Didactic to Constructive: Turning Expert Solutions into Learnable Reasoning
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EGMO-2012-notes_1
|
Let $ABC$ be a triangle with circumcenter $O$.
The points $D$, $E$, $F$ lie in the interiors
of the sides $BC$, $CA$, $AB$ respectively,
such that $\ol{DE} \perp \ol{CO}$ and $\ol{DF} \perp \ol{BO}$.
Let $K$ be the circumcenter of triangle $AFE$.
Prove that the lines $\ol{DK}$ and $\ol{BC}$ are perpendicular.
|
First, note
$\dang EDF = 180\dg - \dang BOC = 180\dg - 2A$,
so $\dang FDE = 2A$.
\begin{center}
\begin{asy}
pair A = Drawing("A", dir(110), dir(110));
pair B = Drawing("B", dir(210), dir(210));
pair C = Drawing("C", dir(330), dir(330));
draw(A--B--C--cycle);
pair O = Drawing("O", origin, dir(45));
pair D = Drawing("D", 0.4*B+0.6*C, dir(-90));
pair F = Drawing("F", extension(A,B,D,foot(D,O,B)), dir(-C));
pair E = Drawing("E", extension(A,C,D,foot(D,O,C)), dir(-B));
pair K = Drawing("K", circumcenter(A,E,F), dir(90));
draw(E--D--F--cycle);
draw(E--K--F, dashed);
draw(B--O--C);
draw(D--K);
draw(circumcircle(D,E,F), dotted);
\end{asy}
\end{center}
Observe that $\dang FKE = 2A$ as well; hence $KFDE$ is cyclic.
Hence
\begin{align*}
\dang KDB &= \dang KDF + \dang FDB \\
&= \dang KEF + \left( 90\dg - \dang DBO \right) \\
&= (90\dg-A) + (90\dg-(90\dg-A)) \\
&= 90\dg.
\end{align*}
and the proof ends here.
|
EGMO-2012-notes_2
|
Let $n$ be a positive integer.
Find the greatest possible integer $m$, in terms of $n$,
with the following property: a table with $m$ rows and $n$ columns can be
filled with real numbers in such a manner that for any
two different rows $\left[ a_1, a_2, \dots, a_n \right]$
and $\left[ b_1, b_2, \dots, b_n \right]$ the following holds:
\[\max\left( {\left| {{a_1} - {b_1}} \right|,\left| {{a_2} - {b_2}} \right|,
\dots,\left| {{a_n} - {b_n}} \right|} \right) = 1. \]
|
\paragraph{Answer.} The largest $m$ is $m(n) = 2^n$.
\paragraph{Construction.} Choose the table with all $2^n$ binary vectors.
\paragraph{Bound.}
We will proceed by induction on $n$, with the base case $n=1$ being immediate.
Take the first column, and assume the entries lie in some interval $I = [r, r+1]$.
Then the rows with first entry equal to $r$
form an $(n-1)$-table, hence at most $m(n-1)$ of them.
The rows with first entry not equal to $r$
also form an $(n-1)$-table, hence at most $m(n-1)$ of them.
So
\[ m(n) \le \underbrace{m(n-1)}_{\text{first entry $r$}}
+ \underbrace{m(n-1)}_{\text{first entry not $r$}} = 2m(n-1) \]
and we're done by induction.
|
EGMO-2012-notes_3
|
Solve over $\RR$ the functional equation
\[ f\left( yf(x + y) + f(x) \right) = 4x + 2yf(x + y). \]
|
The only solution is $f(x) \equiv 2x$ which obviously works.
Let $P(x,y)$ be the given condition. Then:
\begin{itemize}
\ii Note $P(x,0) \implies f(f(x)) = 4x$; in particular $f$ is bijective.
\begin{itemize}
\ii \dots This also implies $f(4x) = f(f(f(x))) = 4f(x)$.
\ii Taking $x=0$ gives $f(0) = 4f(0) \implies f(0) = 0$.
\end{itemize}
\ii Now $P(0,2) \implies f(2f(2)) = 4f(2) = f(8) \implies f(2) = 4$.
\ii Then $P(0,1) \implies f(f(1)) = 2f(1) \implies 4 = 2f(1) \implies f(1)=2$.
\ii Finally, $P(x,1-x)$ gives
\[ f\left( 2(1-x) + f(x) \right) = 4x + 4(1-x) = 4. \]
Since $f$ is a bijection and $f(2) = 4$, this means $2(1-x)+f(x) = 2$,
so $f(x) = 2x$ as desired.
\end{itemize}
|
EGMO-2012-notes_4
|
A set $A$ of integers is called \emph{sum-full} if $A \subseteq A + A$.
A set $A$ of integers is said to be \emph{zero-sum-free} if $0$
is the only integer that cannot be expressed as the
sum of the elements of a finite nonempty subset of $A$.
Does there exist a sum-full zero-sum-free set of integers?
|
The answer is YES, such a set exists, and is given by
\[ A = \left\{ 1, -2, 3, -5, 8, -13, 21, -34, 55, \dots \right\}. \]
To prove every number can be expressed as a
(possibly empty) subset of $A$, we have the following stronger claim.
\begin{claim*}
Let $N$ be an integer (possibly zero or negative).
Let $F$ denote the smallest Fibonacci number
which is strictly greater than $|N|$.
Then there exists a representation of $N$ as the sum of
a (possibly empty) subset of $A$,
where no element used has absolute value greater than $N$.
\end{claim*}
\begin{proof}
By induction on $|N|$.
Omitted.
\end{proof}
On the other hand, the \emph{Zeckendorf representation theorem}
implies that it's not possible to express $0$
as a nonempty subset;
such a representation would amount to an identity of the form
\[ F_{i_1} + F_{i_2} + \dots = F_{j_1} + F_{j_2} + \dots \]
where no terms on either side are consecutive,
and both sides have no common terms; and this is impossible.
|
EGMO-2012-notes_5
|
The numbers $p$ and $q$ are prime and satisfy
\[ \frac{p}{p+1} + \frac{q+1}{q} = \frac{2n}{n+2} \]
for some positive integer $n$.
Find all possible values of $q-p$.
|
We clean up the numerators as
\[ \left( 1-\frac{1}{p+1} \right) + \left( 1 + \frac{1}{q} \right) = 2 - \frac{4}{n+2} \]
which simplifies as
\[ \frac{4}{n+2} = \frac{1}{p+1} - \frac{1}{q} \]
which reduces as
\[ \frac{4}{n+2} = \frac{q-p-1}{q(p+1)} \]
At this point $n$ is a free parameter.
So, this motivates us to rewrite the equation as
\[ n+2 = \frac{4q(p+1)}{(q-p)-1}. \]
This implies that $q-p-1 \mid 4q(p+1)$.
Since $n$ is positive, we have $0 < q-p-1 < q$.
Because $q$ is prime, this implies $q-p-1 \mid 4(p+1)$.
But now $q-p-1 \mid 4(p+1) + 4(q-p-1) = 4q$.
Again, $q$ is prime, so $q-p-1 \mid 4$.
This implies $q-p \in \left\{ 2,3,5 \right\}$.
It is easy to construct working examples: say $(p,q)$ is $(3,5)$, $(2,5)$ and $(2,7)$.
|
EGMO-2012-notes_6
|
There are infinitely many people registered on the social network Mugbook.
Some pairs of (different) users are registered as friends, but each person has only finitely many friends.
Every user has at least one friend.
(Friendship is symmetric; that is, if $A$ is a friend of $B$, then $B$ is a friend of $A$.)
Each person is required to designate one of their friends as their best friend.
If $A$ designates $B$ as her best friend, then (unfortunately)
it does not follow that $B$ necessarily designates $A$ as her best friend.
Someone designated as a best friend is called a $1$-best friend.
More generally, if $n> 1$ is a positive integer,
then a user is an $n$-best friend provided that they have been designated
the best friend of someone who is an $(n-1)$-best friend.
Someone who is a $k$-best friend for every positive integer $k$ is called popular.
\begin{enumerate}
\ii[(a)] Prove that every popular person is the best friend of a popular person.
\ii[(b)] Show that if people can have infinitely many friends,
then it is possible that a popular person is not the best friend of a popular person.
\end{enumerate}
|
First note the following statement is true by induction:
\begin{claim*}
Someone who is an $n$-best friend is also a $k$-best friend for all $1 \le k < n$.
\end{claim*}
\begin{proof}
Induction on $(k,n)$.
\end{proof}
For part (a), suppose Dan is a popular person.
Let the friends of Dan be $P_1$, $P_2$, \dots, $P_m$.
For each $n$, one of the $P_i$'s must be an $n$-best friend.
By pigeonhole, one of the $P_i$'s is an $n$-best friend for infinitely many $n \ge 1$.
Hence be the claim above they are popular too.
For part (b), consider a social graph defined by having Dan ($D$)
and the following chains of best friends:
\begin{align*}
D &\longleftarrow P_{11} \\
D &\longleftarrow P_{21} \longleftarrow P_{22} \\
D &\longleftarrow P_{31} \longleftarrow P_{32} \longleftarrow P_{33} \\
D &\longleftarrow P_{41} \longleftarrow P_{42} \longleftarrow P_{43} \longleftarrow P_{44} \\
&\vdotswithin{\longleftarrow}
\end{align*}
Then Dan is the only popular person in the entire network.
This gives the construction promised.
|
EGMO-2012-notes_7
|
Let $ABC$ be an acute-angled triangle with circumcircle $\Gamma$ and orthocenter $H$. Let $K$ be a point of $\Gamma$ on the other side of $\ol{BC}$ from $A$. Let $L$ be the reflection of $K$ across $\ol{AB}$, and let $M$ be the reflection of $K$ across $\ol{BC}$. Let $E$ be the second point of intersection of $\Gamma$ with the circumcircle of triangle $BLM$. Show that the lines $KH$, $EM$ and $BC$ are concurrent.
|
\begin{center}
\begin{asy}
size(9cm);
pair A = dir(110);
pair B = dir(210);
pair C = dir(330);
draw(A--B--C--cycle);
pair Ap = -B*C/A;
pair Cp = -A*B/C;
draw(A--Ap);
draw(C--Cp);
draw(unitcircle);
pair K = dir(230);
pair M = reflect(B,C)*K;
pair blah = reflect(A,B)*K;
pair L = blah;
draw(M--K--L, dotted);
pair E = IP(unitcircle, circumcircle(B, L, M));
draw(circumcircle(B, L, M));
pair H = A+B+C;
draw(H--K, red);
draw(E--Ap, red);
draw(E--L, dotted);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$A'$", Ap, dir(Ap));
dot("$C'$", Cp, dir(Cp));
dot("$K$", K, dir(K));
dot("$M$", M, dir(M));
dot("$L$", L, dir(270));
dot("$E$", E, dir(100));
dot("$H$", H, dir(45));
/* Source generated by TSQ */
\end{asy}
\end{center}
The main idea is a ``floating orthocenter'' amidst a large number of reflections.
This motivates us to reflect $H$ over $\ol{BC}$ to $A'$
and $H$ over $\ol{AB}$ to $C'$.
Then the problem is equivalent to trying to prove that
the points $E$, $M$, $A'$ are collinear.
Let $E'$ be the second intersection of line $A'M$ with $\Gamma$.
First, we claim that $E'$, $C'$, $L$ are collinear.
Note that and \[ \angle LC'B = \angle KHB = \angle MA'B = \angle E'A'B \] (using the reflections).
From here we get $\angle LC'B + \angle BC'E' = 180^{\circ}$, and
hence the collinearity is proved.
On the other hand,
\[ \angle LBM =
2\left( \angle KBA - KBC \right)
= 2 \angle B \]
and
\[ \angle C'E'A' = 180 -2 \angle B. \]
So $L$, $B$, $M$, $E'$ are cyclic as desired.
|
EGMO-2012-notes_8
|
A word is a finite sequence of letters from some alphabet.
A word is \emph{repetitive} if it is a concatenation of at least two identical subwords
(for example, $ababab$ and $abcabc$ are repetitive, but $ababa$ and $aabb$ are not).
Prove that if a word has the property that swapping any two adjacent letters
makes the word repetitive, then all its letters are identical.
(Note that one may swap two adjacent identical letters, leaving a word unchanged.)
\end{enumerate}
|
A word with the property that swapping any two adjacent letters
makes the word repetitive, but for which all letters the same,
will called a \emph{unicorn}.
The problem asks to show there are no unicorns.
So assume for contradiction there exists a unicorn.
We proceed in three steps.
\paragraph{Step 1. Reduction to binary alphabet.}
In the alphabet, choose one letter (that appears in the unicorn) to be $0$
and replace all the letters with $1$.
The resulting word is still a unicorn on the two symbols $\{0,1\}$.
In this way we only need to consider the problem for the alphabet $\Sigma = \{0,1\}$.
\paragraph{Step 2. Unicorns are repetitive themselves.}
So let $w \in \Sigma^n$ be a unicorn.
It's easy to see that $w = 010101\dotsm$ is not a unicorn;
if we swap the first two fits, the resulting word
$100101\dotsm$ has a doubled $00$ that appears only once, so it is not a unicorn.
Similarly, $w = 101010\dotsm$ is not a unicorn.
Hence, we may assume our unicorn $w$ has at least two adjacent bits equal.
This means that $w$ is itself repetitive.
\paragraph{Step 3. Main argument.}
From now on let $w[i]$ denote the $i$\ts{th} letter of $w$.
Fix an index $1 \le k < n/2$ for which $w[k] \neq w[k+1]$.
Let $w'$ be the word that results from swapping the unequal bits,
so $w'[k] = w[k+1]$ and $w'[k+1] = w[k]$.
Now $w$ is repetitive, and so is $w'$;
so there should exist integers $p \le n/2$ and $q \le n/2$ dividing $n$ (the \emph{periods})
such that the identities
\[
w[i] = w[i+p] \quad \forall 1 \le i \le n-p; \qquad
w'[i] = w'[i+q] \quad \forall 1 \le i \le n-q.
\]
We may assume $p$ and $q$ are different and don't divide each other.
We will now produce a contradiction in two cases.
\begin{itemize}
\ii Consider the case $\gcd(p,q) = 1$.
Then, the number of $1$'s in $w$ ought to be a multiple of $n/p$,
because $w$ consists of $n/p$ repetitions of some word.
Similarly, the number of $1$'s in $w'$ ought to be a multiple of $n/q$.
But these numbers are equal.
So the number of $1$'s is a multiple of both $n/p$ and $n/q$.
When $\gcd(p,q) = 1$, this can only occur if the numbers $1$'s is divisible by $n$,
i.e.\ $w = 0 \dots 0$ or $w = 1 \dots 1$.
Then $w$ is not in fact a unicorn.
\ii Otherwise, let $d = \gcd(p,q) > 1$.
Choose the index $i = k$ to start.
In a \emph{move} we update the index as follows:
\begin{itemize}
\ii If $i + p < n$, replace $i$ with $i + p$.
\ii Otherwise, replace $i$ with $i - q$.
\ii Repeat this until $i \equiv k \pmod q$ again for the first time.
\end{itemize}
This process is well defined since $p$ and $q$ are different and less than $n/2$.
This creates a sequence
\[ k = i_0, i_1, i_2, \dots, i_m \]
with the property that the values of $w$ and $w'$ at all indices
except the ends are equal.
(The assumption $d > 1$ means that all $i_\bullet$ are $k \pmod d$,
so this sequence never contains $k+1$.)
Following the chain then creates a contradiction,
because we started at $w[k]$ and ended at $w'[i_m] = w'[k] \neq w[k]$ following only equalities.
\end{itemize}
A cartoon is shown below for $p = 9$, $q = 6$, $n = 18$, $k = 2$.
\begin{center}
\begin{asy}
size(12cm);
for (int i=1; i<=18; ++i) {
dot("$" + ((string)(i)) + "$", (i,0), dir(-90));
}
void draw_arr(int x, int y, pen p) {
draw((x,0)..((x+y)/2, (y-x)/5)..(y,0), p, EndArrow(TeXHead), Margins);
}
draw_arr(2,11, deepgreen);
draw_arr(11,5, red);
draw_arr(5,14, deepgreen);
draw_arr(14,8, red);
draw_arr(8,2, red);
label("$w$", (0,-2), blue);
label("$w'$", (0,-3), blue);
draw(box((0.6,-1.6), (9.4,-2.4)), deepgreen);
draw(box((9.6,-1.6), (18.4,-2.4)), deepgreen);
draw(box((0.6,-2.6), (6.4,-3.4)), red);
draw(box((6.6,-2.6), (12.4,-3.4)), red);
draw(box((12.6,-2.6), (18.4,-3.4)), red);
label("$\mathbf{0}$", (2,-2), blue);
label("$\mathbf{1}$", (3,-2), blue);
label("$\mathbf{1}$", (2,-3), blue);
label("$\mathbf{0}$", (3,-3), blue);
label("$0$", (11,-2), blue);
label("$0$", (11,-3), blue);
label("$0$", (5,-2), blue);
label("$0$", (5,-3), blue);
label("$0$", (14,-2), blue);
label("$0$", (14,-3), blue);
label("$0$", (8,-2), blue);
label("$0$", (8,-3), blue);
\end{asy}
\end{center}
\begin{remark*}
[The smart solution]
An alternative clever way to do this last part is using roots of unity.
The idea is to show that in our binary language,
if $S$ is the subset of indices with $1$'s among $\{1, \dots, n\}$,
then every repetitive word satisfies
\[ \sum_{s \in S} \exp\left( \frac{2\pi n}{s} \right) = 0. \]
(The converse is not true.)
Hence a repetitive unicorn cannot exist;
swapping \emph{any} two adjacent indices with unequal bits makes the sum change.
(The proof earlier shows this too, since we only used one value of $k$.)
\end{remark*}
|
EGMO-2013-notes_1
|
The side $BC$ of the triangle $ABC$
is extended beyond $C$ to $D$ so that $CD = BC$.
The side $CA$ is extended beyond $A$ to $E$ so that $AE = 2CA$.
Prove that if $AD=BE$ then the triangle $ABC$ is right-angled.
|
Let ray $DA$ meet $\ol{BE}$ at $M$.
Consider the triangle $EBD$.
Since the point lies on median $\ol{EC}$, and $EA = 2AC$,
it follows that $A$ is the centroid of $\triangle EBD$.
\begin{center}
\begin{asy}
pair A = dir(50);
pair B = dir(180);
pair C = dir(0);
draw(A--B--C--cycle, black+1.4);
pair D = 2*C-B;
pair E = 3*A-2*C;
draw(C--D);
draw(A--E);
draw(B--E, dotted);
draw(A--D, dotted);
pair M = extension(A, D, B, E);
draw(B--E--D--M);
dot("$A$", A, dir(A));
dot("$B$", B, dir(-90));
dot("$C$", C, dir(-90));
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
dot("$M$", M, dir(M));
/* TSQ Source:
A = dir 50
B = dir 180 R-90
C = dir 0 R-90
A--B--C--cycle black+1.4
D = 2*C-B
E = 3*A-2*C
C--D
A--E
B--E dotted
A--D dotted
M = extension A D B E
B--E--D--M
*/
\end{asy}
\end{center}
So $M$ is the midpoint of $\ol{BE}$.
Moreover $MA = \frac 12 AD = \frac 12 BE$; so $MA = MB = ME$ and hence $\triangle ABE$ is inscribed in a circle with diameter $\ol{BE}$.
Thus $\angle BAE = 90\dg$, so $\angle BAC = 90\dg$.
|
EGMO-2013-notes_2
|
Determine all integers $m$ for which the $m \times m$ square can be dissected
into five rectangles, the side lengths of which are
the integers $1$, $2$, \dots, $10$ in some order.
|
The answer is that this is possible if and only if $m = 11$ or $m = 13$.
\paragraph{Constructions for $m=11$ and $m=13$.}
See the figures below.
\begin{center}
\begin{asy}
size(12cm);
picture p11, p13;
pen b = black+2.5;
filldraw(p11, box((0,0), (11,11)), palegreen, b);
filldraw(p11, box((0,0), (8,5)), palecyan, b);
filldraw(p11, box((7,9), (11,11)), palecyan, b);
filldraw(p11, box((8,0), (11,9)), palered, b);
filldraw(p11, box((0,5), (7,11)), palered, b);
for (int i=0; i<11; ++i) {
for (int j=0; j<11; ++j) {
draw(p11, shift(i,j)*unitsquare, black);
}
}
add(p11);
filldraw(p13, box((0,0), (13,13)), palegreen, b);
filldraw(p13, box((0,0), (8,3)), palecyan, b);
filldraw(p13, box((6,4), (13,13)), palecyan, b);
filldraw(p13, box((8,0), (13,4)), palered, b);
filldraw(p13, box((0,3), (6,13)), palered, b);
for (int i=0; i<13; ++i) {
for (int j=0; j<13; ++j) {
draw(p13, shift(i,j)*unitsquare, black);
}
}
add(shift(13,-1)*p13);
\end{asy}
\end{center}
\paragraph{Proof the task is impossible unless $11 \le m \le 13$.}
The total area of the rectangles is of the form $A = a_1 b_1 + \dots + a_5 b_5$
for $(a_1, \dots, b_5)$ a permutation of $(1, \dots, 10)$.
By applying the rearrangement inequality repeatedly one can prove that
\[ A \le 1 \cdot 2 + 3 \cdot 4 + 5 \cdot 6 + 7 \cdot 8 + 9 \cdot 10 = 190. \]
Analogously, we can get a matching lower bound
\[ A \ge 1 \cdot 10 + 2 \cdot 9 + 3 \cdot 8 + 4 \cdot 7 + 5 \cdot 6 = 110. \]
Since $A = m^2$ for integer $m$, it follows $11 \le m \le 13$.
\paragraph{Proof the task is impossible for $m=12$.}
It remains to rule out such a tiling for a $12 \times 12$ square.
We start with the following structure claim:
\begin{claim*}
Any tiling for $m > 10$ must either be the spiral pattern below or a reflection of it.
\begin{center}
\begin{asy}
size(3cm);
draw(box((0,0),(7,7)), black+1.5);
draw((0,2)--(5,2), black+1);
draw((5,0)--(5,4), black+1);
draw((3,4)--(7,4), black+1);
draw((3,7)--(3,2), black+1);
\end{asy}
\end{center}
\end{claim*}
\begin{proof}
Label the square $ABCD$.
Because $m > 10$, each corner of the square must be part of a different rectangle.
Consider the rectangle covering $A$, and let the vertex opposite $A$ be $A'$.
Define $B'$, $C'$, $D'$ similarly.
We contend that no two of $A'$, $B'$, $C'$, $D'$ coincide.
Indeed, if $A' = C'$, then we get the figure on the left where the two
purple rectangles are part of the tiling.
The remaining two ``quadrants'' must be dissected into three rectangles,
but there is general a quadrant cannot be divided into either $2$ or $3$ rectangles
in aw way that doesn't create two equal lengths.
\begin{center}
\begin{asy}
size(7cm);
picture pl;
draw(pl, box((0,0),(7,7)), black+1.5);
label(pl, "$A$", (0,0), dir(225));
label(pl, "$B$", (7,0), dir(315));
label(pl, "$C$", (7,7), dir(45));
label(pl, "$D$", (0,7), dir(135));
dotfactor *= 2;
picture pr = rotate(0)*pl;
filldraw(pl, box((0,0), (5,3)), mediummagenta, black+1.5);
filldraw(pl, box((7,7), (5,3)), mediummagenta, black+1.5);
filldraw(pr, box((0,0), (5,3)), mediummagenta, black+1.5);
filldraw(pr, box((5,3), (7,0)), mediummagenta, black+1.5);
label(pl, "$A'=C'$", (5,3), dir(135));
label(pr, "$A'=B'$", (5,3), dir(90));
dot(pl, (5,3));
dot(pr, (5,3));
add(shift(0,0)*pl);
add(shift(11,0)*pr);
\end{asy}
\end{center}
Meanwhile, if $A' = B'$, the two purple rectangles
already obviously have a common dimension (right figure).
Hence $A'$, $B'$, $C'$, $D'$ are all different points.
They must each be part of some rectangle not using the corner.
Moreover, every rectangle in the dissection has at least one vertex
not on the boundary of the square.
So the fifth rectangle must be $A'B'C'D'$ exactly.
\end{proof}
Now on to the proof that $m = 12$ is impossible.
Assume for contradiction a construction exists and refer to the spiral figure above.
Note that because $12 > 1 + 10$,
the side length $1$ cannot belong to any edge of the outer $12 \times 12$ square.
This means that $1$ must be a side length of the center rectangle.
Moreover, because the outer square has perimeter $48$,
and we know $1+2+\dots+10 = 55$, the center rectangle has perimeter $7$.
In other words, the inner rectangle must be $1 \times 6$ exactly.
\begin{center}
\begin{asy}
size(5cm);
draw(box((0,0),(12,12)), black+1.5);
draw((0,2)--(5,2), black+1);
draw((5,0)--(5,8), black+1);
draw((4,8)--(12,8), black+1);
draw((4,12)--(4,2), black+1);
label("$6$", (5,4), dir(0));
label("$1$", (4.5,8), dir(90));
\end{asy}
\end{center}
Hence, we would need to partition the remaining numbers $\{2,3,4,5,7,8,9,10\}$
into eight pairs such that two pairs differ by $1$ and differ by $6$
(in the picture, the paired numbers are on opposite sides of the square).
Note that $7$ must be paired with $8$ (since $7 \pm 6$ and $7-1$ are not available)
and $5$ must be paired with $4$ (since $5 \pm 6$ and $5+1$ are not available).
But then the remaining pairs need to have difference $6$,
and there is no way to form two such pairs from $\{2,3,9,10\}$.
This produces the desired contradiction.
|
EGMO-2013-notes_3
|
Let $n$ be a positive integer.
\begin{enumerate}
\ii[(a)] Prove that there exists a set $S$ of $6n$ positive integers
such that the least common multiple of any two is at most $32n^2$.
\ii[(b)] Show that every set $T$ of $6n$ positive integers
contains two elements with least common multiple exceeding $9n^2$.
\end{enumerate}
|
For part (a), let
\[ S = \{1, 2, \dots, 4n\} \cup \{4n+2, 4n+4, \dots, 8n\} \]
which works.
For (b),
the main idea is as follows.
Let $x_0 < x_1 < \dots < x_m$ be these integers.
Assume for contradiction the LCM's are less than $L$.
The main idea is the estimate
\[ L \ge \frac{x_i x_{i+1}}{\gcd(x_i, x_{i+1})} \ge \frac{x_i x_{i+1}}{x_{i+1} - x_i} \]
whence
\[ \frac{1}{L} \le \frac{1}{x_i} - \frac{1}{x_{i+1}}. \]
so taking the sum starting from any $k$ gives
\[ \frac{m+1-k}{L} \le \frac{1}{x_k} - \frac{1}{x_m} < \frac 1{x_k}. \]
For $(k,m) = (3n,6n)$ and $L = 9n^2$ this is a contradiction (since $x_k \ge 3n$).
\begin{remark*}
China TST 2007/6 follows the same idea.
\end{remark*}
|
EGMO-2013-notes_4
|
Find all positive integers $a$ and $b$ for which there are three consecutive integers at which the polynomial $P(n) = \frac1b(n^5+a)$ takes integer values.
|
The answer is
\begin{itemize}
\ii Either $b = 1$; or
\ii $a \equiv \pm 1 \pmod{11}$ and $b = 11$ (and $a > 0$).
\end{itemize}
We consider only the case $b>1$.
Suppose that $P$ is a prime power dividing $b$.
Evidently $P$ is not even.
Then there exists an $n$ for which
\[ n^5 \equiv (n+1)^5 \equiv (n-1)^5 \pmod{P}. \]
Therefore,
\[ 0 \equiv (n+1)^5 + (n-1)^5 - 2n^5 = 20n^3+10n \pmod{P}. \]
Since $n \not\equiv 0 \pmod{p}$ for obvious reasons, we obtain that
\[ n^2 \equiv -\tfrac{1}{2} \pmod{p}. \]
Then,
\begin{align*}
0 &\equiv 2((n+1)^5 - (n-1)^5) \pmod{P} \\
&= 20n^4+40n^2+4 \\
&\equiv 20 \cdot -\frac{1}{4} + 40 \cdot -\frac{1}{2} + 4 \pmod{P} \\
&= -11.
\end{align*}
This implies $P = 11$.
Since $P$ can be any prime power dividing $b$, this forces $b=11$.
In other words, $b \in \left\{ 1,11 \right\}$.
Finally, observe that
\[ 3^5 \equiv 4^5 \equiv 5^5 \equiv 1 \pmod{11} \]
and \[ (-3)^5 \equiv (-4)^5 \equiv (-5)^5 \equiv -1 \pmod{11} \]
are the only such triples modulo $11$.
Therefore the solution set is the one claimed above.
|
EGMO-2013-notes_5
|
Let $\Omega$ be the circumcircle of the triangle $ABC$.
The circle $\omega$ is tangent to the sides $AC$ and $BC$,
and it is internally tangent to the circle $\Omega$ at the point $P$.
A line parallel to $AB$ intersecting the
interior of triangle $ABC$ is tangent to $\omega$ at $Q$.
Prove that $\angle ACP = \angle QCB$.
|
First, let us extend $\ol{AQ}$ to meet $\ol{BC}$ at $Q_1$.
By homothety, we see that $Q_1$ is just the
contact point of the $A$-excircle with $\ol{BC}$.
\begin{center}
\begin{asy}
size(7cm);
pair A = Drawing("A", dir(100), dir(100));
pair B = Drawing("B", dir(210), dir(190));
pair C = Drawing("C", dir(330), dir(350));
pair I = incenter(A,B,C);
pair Ia = 2*dir(-90)-I;
pair Ib = 2*dir(35)-I;
pair Ic = 2*dir(155)-I;
draw(unitcircle);
pair E = dir(90);
pair T = Drawing("P", 2*foot(origin,I,E)-E, dir(I-E));
pair K = extension(A,B,T,midpoint(I--Ic));
pair L = extension(A,C,T,midpoint(I--Ib));
path w = Drawing(circumcircle(T,K,L));
Drawing("I_A", Ia, dir(-90));
pair Q1 = Drawing("Q_1", foot(Ia, B, C), dir(45));
pair Q = Drawing("Q", IP(A--Q1, w), dir(45));
draw(arc(Ia, abs(Ia-Q1), -20, 200));
draw(B--C);
draw(A--foot(Ia, A, B));
draw(A--foot(Ia, A, C));
draw(T--A--Q1);
draw( (Q-0.8)--(Q+0.8) );
\end{asy}
\end{center}
Now let us perform an inversion around $A$
with radius $\sqrt{AB \cdot AC}$ followed by
a reflection around the angle bisector;
call this map $\Psi$.
Note that $\Psi$ fixes $B$ and $C$.
Moreover it swaps $\ol{BC}$ and $(ABC)$.
Hence, this map swaps the $A$-excircle
with the $A$-mixtilinear incircle $\omega$.
Hence $\Psi$ swaps $P$ and $Q_1$.
It follows that $\ol{AP}$ and $\ol{AQ_1}$
are isogonal with respect to $\angle BAC$,
meaning $\angle BAP = \angle CAQ_1$.
Since $\angle CAQ = \angle CAQ_1$ we are done.
|
EGMO-2013-notes_6
|
Snow White and the Seven Dwarves are living in their house in the forest.
On each of 16 consecutive days, some of the dwarves worked in the diamond mine
while the remaining dwarves collected berries in the forest.
No dwarf performed both types of work on the same day.
On any two different (not necessarily consecutive) days,
at least three dwarves each performed both types of work.
Further, on the first day, all seven dwarves worked in the diamond mine.
Prove that, on one of these 16 days, all seven dwarves were collecting berries.
\end{enumerate}
|
Let $Q_n$ denote the vector space $\left\{ 0,1 \right\}^n$.
For a vector $v$, let $v[i]$ denote the $i$th component.
We may identify each day with a vector $v_k$,
where $v_k[i] = 0$ if dwarf $i$ worked in the diamond mine,
and $v_k[i] = 1$ otherwise.
Let $V = \left\{ v_1, v_2, \dots, v_{16} \right\}$
be the subset of $Q_7$ in the problem,
and assume $v_{16} = 0000000$.
We first prove the following:
\begin{lemma*}
Exactly two vectors start with each of $000$, $001$, \dots, $111$.
Similar statements hold for any choice of three indices.
\end{lemma*}
\begin{proof}
If three vectors start with $000$ (say)
then we run into problems.
\end{proof}
Now we know that $v_{16}$ is the all-null vector.
Ignoring that vector (and hence considering just the first $15$ vectors),
let $n_i$ be the number of $1$'s in $v_i$, where $i = 1, 2, \dots, 15$.
\begin{claim*}
We have
\begin{align*}
\sum_{i=1}^{15} \binom{n_i}{1} &= \frac{16}{2} \binom{7}{1} = 56 \\
\sum_{i=1}^{15} \binom{n_i}{2} &= \frac{16}{2^2} \binom{7}{2} = 84 \\
\sum_{i=1}^{15} \binom{n_i}{3} &= \frac{16}{2^3} \binom{7}{3} = 70.
\end{align*}
\end{claim*}
\begin{proof}
This follows using the lemma and double counting.
For example, the third equation is counting
the number of quadruples $(i, j_1, j_2, j_3)$
such that the $i$th string has a $1$
in the $j_1$th, $j_2$th, $j_3$th position, where $j_1 < j_2 < j_3$.
The left-hand side counts the number when summed by $i$;
the right-hand side counts the number according
to the $\binom73$ choices of $(j_1, j_2, j_3)$.
\end{proof}
We may then rewrite the lemma as
\begin{align*}
\sum_{i=1}^{15} n_i &= 56 \\
\sum_{i=1}^{15} n_i^2 &= 2 \cdot 84 + 56 = 224 \\
\sum_{i=1}^{15} n_i^3 &= 6 \cdot 70 + 3 \cdot 224 - 2 \cdot 56 = 980.
\end{align*}
Now remark that $(n_i-3)(n_i-4)(n_i-7) \le 0$ for each integer $1 \le n_i \le 7$.
We compute
\begin{align*}
0 &\ge \sum_{i=1}^{15} (n_i-3)(n_i-4)(n_i-7) \\
&= \sum_{i=1}^{15} n_i^3 - 14n_i^2 + 61n_i - 84 \\
&= 1 \cdot 980 - 14 \cdot 224 + 61 \cdot 56 - 15 \cdot 84 \\
&= 980 - 3136 + 3416 - 1260 \\
&= 0.
\end{align*}
This implies that $n_i \in \left\{ 3,4,7 \right\}$ for each $i$.
If $n_i \neq 7$ for any $i$ then we have
\[ 224 = \sum n_i^2 \equiv 15 \cdot 2 \not\equiv 0 \pmod{7} \]
which is impossible.
This means that $n_i = 7$ for some $i$, and therefore, $1111111 \in V$, as desired.
\begin{remark*}
Up to permutation there turns out to be a unique set of $16$ vectors.
Xinyang Chen gives the following compact description of the unique solution:
\begin{itemize}
\ii $0000000$ and $1111111$, of course
\ii $1101000$ and its six \emph{cyclic shifts} $0110100$, $0011010$, etc.
\ii The complements of the above seven strings (i.e.\ $0010111$, $1001011$, etc.).
\end{itemize}
To write out all sixteen explicitly as a TL;DR:
\begin{align*}
\Big\{ 0000000, 1111111,
& 1101000, 0110100, 0011010, 0001101, 1000110, 0100011, 1010001 \\
& 0010111, 1001011, 1100101, 1110010, 0111001, 1011100, 0101110 \Big\}.
\end{align*}
\end{remark*}
|
EGMO-2014-notes_1
|
Determine all real constants $t$ such that whenever $a$, $b$ and $c$ are the lengths of sides of a triangle, then so are $a^2+bct$, $b^2+cat$, $c^2+abt$.
|
\paragraph{Answer.}
We will show the answer is exactly \[ 2/3 \le t \le 2. \]
\paragraph{Proof.}
Apply the substitution $a=y+z$, $b=z+x$, $c=x+y$.
We need the inequality
\begin{align*}
a^2 + bct & < b^2 + ca t + c^2 + ab t \\
(y+z)^2 + (x+y)(x+z) t &< (x+y)^2+(x+z)^2 + (y+z)(2x+y+z) t \\
\iff [x^2-(xy+xz+yz+y^2+z^2)]t &< 2x^2+2xy+2xz-2yz \\
\end{align*}
to be true for all $(x,y,z) \in \RR_{>0}^3$
(checking only one inequality is enough by symmetry).
Writing this is a quadratic in $x$, we want
\[ Q(x) \coloneq (2-t)x^2 + (2+t)(y+z)x + ((y^2+yz+z^2)t - 2yz) > 0. \]
\begin{claim*}
Except when $t=2$,
the quadratic $Q$ always has two real roots.
\end{claim*}
\begin{proof}
The discriminant is
\begin{align*}
D &= [(2+t)(y+z)]^2 - 4(2-t)\left( (y^2+yz+z^2)t - 2yz \right) \\
&= \left( (2+t)^2 - 4(2-t)t \right)(y^2+z^2)
+ \left( 2(2+t)^2-4(2-t)(t-2) \right) yz \\
&= \left( 5t^2-4t+4 \right)(y^2+z^2) + \left( 6t^2-8t+24 \right) yz > 0
\end{align*}
which means $Q$ always has two real roots
(aside for the single exceptional case $t=2$,
in which $Q$ is not a quadratic).
\end{proof}
Now, we make a few closing observations.
\begin{itemize}
\ii It is clear we need $t \le 2$,
since a negative leading coefficient will
cause the inequality to fail for $x \gg y,z$.
\ii The case $t=2$ obviously has $Q > 0$ always.
\ii For $t < 2$, as $Q$ always has two real roots,
the assertion is true if and only if all coefficients
of $Q$ are nonnegative.
\ii When $t \ge \frac23$,
we have $(2+t)(y+z) > 0$ obviously, and from
\[ \frac{y^2+yz+z^2}{3} \ge yz \]
the constant coefficient is nonnegative as well.
Thus when $\boxed{\frac23 \le t \le 2}$
we indeed have $Q(x) > 0$ for $x,y,z > 0$.
\ii If $t < \frac23$,
then by letting $y=z$ fails the condition.
\end{itemize}
|
EGMO-2014-notes_2
|
Let $D$ and $E$ be points in the interiors of sides $AB$ and $AC$,
respectively, of a triangle $ABC$, such that $DB = BC = CE$.
Let the lines $CD$ and $BE$ meet at $F$.
Prove that the incenter $I$ of triangle $ABC$,
the orthocenter $H$ of triangle $DEF$ and
the midpoint $M$ of the arc $BAC$ of the circumcircle of triangle $ABC$
are collinear.
|
\paragraph{First solution (Cynthia Du).}
Let $BI$ and $CI$ meet the circumcircle again at $M_B$, $M_C$.
Observe that we have the spiral congruence
\[ \triangle MDB \sim \triangle MEC \]
from $\dang MBD = \dang MBA = \dang MCA = \dang MCE$
and $BD = EC$, $BM = CM$.
That is, $M$ is the Miquel point of $BDEC$.
\begin{center}
\begin{asy}
pair A = dir(110);
pair B = dir(240);
pair C = dir(300);
draw(A--B--C--cycle, heavycyan);
filldraw(unitcircle, opacity(0.1)+lightcyan, heavycyan);
pair I = incenter(A, B, C);
pair D = -C+2*foot(C, I, B);
pair E = -B+2*foot(B, I, C);
pair M_B = circumcenter(A, I, C);
pair M_C = circumcenter(A, I, B);
pair M = dir(90);
filldraw(M--D--B--cycle, opacity(0.1)+lightgreen, heavygreen);
filldraw(M--E--C--cycle, opacity(0.1)+lightgreen, heavygreen);
draw(B--M_B, heavycyan);
draw(C--M_C, heavycyan);
pair T = extension(M, E, B, I);
pair S = extension(M, D, C, I);
draw(circumcircle(D, I, E), lightred+dashed);
draw(D--S, heavygreen);
draw(E--T, heavygreen);
draw(D--I--E, lightred);
draw(C--D, lightgreen);
draw(B--E, lightgreen);
pair F = extension(C, D, B, E);
pair H = orthocenter(D, E, F);
draw(D--H--E, heavycyan);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$I$", I, dir(I));
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
dot("$M_B$", M_B, dir(M_B));
dot("$M_C$", M_C, dir(M_C));
dot("$M$", M, dir(M));
dot("$T$", T, dir(T));
dot("$S$", S, dir(S));
dot("$F$", F, dir(F));
dot("$H$", H, dir(90));
/* TSQ Source:
A = dir 110
B = dir 240
C = dir 300
A--B--C--cycle heavycyan
unitcircle 0.1 lightcyan / heavycyan
I = incenter A B C
D = -C+2*foot C I B
E = -B+2*foot B I C
M_B = circumcenter A I C
M_C = circumcenter A I B
M = dir 90
M--D--B--cycle 0.1 lightgreen / heavygreen
M--E--C--cycle 0.1 lightgreen / heavygreen
B--M_B heavycyan
C--M_C heavycyan
T = extension M E B I
S = extension M D C I
circumcircle D I E lightred dashed
D--S heavygreen
E--T heavygreen
D--I--E lightred
C--D lightgreen
B--E lightgreen
F = extension C D B E
H = orthocenter D E F R90
D--H--E heavycyan
*/
\end{asy}
\end{center}
Let $T = \ol{ME} \cap \ol{BI}$ and $S = \ol{MD} \cap \ol{CI}$.
First, since $\ol{BI}$ is the perpendicular bisector of $\ol{CD}$
we have that
\[ \dang DIT = \dang CIT = \dang CIB = 90\dg - \half \angle A
= \dang MCB = \dang MED = \dang TED \]
and so $D$, $I$, $T$, $E$ is cyclic.
Similarly $S$ lies on this circle too.
But $\dang SDE = \dang EDM = \dang MED = \dang TED$
so in fact $\ol{ST} \parallel \ol{DE}$ (isosceles trapezoid).
Then $\triangle IST$ and $\triangle HDE$ are homothetic,
so $\ol{IH}$, $\ol{DS}$, and $\ol{ET}$ concur (at $M$).
\paragraph{Second solution (Evan Chen).}
Observe that we have the spiral congruence
\[ \triangle MDB \sim \triangle MEC \]
from $\dang MBD = \dang MBA = \dang MCA = \dang MCE$
and $BD = EC$, $BM = CM$.
That is, $M$ is the Miquel point of $BDEC$.
\begin{center}
\begin{asy}
pair A = dir(110);
pair B = dir(240);
pair C = dir(300);
draw(A--B--C--cycle, heavycyan);
filldraw(unitcircle, opacity(0.1)+lightcyan, heavycyan);
pair I = incenter(A, B, C);
pair D = -C+2*foot(C, I, B);
pair E = -B+2*foot(B, I, C);
pair M = dir(90);
pair F = extension(C, D, B, E);
pair X = midpoint(B--D);
pair Y = midpoint(C--E);
draw(X--M--Y, red+dotted);
pair H = orthocenter(D, E, F);
draw(C--D, lightgreen);
draw(B--E, lightgreen);
draw(I--M, red+dotted);
draw(CP(X, B), red+dashed);
draw(CP(Y, C), red+dashed);
filldraw(B--D--E--C--cycle, opacity(0.1)+yellow, lightgreen);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$I$", I, dir(I));
dot("$D$", D, dir(120));
dot("$E$", E, dir(75));
dot("$M$", M, dir(M));
dot("$F$", F, dir(F));
dot("$X$", X, dir(A+B));
dot("$Y$", Y, dir(A+C));
dot("$H$", H, dir(90));
/* TSQ Source:
A = dir 110
B = dir 240
C = dir 300
A--B--C--cycle heavycyan
unitcircle 0.1 lightcyan / heavycyan
I = incenter A B C
D = -C+2*foot C I B R120
E = -B+2*foot B I C R75
M = dir 90
F = extension C D B E
X = midpoint B--D RA+B
Y = midpoint C--E RA+C
X--M--Y red dotted
H = orthocenter D E F R90
C--D lightgreen
B--E lightgreen
I--M red dotted
CP X B red dashed
CP Y C red dashed
B--D--E--C--cycle 0.1 yellow / lightgreen
*/
\end{asy}
\end{center}
Let $X$ and $Y$ be the midpoints of $\ol{BD}$ and $\ol{CE}$.
Then $MX = MY$ by our congruence.
Consider now the circles with diameters $\ol{BD}$ and $\ol{CE}$.
We now claim that $H$, $I$, $M$ all lie on the radical axis
of these circles.
Note that $I$ is the orthocenter of $\triangle BFC$
and $H$ is the orthocenter of $\triangle DEF$,
so this follows from the so-called Steiner line of $BCDE$
(perpendicular to Gauss line $\ol{XY}$).
For $M$, we observe $MX^2 - XB^2 = MY^2 - YC^2$ thus completing the proof.
\paragraph{Third solution (homothety, official solution).}
Extend $DH$ and $EH$ to meet $BI$ and $CI$ at $D_1$ and $E_1$.
Note $DD_1 \perp BE$, $CI \perp BE$, so $DD_1 \parallel CI$.
Similarly $EE_1 \parallel BI$.
So $HE_1ID_1$.
\begin{center}
\begin{asy}
size(9cm);
pair A = dir(100);
pair B = dir(240);
pair C = dir(300);
pair I = incenter(A, B, C);
pair D = -C+2*foot(C, B, I);
pair E = -B+2*foot(B, C, I);
pair F = extension(B, E, C, D);
pair H = orthocenter(D, E, F);
draw(A--B--C--cycle);
draw(B--E);
draw(C--D);
pair D_1 = extension(D, H, B, I);
pair E_1 = extension(E, H, C, I);
pair D_2 = dir(20);
pair E_2 = dir(170);
draw(unitcircle);
pair M = D_2+E_2-I;
draw(circumcircle(B, C, D_1));
draw(B--D_2--M--E_2--C);
draw(E_1--E, dotted);
draw(M--I, dotted);
draw(D_1--D, dotted);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$I$", I, dir(I));
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
dot("$H$", H, dir(90));
dot("$D_1$", D_1, dir(100));
dot("$E_1$", E_1, dir(80));
dot("$D_2$", D_2, dir(D_2));
dot("$E_2$", E_2, dir(E_2));
dot("$M$", M, dir(M));
/* Source generated by TSQ
A = dir 100
B = dir 240
C = dir 300
I = incenter A B C
D = -C+2*foot C B I
E = -B+2*foot B C I
F = extension B E C D
H = orthocenter D E F R90
A--B--C--cycle
B--E
C--D
D_1 = extension D H B I R100
E_1 = extension E H C I R80
D_2 = dir 20
E_2 = dir 170
unitcircle
M = D_2+E_2-I
circle B C D_1
B--D_2--M--E_2--C
E_1--E dotted
M--I dotted
D_1--D dotted
*/
\end{asy}
\end{center}
Angle chase to show that $B$, $E_1$, $F$, $C$ are cyclic --
$\angle DCE_1 = \angle DCI$ is computable in terms of $ABC$
and \[ \angle E_1BF = \angle E_1BE = \angle E_1EB = \angle HEF = \angle HDF = \angle HDC = \angle DCE_1 = \angle FCE_1. \]
Thus $B$, $D_1$, $F$, $C$ are also cyclic.
So $B$, $D_1$, $E_1$, $C$ are cyclic.
Extend $BI$ and $CI$ to meet the circumcircle again at $D_2$ and $E_2$.
Direct computation gives that $ME_2ID_2$ is also a parallelogram.
We also get $E_1D_1$ is parallel to $E_2D_2$
(both are antiparallel to $BC$ through $\angle BIC$).
So we have homothetic paralellograms and that finishes the problem.
|
EGMO-2014-notes_3
|
We denote the number of positive divisors of a positive integer $m$ by $d(m)$
and the number of distinct prime divisors of $m$ by $\omega(m)$.
Let $k$ be a positive integer.
Prove that there exist infinitely many positive integers $n$
such that $\omega(n) = k$ and
$d(n)$ does not divide $d(a^2+b^2)$
for any positive integers $a, b$ satisfying $a + b = n$.
|
Let $n = 2^{p-1}t$, where $t \equiv 5 \pmod 6$, $\omega(t) = k-1$,
and $p \gg t$ is a sufficiently large prime.
Let $a+b=n$ and $a^2+b^2=c$.
We claim that $p \nmid d(c)$,
which solves the problem since $p \mid d(n)$.
First, note that $3 \nmid a^2+b^2$, since $3 \nmid n$.
Next, note that $c < 2n^2 < 5^{p-1}$ (since $p \gg t$)
so no exponent of an odd prime in $c$ exceeds $p-2$.
Moreover, $c < 2^{3p-1}$.
So, it remains to check that $\nu_2(c) \notin \{p-1, 2p-1\}$.
On the one hand, if $\nu_2(a) < \nu_2(b)$,
then $\nu_2(a) = p-1$ and $\nu_2(c) = 2\nu_2(a) = 2p-2$.
On the other hand, if $\nu_2(a) = \nu_2(b)$
then $\nu_2(a) \le p-2$, and $\nu_2(c) = 2\nu_2(a)+1$
is odd and less than $2p-1$.
\begin{remark*}
Personally, I find the condition to be artificial,
but the construction is kind of fun.
I also think the scores on this problem during the real contest
are low mostly because of the difficulty of problem 2.
\end{remark*}
|
EGMO-2014-notes_4
|
Determine all positive integers $n\geq 2$ for which there exist integers
$x_1$, $x_2$, \dots, $x_{n-1}$ satisfying the condition that
if $0<i<n$, $0<j<n$, $i\neq j$ and $n$ divides $2i+j$, then $x_i<x_j$.
|
The answer is $n = 2^k$ and $n = 3 \cdot 2^k$, for each $k \ge 0$ (excluding $n=1$).
We work with the set $S = \{1, 2, \dots, n-1\} \bmod n$
of nonzero residues modulo $n$ instead.
We define the relation $\prec$ on $S$ to mean that
$2i + j \equiv 0 \pmod n$ and $i \neq j$, for $i,j \in S$.
Then the problem satisfies the conditions if and only if $\prec$ has no cycles,
i.e.\ $\prec$ imposes a partial order on $S$.
The existence of a cycle for $\prec$ is equivalent to some choice of $t_1 \in S$
and an integer $m \ge 2$ such that
\[ t_1 \prec t_2 \prec \dots \prec t_m \prec t_1. \]
Unwinding the definition, this is equivalent to two conditions:
\begin{itemize}
\ii We need $t_i \not\equiv t_{i+1} \pmod n$ for $i=1,\dots,m$ (where $t_{m+1}=t_1$).
This is equivalent to
\[ 3 \cdot 2^{i-1} \cdot t_1 \equiv 0 \pmod n \qquad (\heartsuit). \]
\ii For $t_m \prec t_1$ to be true, we need
\[ (-2)^m t_1 \equiv t_1 \pmod n \iff \left( (-2)^m-1 \right) t_1 \equiv 0 \pmod n.
\qquad (\spadesuit) \]
\end{itemize}
We now analyze three cases:
\begin{itemize}
\ii Let $n = 2^k$.
Suppose for contradiction some cycle exists.
Then $(-2)^m-1$ is coprime to $n$,
so $(\spadesuit)$ would imply $t_1 \equiv 0 \pmod n$, contradiction.
\ii Let $n = 3 \cdot 2^k$.
Suppose for contradiction some cycle exists.
If $(\spadesuit)$ holds for some $m$, then $2^k \mid t_1$,
so the only possibility is that $t_1 \equiv \pm 2^k \pmod{n}$
and $3 \mid (-2)^m-1$.
However, in that case $(\heartsuit)$ is violated for $i=1$, contradiction.
\ii Suppose $n$ had an $n$ has an odd divisor $d \mid n$ and $d \ge 5$.
Then taking $t_1 = n/d$ and $m = \varphi(d)$, the equation $(\spadesuit)$ is true.
Moreover, $(\heartsuit)$ is true because there is at least one odd prime $p$
with $\nu_p(n) > \nu_p(3t_1) = \nu_p(3n/d)$ (since $d \ge 5$ is odd).
So indeed it's possible to construct a cycle.
\end{itemize}
Thus these are all the answers and the only answers.
|
EGMO-2014-notes_5
|
Let $n$ be a positive integer.
We have $n$ boxes where each box contains a non-negative number of pebbles.
In each move we are allowed to take two pebbles from a box we choose,
throw away one of the pebbles and put the other pebble in another box we choose.
An initial configuration of pebbles is called \emph{solvable} if it is possible
to reach a configuration with no empty box, in a finite (possibly zero) number of moves.
Determine all initial configurations of pebbles which are not solvable,
but become solvable when an additional pebble is added to a box,
no matter which box is chosen.
|
The point of the problem is to characterize all the solvable configurations.
We claim that it is given by the following:
\begin{claim*}
A configuration $(a_1, \dots, a_n)$ is solvable if and only if
\[ \sum_{i=1}^n \left\lceil \frac{a_i}{2} \right\rceil \ge n. \]
\end{claim*}
\begin{proof}
The proof is by induction on the number of stones.
If there are fewer than $n$ stones there is nothing to prove.
Now assume there are at least $n$ stones,
and let $S = \sum \left\lceil a_i/2 \right\rceil$.
Then:
\begin{itemize}
\ii If $S < n$, this remains true after any operation,
so by induction the configuration is not solvable.
\ii Suppose $S \ge n$, and also that there is an empty box (else we are already done).
Then there must be some box with at least two stones.
In that case, using those two stones to fill the empty box
does not change the value of $S$,
but decreases the total number of stones by one, as desired.
\qedhere
\end{itemize}
\end{proof}
From here we may then extract the answer to the original problem:
we require all $a_i$ to be even and $\sum a_i = 2n-2$.
\begin{remark*}
It should be unsurprising that a criteria of this form exists,
since (1) intuitively, one loses nothing by filling empty boxes as soon as possible,
and then ignoring boxes with one pebble in them,
(2) the set of configurations is a graded partially ordered set,
so one can inductively look at small cases.
\end{remark*}
|
EGMO-2014-notes_6
|
Solve over $\RR$ the functional equation
\[ f(y^2+2xf(y)+f(x)^2) = (y+f(x)) (x+f(y)). \]
\end{enumerate}
|
A key motivation throughout the problem
is that the left-hand side is asymmetric
while the right-hand side is symmetric.
Thus any time we plug in two values for $x$ and $y$
we will also plug in the opposite pair.
Once $f$ is injective this will basically kill the problem.
First, we prove the following.
\begin{lemma*}
There is a unique $z \in \RR$ such that $f(z) = 0$.
\end{lemma*}
\begin{proof}
Clearly by putting $y=-f(x)$ such $z$ exists.
Now, suppose $f(u) = f(v) = 0$.
Then:
\begin{itemize}
\ii Plug $(x,y) = (u,v)$ gives $f(v^2) = uv$.
\ii Plug $(x,y) = (v,u)$ gives $f(u^2) = uv$.
\ii Plug $(x,y) = (u,u)$ gives $f(u^2) = u^2$.
\ii Plug $(x,y) = (v,v)$ gives $f(v^2) = v^2$.
\end{itemize}
Consequently $u^2 = uv = v^2$ which yields $u = v$.
\end{proof}
Next let $(x,y) = (z,0)$ and $(x,y) = (0,z)$ to get
\begin{align*}
& f\left( 2zf(0) \right)
= f\left( z^2 + f(0)^2 \right) = 0 \\
\implies & 2z f(0) = z^2 + f(0)^2 = z \\
\implies &\boxed{f(0) = z \in \left\{ 0, \half \right\}}.
\end{align*}
We now set to prove:
\begin{lemma*}
The function $f$ is injective.
\end{lemma*}
\begin{proof}
By putting $(x,y)=(x,z)$ and $(x,y)=(z,x)$ we get
\[ f\left( f(x)^2 + z^2 \right) = f\left( 2zf(x) + x^2 \right) = x(z+f(x)). \]
Now suppose $f(x_1) = f(x_2)$ but $x_1 \neq x_2$.
This can only happen if $f(x_1) = f(x_2) = -z$.
And now \[ f(x_i)^2 + z^2 = 2zf(x_i) + x_i^2 = z \qquad i = 1, 2. \]
Solving, we have $x_i = \pm 1$, $z = \half$,
(since $z = 0$ is not permissible).
Thus we have ``almost injectivity''.
Now plug in $(x,y) = (-1,0)$ and $(x,y) = (0,-1)$ in
the original and equate in order to obtain
$f(-\frac34) = f(\frac54)$, which contradicts the work above.
\end{proof}
Finally we may use the symmetry trick in full to obtain
\[ y^2 + 2xf(y) + f(x)^2 = x^2 + 2yf(x) + f(y)^2. \qquad(\heartsuit) \]
In particular, by setting $y=0$ we obtain
\[ f(x)^2 = \left( z - x \right)^2. \]
Two easy cases remain:
\begin{itemize}
\ii In the $z=0$ case simply note
that $(\heartsuit)$ gives
$2xf(y) = 2yf(x)$, so for $x \neq 0$
the value $f(x)/x$ is constant
and hence $f(x) \equiv \pm x$ follows.
\ii In the $z=\half$ case $(\heartsuit)$ becomes
$ \left( 2f(y)+1 \right)x = \left( 2f(x)+1 \right)y $
and hence we're done again by the same reasoning.
\end{itemize}
|
EGMO-2015-notes_1
|
Let $\triangle ABC$ be an acute-angled triangle,
and let $D$ be the foot of the altitude from $C$.
The angle bisector of $\angle ABC$ intersects $CD$ at $E$ and
meets the circumcircle $\omega$ of $\triangle ADE$ again at $F$.
If $\angle ADF = 45^{\circ}$, show that $CF$ is tangent to $\omega$.
|
\begin{center}
\begin{asy}
pair A = dir(0);
pair B = dir(180);
pair F = dir(65);
pair D = extension(A, B, F, dir(205));
pair K = F*F;
pair C = extension(D, D+dir(90), B, K);
pair E = extension(B, F, C, D);
filldraw(unitcircle, palecyan+opacity(0.2), heavyblue);
filldraw(circumcircle(A, D, E), palered+opacity(0.2), red+dotted);
filldraw(CP(E,D), pink+opacity(0.2), magenta);
draw(arc(F,abs(E-F),170,310), orange+dashed);
draw(K--A--B--C--D, deepcyan);
draw(C--A, deepcyan+dashed);
draw(B--F--A, deepgreen);
draw(F--D, deepgreen);
draw(C--F, mediumred);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$F$", F, dir(F));
dot("$D$", D, dir(225));
dot("$K$", K, dir(K));
dot("$C$", C, dir(C));
dot("$E$", E, dir(165));
\end{asy}
\end{center}
Let $BC$ meet the circle with diameter $\ol{AB}$ at $K$.
By the conditions of the problem, we have $FK = FE = FA$.
Thus $E$ is the incenter of $\triangle KBA$.
By angle chasing, we can now show that
\[ \angle KFE = 90^{\circ} - \frac12 \angle KEF = \angle BCD, \]
so $KCFE$ is cyclic and thus
\[ \angle CKE = 135^{\circ} \implies \angle CFE = 45^{\circ} \]
as needed.
\begin{remark*}
One can also realize $CKEF$ is cyclic by noting
\[ BE \cdot BF = BD \cdot BA = BK \cdot BC. \]
\end{remark*}
\begin{remark*}
Another approach is to use barycentric coordinates on $\triangle ABK$.
Letting $a=BK$, $b=AK$, $c=AB$ we have $E = (a:b:c)$, $D = (s-b:s-a:0)$, and
\[ F = (a(a+c) : -b^2 : c(a+c))
= (a(a+c) : a^2-c^2 : c(a+c))
= (a : a-c : c). \]
\end{remark*}
|
EGMO-2015-notes_3
|
Let $n, m$ be integers greater than $1$,
and let $a_1, a_2, \dots, a_m$ be positive integers
not greater than $n^m$.
Prove that there exist integers $b_1, b_2, \dots, b_m$ not greater than $n$ such that
\[ \gcd(a_1 + b_1, a_2 + b_2, \dots, a_m + b_m) < n. \]
|
In fact, we will prove that it's possible to choose $b_i \in \{0,1\}$!
Assume not, and all GCD's are at least $n$.
Consider the choices:
\begin{itemize}
\ii $b_1 = \dots = b_m = 0$.
\ii $b_1 = b_2 = \dots = b_{k-1} = b_{k+1} = b_m = 0$ and $b_k = 1$,
for some $2 \le k \le m$.
\end{itemize}
This generates $m$ gcd's, say $g_1$, \dots, $g_m$.
Each divides $a_1$.
Moreover, they are pairwise coprime, s
\[ a_1 \ge \prod_i g_i = n(n+1) \dots + (n+m-1) > n^m \]
which is impossible.
|
EGMO-2015-notes_4
|
Determine whether or not there exists an infinite sequence
$a_1$, $a_2$, \dots\ of positive integers satisfying
\[ a_{n+2} = a_{n+1} + \sqrt{a_{n+1}+a_{n}} \]
for every positive integer $n$.
|
In fact, we will show the following stronger result:
the largest $N$ for which one can find $(a_1, \dots, a_N)$ satisfying
(for all $1 \le n \le N-2$) is actually $N = 5$.
This largest $N$ is obtained
for example by $(a_1,a_2,a_3,a_4,a_5) = (477,7,29,35,43)$.
\begin{remark*}
Basically, the idea is to choose $a_3$ first; then as long as the number
\[ s \coloneq \sqrt{a_3 + a_4} = \sqrt{2a_3 + \sqrt{a_2 + a_3}} \]
is as integer, the rest of the sequence can be chosen to have integer values.
Unfortunately, $a_1$ may turn out to be negative in this situation.
But if one experiments with numbers, it should be possible to ensure $a_2 < a_3$,
and after this no further obstructions arise.
For example, if one makes the arbitrary starting choice $s = 10$,
then choosing $a_3 = 46$ gives the nice choice $a_2 = 18$, thus $a_1 = 766$.
(We picked this number so that $2a_3 + \sqrt{a_3}$ was just under $100$.)
Meanwhile, moving forward, $a_4 = 46 + \sqrt{46+18} = 54$
and $a_5 = 54 + \sqrt{46+54} = 64$.
Hence $(a_1,a_2,a_3,a_4,a_5) = (766, 18, 46, 54, 64)$ is another example.
\end{remark*}
Let
\[ x_n \coloneq a_{n+1}-a_n = \sqrt{a_{n}+a_{n-1}}. \]
We will rewrite everything in terms of the $(x_n)$.
Since the $(a_n)$ are strictly increasing for $n \ge 2$
so are the $(x_n)$ when $n \ge 3$.
For $n \ge 2$ observe that
\[ x_{n+1}^2 - x_n^2 = a_{n+1} - a_{n-1}
= \left( a_{n+1}-a_{n} \right) + \left( a_{n}-a_{n-1} \right) = x_n + x_{n-1} \]
thus
\[ x_{n+1}-x_n = \frac{x_n+x_{n-1}}{x_{n+1}+x_n}. \]
But suppose $n \ge 4$.
Then since the $x_n$ are supposed to be strictly increasing, the right-hand side is $< 1$.
Yet $x_{n+1} - x_n \ge 1$ as well, which is a contradiction.
Thus it is impossible to have six terms in the sequence.
|
EGMO-2015-notes_5
|
Let $m, n$ be positive integers with $m > 1$.
Anastasia partitions the integers $1, 2, \dots , 2m$ into $m$ pairs.
Boris then chooses one integer from each pair
and finds the sum of these chosen integers.
Prove that Anastasia can select the pairs
so that Boris cannot make his sum equal to $n$.
|
Overall the idea is to try find a few constructions which eliminate most of the cases,
then clean out the last few ones leftover.
% (It's like using reavers on zerglings:
% knock out 90\% of the lings with a couple scarabs
% and then have the zealots finish the rest.)
\begin{itemize}
\ii If $n \notin [m^2, m^2+m]$, then use the construction
\[
\begin{array}{ccccc}
1 & 3 & \dots & 2m-3 & 2m-1 \\
2 & 4 & \dots & 2m-2 & 2m
\end{array}
\]
\ii If $n \not\equiv 1 + 2 + \dots + m = \half m(m+1) \pmod m$, use the construction
\[
\begin{array}{ccccc}
1 & 2 & \dots & m-1 & m \\
m+1 & m+2 & \dots & 2m-1 & 2m
\end{array}
\]
\end{itemize}
Henceforth, assume $m \ge 4$ (smaller cases can be dispensed with by hand).
\begin{itemize}
\ii Assume $m$ is odd, and either $n = m^2$ and $n = m^2 + m$.
Use the construction
\[
\begin{array}{ccccc}
1 & 2 & \dots & m-1 & m \\
m+2 & m+3 & \dots & 2m & m+1
\end{array}.
\]
The possible values of this modulo $m+1$ are
\[ \tfrac 12 m(m+1) + \{0,1\} \equiv \tfrac12(m+1) + \{0,1\} \pmod{m+1} \]
since $m$ is odd.
But $m^2$ and $m^2+m$ leave residues $1$ and $2$ modulo $m$, done.
\ii Assume $m$ is even (so $m+1$ is odd), and $n = m^2 + \half m \equiv \half \pmod{m+1}$.
The same pairing as before has possible residues
\[ \tfrac 12m (m+1) + \{0,1\} \equiv \{0,1\} \pmod{m+1}. \]
This completes the proof.
\end{itemize}
|
EGMO-2015-notes_6
|
Let $H$ be the orthocenter and $G$ be the centroid of
acute-angled triangle $ABC$ with $AB \neq AC$.
The line $AG$ intersects the circumcircle
of $ABC$ at $A$ and $P$.
Let $P'$ be the reflection of $P$ in the line $BC$.
Prove that $\angle CAB = 60$ if and only if $HG = GP'$.
\end{enumerate}
|
The following complex numbers solution was given by Stefan Tudose.
First
\[ \frac{pa(b+c)-bc(p+a)}{pa-bc}
= \frac{b+c}{2} \implies p
= -\frac{2bc-ab-ac}{bc(2a-b-c)}. \]
Then
\[ p' = b+c-bc\ol p = \frac{ab+ac-b^2-c^2}{2a-b-c}. \]
Now let $D$ be the midpoint of $\ol{HP'}$.
Then
\begin{align*}
d = \frac{h+p'}{2} & = \frac{a^2-b^2-c^2+ab+ac-bc}{2a-b-c} \\
h-p' &= \frac{2(a^2-bc)}{2a-b-c} \\
g-d &= \frac{2b^2+2c^2-a^2+bc-2ab-2ac}{3(2a-b-c)}
\end{align*}
Then $HG = GP' \iff \ol{GD} \perp \ol{HP'}$ and we have
\[
\frac{g-d}{h-p'}
= \frac{2b^2+2c^2-a^2+bc-2ab-2ac}{6(a^2-bc)}
\]
which we would like to be pure imaginary.
However the negative conjugate equals
\[
\ol{\left( \frac{g-d}{h-p'} \right)}
= -\frac{2a^2c^2+2a^2b^2-b^2c^2+a^2bc-2abc^2-2ab^2c}{6bc(bc-a^2)}.
\]
Expanding, the condition we have becomes
\[ b^3c+bc^3+b^2c^2=a^2bc+a^2c^2+a^2b^2
\iff (b^2+bc+c^2)(a^2-bc) = 0. \]
Now $b^2+bc+c^2=0 \iff \angle A = 60\dg$ as desired.
\begin{remark*}
One can also proceed by $|g-h| = |g-p'|$
which is longer but ultimately the same calculation.
\end{remark*}
\begin{remark*}
The condition $AB \neq AC$ is not cosmetic;
it cannot be dropped from the problem condition.
This is reflected in the presence of $a^2-bc$ factor.
\end{remark*}
|
EGMO-2016-notes_1
|
Let $n$ be an odd positive integer,
and let $x_1$, $x_2$, \dots, $x_n$ be nonnegative real numbers.
Show that
\[ \min(x_i^2+x_{i+1}^2) \le \max(2x_jx_{j+1}) \]
where $1\le i,j\le n$ and $x_{n+1}=x_1$.
|
It suffices to exhibit a single pair $(i,j)$ where
\[ x_i^2 + x_{i+1}^2 \le 2 x_j x_{j+1}. \]
Since $n$ is odd we can find three adjacent $x_k$'s,
call them $a$, $b$, and $c$ (with $b$ in middle) such that $a \ge b \ge c$.
Then indeed
\[ 2ab \ge b^2+c^2 \]
so we're done.
\begin{remark*}
When $n$ is even, counterexamples to the problem statement alternate up and down,
such as $(1,100,1,100,1,100,\dots)$.
This provides a reason why one should consider $a \ge b \ge c$ to exploit
the parity of $n$.
\end{remark*}
|
EGMO-2016-notes_2
|
Let $ABCD$ be a cyclic quadrilateral,
and let diagonals $AC$ and $BD$ intersect at $X$.
Let $C_1,D_1$ and $M$ be the midpoints of segments $CX$, $DX$ and $CD$, respectively.
Lines $AD_1$ and $BC_1$ intersect at $Y$, and line $MY$ intersects diagonals $AC$ and $BD$
at different points $E$ and $F$, respectively.
Prove that line $XY$ is tangent to the circle through $E$, $F$ and $X$.
|
We present two approaches.
\paragraph{First approach through isogonality lemma.}
Note $ABC_1D_1$ is cyclic.
By the ``isogonality lemma'' applied to $\triangle YC_1D_1$,
it follows that $YX$ and $YM$ are isogonal
with respect to $\triangle YC_1D_1$.
Then
\[ \angle EXY = \angle XC_1Y + \angle C_1YX
= \angle AD_1X + \angle MYA = \angle YFX \]
and we're done (angles are directed).
\begin{remark*}
Note that the points $C$ and $D$ can be more or less deleted immediately.
The ``isogonality lemma'' people has mentioned has appeared in other places too.
In addition to the Taiwan TST 2014 round 2 problem 6, it also shows itself in
ELMO 2012/5, % http://aops.com/community/c6h486936p2728469
BrMO2 2013/2, % http://www.aops.com/community/c6h518987
SL 2009 G4, SL 2012 G2.
\end{remark*}
\paragraph{Solution with complex numbers, by Sanjana Das.}
Quadrilateral $ABC_1D_1$ is also cyclic ($\dang AC_1D_1 = \dang ACD = \dang ABD_1$).
We want $\dang YXE = \dang XFE$,
or \[\dang(XY, AC) = \dang(BD, MY).\]
Now use complex numbers with unit circle $(ABC_1D_1)$
(using $c$ and $d$ to refer to $C_1$ and $D_1$ for berevity).
We want to check that $\frac{(x - y)(m - y)}{(a - c)(b - d)}$ is real.
First, if $Z = AB \cap C_1D_1$, then $XY \perp OZ$ where $O$ is the
circumcenter of $(ABC_1D_1)$ by Brocard's Theorem, so we want to show
\[ \frac{(ab(c + d) - cd(a + b))\cdot (y - m)}
{(a - c)(b - d)(ab - cd)} \in i\RR.\]
Now we have $m = c + d - x$, which means
\begin{align*}
y - m &= x + y - c - d \\
&= \frac{ac(b +d) - bd(a +c)}{ac - bd} + \frac{ad(b + c) - bc(a+ d)}{ad - bc} - c - d \\
&= \frac{ab(2acd + 2bcd + c^3 + d^3 - ac^2 - ad^2 - bc^2 - bd^2 - c^2d - cd^2)}{(ac - bd)(ad - bc)} \\
&= -\frac{ab(a + b - c - d)(c - d)^2}{(ac - bd)(ad - bc)}.
\end{align*}
This means our final expression is
\[-\frac{ab(ab(c + d) - cd(a + b))(a + b - c - d)(c - d)^2}{(ac - bd)(ad - bc)(a - c)(b - d)(ab - cd)},\]
which is visibly the negative of its conjugate and is therefore imaginary.
|
EGMO-2016-notes_3
|
Let $m$ be a positive integer. Consider a $4m \times 4m$ array of square unit cells.
Two different cells are \emph{related} to each other
if they are in either the same row or in the same column.
No cell is related to itself.
Some cells are colored blue, such that every cell is
related to at least two blue cells.
Determine the minimum number of blue cells.
|
We'll prove that the answer is at least $6m$.
The construction is to take
diagonal copies of the matrix
\[ \begin{bmatrix}
& 1 & 1 & 1 \\
1 &&& \\
1 &&& \\
1 &&&
\end{bmatrix}. \]
We present two proofs this is optimal.
\paragraph{Bipartite graph (Kevin Sun).}
Consider the bipartite subgraph
\[ H \subseteq K_{4m,4m} \]
with vertices being rows/columns,
and edges corresponding blue squares.
Assume for contradiction the are fewer than $6m$ edges.
Then there are at least $|V(H)|-|E(H)|=8m-6m=2m$
connected components in $H$,
so some connected component has at most three vertices.
We consider two cases:
\begin{itemize}
\ii Note that if there are any isolated vertices $v$
(connected components with no edges),
then it follows $H$ needs at least $8m$ edges,
because every vertex on the opposite shore from $v$ has
blue-degree at least $2$.
\ii Thus assume no isolated vertices exist.
On the other hand, every blue edge
is incident to at least two more blue edges,
so no connected component can have fewer than three edges.
As the connected components are bipartite,
it follows each connected component has at least four vertices,
which is impossible.
\end{itemize}
\paragraph{Brute force (mine).}
%I'll admit that unlike P1 and P2 (which I solved in about five minutes each),
%there is \emph{no way} I would have gotten this problem under time pressure:
%I had to stay up all night (despite a sore throat) to solve it.
%Also, this solution (linear programming after optimization)
%seems too silly to be true.
We prove that for a general $n \times n$ board,
a good configuration has at least $\frac32n$ blue squares.
Consider any minimal good configuration
with $k$ squares, and assume $k \le \frac 32 n$
(and we will prove $k \ge \frac32n$).
We observe that no column and row can be empty in a good
configuration with $k \le 2n$ squares.
Now, assume there are $x_i$ columns with exactly $i$ blue squares,
and $a_i$ rows with exactly $i$ blue squares (for $i = 1,\dots,n$.)
Permute the board to arrive at the following:
\[
\begin{array}{r|c|c|c|}
\multicolumn{1}{c}{\ }
& \multicolumn{1}{c}{x_1\text{ cols}}
& \multicolumn{1}{c}{x_2\text{ cols}}
& \multicolumn{1}{c}{x_3+x_4+\dots+x_n\text{ cols}} \\\cline{2-4}
a_1 \text{ rows} & 0 & 0 & a_1 \\ \cline{2-4}
a_2 \text{ rows} & 0 & P & Q \\ \cline{2-4}
&&& \\[1em]
a_3+\dots+a_n & x_1 & R & S \\
\text{rows} &&& \\[1em]\cline{2-4}
\end{array}
\]
In each subgrid, the letter denotes the number of blue squares in that region.
Consequently, we have $P+Q=2a_2$, $P+R=2x_2$,
and $a_1+Q+S = 3x_3+4x_4+\dots+nx_n$.
We also know that $n = \sum a_i = \sum x_i$ and $k = \sum ia_i = \sum ix_i$.
Without loss of generality we assume that $a_1 \le x_1$.
Then, it would suffice to prove that
\[ \sum ia_i \ge \frac32 \sum a_i \iff a_1 \le a_2 + 3a_3 + 5a_4 + \dots. \]
We observe that
\begin{align*}
0 &\le S = 3x_3 + \dots + nx_n - a_1 - Q \\
&= \left( a_1+2a_2+\dots+na_n \right) - (x_1+2x_2) - (a_1-Q) \\
\implies x_1 &\le (3a_3+4a_4+\dots+na_n) - (Q - 2a_2 + 2x_2) \\
&= (3a_3+4a_4+\dots+na_n) - R \\
&\le 3a_3 + 4a_4 + \dots + na_n \\
&\le a_2 + 3a_3 + 5a_4 + \dots + (2n-3)a_n.
\end{align*}
Since we assumed $a_1 \le x_1$, we are done.
% http://aops.com/community/c6t30365f6h1226675_related_blue_squares_in_4m_by_4m_board
|
EGMO-2016-notes_4
|
Two circles $\omega_1$ and $\omega_2$, of equal radius
intersect at different points $X_1$ and $X_2$.
Consider a circle $\omega$ externally tangent to $\omega_1$ at $T_1$
and internally tangent to $\omega_2$ at point $T_2$.
Prove that lines $X_1T_1$ and $X_2T_2$ intersect at a point lying on $\omega$.
|
We present two solutions, one by homothety and another by inversion.
\begin{center}
\begin{asy}
/*
Converted from GeoGebra by User:Azjps using Evan's magic cleaner
https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py
*/
size(8cm);
pen zzttqq = rgb(0.6,0.2,0.);
pen dcrutc = rgb(0.86274,0.07843,0.23529);
pen aqaqaq = rgb(0.62745,0.62745,0.62745);
pair X_1 = (0.5,0.62449), X_2 = (0.5,-0.62449), M = (0.5,0.), T_2 = (1.35807,0.71538), T_1 = (0.76813,0.22354);
draw(circle((0.,0.), 0.8), linewidth(0.6));
draw(circle((1.,0.), 0.8), linewidth(0.6));
filldraw(circle((1.17050,0.34064), 0.41906), paleyellow, brown);
draw(X_2--T_2, linewidth(0.6) + linetype("2 2") + dcrutc);
draw(X_1--(0.90858,0.01351), linewidth(0.6) + linetype("2 2") + dcrutc);
draw(X_1--X_2, linewidth(0.6) + aqaqaq);
draw(M--T_2, gray);
dot("$X_1$", X_1, dir(90));
dot("$X_2$", X_2, dir(-90));
dot("$M$", M, dir(180));
dot("$T_2$", T_2, dir((1.076, 2.708)));
dot("$T_1$", T_1, dir(170));
\end{asy}
\end{center}
\paragraph{First solution (homothety).}
Consider the composition of homotheties
\[ \omega_1 \xrightarrow{T_1} \omega \xrightarrow{T_2} \omega_2. \]
This is a negative homothety, and since $\omega_1$ and $\omega_2$ have equal radius,
it follows that the composition is just reflection
about the midpoint of $\ol{X_1X_2}$.
In particular, it sends $X_1$ to $X_2$, and thus $\ol{X_1T_1} \cap \ol{X_2T_2}$
is just the image of $X_1$ under the first homothety.
\paragraph{Second solution (inversion).}
Invert around $X_1$. Then we have the following:
\begin{itemize}
\ii $\omega_1^\ast$ and $\omega_2^\ast$ are lines intersecting at $X_2^\ast$,
and $X_1$ is a point on the external angle bisector of the two lines.
\ii The circle $\omega^\ast$ becomes tangent to the two lines
(in a region adjacent to $X_1$).
\ii Let $P = \ol{T_1^\ast X_1} \cap \omega^\ast$.
Then one wishes to show $X_1 X_2^\ast P T_2^\ast$ are concyclic.
\end{itemize}
To show this, notice that
\[ \dang T_2^\ast X_2^\ast X_1 = \dang X_2^\ast T_2^\ast T_1^\ast
= \dang T_2^\ast P T_1^\ast = \dang T_2^\ast P X_1. \]
|
EGMO-2016-notes_6
|
Let $S$ be the set of all positive integers $n$ such that $n^4$ has a divisor
in the range $n^2 + 1$, $n^2+2$, \dots, $n^2 + 2n$.
Prove that there are infinitely many elements of $S$ of
each of the forms $7m$, $7m+1$, $7m+2$, $7m+5$, $7m+6$
and no elements of $S$ of the form $7m+3$ and $7m+4$,
where $m$ is an integer.
\end{enumerate}
|
Start from the implication
\[ n^2 + k \mid n^4 \iff n^2 + k \mid k^2. \]
Since $1 \le k \le 2n$, in fact the quotient $\frac{k^2}{n^2+k}$
can only take values from $1$ to $3$.
In other words, $S$ is the set of integers $n$ for which at least one equation
\begin{align*}
n^2 + k &= k^2 \\
2(n^2 + k) &= k^2 \\
3(n^2 + k) &= k^2
\end{align*}
has at least one solution $1 \le k \le 2n$.
The first equation has no solutions with $k \ge 1$
since we can put $(k-1)^2 < n^2 < k^2$.
The other two are Pell equations,
and one can check that $n^2 \equiv 2 \pmod 7$ has no solutions at all
for $k \pmod 7$ in either case.
The assertion about infinitely many solutions then follows by
using the Pell recursion, and taking modulo $7$.
|
EGMO-2017-notes_1
|
Let $ABCD$ be a convex quadrilateral
with $\angle DAB = \angle BCD = 90\dg$ and $\angle ABC > \angle CDA$.
Let $Q$ and $R$ be points on segments $BC$ and $CD$,
respectively, such that line $QR$ intersects lines $AB$ and $AD$
at points $P$ and $S$, respectively.
It is given that $PQ=RS$.
Let the midpoint of $BD$ be $M$ and the midpoint of $QR$ be $N$.
Prove that the points $M$, $N$, $A$ and $C$ lie on a circle.
|
The condition is equivalent to $N$ being the midpoint
of both $\ol{PS}$ and $\ol{QR}$ simultaneously.
(Thus triangles $BAD$ and $BCD$ play morally dual roles.)
\begin{center}
\begin{asy}
import graph; size(12cm);
pen zzttqq = rgb(0.6,0.2,0.); pen cqcqcq = rgb(0.75294,0.75294,0.75294);
draw((-1.83576,5.20298)--(-4.,1.5)--(-0.44779,-2.69232)--(4.5,1.5)--cycle, linewidth(0.6) + zzttqq);
draw(circle((0.25,1.5), 4.25), linewidth(0.6));
draw((-5.40339,-0.90118)--(6.45435,0.35776), linewidth(0.6));
draw((-1.83576,5.20298)--(-4.,1.5), linewidth(0.6) + zzttqq);
draw((-4.,1.5)--(-0.44779,-2.69232), linewidth(0.6) + zzttqq);
draw((-0.44779,-2.69232)--(4.5,1.5), linewidth(0.6) + zzttqq);
draw((4.5,1.5)--(-1.83576,5.20298), linewidth(0.6) + zzttqq);
draw((-4.,1.5)--(-5.40339,-0.90118), linewidth(0.6));
draw((4.5,1.5)--(6.45435,0.35776), linewidth(0.6));
dot("$D$", (4.5,1.5), dir(40));
dot("$B$", (-4.,1.5), dir(180));
dot("$M$", (0.25,1.5), dir(90));
dot("$A$", (-1.83576,5.20298), dir(130));
dot("$C$", (-0.44779,-2.69232), dir(270));
dot("$Q$", (-2.24919,-0.56630), dir(80));
dot("$R$", (2.67884,-0.04308), dir(135));
dot("$P$", (-5.40339,-0.90118), dir(270));
dot("$S$", (6.45435,0.35776), dir(270));
dot("$N$", (0.21482,-0.30469), dir(80));
\end{asy}
\end{center}
The rest is angle chasing.
We have
\begin{align*}
\dang ANC &= \dang ANP + \dang QNC \\
&= 2\dang ASP + 2\dang QRC \\
&= 2 \dang DSR + 2 \dang DRS = 2 \dang RDS \\
&= 2 \dang ADC = \dang AMC.
\end{align*}
|
EGMO-2017-notes_2
|
Find the smallest positive integer $k$ for which
there exists a coloring of the positive integers
$\ZZ_{>0}$ with $k$ colors and a function $f \colon \ZZ_{>0} \to \ZZ_{>0}$
with the following two properties:
\begin{enumerate}[(i)]
\ii For all positive integers $m$, $n$ of the same color,
$f(m+n)=f(m)+f(n)$.
\ii There are positive integers $m$, $n$
such that $f(m+n) \neq f(m)+f(n)$.
\end{enumerate}
|
Answer: $k=3$.
Construction for $k=3$:
let
\[ f(n) =
\begin{cases} n/3 & n\equiv 0 \pmod 3 \\
n & \text{else} \end{cases} \]
and color the integers modulo $3$.
Now we prove that for $k=2$ a function $f$ obeying (i) must be linear,
even if $f \colon \ZZ_{>0} \to \RR_{>0}$.
Call the colors blue/red and WLOG $f(1) = 1$.
First, we obviously have:
\begin{claim*}
$f(2n) = 2f(n)$ for every $n$.
\end{claim*}
Now we proceed by induction in the following way.
Assume that $f(1) = 1$, $f(2) = 2$, \dots, $f(2n) = 2n$.
For brevity let $m = 2n+1$ be red
and assume for contradiction that $f(m) \neq m$.
The proof now proceeds in four steps.
First:
\begin{itemize}
\ii The number $m-2$ must be blue.
Indeed if $m-2$ was red we would have $f(2m-2) = f(m) + f(m-2)$
which is a contradiction as $f(2m-2) = 2f(m-1) = 2m-2$ and $f(m-2)=m-2$.
\ii The number $2$ must be red.
Indeed if it was blue then $f(m) = f(2) + f(m-2) = m$ contradiction.
\end{itemize}
Observe then that $f(m+2) = f(m)+2$ since $m$ and $2$ are both red.
Now we consider two cases:
\begin{itemize}
\ii If $m+2$ is red, then
$f(2m+2) = f(m+2) + f(m) = (f(m) + 2) + f(m)$.
But $f(2m+2) = f(4n+4) = 4f(n+1) = 2m+2$, contradiction.
\ii If $m+2$ is blue,
then $2f(m) = f(2m) = f(m+2) + f(m-2) = f(m) + 2 + (m-2)$.
So then $f(m) = m$ again a contradiction.
\end{itemize}
So $f(m) = m$, which completes the induction.
|
EGMO-2017-notes_3
|
There are $2017$ lines in the plane such that
no three of them go through the same point.
Turbo the snail sits on a point on exactly one of the lines
and starts sliding along the lines in the following fashion:
she moves on a given line until she reaches an intersection of two lines.
At the intersection, she follows her journey
on the other line turning left or right,
alternating her choice at each intersection point she reaches.
She can only change direction at an intersection point.
Can there exist a line segment through which she passes
in both directions during her journey?
|
\textbf{Color the regions} of the plane black and white in alternating colors.
Then:
\begin{claim*}
Turbo will always move in one orientation around black regions
and the other orientation around white regions.
\end{claim*}
This completes the proof, even if Turbo may pick
which of left/right she follows each time!
\begin{remark*}
An example of a cyclic path:
take a pentagon and extend its sides to form a star.
Then Turbo can trace around the exterior of the star.
\end{remark*}
|
EGMO-2017-notes_4
|
Let $n\geq1$ be an integer and let $t_1<t_2<\dots<t_n$ be positive integers.
In a group of $t_n+1$ people, some games of chess are played.
Two people can play each other at most once.
Prove that it is possible for the following two conditions to hold at the same time:
\begin{enumerate}[(i)]
\ii The number of games played by each person is one of $t_1,t_2,\dots,t_n$.
\ii For every $i$ with $1\leq i\leq n$,
there is someone who has played exactly $t_i$ games of chess.
\end{enumerate}
|
Phrased in graph theory,
the problem asks to produce a simple graph $G$ on $t_n+1$ vertices
such that all degrees are in the set $\{t_1, \dots, t_n\}$
and each degree appears at least once.
We proceed by induction on $n$.
If $n = 1$, take a clique on $t_1+1$ vertices.
For $n = 2$, take a clique on $t_1$ vertices and
an empty graph on $t_2 + 1 - t_1$ vertices, and join them all together.
For the inductive step:
\begin{itemize}
\ii Take an example for the $(n-2)$ tuple $(t_2-t_1, \dots, t_{n-1}-t_1)$,
which has $t_{n-1} - t_1 + 1$ vertices.
\ii Then add in $t_n - t_{n-1}$ isolated vertices.
\ii Finally add in $t_1$ universal vertices.
\end{itemize}
\begin{remark*}
The universal vertices are ``forced'',
so the only parameter is the number of universal vertices to add in.
\end{remark*}
|
EGMO-2017-notes_5
|
An $n$-tuple $(a_1,a_2,\dots,a_n)$ of positive integers is \emph{expensive} if
\[ (a_1+a_2)(a_2+a_3)\dots(a_{n-1}+a_n)(a_n+a_1) = 2^{2k-1} \]
for some positive integer $k$.
\begin{enumerate}
\ii[(a)] Find all integers $n \ge 2$ for which there exists an expensive $n$-tuple.
\ii[(b)] Prove that each odd integer $m \ge 1$
appears in an expensive $n$-tuple for some $n \ge 2$.
\end{enumerate}
|
For part (a), the answer is odd $n$,
with construction given by setting $a_1 = \dots = a_n = 1$.
We show by induction that even $n$ all fail,
with $n=2$ being plain.
Construct a regular $n$-gon whose vertices are labeled $a_i$
and edges are labeled with $a_i + a_j$.
\begin{center}
\begin{asy}
size(5cm);
int[] x = {1,3,13,3,1,7,9,7,1,1}; // repeat last guy
for (int i=0; i<9; ++i) {
dot( (string) x[i], dir(40*i), dir(40*i), darkcyan);
draw(dir(40*i)--dir(40*i+40), darkcyan);
label( (string) (x[i] + x[i+1]),
midpoint(dir(40*i)--dir(40*i+40)), dir(40*i+20), red);
}
\end{asy}
\end{center}
Then we observe that it's impossible for an edge to exceed
both of its neighbors,
since if $a+b = 2^e$ then $\min(a+1, b+1) > 2^{e-1}$.
Consequently we can take two adjacent edges $a+b = b+c$
with the same label; hence $a = c$.
Then delete $b$, $c$ and induct down.
For part (b), for odd integers $m$, let $f(m)$ denote $2^k - m$
where $2^k$ is the smallest power of two exceeding $m$.
Note $f(m) \le m$ with equality if and only if $m = 1$.
So repeatedly apply $f$ until we wrap around;
this gives twice a square of a power of two.
Example when $m = 13$:
\begin{center}
\begin{asy}
size(4cm);
int[] x = {13,3,1,1,3,13}; // repeat last guy
for (int i=0; i<5; ++i) {
dot( (string) x[i], dir(72*i), dir(72*i), darkcyan);
draw(dir(72*i)--dir(72*i+72), darkcyan);
label( (string) (x[i] + x[i+1]),
midpoint(dir(72*i)--dir(72*i+72)), dir(72*i+36), red);
}
\end{asy}
\end{center}
\begin{remark*}
By applying the process of (a) in reverse,
one essentially finds an inductive characterization
of all expensive $n$-tuples.
\end{remark*}
|
EGMO-2017-notes_6
|
Let $ABC$ be an acute scalene triangle.
The reflections of the centroid $G$ and the circumcenter $O$ of $ABC$ in its sides $BC$, $CA$, $AB$
are denoted by $G_1$, $G_2$, $G_3$ and $O_1$, $O_2$, $O_3$, respectively.
Show that the circumcircles of triangles $G_1G_2C$, $G_1G_3B$, $G_2G_3A$,
$O_1O_2C$, $O_1O_3B$, $O_2O_3A$ and $ABC$ have a common point.
\end{enumerate}
|
Here is an approach with complex numbers.
Let $P$ be an arbitrary point.
Let $P_B$ and $P_C$ be the reflections of $P$ across $AB$ and $AC$
and let $Q_B$ and $Q_C$ be the second intersections of
lines $AP_B$ and $AP_C$ with the circumcircle.
Then we will compute the intersection of
$(A P_B P_C)$ and $(A Q_B Q_C) \equiv (ABC)$.
We have $p_B = a+c-ac\ol p$, $p_C = a+b-ab\ol p$.
To compute $q_B$ note that
\begin{align*}
a+q_B &= p_B + a q_B \ol{p_B} \\
&= a+c-ac\ol p + aq_B \left( 1/a + 1/c - p/ac \right) \\
\implies ac\ol p - c &= (a/c-p/c) q_B \\
\implies q_B &= c^2 \frac{a\ol p -1}{a-p}.
\end{align*}
Similarly $q_C = b^2 \frac{a \ol p - 1}{a-p}$.
Then, the desired intersection is
\begin{align*}
\frac{p_B q_C - p_C q_B}{p_B - p_C + q_C - q_B}
&= \frac{ \left( \frac{a \ol p -1}{a-p} \right)
\left( b^2(a+c-ac \ol p) - c^2 (a+b-ab\ol p) \right)}%
{(b-c)(a \ol p-1) + (b^2-c^2) \cdot \frac{a \ol p - 1}{a-p}} \\
&= \frac{b^2(a+c-ac \ol p) - c^2(a+b-ab\ol p)}%
{(b-c)(a-p) + (b^2-c^2)} \\
&= \frac{(b-c)(a(b+c)+bc) - (b-c) abc \ol p}{(a-p)+(b+c)} \\
&= \frac{ab+bc+ca - abc \ol p}{a+b+c-p}
\end{align*}
which is in any case symmetric in $a$, $b$, $c$.
Moreover taking $p = \frac13(a+b+c)$ and $p=0$
give the same numbers (and indeed any $p$ on the Euler line).
|
EGMO-2018-notes_1
|
Let $ABC$ be a triangle with $CA=CB$ and $\angle{ACB}=120^{\circ}$,
and let $M$ be the midpoint of $AB$.
Let $P$ be a variable point of the circumcircle of $ABC$,
and let $Q$ be the point on the segment $CP$ such that $QP = 2QC$.
It is given that the line through $P$ and perpendicular to $AB$
intersects the line $MQ$ at a unique point $N$.
Prove that there exists a fixed circle such that $N$
lies on this circle for all possible positions of $P$.
|
Since $\vec N = 3\vec Q - 2\vec M$,
it suffices to show $Q$ moves on a fixed circle.
That fixed circle is the image of $(CAB)$ under a homothety at $C$ with ratio $1/3$,
so we are done.
\begin{center}
\begin{asy}
pair C = dir(90);
pair A = dir(150);
pair B = dir(30);
pair M = midpoint(A--B);
pair P = dir(-40);
pair Q = (2*C+P)/3;
pair N = extension(M, Q, P, P+C);
draw(unitcircle);
draw(A--B--C--cycle, dotted);
draw(C--P--N--M--cycle);
dot("$C$", C, dir(C));
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$M$", M, dir(270));
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(225));
dot("$N$", N, dir(N));
/* TSQ Source:
C = dir 90
A = dir 150
B = dir 30
M = midpoint A--B R270
P = dir -40
Q = (2*C+P)/3 R225
N = extension M Q P P+C
unitcircle
A--B--C--cycle dotted
C--P--N--M--cycle
*/
\end{asy}
\end{center}
\begin{remark*}
Note that points $A$ and $B$ are superfluous,
and the choice of constant $2$ or the definition of $M$
is also arbitrary (only the fact that $\ol{CM} \parallel \ol{PN}$ matters).
Moreover, the condition $\angle ACB = 120^{\circ}$ is never used either.
As a result, I did not find this problem very inspiring.
\end{remark*}
|
EGMO-2018-notes_2
|
Consider the set
\[ A = \left\{ 1 + \frac 1k : k = 1, 2, 3, \dots \right\}. \]
For every integer $x \ge 2$, let $f(x)$ denote
the minimum integer such that $x$ can be written as the product
of $f(x)$ elements of $A$ (not necessarily distinct).
Prove that there are infinitely many pairs of integers $x \ge 2$
and $y \ge 2$ for which \[ f(xy) < f(x) + f(y). \]
|
One of many constructions: let $n = 2^e+1$
for $e \equiv 5 \pmod{10}$
and let $x = 11$, $y = n/11$ be our two integers.
We prove two lemmas:
\begin{claim*}
For any $m \ge 2$ we have
$f(m) \ge \left\lceil \log_2 m \right\rceil$.
\end{claim*}
\begin{proof}
This is obvious.
\end{proof}
It follows that $f(n) = e+1$, since $n = \frac{n}{n-1} \cdot 2^e$.
\begin{claim*}
$f(11) = 5$.
\end{claim*}
\begin{proof}
We have $11 = \frac{33}{32} \cdot \frac 43 \cdot 2^3$.
So it suffices to prove $f(11) > 4$.
Note that a decomposition of $11$ must
contain a fraction at most $\frac{11}{10} = 1.1$.
But $2^3 \cdot 1.1 = 8.8 < 11$, contradiction.
\end{proof}
To finish, note that
\[ f(11) + f(n/11) \ge 5 + \log_2(n/11) = 1 + \log_2(16n/11) > 1 + e = 1 + f(n). \]
\begin{remark*}
Most solutions seem to involve picking $n$ such that
$f(n)$ is easy to compute.
Indeed, it's hard to get nontrivial lower bounds other than the log,
and even harder to actually come up with complicated constructions.
It might be said the key to this problem
is doing as little number theory as possible.
\end{remark*}
|
EGMO-2018-notes_3
|
The $n$ contestants of EGMO are named $C_1$, $C_2$, \dots, $C_n$.
After the competition,
they queue in front of the restaurant according to the following rules.
\begin{itemize}
\ii The Jury chooses the initial order of the contestants in the queue.
\ii Every minute, the Jury chooses an integer $i$ with $1 \leq i \leq n$.
\begin{itemize}
\ii If contestant $C_i$ has at least $i$ other contestants in front of her,
she pays one euro to the Jury and moves forward in the queue by exactly $i$ positions.
\ii If contestant $C_i$ has fewer than $i$ other contestants in front of her,
the restaurant opens and the process ends.
\end{itemize}
\end{itemize}
For every $n$, prove that this process must terminate
and determine the maximum number of euros
that the Jury can collect by cunningly choosing
the initial order and the sequence of moves.
|
The maximum money is
$1 + 3 + 7 + \dots + (2^{n-1}-1) = 2^n-n-1$,
which is finite.
Call the 1-euro process a jump,
and let $x_i$ denote the number of times that $C_i$ jumps.
Note that:
\begin{itemize}
\ii Whenever $C_i$ jumps it must jump over some $C_j$ with $j > i$.
\ii Contestant $C_i$ can jump over a $C_j$ with $j > i$ at most $1 + x_j$ times.
\end{itemize}
Now, we have $x_n = 0$ and in general,
\begin{align*}
x_{n-1} &\le 1 + x_n \le 1 \\
x_{n-2} &\le (1 + x_{n-1}) + (1 + x_{n-2}) \le 3 \\
x_{n-3} &\le (1 + x_{n-1}) + (1 + x_{n-2}) + (1 + x_{n-3}) \le 7
\end{align*}
and so on, which gives the bound.
The construction is inductive; here is the example for $n = 3$,
with the food towards the right:
\[
\begin{array}{ccc}
C_1 & C_2 & C_3 \\
C_2 & C_1 & C_3 \\
C_2 & C_3 & C_1 \\
C_3 & C_1 & C_2 \\
C_3 & C_2 & C_1
\end{array}
\]
In general, we can start the contestants in reverse order,
apply inductive hypothesis to $C_1$ through $C_{n-1}$ to flip their order,
then have $C_1$, $C_2$, $\dots$, $C_{n-1}$ jump over $C_n$,
then repeat.
This gives a construction with value $a_1 = 0$
and $a_n = 2a_{n-1} + (n-1)$ which is the same as the bound.
|
EGMO-2018-notes_4
|
Let $n \ge 3$ be an integer.
Several non-overlapping dominoes are placed on an $n \times n$ board.
The \emph{value} of a row or column is the number of dominoes
that cover at least one cell of that row or column.
A domino configuration is called \emph{balanced} if there exists
some $k \ge 1$ such that every row and column has value $k$.
Prove that a balanced configuration exists for every $n \ge 3$
and find the minimum number of dominoes needed
in such a configuration.
|
The answer is $2n/3$ when $n \equiv 0 \pmod 3$, and $2n$ otherwise.
\paragraph{Proof this is best possible.}
To prove these are best possible, assume there are $d$ dominoes.
\begin{claim*}
In any balanced configuration, we always have $k \cdot 2n = 3d$.
\end{claim*}
\begin{proof}
Consider counting the number of ordered pairs
\[ (\text{row or column},
\text{domino touching at least one cell in that row or column}). \]
On the one hand, this is equal to $k \cdot 2n$,
because there are $2n$ choices for the row and column,
and by the balanced hypothesis, there are $k$ dominoes for that row or column.
On the other hand, it must be equal to $3d$,
because for each domino there are either one column and two rows it touches,
or two columns and one row.
Hence the result.
\end{proof}
Written another way, we have
\[ d = 2n \cdot \frac{k}{3}. \]
Since $k \ge 1$, the first few possible values of $d$ are $2n/3$, $4n/3$, $2n$, \dots.
So we get a lower bound by taking the first integer in this sequence.
\paragraph{Constructions.}
When $n \equiv 0 \pmod 3$, one takes block-diagonal
copies of the following $3 \times 3$ square with $k = 1$.
\[
\begin{bmatrix}
A & A & \\
& & B \\
& & B
\end{bmatrix}.
\]
On the other hand, we give below construction for $4 \le n \le 7$
which have $k = 3$ and $2n$ dominoes.
By taking block-diagonal copies of these, we obtain a $k = 3$ construction
using $2n$ dominoes for any value of $n \ge 4$.
\[
\begin{bmatrix}
A & A & B & C \\
D & D & B & C \\
W & X & Y & Y \\
W & X & Z & Z
\end{bmatrix}
\qquad
\begin{bmatrix}
A & A & B & B & C \\
H & X & & & C \\
H & X & & & D \\
G & & Y & Y & D \\
G & F & F & E & E
\end{bmatrix}
\]
\[
\begin{bmatrix}
A & A & B & C & & \\
D & D & B & C & & \\
& & W & W & Y & Z \\
& & X & X & Y & Z \\
P & Q & & & R & R \\
P & Q & & & S & S \\
\end{bmatrix}
\qquad
\begin{bmatrix}
A & A & B & B & & & C \\
& W & W & & & X & C \\
& & P & & & X & D \\
H & & P & & & & D \\
H & Z & & Q & Q & & \\
G & Z & & & Y & Y & \\
G & & & F & F & E & E
\end{bmatrix}
\]
\begin{remark*}
Most of the difficulty of the problem is the construction for $n \in \{5,7\}$.
\end{remark*}
|
EGMO-2018-notes_5
|
Let $\Gamma $ be the circumcircle of triangle $ABC$.
A circle $\Omega$ is tangent to the line segment $AB$
and is tangent to $\Gamma$ at a point
lying on the same side of the line $AB$ as $C$.
The angle bisector of $\angle BCA$ intersects $\Omega$
at two different points $P$ and $Q$. Prove that $\angle ABP = \angle QBC$.
|
If we let $M$ denote the midpoint of arc $\widehat{AB}$
then the inversion at $M$ with radius $MA = MB$ fixes $\Omega$,
so it swaps $P$ and $Q$, thus
\[ \dang MPB = \dang QBM. \]
\begin{center}
\begin{asy}
pair A = dir(210);
pair B = dir(330);
pair C = dir(110);
pair T = dir(20);
pair M = dir(270);
pair S = extension(T, M, A, B);
pair O = extension(origin, T, S, S+M);
pair P = IP(CP(O, T), C--M);
pair Q = OP(CP(O, T), C--M);
filldraw(unitcircle, opacity(0.1)+lightcyan, deepcyan);
draw(C--M, deepcyan);
draw(A--B--C--cycle, deepcyan);
filldraw(CP(O, T), opacity(0.1)+lightgreen, deepgreen);
draw(A--P--B--Q--cycle, lightred);
draw(M--T, blue+dotted);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$T$", T, dir(T));
dot("$M$", M, dir(M));
dot("$S$", S, dir(S));
dot("$P$", P, dir(140));
dot("$Q$", Q, dir(Q));
/* TSQ Source:
A = dir 210
B = dir 330
C = dir 110
T = dir 20
M = dir 270
S = extension T M A B
O := extension origin T S S+M
P = IP CP O T C--M R140
Q = OP CP O T C--M
unitcircle 0.1 lightcyan / deepcyan
C--M deepcyan
A--B--C--cycle deepcyan
CP O T 0.1 lightgreen / deepgreen
A--P--B--Q--cycle lightred
M--T blue dotted
*/
\end{asy}
\end{center}
But
\begin{align*}
\dang MPB &= \dang MCB + \dang CBP \\
\dang QBM &= \dang ABM + \dang QBA
\end{align*}
implying the desired isogonality,
since $\dang ABM = \dang ACM = \dang MCB$.
|
EGMO-2018-notes_6
|
Fix a real number $0 < t < \half$.
\begin{enumerate}
\ii [(a)]
Prove that there exists a positive integer $n$
such that for every set $S$ of $n$ positive integers,
the following holds:
there exist distinct $x, y \in S$ and \emph{nonnegative} integer $m \ge 0$
such that $|x-my| \le ty$.
\ii [(b)]
Determine whether there exists an infinite set $S$
of positive integers such that the following holds:
for any distinct $x, y \in S$ and \emph{positive} integer $m > 0$,
we have $|x-my| > ty$.
\end{enumerate}
\end{enumerate}
|
\paragraph{Solution to (a).}
Assume not. Let $S = \left\{ s_1 < \dots < s_n \right\}$.
Consider
\[ 1 > \frac{s_1}{s_2} > \frac{s_1}{s_3} > \dots > \frac{s_1}{s_n} > t. \]
Note that two of the fractions above are within a factor of
$t^{1/(n-1)}$ of each other; taking $n$ large enough so that
$t^{1/(n-1)} \ge 1-t$ gives the conclusion.
\paragraph{Solution to (b).}
Yes, such a set $S$ exists.
To construct it, we employ a greedy algorithm.
Let $N$ be a large integer such that $t < 1/2-1/N$.
We define $S = \left\{ s_1 < s_2 < \dots \right\}$ inductively as follows.
\begin{itemize}
\ii First, let $s_1$ be any prime number exceeding $N$.
\ii Then, given $s_1$, \dots, $s_k$,
we let $s_{k+1}$ equal a prime number which is greater than $2s_k$,
and is congruent to $\tfrac{s_i-1}{2} \pmod{s_i}$ for $i = 1, \dots, k$.
This is possible by Chinese remainder theorem and Dirichlet theorem.
\end{itemize}
By construction, this works: if $i < j$ then $s_i / s_j < 1/2$
while $s_j / s_i$ has fractional part within $s_i^{-1}$ of $1/2$.
|
EGMO-2019-notes_1
|
Find all triples $(a, b, c)$ of real numbers such that $ab + bc + ca = 1$ and
\[ a^2b + c = b^2c + a = c^2a + b. \]
|
Answer: $(\pm 1/\sqrt3, \pm 1/\sqrt3, \pm1/\sqrt3)$
where the signs correspond,
and $(\pm1, \pm1, 0)$ and permutations where the signs correspond.
These work and we prove that is all.
We begin by eliminating the condition
via homogenization:
the first equality now reads
\begin{align*}
a^2b+c\left[ ab+bc+ca \right] &= b^2c+a\left[ ab+bc+ca \right] \\
\iff c^2(b+a) &= c(b^2+a^2) \\
\iff c&=0 \text{ or } a^2+b^2 = c(a+b).
\end{align*}
Cyclic variations hold.
So we have two cases.
\begin{itemize}
\ii If any of the variables is zero,
say $a = 0$, then the other two are nonzero.
So from $b^2 = bc$ we get $b = c$ giving
$(\pm1, \pm1, 0)$.
\ii Now assume all three variables are nonzero,
so $a^2+b^2 = c(a+b)$.
If we sum cyclically we get
\[ 2(a^2+b^2+c^2) = 2(ab+bc+ca)
\iff (a-b)^2 + (b-c)^2 + (c-a)^2 = 0 \]
which forces $a=b=c$ and gives the last solution.
\end{itemize}
|
EGMO-2019-notes_2
|
Let $n$ be a positive integer.
Dominoes are placed on a $2n \times 2n$ board
in such a way that every cell of the board is
(orthogonally) adjacent to exactly one cell covered by a domino.
For each $n$, determine the largest number of dominoes
that can be placed in this way.
|
The answer is $\binom{n+1}{2}$ and a construction is shown below.
For each domino, its \emph{aura} consists of all the cells
which are adjacent to a cell of the domino.
There are up to eight squares in each aura,
but some auras could be cut off by the boundary of the board,
which means that there could be as few as five squares.
So we want to estimate how many auras get cut off.
We denote by $a$, $b$, $c$, $k$ the number of auras
with $5$, $6$, $7$, $8$ cells on the boundary,
so we want an upper bound on $a+b+c+k$.
Note also that $\boxed{a \le 4}$
since such a cell uses a corner of the grid.
\begin{center}
\begin{asy}
size(11cm);
path aura = (1,0)--(2,0)--(2,1)--(3,1)--(3,3)--(2,3)--(2,4)--(1,4)--(1,3)--(0,3)--(0,1)--(1,1)--cycle;
picture aura_up(pen fill) {
picture pic = new picture;
fill(pic, aura, fill);
draw(pic, (1,1)--(2,1), lightgray );
draw(pic, (1,3)--(2,3), lightgray );
draw(pic, (0,2)--(3,2), lightgray );
draw(pic, (1,1)--(1,3), lightgray );
draw(pic, (2,1)--(2,3), lightgray );
draw(pic, aura, blue+1.25);
return pic;
}
picture aura_right(pen fill) {
return shift(0,3) * rotate(-90) * aura_up(fill);
}
add(shift(-1,-1)*aura_up(palegreen));
add(shift(-1,3)*aura_up(palegreen));
add(shift(-1,7)*aura_up(palegreen));
add(shift(1,1)*aura_up(palegreen));
add(shift(1,5)*aura_up(palegreen));
add(shift(3,3)*aura_up(palegreen));
add(shift(6,3)*aura_up(palegreen));
add(shift(8,1)*aura_up(palegreen));
add(shift(8,5)*aura_up(palegreen));
add(shift(10,-1)*aura_up(palegreen));
add(shift(10,3)*aura_up(palegreen));
add(shift(10,7)*aura_up(palegreen));
add(shift(4,6)*aura_right(palered));
add(shift(2,8)*aura_right(palered));
add(shift(6,8)*aura_right(palered));
/*
add(shift(0,10)*aura_right(palered));
add(shift(4,10)*aura_right(palered));
add(shift(8,10)*aura_right(palered));
*/
add(shift(4,1)*aura_right(palered));
add(shift(2,-1)*aura_right(palered));
add(shift(6,-1)*aura_right(palered));
add(shift(0,-3)*aura_right(palered));
add(shift(4,-3)*aura_right(palered));
add(shift(8,-3)*aura_right(palered));
draw(shift(0,-2)*scale(12)*unitsquare, black+2);
path cell = shift(16,-2)*unitsquare;
for (int i = 0; i <= 11; ++i) {
for (int j = 0; j <= 11; ++j) {
if (max(abs(i-5.5), abs(j-5.5)) == 5.5)
filldraw(shift(i,j)*cell, paleblue, gray);
/*
else if (max(abs(i-5.5), abs(j-5.5)) == 3.5)
filldraw(shift(i,j)*cell, paleblue, gray);
else if (max(abs(i-5.5), abs(j-5.5)) == 1.5)
filldraw(shift(i,j)*cell, paleblue, gray);
*/
else draw(shift(i,j)*cell, gray);
}
}
\end{asy}
\end{center}
The big observation about the auras on the edge:
\begin{claim*}
The auras of type $a$, $b$, $c$ have
$4$, $4$, and $3$-$4$ cells on the boundary of the grid, respectively.
(The boundary is the $4(2n-1)$ cells touching an edge of the board.)
\end{claim*}
Consequently, we have a bound
$4a + 4b + 3c \le 4(2n-1)$.
On the other hand, we obviously have
$5a + 6b + 7c + 8k = 4n^2$.
Therefore,
\begin{align*}
4n^2 + 2(2n-1) &\ge (5a+6b+7c+8k) + (2a+2b+1.5c) \\
&= 8(a+b+c+k) + 0.5c - a \\
&\ge 8(a+b+c+k) + 0 - 4 \\
\implies \frac{n(n+1)}{2} + \frac 14 &\ge a+b+c+k
\end{align*}
which implies the desired bound after taking the floor of left-hand side.
\begin{remark*}
In fact IMO 1999/3 shows the reverse bound:
we now give a proof that every tiling in this problem
has \emph{exactly} $\half n (n+1)$ dominoes.
Color the board as shown below into ``rings''.
\begin{center}
\begin{asy}
size(5cm);
path cell = unitsquare;
for (int i = 0; i <= 11; ++i) {
for (int j = 0; j <= 11; ++j) {
if (max(abs(i-5.5), abs(j-5.5)) == 5.5)
filldraw(shift(i,j)*cell, paleblue, gray);
else if (max(abs(i-5.5), abs(j-5.5)) == 3.5)
filldraw(shift(i,j)*cell, paleblue, gray);
else if (max(abs(i-5.5), abs(j-5.5)) == 1.5)
filldraw(shift(i,j)*cell, paleblue, gray);
else draw(shift(i,j)*cell, gray);
}
}
\end{asy}
\end{center}
Every aura covers exactly four blue cells. Done.
\end{remark*}
|
EGMO-2019-notes_3
|
Let $ABC$ be a triangle such that $\angle CAB > \angle ABC$, and let $I$ be its incenter.
Let $D$ be the point on segment $BC$ such that $\angle CAD = \angle ABC$.
Let $\omega$ be the circle tangent to $AC$ at $A$ and passing through $I$.
Let $X$ be the second point of intersection of $\omega$ and the circumcircle of $ABC$.
Prove that the angle bisectors of $\angle DAB$ and $\angle CXB$
intersect at a point on line $BC$.
|
Here is a cross-ratio/trig solution:
rare for me to do one of these,
but this problem called out to me that way.
As usual, let $\alpha = \angle BAC$, etc.
Let $T$ be the foot of the bisector of $\angle BAD$ on $\ol{BD}$,
so that
\[ \frac{TB}{TC} = \frac{AB \sin \angle BAT}{AC \sin \angle CAT}
= \frac{AB \sin \frac{\alpha-\beta}{2}}{AC \sin \frac{\alpha+\beta}{2}}. \]
Also, call $(ABC)$ by $\Gamma$ and let ray $XI$ meet $\Gamma$ again at $W$,
meaning that $\angle WXA = \angle IXA = \angle IAC = \frac{\beta}{2}$,
since we assume $\ol{AC}$ was tangent to $(IAX)$.
Thus arc $\widehat{AW}$ also has measure $\beta$.
\begin{center}
\begin{asy}
pair A = dir(50);
pair B = dir(210);
pair C = dir(330);
pair I = incenter(A, B, C);
pair M_a = circumcenter(B, I, C);
pair M_b = circumcenter(C, I, A);
pair M_c = circumcenter(A, I, B);
pair W = A*B/M_a;
pair X = -W+2*foot(origin, I, W);
pair D = extension(B, C, A, C*C/A);
filldraw(unitcircle, opacity(0.1)+lightcyan, lightblue);
draw(A--B--C--cycle, lightblue);
draw(X--W, lightred);
draw(B--M_b, lightred);
draw(C--M_c, lightred);
draw(A--M_a, lightred);
filldraw(W--M_b--A--M_c--cycle, opacity(0.1)+yellow, deepgreen);
filldraw(X--B--M_a--C--cycle, opacity(0.1)+yellow, deepgreen);
draw(A--D, lightblue);
pair T = extension(A, incenter(A, B, D), B, D);
draw(X--T--A, dashed+brown);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$I$", I, dir(I));
dot("$M_a$", M_a, dir(M_a));
dot("$M_b$", M_b, dir(M_b));
dot("$M_c$", M_c, dir(M_c));
dot("$W$", W, dir(W));
dot("$X$", X, dir(X));
dot("$D$", D, dir(D));
dot("$T$", T, dir(T));
/* TSQ Source:
A = dir 50
B = dir 210
C = dir 330
I = incenter A B C
M_a = circumcenter B I C
M_b = circumcenter C I A
M_c = circumcenter A I B
W = A*B/M_a
X = -W+2*foot origin I W
D = extension B C A C*C/A
unitcircle 0.1 lightcyan / lightblue
A--B--C--cycle lightblue
X--W lightred
B--M_b lightred
C--M_c lightred
A--M_a lightred
W--M_b--A--M_c--cycle 0.1 yellow / deepgreen
X--B--M_a--C--cycle 0.1 yellow / deepgreen
A--D lightblue
T = extension A incenter A B D B D
X--T--A dashed brown
*/
\end{asy}
\end{center}
Now,
\begin{align*}
\frac{XB}{XC} &= -(BC;XM_a)_\Gamma \overset{I}{=} -(M_b M_c; W A)_\Gamma \\
&= -\frac{\sin \half \mathrm{m}\widehat{M_b W}}{\sin \half \mathrm{m}\widehat{M_c W}}
\div \frac{\sin \half \mathrm{m}\widehat{M_b A}}{\sin \half \mathrm{m}\widehat{M_c A}}
= \frac{\sin\frac{\alpha-\beta}{2}}
{\sin\frac{\alpha +\gamma}{2}}
\div \frac{\sin \frac{\beta}{2}}{\sin \frac{\gamma}{2}} \\
\end{align*}
Therefore,
\[ \frac{XB}{XC} \div \frac{TB}{TC}
= \frac{\sin\frac{\alpha+\beta}{2} \sin\frac{\gamma}{2}}%
{\sin\frac{\alpha+\gamma}{2} \sin\frac{\beta}{2}}
\div \frac{AB}{AC} \\
= \frac{2\cos \frac{\gamma}{2} \sin \frac{\gamma}{2}}%
{2\cos \frac{\beta}{2} \sin \frac{\beta}{2}} \div \frac{AB}{AC} = 1 \]
the last step being the law of sines $AB/AC = \sin\gamma/\sin\beta$.
|
EGMO-2019-notes_4
|
Let $ABC$ be a triangle with incenter $I$.
The circle through $B$ tangent to $AI$ at $I$
meets side $AB$ again at $P$.
The circle through $C$ tangent to $AI$ at $I$
meets side $AC$ again at $Q$.
Prove that $PQ$ is tangent to the incircle of $ABC$.
|
Let $E$ and $F$ be the tangency points of the incircle on $AC$ and $AB$.
By angle chasing,
\[ \angle PIF = \angle AIF - \angle AIP
= \left( 90\dg - \half \angle A \right) - \half \angle B = \half \angle C. \]
Similarly, $\angle EIQ = \half \angle B$.
\begin{center}
\begin{asy}
pair A = dir(110);
pair B = dir(210);
pair C = dir(330);
pair I = incenter(A, B, C);
pair D = foot(I, B, C);
pair E = foot(I, C, A);
pair F = foot(I, A, B);
pair _P = abs(A-I)*abs(A-I)/abs(A-B) * dir(B-A) + A;
pair _Q = abs(A-I)*abs(A-I)/abs(A-C) * dir(C-A) + A;
pair P = _P;
pair Q = _Q;
filldraw(incircle(A, B, C), opacity(0.1)+blue, blue);
filldraw(A--B--C--cycle, opacity(0.1)+blue, blue);
draw(P--Q, red);
draw(P--I--F, deepgreen);
draw(E--I--Q, deepgreen);
pair T = foot(I, P, Q);
draw(I--T, red+dashed);
draw(A--I, lightblue);
draw(B--I--C, lightblue);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$I$", I, dir(270));
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(Q));
dot("$T$", T, dir(T));
/* --------------------------------+
| TSQX: by CJ Quines and Evan Chen |
| https://github.com/vEnhance/tsqx |
+----------------------------------+
A = dir 110
B = dir 210
C = dir 330
I 270 = incenter A B C
D = foot I B C
E = foot I C A
F = foot I A B
!pair _P = abs(A-I)*abs(A-I)/abs(A-B) * dir(B-A) + A;
!pair _Q = abs(A-I)*abs(A-I)/abs(A-C) * dir(C-A) + A;
P = _P
Q = _Q
incircle A B C / 0.1 blue / blue
A--B--C--cycle / 0.1 blue / blue
P--Q / red
P--I--F / deepgreen
E--I--Q / deepgreen
T = foot I P Q
I--T / red dashed
A--I / lightblue
B--I--C / lightblue
*/
\end{asy}
\end{center}
Let $T_p$, $T_q$ denote the second tangency of $P$, $Q$ to the incircle.
Then $\angle E I T_q = \angle B$ and $\angle T_p I F = \angle C$.
Since $\angle E I F = \angle B + \angle C$, it follows $T_p = T_q$.
\begin{remark*}
Notice we really only need $\angle PIQ = 90\dg - \half \angle A$
by essentially the same argument.
\end{remark*}
|
EGMO-2019-notes_5
|
Let $n\ge 2$ be an integer,
and let $a_1, a_2, \dots , a_n$ be positive integers.
Show that there exist positive integers $b_1, b_2, \dots, b_n$
satisfying the following three conditions:
\begin{enumerate}[(a)]
\ii $a_i\le b_i$ for $i=1, 2, \dots, n$;
\ii the remainders of $b_1$, $b_2$, \dots, $b_n$
on division by $n$ are pairwise different,
\ii $b_1 + \dots + b_n \le n \left( \frac{n-1}{2}
+ \left\lfloor \frac{a_1 + \dots + a_n}{n} \right\rfloor \right)$.
\end{enumerate}
|
Note that if $a_i > n$,
we can replace $a_i$ with $a_i-n$ and $b_i$ with $b_i-n$,
and nothing changes.
So we may as well assume $a_i \in \{1, \dots, n\}$ for each $i$.
We choose our $b_i$'s in the following way.
Draw an $n \times n$ grid and in the $i$th
column fill in the bottom $a_i-1$ cells red.
We can select $b_i$ by marking $n$ cells,
one in each row or column.
If we chose $j$th lowest row in the $i$th column,
then we would set $b_i = j$ on non-red cells
and $b_i = j + n$ on red cells.
In this way, define the \emph{penalty} $p$ as
the number of selected cells which are red.
Then
\[ b_1 + \dots + b_n = (1+2+\dots+n) + n \cdot p
= n \cdot \frac{n-1}{2} + n \cdot (p+1). \]
and we seek to minimize the penalty $p$.
\begin{center}
\begin{asy}
size(8cm);
fill( (1,0)--(1,2)--(2,2)--(2,3)--(4,3)--(4,4)--(5,4)--(5,0)--cycle, palered);
for (int i=0; i<=5; ++i) {
draw((i,0)--(i,5), gray);
draw((0,i)--(5,i), gray);
}
draw(scale(5)*unitsquare);
label("$b_i \equiv 1 \pmod 5$", (0,0.5), dir(180), lightblue);
label("$b_i \equiv 2 \pmod 5$", (0,1.5), dir(180), lightblue);
label("$b_i \equiv 3 \pmod 5$", (0,2.5), dir(180), lightblue);
label("$b_i \equiv 4 \pmod 5$", (0,3.5), dir(180), lightblue);
label("$b_i \equiv 5 \pmod 5$", (0,4.5), dir(180), lightblue);
label("$a_1-1$", (0.5,0), dir(-90), lightred);
label("$a_2-1$", (1.5,0), dir(-90), lightred);
label("$a_3-1$", (2.5,0), dir(-90), lightred);
label("$a_4-1$", (3.5,0), dir(-90), lightred);
label("$a_5-1$", (4.5,0), dir(-90), lightred);
label("$b_1$", (0.5, 3.5), blue);
label("$b_2$", (1.5, 2.5), blue);
label("$b_3$", (2.5, 4.5), blue);
label("$b_4$", (3.5, 0.5), blue);
label("$b_5$", (4.5, 1.5), blue);
\end{asy}
\end{center}
But the expected penalty of a \emph{random} permutation
is the red area divided by $n$, which is \[ \mathbb E[p] = \frac{(a_1-1) + \dots + (a_n-1)}{n} \]
and so there exists a choice for which
the penalty is at most $\left\lfloor \mathbb E[p] \right\rfloor$.
This gives the required result.
\begin{remark*}
The visual aid can be excised from the solution;
which can then be rewritten more tersely as follows.
After assuming $1 \le a_i \le n$ for each $n$,
pick a uniformly random permutation $\sigma$ on
$\{1, \dots, n\}$ and define
\[ b_i =
\begin{cases}
n + \sigma(i) & a_i > \sigma(i) \\
\sigma(i) & \text{otherwise}.
\end{cases}
\]
As before $\mathbb E[e_\sigma] = \sum_{i=1}^n \frac{a_i-1}{n}$
and the rest is the same.
\end{remark*}
|
EGMO-2019-notes_6
|
On a circle, Alina draws $2019$ chords, the endpoints of which are all different.
A point is considered marked if it is either
\begin{enumerate}[(i)]
\ii one of the $4038$ endpoints of a chord; or
\ii an intersection point of at least two chords.
\end{enumerate}
Of the $4038$ points meeting criterion (i),
Alina labels $2019$ points with a $0$ and the other $2019$ points with a $1$.
She labels each point meeting criterion (ii)
with an arbitrary integer (not necessarily positive).
Along each chord, Alina considers the segments connecting two consecutive marked points.
(A chord with $k$ marked points has $k-1$ such segments.)
She labels each such segment in yellow with the sum of the labels of its two endpoints
and in blue with the absolute value of their difference.
Alina finds that the $N + 1$ yellow labels take each value $0, 1, \dots, N$ exactly once.
Show that at least one blue label is a multiple of $3$.
\end{enumerate}
|
Only the labels mod $3$ matter at all.
Assume for contradiction no blue labels are divisible by $3$.
Let $e_{ij}$ denote the number of segments
joining $i \pmod 3$ to $j \pmod 3$.
By double-counting (noting that points in (ii) are
counted an even number of times
but points in (i) are counted once)
we derive that
\begin{align*}
e_{01} + e_{02} &\equiv 2019 \pmod 2 \\
e_{01} + e_{12} &\equiv 2019 \pmod 2 \\
e_{02} + e_{12} &\equiv 0 \pmod 2
\end{align*}
which gives \[ e_{02} \equiv e_{12} \equiv 1 - e_{01} \pmod 2. \]
However, one can check this is incompatible
with the hypothesis that the yellow labels
are $0$, $1$, \dots, $N$.
\begin{remark*}
In addition, we can replace the points/segments
by any graph $G$ for which there are
\begin{itemize}
\ii an odd number of leaves (or just odd-degree vertices) labeled $0$,
\ii an odd number of leaves (or just odd-degree vertices) labeled $1$,
\ii and the remaining vertices have even degree.
\end{itemize}
Thus the geometry of the problem is smoke and mirrors, too.
\end{remark*}
|
EGMO-2020-notes_1
|
The positive integers $a_0$, $a_1$, $a_2$, \dots, $a_{3030}$ satisfy
\[ 2a_{n + 2} = a_{n + 1} + 4a_n \text{ for } n = 0, 1, 2, \dots, 3028. \]
Prove that at least one of the numbers
$a_0$, $a_1$, $a_2$, \dots, $a_{3030}$ is divisible by $2^{2020}$.
|
The idea is this:
\begin{itemize}
\ii All terms $a_0$, \dots, $a_{3030}$ are integers
(divisible by $2^0=1$);
\ii Hence all terms $a_1$, \dots, $a_{3029}$ are divisible by $2$,
\ii Hence all terms $a_1$, \dots, $a_{3028}$ are divisible by $4$,
\ii Hence all terms $a_2$, \dots, $a_{3027}$ are divisible by $8$,
\ii \dots and so on.
\end{itemize}
The $2021$st item in this list reads as $2^{2020} \mid a_{1010}$.
One can also phrase this with induction.
Replace $1010$ by $N$ in the obvious way and proceed by induction on $N \ge 0$
with the base case $N = 0$ being vacuous.
Notice that for any index $k \neq 0, 3N-1, 3N$ we have
\[ a_k = 2a_{k+1} - 4a_{k-1} = 2(2a_{k+2}-4a_k)-4a_{k-1}
= 4(a_{k+2}-2a_k-a_{k-1}) \]
so it follows that $(a_1, a_2, \dots, a_{3N-2})$
are all divisible by $4$ and satisfy the same relations.
But then $(a_1/4, a_2/4, \dots, a_{3N-2}/4)$ has length $3(N-1)+1$
and so one of them is divisible by $4^{N-1}$;
hence some term of our original sequence is divisible by $4^N$.
|
EGMO-2020-notes_2
|
Find all lists $(x_1, x_2, \dots, x_{2020})$ of non-negative real numbers
such that the following three conditions are all satisfied:
\begin{itemize}
\ii $x_1 \le x_2 \le \dots \le x_{2020}$;
\ii $x_{2020} \le x_1 + 1$;
\ii there is a permutation $(y_1, y_2, \dots, y_{2020})$
of $(x_1, x_2, \dots, x_{2020})$ such that
\[ \sum_{i = 1}^{2020} ((x_i + 1)(y_i + 1))^2 = 8 \sum_{i = 1}^{2020} x_i^3. \]
\end{itemize}
|
The main point of the problem is to prove an inequality.
\begin{claim*}
If $a$ and $b$ are real numbers with $|a-b| \le 1$,
then \[ (a+1)^2(b+1)^2 \geq 4(a^3+b^3). \]
Moreover, equality holds only when $\{a,b\}=\{1,0\}$ or $\{a,b\}=\{2,1\}$.
\end{claim*}
\begin{proof}
Write
\begin{align*}
a^3+b^3 &= (a+b)(a^2-ab+b^2) = (a+b)[(a-b)^2 + ab] \\
&\le (a+b)(1+ab) \\
&\le \left( \frac{(a+b)+(1+ab)}{2} \right)^2 = \frac{(a+1)^2(b+1)^2}{4}. \qedhere
\end{align*}
\end{proof}
From this it follows that $(x_n)_n$
should be either $(1,\dots,1,2,\dots,2)$ or $(0,\dots,0,1,\dots,1)$.
\begin{remark*}
There is a sense in which the problem \emph{must} be an inequality,
because given a \emph{fixed} permutation of the indices
of $y_i$ by $x_i$, one gets an equation which \emph{a priori}
ought to cut out a codimension $1$ surface in $\RR^{2020}$.
So the fact that the answer set appears finite
is an indication that some inequality at play here.
\end{remark*}
|
EGMO-2020-notes_3
|
Let $ABCDEF$ be a convex hexagon such that $\angle A = \angle C = \angle E$ and $\angle B = \angle D = \angle F$
and the (interior) angle bisectors of $\angle A$, $\angle C$, and $\angle E$ are concurrent.
Prove that the (interior) angle bisectors of $\angle B$, $\angle D$, and $\angle F$ must also be concurrent.
|
In general, if hexagon $ABCDEF$ has $\angle A = \angle C = \angle E$ and $\angle B = \angle D = \angle F$,
then its sides can be extended to form two
equilateral triangles $PQR$ and $XYZ$, as shown.
\begin{center}
\begin{asy}
size(5cm);
pair P = dir(90);
pair Q = dir(210);
pair R = dir(330);
pair O = (0.1,0.2);
pair z = dir(-40);
pair Y = O+z*(P-O);
pair X = O+z*(R-O);
pair Z = O+z*(Q-O);
filldraw(P--Q--R--cycle, opacity(0.1)+lightblue, blue);
filldraw(X--Y--Z--cycle, opacity(0.1)+lightcyan, deepcyan);
pair A = extension(Q, P, Y, Z);
pair B = extension(P, R, Y, Z);
pair C = extension(P, R, X, Y);
pair D = extension(R, Q, X, Y);
pair E = extension(R, Q, X, Z);
pair F = extension(Q, P, X, Z);
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(Q));
dot("$R$", R, dir(R));
dot("$Y$", Y, dir(Y));
dot("$X$", X, dir(X));
dot("$Z$", Z, dir(Z));
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
\end{asy}
\end{center}
The problem is solved upon proving the following claim.
\begin{claim*}
The angle bisectors of $\angle A$, $\angle C$, $\angle E$ are concurrent if and only if
the unique spiral similarity sending $PQR$ to $YZX$ is a rotation;
equivalently, the two triangles are congruent.
\end{claim*}
\begin{center}
\begin{asy}
pair P = dir(90);
pair Q = dir(210);
pair R = dir(330);
pair O = (0.1,0.2);
pair z = dir(-40);
pair Y = O+z*(P-O);
pair X = O+z*(R-O);
pair Z = O+z*(Q-O);
filldraw(P--Q--R--cycle, opacity(0.1)+lightblue, blue);
filldraw(X--Y--Z--cycle, opacity(0.1)+lightcyan, deepcyan);
pair A = extension(Q, P, Y, Z);
pair B = extension(P, R, Y, Z);
pair C = extension(P, R, X, Y);
pair D = extension(R, Q, X, Y);
pair E = extension(R, Q, X, Z);
pair F = extension(Q, P, X, Z);
draw(circumcircle(O, A, C), deepgreen);
draw(circumcircle(O, C, E), deepgreen);
draw(circumcircle(O, E, A), deepgreen);
draw(A--O, red);
draw(C--O, red);
draw(E--O, red);
label("$2\theta$", A, 5*dir(40), orange);
label("$2\theta$", C, 5*dir(280), orange);
label("$2\theta$", E, 5*dir(160), orange);
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(Q));
dot("$R$", R, dir(R));
dot("$O$", O, dir(O));
dot("$Y$", Y, dir(Y));
dot("$X$", X, dir(X));
dot("$Z$", Z, dir(Z));
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
/* TSQ Source:
P = dir 90
Q = dir 210
R = dir 330
O = (0.1,0.2)
z := dir -40
Y = O+z*(P-O)
X = O+z*(R-O)
Z = O+z*(Q-O)
P--Q--R--cycle 0.1 lightblue / blue
X--Y--Z--cycle 0.1 lightcyan / deepcyan
A = extension Q P Y Z
B = extension P R Y Z
C = extension P R X Y
D = extension R Q X Y
E = extension R Q X Z
F = extension Q P X Z
circumcircle O A C deepgreen
circumcircle O C E deepgreen
circumcircle O E A deepgreen
A--O red
C--O red
E--O red
!label("$2\theta$", A, 5*dir(40), orange);
!label("$2\theta$", C, 5*dir(280), orange);
!label("$2\theta$", E, 5*dir(160), orange);
*/
\end{asy}
\end{center}
\begin{proof}[Proof of ``if'' direction]
Let $O$ be the center of the rotation.
Then $O$ is equidistant form $\ol{YZ}$ and $\ol{PQ}$ by rotation;
similarly for the other pairs of sides.
So the bisectors meet at $O$.
\end{proof}
\begin{proof}[Proof of ``only if'' direction]
Let $O$ be the concurrency point.
Let $\angle PAB = \angle RCD = \angle QEF = 2\theta$
Since $\angle APC = \angle AYC = 60\dg$ the quadrilateral $PACY$ is cyclic.
But $\angle PAO + \angle PCO = (90\dg+\theta) + (90\dg-\theta) = 180\dg$, so $PAOCY$ is cyclic.
Now $\angle PAO = \angle YCO \implies OP = OY$, and $\angle POY = 2\theta$.
So $O$ is the center of rotation from $\triangle PQR$ to $\triangle YXZ$ with angle $2\theta$.
\end{proof}
|
EGMO-2020-notes_4
|
A permutation of the integers $1, 2, \dots, m$ is called \emph{fresh} if there exists
no positive integer $k < m$ such that the first $k$ numbers
in the permutation are $1, 2, \dots, k$ in some order.
Let $f_m$ be the number of fresh permutations of the integers $1, 2, \dots, m$.
Prove that $f_n \ge n \cdot f_{n - 1}$ for all $n \ge 3$.
|
For every fresh permutation on $(1, 2, \dots, n-1)$
we generate $n$ fresh permutations on $(1, 2, \dots, n)$ in the following way:
\begin{itemize}
\ii Insert $n$ in the $k$th position for $k=1, 2, \dots, n-1$;
\ii Replace $n-1$ with $n$ and append $n-1$ to the end.
\end{itemize}
For example, $3142$ would generate $53142$, $35142$, $31542$, $31452$ and $31524$.
All permutations generated this way are distinct.
Indeed, the only thing to note is that the second type of permutation
yields a non-fresh permutation when $n$ is deleted
(because $n-1$ is at the end, and $n \ge 3$).
This implies the result.
|
EGMO-2020-notes_5
|
Triangle $ABC$ has circumcircle $\Gamma$ and obeys $\angle BCA > 90^{\circ}$.
There is a point $P$ in the interior of the line segment $AB$
such that $PB = PC$ and the length of $PA$ equals the radius of $\Gamma$.
The perpendicular bisector of $\ol{PB}$ intersects $\Gamma$ at the points $D$ and $E$.
Prove $P$ is the incenter of triangle $CDE$.
|
Let $O$ be the center of $\Gamma$ and $M$
the arc midpoint of $\ol{DE}$.
\begin{claim*}
Quadrilateral $APMO$ is a rhombus.
\end{claim*}
\begin{proof}
Since $PA = MO$ and both are perpendicular to $\ol{DE}$,
it follows $APMO$ is a parallelogram.
In fact though, because $AO = MO$, we get the rhombus.
\end{proof}
\begin{center}
\begin{asy}
size(6cm);
pair B = dir(0);
pair C = dir(180);
pair D = dir(37);
pair A = D*D;
pair E = dir(60)*A;
pair F = dir(-60)*A;
pair O = origin;
pair J = A+O-D;
filldraw(unitcircle, opacity(0.1)+lightcyan, heavycyan);
filldraw(A--B--C--cycle, opacity(0.1)+lightcyan, heavycyan);
draw(A--D, red);
draw(O--J, red);
filldraw(C--E--F--cycle, opacity(0.1)+green, heavygreen);
filldraw(A--E--O--F--cycle, opacity(0.1)+lightblue, blue);
draw(E--F, blue);
draw(A--O, blue);
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
dot("$A$", A, dir(A));
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
dot("$O$", O, dir(-90));
dot("$J$", J, dir(J));
label("IMO 2002/2", dir(-90), dir(-90));
\end{asy}
\qquad
\begin{asy}
size(6cm);
pair C = dir(180);
pair K = dir(37);
pair M = K*K;
pair D = dir(60)*M;
pair E = dir(-60)*M;
pair O = origin;
pair P = M+O-K;
filldraw(unitcircle, opacity(0.1)+lightcyan, heavycyan);
filldraw(C--D--E--cycle, opacity(0.1)+green, heavygreen);
filldraw(M--D--O--E--cycle, opacity(0.1)+lightblue, blue);
draw(D--E, blue);
draw(M--O, blue);
draw(C--P, deepcyan);
draw(P--M, blue+dotted);
pair A = P+O-M;
pair B = -A+2*foot(O, P, A);
draw(A--B--O, orange);
dot("$C$", C, dir(C));
dot("$M$", M, dir(M));
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
dot("$O$", O, dir(-90));
dot("$P$", P, dir(P));
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
label("EGMO 2020/5", dir(-90), dir(-90));
/* TSQ Source:
C = dir 180
K := dir 37
M = K*K
D = dir(60)*M
E = dir(-60)*M
O = origin R-90
P = M+O-K
unitcircle 0.1 lightcyan / heavycyan
C--D--E--cycle 0.1 green / heavygreen
M--D--O--E--cycle 0.1 lightblue / blue
D--E blue
M--O blue
C--P deepcyan
P--M blue dotted
A = P+O-M
B = -A+2*foot O P A
A--B--O orange
*/
\end{asy}
\end{center}
Since $PC = PB$, and $PA = PM$,
it follows that $C$, $P$, $M$ are collinear now.
\begin{claim*}
We have $MP = MD = ME$ equal to the radius of $\Gamma$.
\end{claim*}
\begin{proof}
Note $PBMO$ is an isosceles trapezoid.
Since $\ol{DE}$ is the perpendicular bisector of $\ol{PB}$,
it is the perpendicular bisector of $\ol{OM}$ too.
Hence $MDOE$ is a rhombus (with $60\dg$ internal angles), the end.
\end{proof}
Since $MP = MD = ME$ we are done by Fact 5.
\begin{remark*}
The same figure appears in IMO 2002/2,
drawn on the left for contrast.
\end{remark*}
|
EGMO-2020-notes_6
|
Find all integers $m > 1$ for which the sequence $(a_n)_{n\ge1}$
defined recursively by
\[ a_{n+2} = m(a_{n+1} + a_n) - a_{n-1} \]
with initial conditions $a_1 = a_2 = 1$ and $a_3 = 4$
contains only perfect squares.
\end{enumerate}
|
The answer is $m=2$ and $m=10$.
The verification they work is left as an exercise.
Now compute the first several terms:
\begin{align*}
a_1 &= 1 \\
a_2 &= 1 \\
a_3 &= 4 \\
a_4 &= 5m-1 \\
a_5 &= 5m^2+3m-1 \\
a_6 &= 5m^3+8m^2-2m-4.
\end{align*}
We will now prove:
\begin{claim*}
If $a_4 \cdot a_6$ is a square then $m \in \{2,10\}$.
\end{claim*}
\begin{proof}
A computation gives
\begin{align*}
16 a_4 a_6
&= 400 m^4 + 560 m^3 - 288 m^2 - 288 m + 64 \\
&= \left(20m^2+14m-\frac{121}{10} \right)^2
+ \frac{508}{10}m - \frac{8241}{100}.
\end{align*}
Let $A = 200m^2+140m-121 \equiv -1 \pmod{20}$.
Then $42A+441 > 5080m-8241$ for all $m$ and hence
\[ A^2 < \underbrace{A^2+5080m-8241}_{=1600a_3a_5} < (A+21)^2. \]
But the inner term is the square of a multiple of $20$
so it must equal $(A+1)^2$.
Thus, we have
\[ 2(\underbrace{200m^2+140m-121}_{=A})+1 = 5080m-8241 \implies m \in \{2,10\} \]
as desired.
\end{proof}
\begin{remark*}
In general, if $f(x) \in \ZZ[x]$ has even degree
and the leading coefficient is a square,
then $f(x)$ should be a square finitely often
(unless $f$ is itself the square of a polynomial).
The proof proceeds along the same lines,
by bounding $f$ between two squares.
See China TST 2001 for example
which asked students to determine all $x$ for which
\[ f(x) = x^6 + 15x^5 + 85x^4 + 225x^3 + 274x^2 + 120x + 1 \]
was equal to a perfect square.
\end{remark*}
|
EGMO-2021-notes_1
|
The number $2021$ is fantabulous.
For any positive integer $m$,
if any element of the set $\{m, 2m+1, 3m\}$ is fantabulous,
then all the elements are fantabulous.
Does it follow that the number $2021^{2021}$ is fantabulous?
|
Write $a \iff b$ to mean $a$ fantabulous iff $b$ fantabulous.
Notice that for any integer $n$, we have
\[ 2n \iff 4n+1 \iff 12n+3 \iff 6n+1 \iff 3n \iff n. \]
Together with the given $2n+1 \iff n$,
it follows that if any integer is fantabulous then all of them are.
|
EGMO-2021-notes_2
|
Find all functions $f \colon \QQ \to \QQ$
such that the equation
\[f(xf(x)+y) = f(y) + x^2\]
holds for all rational numbers $x$ and $y$.
|
The answers are $f(x) \equiv +x$
and $f(x) \equiv -x$ which work.
To show they are the only ones,
we follow an approach similar to \texttt{SnowPanda}.
\begin{claim*}
If $f(z) = 0$ then $z = 0$.
In other words, $f$ has at most one root, at $0$.
\end{claim*}
\begin{proof}
Take $P(z,0)$.
\end{proof}
\begin{claim*}
The function $f$ is linear.
\end{claim*}
\begin{proof}
Let $a,b \in \QQ$ be any two nonzero rational numbers, so $f(a), f(b) \neq 0$.
Choose nonzero integers $m$ and $n$ such that
\[ \frac nm = \frac{af(a)}{bf(b)}. \]
Then for any $y \in \QQ$ we have
\begin{align*}
f(y + maf(a)) &= f(y) + m \cdot a^2 \\
f(y + nbf(b)) &= f(y) + n \cdot b^2.
\end{align*}
The two left-hand sides were equal by construction, so we get
\[ \frac{af(a)}{bf(b)} = \frac nm = \frac{a^2}{b^2}. \]
Thus $f(a) / a = f(b) / b$, as needed.
\end{proof}
Once $f$ is linear we here quickly recover the solution set.
|
EGMO-2021-notes_3
|
Let $ABC$ be a triangle with an obtuse angle at $A$.
Let $E$ and $F$ be the intersections of the external bisector of angle $A$
with the altitudes of $ABC$ through $B$ and $C$ respectively.
Let $M$ and $N$ be the points on the segments $EC$ and $FB$ respectively
such that $\angle EMA = \angle BCA$ and $\angle ANF = \angle ABC$.
Prove that the points $E$, $F$, $N$, $M$ lie on a circle.
|
Call the altitudes $\ol{BZ}$ and $\ol{CY}$,
and $H$ the orthocenter.
Let $W$ be the midpoint of $\ol{BC}$.
Then according to
\textbf{\href{https://aops.com/community/c6h89098p519896}{IMO Shortlist 2005 G5}},
the line $AW$ is concurrent
with $(HYZ)$, $(HEF)$, $(HBC)$ at a point $Q$.
\begin{center}
\begin{asy}
import graph; size(9.98895cm);
pen ffxfqq = rgb(1.,0.49803,0.); pen yqqqqq = rgb(0.50196,0.,0.); pen zzttqq = rgb(0.6,0.2,0.); pen qqwuqq = rgb(0.,0.39215,0.);
pair H = (0.2,3.2), B = (-1.,0.2), C = (2.2,0.2), A = (0.2,1.), Y = (-0.55862,1.30344), Z = (1.21538,1.67692), F = (1.57146,1.14279), Q = (-0.68,2.76), M = (0.87710,0.51931);
draw(H--B--C--cycle, linewidth(0.6) + zzttqq);
draw(circle((0.34465,1.81068), 1.39682), linewidth(0.6) + ffxfqq);
draw(circle((0.2,2.1), 1.1), linewidth(0.6) + yqqqqq);
draw(circle((0.6,1.3), 1.94164), linewidth(0.6) + yqqqqq);
draw(H--B, linewidth(0.6) + zzttqq);
draw(B--C, linewidth(0.6) + zzttqq);
draw(C--H, linewidth(0.6) + zzttqq);
draw(B--Z, linewidth(0.6));
draw(C--Y, linewidth(0.6));
draw(B--F, linewidth(0.6) + blue);
draw(C--(-0.71824,0.90439), linewidth(0.6) + blue);
draw((0.6,0.2)--Q, linewidth(0.6) + green);
draw(circle((-1.,1.5), 1.3), linewidth(0.6) + qqwuqq);
draw((-0.71824,0.90439)--F, linewidth(0.6) + yqqqqq);
dot("$H$", H, dir((-5.899, 6.902)));
dot("$B$", B, dir((-8.675, -9.665)));
dot("$C$", C, dir((5.858, -4.103)));
dot("$A$", A, dir((1.801, 3.594)));
dot("$Y$", Y, dir((-13.028, -3.649)));
dot("$Z$", Z, dir((6.784, -1.212)));
dot("$E$", (-0.71824,0.90439), dir((-10.327, -2.673)));
dot("$F$", F, dir((7.966, -3.412)));
dot("$Q$", Q, dir((-8.162, 8.978)));
dot("$N$", (-0.15969,0.50808), dir((1.837, -9.672)));
dot("$W$", (0.6,0.2), dir((-1.409, -11.804)));
dot("$M$", M, dir((-1.312, -9.939)));
\end{asy}
\end{center}
Since $WA \cdot WQ = WB^2$,
it follows that $(AQB)$ is tangent to $\ol{BC}$,
ergo $N \in (AQB)$.
Then
\[ \dang QNF = \dang QNB = \dang QAB = \dang QAZ = \dang QHZ = \dang QHF \]
and hence $N$ lies on $(HQEF)$.
Similarly, so does $M$.
|
EGMO-2021-notes_4
|
Let $ABC$ be a triangle with incenter $I$
and let $D$ be an arbitrary point on the side $BC$.
Let the line through $D$ perpendicular to $BI$ intersect $CI$ at $E$.
Let the line through $D$ perpendicular to $CI$ intersect $BI$ at $F$.
Prove that the reflection of $A$ across the line $EF$ lies on the line $BC$.
|
We begin as follows:
\begin{claim*}
$AEIF$ is cyclic.
\end{claim*}
\begin{proof}
Let $X = \ol{CA} \cap \ol{DF}$.
Then
\begin{align*}
\dang FXA &= \dang DXC = \dang CDX = 90\dg - \half C
= \dang AIB = \dang FIA \\
\dang EXF &= \dang EXD = \dang XDE = \dang FDE = \dang EIF
\end{align*}
which implies $AFXI$ and $FXIE$ are cyclic, respectively.
\end{proof}
\begin{center}
\begin{asy}
pair A = dir(110);
pair B = dir(210);
pair C = dir(330);
pair D = 0.7*C+0.3*B;
pair I = incenter(A, B, C);
pair E = extension(C, I, D, foot(D, B, I));
pair F = extension(B, I, D, foot(D, C, I));
filldraw(A--B--C--cycle, opacity(0.1)+lightcyan, deepcyan);
draw(B--F, lightblue);
draw(C--I--A, lightblue);
filldraw(D--E--F--cycle, opacity(0.1)+lightred, red);
filldraw(circumcircle(A, E, F), opacity(0.1)+yellow, deepgreen+dashed);
pair Y = extension(D, E, A, B);
pair X = extension(D, F, A, C);
draw(D--Y, red);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(270));
dot("$I$", I, dir(250));
dot("$E$", E, dir(225));
dot("$F$", F, dir(F));
dot(Y);
dot("$X$", X, dir(X));
/* TSQ Source:
A = dir 110
B = dir 210
C = dir 330
D = 0.7*C+0.3*B R270
I = incenter A B C R250
E = extension C I D foot D B I R225
F = extension B I D foot D C I
A--B--C--cycle 0.1 lightcyan / deepcyan
B--F lightblue
C--I--A lightblue
D--E--F--cycle 0.1 lightred / red
circumcircle A E F 0.1 yellow / deepgreen dashed
Y .= extension D E A B
X = extension D F A C
D--Y red
*/
\end{asy}
\end{center}
Now, the dilation of the Simson line of $A$
to $\triangle IEF$ should be collinear,
but the reflections of $A$ about lines $BI$ and $CI$
lie on line $BC$ by definition.
This solves the problem.
|
EGMO-2021-notes_5
|
A plane has a special point $O$ called the origin.
Let $P$ be a set of $2021$ points in the plane such that
\begin{itemize}
\ii no three points in $P$ lie on a line and
\ii no two points in $P$ lie on a line through the origin.
\end{itemize}
A triangle with vertices in $P$ is \emph{fat}
if $O$ is strictly inside the triangle.
Find the maximum number of fat triangles.
|
For every pair of points $P$ and $Q$,
draw a directed arrow $P \to Q$ if $\angle POQ$ is labeled clockwise.
This gives a tournament on $2021$ vertices,
and a triangle is fat if its three edges form a directed cycle.
Consequently, by
\href{https://aops.com/community/c6h130647p740398}{\textbf{Canada 2006/4}},
there are at most $\frac16 n(n+1)(2n+1)$
fat triangles, where $n = 1010$.
Equality occurs if the set of points are the vertices of a regular
$2021$-gon containing $O$.
|
EGMO-2021-notes_6
|
Does there exist a nonnegative integer $a$ for which the equation
\[\left\lfloor\frac{m}{1}\right\rfloor
+ \left\lfloor\frac{m}{2}\right\rfloor
+ \left\lfloor\frac{m}{3}\right\rfloor
+ \dotsb + \left\lfloor\frac{m}{m}\right\rfloor = n^2 + a \]
has more than one million different solutions
$(m, n)$ where $m$ and $n$ are positive integers?
\end{enumerate}
|
The answer is yes.
In fact, we prove the following general version:
\begin{claim*}
Consider any function $f \colon \NN \to \NN$
satisfying $f(m) = o(m^2)$.
Then for some $a$, the equation
\[ f(m) = n^2 + a \]
has at least 1000000 solutions $(m,n)$.
\end{claim*}
\begin{proof}
We consider triples $(m,n,a)$ satisfying the equation
with the additional property that
\[ n = \left\lfloor \sqrt{f(m)} \right\rfloor^2
\implies a = f(m) - n^2 \in [0, 2\sqrt{f(m)}]. \]
Now, let $M$ be large enough that
$\max_{m=1}^M f(m) < \frac{M^2}{2 \cdot 10^{12}}$.
Then there are $M$ triples,
but $a < \frac{M}{10^6}$ in each triple.
So some $a$ appears $10^6$ times.
\end{proof}
|
IMO-1997-notes_1
|
In the plane there is an infinite chessboard.
For any pair of positive integers $m$ and $n$,
consider a right-angled triangle with vertices at lattice points
and whose legs, of lengths $m$ and $n$, lie along edges of the squares.
Let $S_1$ be the total area of the black part of the triangle
and $S_2$ be the total area of the white part.
Let $f(m,n) = | S_1 - S_2 |$.
\begin{enumerate}[(a)]
\ii Calculate $f(m,n)$ for all positive integers $m$ and $n$
which are either both even or both odd.
\ii Prove that $f(m,n) \leq \frac 12 \max \{m,n\}$ for all $m$ and $n$.
\ii Show that there is no constant $C$
such that $f(m,n) < C$ for all $m$ and $ n$.
\end{enumerate}
|
In general, we say the \emph{discrepancy} of a region in the plane
equals its black area minus its white area.
We allow negative discrepancies,
so discrepancy is additive and $f(m,n)$ equals the absolute value
of the discrepancy of a right triangle with legs $m$ and $n$.
For (a), the answers are $0$ and $1/2$ respectively.
To see this, consider the figure shown below.
\begin{center}
\begin{asy}
size(8cm);
pair A = (0,5);
pair B = (9,0);
pair M = midpoint(A--B);
for (int i=0; i<=5; ++i) {
draw( (0,i)--(9,i), gray );
}
for (int j=0; j<=9; ++j) {
draw( (j,0)--(j,5), gray );
}
dot("$M$", M, dir(50));
dot("$A$", A, dir(90));
dot("$B$", B, dir(0));
dot("$C$", (0,0), dir(180));
filldraw(A--B--(0,0)--cycle, opacity(0.1)+yellow, black+1.5);
pair P = (0,2.5);
pair Q = (9,2.5);
dot("$P$", P, dir(180));
dot("$Q$", Q, dir(0));
draw(P--Q--B, blue+1.5);
\end{asy}
\end{center}
Notice that triangles $APM$ and $BQM$ are congruent,
and when $m \equiv n \pmod 2$, their colorings actually coincide.
So, the discrepancy of the triangle
is exactly equal to the discrepancy of $CPQB$, which is an $m \times n/2$
rectangle and hence equal to $0$ or $1/2$ according to parity.
For (b), note that a triangle with legs $m$ and $n$, with $m$ even and $n$ odd,
can be dissected into one right triangle with legs $m$ and $n-1$
plus a thin triangle of area $1/2$ which has height $m$ and base $1$.
The former region has discrepancy $0$ by (a),
and the latter region obviously has discrepancy at most its area of $m/2$,
hence $f(m,n) \le m/2$ as needed.
(An alternative slower approach, which requires a few cases,
is to prove that two adjacent columns have at most discrepancy $1/2$.)
For (c), we prove:
\begin{claim*}
For each $k \ge 1$, we have
\[ f(2k, 2k+1) = \frac{2k-1}{6}. \]
\end{claim*}
\begin{proof}
An illustration for $k=2$ is shown below,
where we use $(0,0)$, $(0,2k)$, $(2k+1,0)$ as the three vertices.
\begin{center}
\begin{asy}
size(8cm);
fill( (0,4)--(5,0)--(5,4)--cycle, palered );
draw(box( (0,0), (5,4) ), black);
fill( (0,3)--(1,3)--(1,3.2)--(0,4)--cycle, gray);
fill( (1,2)--(2,2)--(2,2.4)--(1.25,3)--(1,3)--cycle, gray);
fill( (2,1)--(3,1)--(3,1.6)--(2.50,2)--(2,2)--cycle, gray);
fill( (3,0)--(4,0)--(4,0.8)--(3.75,1)--(3,1)--cycle, gray);
fill(shift(1,0)*unitsquare, gray);
fill(shift(0,1)*unitsquare, gray);
for (int i=1; i<4; ++i) {
draw( (0,i)--(5,i), gray );
}
for (int i=1; i<5; ++i) {
draw( (i,0)--(i,4), gray );
}
draw( (0,4)--(5,0)--(0,0)--cycle, blue+2 );
\end{asy}
\end{center}
WLOG, the upper-left square is black, as above.
The $2k$ small white triangles just below the diagonal have area sum
\[ \frac12 \cdot \frac{1}{2k+1} \cdot \frac{1}{2k}
\left[ 1^2 + 2^2 + \dots + (2k)^2 \right] = \frac{4k+1}{12} \]
The area of the $2k$ black polygons sums just below the diagonal to
\[ \sum_{i=1}^{2k} \left( 1
- \frac12 \cdot \frac{1}{2k+1} \cdot \frac{1}{2k} \cdot i^2 \right)
= 2k - \frac{4k+1}{12} = \frac{20k-1}{12}. \]
Finally, in the remaining $1+2+\dots+2k$ squares,
there are $k$ more white squares than black squares.
So, it follows
\[ f(2k, 2k+1)
= \left\lvert -k + \frac{20k-1}{12} - \frac{4k+1}{12} \right\rvert
= \frac{2k-1}{6}. \]
\end{proof}
|
IMO-1997-notes_2
|
Let $ABC$ be a triangle with $\angle A < \min(\angle B, \angle C)$.
The points $B$ and $C$ divide the circumcircle of the triangle into two arcs.
Let $U$ be an interior point of the arc between $B$ and $C$ which does not contain $A$.
The perpendicular bisectors of $ AB$ and $ AC$ meet the line $AU$ at $V$ and $W$, respectively.
The lines $BV$ and $CW$ meet at $T$.
Show that $AU = TB + TC$.
|
Let $\ol{BTV}$ meet the circle again at $U_1$,
so that $AU_1 UB$ is an isosceles trapezoid.
Define $U_2$ similarly.
\begin{center}
\begin{asy}
pair A = dir(110);
pair B = dir(230);
pair C = dir(310);
pair U = dir(250);
pair U_1 = A*B/U;
pair U_2 = A*C/U;
pair T = extension(B, U_1, C, U_2);
filldraw(unitcircle, opacity(0.1)+lightcyan, blue);
filldraw(A--B--C--cycle, opacity(0.1)+lightred, red);
draw(A--U, deepgreen);
draw(C--U_2, deepgreen);
draw(B--U_1, deepgreen);
draw(B--U_2, red);
draw(U_1--U--U_2, red);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$U$", U, dir(U));
dot("$U_1$", U_1, dir(U_1));
dot("$U_2$", U_2, dir(U_2));
dot("$T$", T, dir(T));
/* TSQ Source:
A = dir 110
B = dir 230
C = dir 310
U = dir 250
U_1 = A*B/U
U_2 = A*C/U
T = extension B U_1 C U_2
unitcircle 0.1 lightcyan / blue
A--B--C--cycle 0.1 lightred / red
A--U deepgreen
C--U_2 deepgreen
B--U_1 deepgreen
*/
\end{asy}
\end{center}
Now from the isosceles trapezoids we get
\[ AU = BU_1 = BT + TU_1 = BT + TC \]
as desired.
|
IMO-1997-notes_3
|
Let $x_1$, $x_2$, \dots, $x_n$ be real numbers satisfying the conditions:
\begin{align*}
|x_1 + x_2 + \dots + x_n| &= 1 \\
|x_i| &\le \frac{n+1}{2} \qquad \text{for } i= 1,2, \dots, n
\end{align*}
Show that there exists a permutation $y_1$, $y_2$, \dots, $y_n$
of $x_1$, $x_2$, \dots, $x_n$ such that
\[ | y_1 + 2 y_2 + \dotsb + n y_n | \leq \frac {n + 1}{2}. \]
|
WLOG $\sum x_i = 1$ (by negating $x_i$) and $x_1 \le x_2 \le \dots \le x_n$.
Notice that
\begin{itemize}
\ii The largest possible value of the sum in question is
\[ A = x_1 + 2x_2 + 3x_3 + \dots + nx_n. \]
while the smallest value is
\[ B = nx_1 + (n-1)x_2 + \dots + x_n. \]
\ii Meanwhile, the \emph{average} value across all permutations is
\[ 1 \cdot \frac1n + 2 \cdot \frac1n + \dots + n \cdot \frac1n = \frac{n+1}{2}. \]
\end{itemize}
Now imagine we transform the sum $A$ to the sum $B$,
one step at a time, by swapping adjacent elements.
Every time we do a swap of two neighboring $u \le v$, the sum decreases by
\[ (iu + (i+1)v) - (iv + (i+1)u) = v-u \le n+1. \]
We want to prove we land in the interval
\[ I = \left[ -\frac{n+1}{2}, \frac{n+1}{2} \right] \]
at some point during this transformation.
But since $B \le \frac{n+1}{2} \le A$ (since $\frac{n+1}{2}$ was the average)
and our step sizes were at most the length of the interval $I$,
this is clear.
|
IMO-1997-notes_4
|
An $n \times n$ matrix whose entries come
from the set $S = \{1, 2, \dots , 2n - 1\}$
is called a \emph{silver} matrix if,
for each $i = 1, 2, \dots , n$,
the $i$-th row and the $i$-th column together
contain all elements of $S$. Show that:
\begin{enumerate}[(a)]
\ii there is no silver matrix for $n = 1997$;
\ii silver matrices exist for infinitely many values of $n$.
\end{enumerate}
|
\paragraph{Solution to (a).}
Define a \emph{cross} to be the union of the $i$th row and $i$th column.
Every cell of the matrix not on the diagonal is contained in exactly two crosses,
while each cell on the diagonal is contained in one cross.
On the other hand, if a silver matrix existed for $n=1997$,
then each element of $S$ appears in all $1997$ crosses.
Since $1997$ is odd, each number $s \in S$ must appear on the diagonal an odd number of times.
(For example, $s$ could appear on the diagonal once and off-diagonal $998$ times,
or on the diagonal three times and off-diagonal $997$ times, etc.)
In particular, each number $s$ appears at least once on the diagonal.
However, $|S| = 3993$ while there are only $1997$ diagonal cells.
This is a contradiction.
\paragraph{Solution to (b).}
We construct a silver matrix $M_e$ for $n = 2^e$ for each $e \ge 1$.
We write the first three explicitly for concreteness:
\begin{align*}
M_1 &= \begin{bmatrix}
1 & 2 \\ 3 & 1
\end{bmatrix} \\
M_2 &= \begin{bmatrix}
{\color{red}1} & {\color{red}2} & 4 & 5 \\
{\color{red}3} & {\color{red}1} & 6 & 7 \\
7 & 5 & {\color{red}1} & {\color{red}2} \\
6 & 4 & {\color{red}3} & {\color{red}1}
\end{bmatrix} \\
M_3 &= \begin{bmatrix}
{\color{red}1} & {\color{red}2} & {\color{red}4} & {\color{red}5} & 8 & 9 & 11 & 12\\
{\color{red}3} & {\color{red}1} & {\color{red}6} & {\color{red}7} & 10 & 15 & 13 & 14 \\
{\color{red}7} & {\color{red}5} & {\color{red}1} & {\color{red}2} & 14 & 12 & 8 & 9 \\
{\color{red}6} & {\color{red}4} & {\color{red}3} & {\color{red}1} & 13 & 11 & 10 & 15 \\
15 & 9 & 11 & 12 & {\color{red}1} & {\color{red}2} & {\color{red}4}
& {\color{red}5} \\
10 & 8 & 13 & 14 & {\color{red}3} & {\color{red}1} & {\color{red}6}
& {\color{red}7} \\
14 & 12 & 15 & 9 & {\color{red}7} & {\color{red}5} & {\color{red}1}
& {\color{red}2} \\
13 & 11 & 10 & 8 & {\color{red}6} & {\color{red}4} & {\color{red}3}
& {\color{red}1} \\
\end{bmatrix}
\end{align*}
The construction is described recursively as follows.
Let
\[
M_e' = \left[
\begin{array}{c|c}
{\color{red}M_{e-1}} & M_{e-1} + (2^e-1) \\ \hline
M_{e-1} + (2^e-1) & {\color{red}M_{e-1}} \\
\end{array}
\right].
\]
Then to get from $M_e'$ to $M_e$,
replace half of the $2^e$'s with $2^{e+1}-1$:
in the northeast quadrant, the even-indexed ones,
and in the southwest quadrant, the odd-indexed ones.
\begin{remark*}
In fact, it turns out silver matrices exist for all even dimensions.
A claimed proof is outlined at \url{https://aops.com/community/p7375020}.
\end{remark*}
|
IMO-1997-notes_5
|
Find all pairs $(a,b)$ of positive integers satisfying
\[ a^{b^2} = b^a. \]
|
The answer is $(1,1)$, $(16,2)$ and $(27,3)$.
We assume $a,b > 1$ for convenience.
Let $T$ denote the set of non perfect powers other than $1$.
\begin{claim*}
Every integer greater than $1$
is uniquely of the form $t^n$ for some $t \in T$, $n \in \NN$.
\end{claim*}
\begin{proof}
Clear.
\end{proof}
Let $a = s^m$, $b = t^n$.
\[ s^{m \cdot (t^n)^2} = t^{n \cdot s^m}. \]
Hence $s = t$ and we have
\[ m \cdot t^{2n} = n \cdot t^m
\implies t^{2n-m} = \frac nm. \]
Let $n = t^e m$ and $2 \cdot t^e m - m = e$, or
\[ e + m = 2t^e \cdot m. \]
We resolve this equation by casework
\begin{itemize}
\ii If $e > 0$, then $2t^e \cdot m > 2e \cdot m > e+m$.
\ii If $e=0$ we have $m=n$ and $m = 2m$, contradiction.
\ii If $e = -1$ we apparently have
\[ \frac{2}{t} \cdot m = m-1 \implies
m = \frac{t}{t-2} \]
so $(t,m) = (3,3)$ or $(t,m) = (4,2)$.
\ii If $e = -2$ we apparently have
\[ \frac{2}{t^2} \cdot m = m - 2
\implies m = \frac{2}{1 - 2/t^2} = \frac{2t^2}{t^2-2}. \]
This gives $(t,m) = (2,2)$.
\ii If $e \le -3$ then let $k = -e \ge 3$, so the equation is
\[ m-k = \frac{2m}{t^k}
\iff m = \frac{k \cdot t^k}{t^k-2}
= k + \frac{2k}{t^k-2}. \]
However, for $k \ge 3$ and $t \ge 2$,
we always have $2k \le t^k - 2$,
with equality only when $(t,k) = (2,3)$;
this means $m=4$, which is not a new solution.
\end{itemize}
|
IMO-1997-notes_6
|
For each positive integer $n$,
let $f(n)$ denote the number of ways of representing $n$
as a sum of powers of 2 with nonnegative integer exponents.
Representations which differ only in the ordering
of their summands are considered to be the same.
For instance, $f(4) = 4$,
because the number $4$ can be represented in the following four ways:
$4$; $2+2$; $2+1+1$; $1+1+1+1$.
Prove that for any integer $n \geq 3$
we have $2^{\frac{n^2}{4}} < f(2^n) < 2^{\frac{n^2}2}$.
\end{enumerate}
|
It's clear that $f$ is non-decreasing.
By sorting by the number of $1$'s we used,
we have the equation
\[ f(N) =
f\left( \left\lfloor \frac N2 \right\rfloor \right)
+ f\left( \left\lfloor \frac N2 \right\rfloor -1 \right)
+ f\left( \left\lfloor \frac N2 \right\rfloor -2 \right)
+ \dots
+ f(1) + f(0). \quad (\bigstar)
\]
\paragraph{Upper bound.}
We now prove the upper bound by induction.
Indeed, the base case is trivial and for the inductive step
we simply use $(\bigstar)$:
\[ f(2^n) = f(2^{n-1}) + f(2^{n-1}-1) + \dots
< 2^{n-1} f(2^{n-1})
< 2^{n-1} \cdot 2^{\frac{(n-1)^2}{2}}
= 2^{\frac{n^2}{2} - \half}.
\]
\paragraph{Lower bound.}
First, we contend that $f$ is convex.
We'll first prove this in the even case
to save ourselves some annoyance:
\begin{claim*}
[$f$ is basically convex]
If $2 \mid a+b$ then
we have $f(2a) + f(2b) \ge 2 f\left( a+b \right)$.
\end{claim*}
\begin{proof}
Since $f(2k+1) = f(2k)$, we will only prove the first equation.
Assume WLOG $a \ge b$ and use
$(\bigstar)$ on all three $f$ expressions here;
after subtracting repeated terms, the inequality then rewrites as
\[ \sum_{(a+b)/2 \le x \le a} f(x)
\ge \sum_{b \le x \le (a+b)/2} f(x). \]
This is true since there are an equal number of terms on each side
and $f$ is nondecreasing.
\end{proof}
\begin{claim*}
For each $1 \le k < 2^{n-1}$, we have
\[ f(2^{n-1} - k) + f(k+1) \ge 2f(2^{n-2}) \]
\end{claim*}
\begin{proof}
Use the fact that $f(2t+1)=f(2t)$ for all $t$
and then apply convexity as above.
\end{proof}
Now we can carry out the induction:
\[ f(2^n) = f(2^{n-1}) + f(2^{n-1}-1) + \dots
> 2^{n-1} f(2^{n-2}) + f(0)
> 2^{n-1} 2^{\frac{(n-2)^2}{4}} = 2^{\frac{n^2}{4}}.
\]
|
IMO-1998-notes_1
|
A convex quadrilateral $ABCD$ has perpendicular diagonals.
The perpendicular bisectors of the sides $AB$ and $CD$ meet
at a unique point $P$ inside $ABCD$.
Prove that the quadrilateral $ABCD$ is cyclic
if and only if triangles $ABP$ and $CDP$ have equal areas.
|
If $ABCD$ is cyclic, then $P$ is the circumcenter,
and $\angle APB + \angle PCD = 180\dg$.
The hard part is the converse.
\begin{center}
\begin{asy}
import graph; size(10cm);
pen zzttqq = rgb(0.6,0.2,0.); pen aqaqaq = rgb(0.62745,0.62745,0.62745); pen cqcqcq = rgb(0.75294,0.75294,0.75294);
draw((2.28154,-9.32038)--(8.62921,-11.79133)--(2.93422,-2.54008)--cycle, linewidth(0.6) + zzttqq);
draw((2.28154,-9.32038)--(-2.,-4.)--(-4.,-12.)--cycle, linewidth(0.6) + zzttqq);
draw((8.62921,-11.79133)--(-2.,-4.), linewidth(0.6));
draw((2.93422,-2.54008)--(-4.,-12.), linewidth(0.6));
draw((8.62921,-11.79133)--(2.93422,-2.54008), linewidth(0.6));
draw((-2.,-4.)--(-4.,-12.), linewidth(0.6));
draw((2.28154,-9.32038)--(5.78172,-7.16571), linewidth(0.6) + blue);
draw((2.28154,-9.32038)--(-3.,-8.), linewidth(0.6) + blue);
draw((2.28154,-9.32038)--(8.62921,-11.79133), linewidth(0.6) + zzttqq);
draw((8.62921,-11.79133)--(2.93422,-2.54008), linewidth(0.6) + zzttqq);
draw((2.93422,-2.54008)--(2.28154,-9.32038), linewidth(0.6) + zzttqq);
draw((2.28154,-9.32038)--(-2.,-4.), linewidth(0.6) + zzttqq);
draw((-2.,-4.)--(-4.,-12.), linewidth(0.6) + zzttqq);
draw((-4.,-12.)--(2.28154,-9.32038), linewidth(0.6) + zzttqq);
draw((2.93422,-2.54008)--(-0.31533,2.73866), linewidth(0.6));
draw((-2.,-4.)--(-0.31533,2.73866), linewidth(0.6));
draw(circle((2.25532,-9.31383), 6.80768), linewidth(0.6) + aqaqaq);
draw((0.51354,-5.84245)--(-3.,-8.), linewidth(0.6) + blue);
draw((0.51354,-5.84245)--(5.78172,-7.16571), linewidth(0.6) + blue);
dot("$D$", (-4.,-12.), dir((-76.113, -27.810)));
dot("$C$", (-2.,-4.), dir((-75.616, 41.734)));
dot("$B$", (2.93422,-2.54008), dir((1.940, 37.397)));
dot("$A$", (8.62921,-11.79133), dir((31.751, -33.422)));
dot("$P$", (2.28154,-9.32038), dir((-11.247, -66.945)));
dot("$M$", (5.78172,-7.16571), dir((39.728, 27.050)));
dot("$N$", (-3.,-8.), dir((-82.402, 23.307)));
dot("$E$", (0.51354,-5.84245), dir((-30.585, 62.531)));
dot("$X$", (-0.31533,2.73866), dir((-23.973, 36.915)));
\end{asy}
\end{center}
Let $M$ and $N$ be the midpoints of $\ol{AB}$ and $\ol{CD}$.
\begin{claim*}
Unconditionally, we have $\dang NEM = \dang MPN$.
\end{claim*}
\begin{proof}
Note that $\ol{EN}$ is the median of right triangle $\triangle ECD$, and similarly for $\ol{EM}$.
Hence $\dang NED = \dang EDN = \dang BDC$, while $\dang AEM = \dang ACB$.
Since $\dang DEA = 90\dg$, by looking at quadrilateral $XDEA$ where $X = \ol{CD} \cap \ol{AB}$,
we derive that $\dang NED + \dang AEM + \dang DXA = 90\dg$, so
\[ \dang NEM = \dang NED + \dang AEM + 90\dg = -\dang DXA = -\dang NXM = -\dang NPM \]
as needed.
\end{proof}
However, the area condition in the problem tells us
\[ \frac{EN}{EM} = \frac{CN}{CM} = \frac{PM}{PN}. \]
Finally, we have $\angle MEN > 90\dg$ from the configuration.
These properties uniquely determine the point $E$:
it is the reflection of $P$ across the midpoint of $MN$.
So $EMPN$ is a parallelogram, and thus $\ol{ME} \perp \ol{CD}$.
This implies $\dang BAE = \dang CEM = \dang EDC$ giving $ABCD$ cyclic.
|
IMO-1998-notes_2
|
In a competition, there are $a$ contestants
and $b$ judges, where $b \ge 3$ is an odd integer.
Each judge rates each contestant as either ``pass'' or ``fail''.
Suppose $k$ is a number such that for any two judges,
their ratings coincide for at most $k$ contestants.
Prove that
\[ \frac ka \ge \frac{b-1}{2b}. \]
|
This is a ``routine'' problem with global ideas.
We count pairs of coinciding ratings,
i.e.\ the number $N$ of tuples
\[(\{J_1, J_2\}, C) \]
of two distinct judges and a contestant
for which the judges gave the same rating.
On the one hand, if we count by the judges,
we have \[ N \le \binom b2 k \]
by he problem condition.
However, if we write $b=2m+1$ (so $m \coloneq \frac{b-1}{2}$), then each contestant $C$
contributes at least $\binom{m}{2} + \binom{m+1}{2} = m^2$ to $N$, and so
\[ N \ge a \cdot \left( \frac{b-1}{2} \right)^2 \]
Putting together the two estimates for $N$ yields the conclusion.
|
IMO-1998-notes_4
|
Determine all pairs $(x,y)$ of positive integers
such that $x^{2}y+x+y$ is divisible by $xy^{2}+y+7$.
|
The answer is $(7k^2,7k)$ for all $k \ge 1$,
as well as $(11,1)$ and $(49,1)$.
We are given $xy^2+y+7 \mid x^2y+x+y$.
Multiplying the right-hand side by $y$ gives
\[ xy^2+y+7 \mid x^2y^2+xy+y^2 \]
Then subtracting $x$ times the left-hand side gives
\[ xy^2+y+7 \mid y^2-7x. \]
We consider cases based on the sign of $y^2=7x$.
\begin{itemize}
\ii If $y^2 > 7x$, then $0 < y^2-7x < xy^2+y+7$,
contradiction.
\ii If $y^2=7x$, let $y = 7k$, so $x = 7k^2$.
Plugging this back in to the original equation reads
\[ 343k^4 + 7k + 7 \mid 343k^5 + 7k^2 + 7k \]
which is always valid, hence these are all solutions.
\ii If $y^2 < 7x$, then $|y^2-7x| \le 7x$,
so $y \in \{1,2\}$.
When $y=1$ we get
\[ x+8 \mid x^2+x+1 \iff x+8 \mid 64-8+1=57.\]
This has solutions $x=11$ and $x=49$.
When $y=2$
\begin{align*}
4x+9 \mid 2x^2+x+2 \\
\implies 4x+9 &\mid 16x^2+8x+16 \\
\implies 4x+9 &\mid 81-18+16 = 79
\end{align*}
which never occurs.
\end{itemize}
|
IMO-1998-notes_5
|
Let $I$ be the incenter of triangle $ABC$.
Let the incircle of $ABC$ touch
the sides $BC$, $CA$, and $AB$ at $K$, $L$, and $M$, respectively.
The line through $B$ parallel to $MK$ meets the lines
$LM$ and $LK$ at $R$ and $S$, respectively.
Prove that angle $RIS$ is acute.
|
Observe that $\triangle MKL$ is acute with circumcenter $I$.
We now present two proofs.
\paragraph{First simple proof (grobber).}
The problem is equivalent to showing $BI^2 > BR \cdot BS$.
But from
\[ \triangle BRK \sim \triangle MKL \sim \triangle BLS \]
we conclude
\[ BR = t \cdot \frac{MK}{ML},
\qquad BS = t \cdot \frac{ML}{MK} \]
where $t = BK = BL$ is the length
of the tangent from $B$.
Hence $BR \cdot BS = t^2$.
Since $BI > t$ is clear, we are done.
\paragraph{Second projective proof.}
Let $N$ be the midpoint of $\ol{KL}$,
and let ray $MN$ meet the incircle again at $P$.
Note that line $\ol{RBS}$ is the polar of $N$.
By Brocard's theorem, lines $MK$ and $PL$ should thus
meet the polar of $N$, so we conclude $R = \ol{MK} \cap \ol{PL}$.
Analogously, $S = \ol{ML} \cap \ol{PK}$.
Again by Brocard's theorem, $\triangle NRS$ is self-polar,
so $N$ is the orthocenter of $\triangle RIS$.
Since $N$ lies between $I$ and $B$ we are done.
|
IMO-1998-notes_6
|
Classify all functions $f \colon \NN \to \NN$
satisfying the identity
\[ f(n^2 f(m)) = m f(n)^2. \]
\end{enumerate}
|
Let $\mathcal P$ be the set of primes,
and let $g \colon \mathcal P \to \mathcal P$ be any involution on them.
Extend $g$ to a completely multiplicative function on $\NN$.
Then $f(n) = d g(n)$ is a solution for any $d \in \NN$
which is fixed by $g$.
It's straightforward to check these all work,
since $g \colon \NN \to \NN$ is an involution on them.
So we prove these are the only functions.
Let $d = f(1)$.
\begin{claim*}
We have $df(n) = f(dn)$ and $d \cdot f(ab) = f(a) f(b)$.
\end{claim*}
\begin{proof}
Let $P(m,n)$ denote the assertion in the problem statement.
Off the bat,
\begin{itemize}
\ii $P(1,1)$ implies $f(d) = d^2$.
\ii $P(n,1)$ implies $f(f(n)) = d^2n$.
In particular, $f$ is injective.
\ii $P(1,n)$ implies $f(dn^2) = f(n)^2$.
\end{itemize}
Then
\begin{align*}
f(a)^2 f(b)^2 &= f(da^2) f(b)^2 & \text{by third bullet}\\
&= f(b^2 f(f(da^2))) & \text{by problem statement} \\
&= f(b^2 \cdot d^2 \cdot da^2) & \text{by second bullet} \\
&= f(dab)^2 & \text{by third bullet} \\
\implies f(a) f(b) &= f(dab).
\end{align*}
This implies the first claim by taking $(a,b) = (1,n)$.
Then $df(a) = f(da)$, and so we actually have
$f(a) f(b) = d f(ab)$.
\end{proof}
\begin{claim*}
All values of $f$ are divisible by $d$.
\end{claim*}
\begin{proof}
We have
\begin{align*}
f(n^2) &= \frac 1d f(n)^2 \\
f(n^3) &= \frac{f(n^2) f(n)}{d} = \frac{f(n)^3}{d^2} \\
f(n^4) &= \frac{f(n^3) f(n)}{d} = \frac{f(n)^4}{d^3}
\end{align*}
and so on,
which implies the result.
\end{proof}
Then, define $g(n) = f(n) / d$.
We conclude that $g$ is completely multiplicative, with $g(1) = 1$.
However, $f(f(n)) = d^2n$ also implies $g(g(n)) = n$,
i.e.\ $g$ is an involution.
Moreover, since $f(d) = d^2$, $g(d) = d$.
All that remains is to check that $g$ must map primes to primes
to finish the description in the problem.
This is immediate; since $g$ is multiplicative and $g(1) = 1$,
if $g(g(p)) = p$ then $g(p)$ can have at most one prime factor,
hence $g(p)$ is itself prime.
\begin{remark*}
The IMO problem actually asked for the least value of $f(1998)$.
But for instruction purposes,
it is probably better to just find all $f$.
Since $1998 = 2 \cdot 3^3 \cdot 37$,
this answer is $2^3 \cdot 3 \cdot 5 = 120$, anyways.
\end{remark*}
|
IMO-1999-notes_1
|
A set $S$ of points from the space will be called
completely symmetric if it has at least three elements
and fulfills the condition that for every two distinct points $A$ and $B$ from $S$,
the perpendicular bisector plane of the segment $AB$ is a plane of symmetry for $S$.
Prove that if a completely symmetric set is finite,
then it consists of the vertices of either a regular polygon,
or a regular tetrahedron or a regular octahedron.
|
Let $G$ be the centroid of $S$.
\begin{claim*}
All points of $S$ lie on a sphere $\Gamma$ centered at $G$.
\end{claim*}
\begin{proof}
Each perpendicular bisector plane passes through $G$.
So if $A,B \in S$ it follows $GA = GB$.
\end{proof}
\begin{claim*}
Consider any plane passing through three or more points of $S$.
The points of $S$ in the plane form a regular polygon.
\end{claim*}
\begin{proof}
The cross section is a circle because we are intersecting a plane with sphere $\Gamma$.
Now if $A$, $B$, $C$ are three adjacent points on this circle,
by taking the perpendicular bisector we have $AB=BC$.
\end{proof}
If the points of $S$ all lie in a plane, we are done.
Otherwise, the points of $S$ determine a polyhedron $\Pi$ inscribed in $\Gamma$.
All of the faces of $\Pi$ are evidently regular polygons,
of the same side length $s$.
\begin{claim*}
Every face of $\Pi$ is an equilateral triangle.
\end{claim*}
\begin{proof}
Suppose on the contrary some face $A_1 A_2 \dots A_n$ has $n > 3$.
Let $B$ be any vertex adjacent to $A_1$ in $\Pi$ other than $A_2$ or $A_n$.
Consider the plane determined by $\triangle A_1 A_3 B$.
This is supposed to be a regular polygon, but arc $A_1 A_3$ is longer than arc $A_1 B$,
and by construction there are no points inside these arcs.
This is a contradiction.
\end{proof}
Hence, $\Pi$ has faces all congruent equilateral triangles.
This implies it is a regular polyhedron --- either
a regular tetrahedron, regular octahedron, or regular icosahedron.
We can check the regular icosahedron fails by
taking two antipodal points as our counterexample.
This finishes the problem.
|
IMO-1999-notes_2
|
Find the least constant $C$ such that for any integer $n > 1$ the inequality
\[\sum_{1 \le i < j \le n} x_i x_j (x_i^2 + x_j^2)
\le C \left( \sum_{1 \le i \le n} x_i \right)^4\]
holds for all real numbers $x_1, \dots, x_n \ge 0$.
Determine the cases of equality.
|
Answer: $C = \frac 18$, with equality when two $x_i$ are equal
and the remaining $x_i$ are equal to zero.
We present two proofs of the bound.
\paragraph{First solution by smoothing.}
Fix $\sum x_i = 1$.
The sum on the left-hand side can be interpreted as
$\sum_{i=1}^n x_i^3 \sum_{j \neq i} x_j = \sum_{i=1}^n x_i^3(1-x_i)$,
so we may rewrite the inequality as:
Then it becomes \[ \sum_i (x_i^3 - x_i^4) \le C. \]
\begin{claim*}
[Smoothing]
Let $f(x) = x^3 - x^4$.
If $u + v \le \frac 34$, then $f(u) + f(v) \le f(0) + f(u+v)$.
\end{claim*}
\begin{proof}
Note that
\begin{align*}
(u^3-u^4)+(v^3-v^4) &\le (u+v)^3-(u+v)^4 \\
\iff uv(4u^2+4v^2+6uv) &\le 3uv(u+v)
\end{align*}
If $u+v\le \frac 34$ this is obvious as $4u^2+4v^2+6uv \le 4(u+v)^2$.
\end{proof}
Observe that if three nonnegative reals have pairwise sums
exceeding $\frac34$ then they have sum at least $\frac 98$.
Hence we can smooth until $n-2$ of the terms are zero.
Hence it follows
\[ C = \max_{a+b=1} (a^3+b^3-a^4-b^4) \]
which is routine computation giving $C = \frac18$.
\paragraph{Second solution by AM-GM (Nairit Sarkar).}
Write
\begin{align*}
\text{LHS}
&\le \left( \sum_{1 \le k \le n} x_k^2 \right)
\left( \sum_{1 \le i < j \le n} x_i x_j \right)
= \half \left( \sum_{1 \le k \le n} x_k^2 \right)
\left( \sum_{1 \le i < j \le n} 2 x_i x_j \right) \\
&\le \half \left( \frac{\sum_k x_k^2 + 2 \sum_{i<j} x_i x_j}{2} \right)^2
= \frac 18 \left( \sum_{1 \le i < n} x_i \right)^4
\end{align*}
as desired.
|
IMO-1999-notes_3
|
Let $n$ be an even positive integer.
Find the minimal number of cells on the $n \times n$ board
that must be marked so that any cell
(marked or not marked) has a marked neighboring cell.
|
For every marked cell, consider the marked cell adjacent
to it; in this way we have a \emph{domino} of two cells.
For each domino, its \emph{aura} consists of all the cells
which are adjacent to a cell of the domino.
There are up to eight squares in each aura,
but some auras could be cut off by the boundary of the board,
which means that there could be as few as five squares.
We will prove that $\half n (n+2)$
is the minimum number of auras
needed to cover the board
(the auras need not be disjoint).
\begin{itemize}
\ii A construction is shown on the left below,
showing that $\half n (n+2)$ is sufficient.
\ii Color the board as shown to the right into ``rings''.
Every aura takes covers exactly (!) four blue cells.
Since there are $2n(n+2)$ blue cells, this implies the lower bound.
\end{itemize}
\begin{center}
\begin{asy}
size(11cm);
path aura = (1,0)--(2,0)--(2,1)--(3,1)--(3,3)--(2,3)--(2,4)--(1,4)--(1,3)--(0,3)--(0,1)--(1,1)--cycle;
picture aura_up(pen fill) {
picture pic = new picture;
fill(pic, aura, fill);
draw(pic, (1,1)--(2,1), lightgray );
draw(pic, (1,3)--(2,3), lightgray );
draw(pic, (0,2)--(3,2), lightgray );
draw(pic, (1,1)--(1,3), lightgray );
draw(pic, (2,1)--(2,3), lightgray );
draw(pic, aura, blue+1.25);
return pic;
}
picture aura_right(pen fill) {
return shift(0,3) * rotate(-90) * aura_up(fill);
}
add(shift(-1,-1)*aura_up(palegreen));
add(shift(-1,3)*aura_up(palegreen));
add(shift(-1,7)*aura_up(palegreen));
add(shift(1,1)*aura_up(palegreen));
add(shift(1,5)*aura_up(palegreen));
add(shift(3,3)*aura_up(palegreen));
add(shift(6,3)*aura_up(palegreen));
add(shift(8,1)*aura_up(palegreen));
add(shift(8,5)*aura_up(palegreen));
add(shift(10,-1)*aura_up(palegreen));
add(shift(10,3)*aura_up(palegreen));
add(shift(10,7)*aura_up(palegreen));
add(shift(4,6)*aura_right(palered));
add(shift(2,8)*aura_right(palered));
add(shift(6,8)*aura_right(palered));
/*
add(shift(0,10)*aura_right(palered));
add(shift(4,10)*aura_right(palered));
add(shift(8,10)*aura_right(palered));
*/
add(shift(4,1)*aura_right(palered));
add(shift(2,-1)*aura_right(palered));
add(shift(6,-1)*aura_right(palered));
add(shift(0,-3)*aura_right(palered));
add(shift(4,-3)*aura_right(palered));
add(shift(8,-3)*aura_right(palered));
draw(shift(0,-2)*scale(12)*unitsquare, black+2);
path cell = shift(16,-2)*unitsquare;
for (int i = 0; i <= 11; ++i) {
for (int j = 0; j <= 11; ++j) {
if (max(abs(i-5.5), abs(j-5.5)) == 5.5)
filldraw(shift(i,j)*cell, paleblue, gray);
else if (max(abs(i-5.5), abs(j-5.5)) == 3.5)
filldraw(shift(i,j)*cell, paleblue, gray);
else if (max(abs(i-5.5), abs(j-5.5)) == 1.5)
filldraw(shift(i,j)*cell, paleblue, gray);
else draw(shift(i,j)*cell, gray);
}
}
\end{asy}
\end{center}
Note that this proves that a partition into disjoint auras
actually always has exactly $\half n (n+2)$ auras,
thus also implying EGMO 2019/2.
|
IMO-1999-notes_4
|
Find all pairs of positive integers $(x,p)$
such that $p$ is a prime and $x^{p-1}$ is a divisor of $(p-1)^{x}+1$.
|
If $p = 2$ then $x \in \{1,2\}$,
and if $p = 3$ then $x \in \{1,3\}$, since this is IMO 1990/3.
Also, $x = 1$ gives a solution for any prime $p$.
We show that there are no other solutions.
Assume $x > 1$ and let $q$ be smallest prime divisor of $x$.
We have $q > 2$ since $(p-1)^x+1$ is odd.
Then
\[ (p-1)^x \equiv -1 \pmod q \implies (p-1)^{2x} \equiv 1 \pmod q \]
so the order of $p-1 \bmod q$ is even and divides $\gcd(q-1,2x) \le 2$.
This means that
\[ p-1 \equiv -1 \pmod q \implies p = q. \]
In other words $p \mid x$ and we get $x^{p-1} \mid (p-1)^{x}+1$.
By exponent lifting lemma, we now have
\[ 0 < (p-1) \nu_{p}(x) \le 1 + \nu_p(x). \]
This forces $p=3$,
which we already addressed.
|
IMO-1999-notes_5
|
Two circles $\Omega_{1}$ and $\Omega_{2}$ touch internally the circle
$\Omega$ in $M$ and $N$ and the center of $\Omega_{2}$ is on $\Omega_{1}$.
The common chord of the circles $\Omega_{1}$ and $\Omega_{2}$
intersects $\Omega$ in $A$ and $B$
Lines $MA$ and $MB$ intersects $\Omega_{1}$ in $C$ and $D$.
Prove that $\Omega_{2}$ is tangent to $CD$.
|
Let $P$ and $Q$ be the centers of $\Omega_1$ and $\Omega_2$.
Let line $MQ$ meet $\Omega_1$ again at $W$,
the homothetic image of $Q$ under $\Omega_1 \to \Omega$.
Meanwhile, let $T$ be the intersection of segment $PQ$
with $\Omega_2$, and let $L$ be its homothetic image on $\Omega$.
Since $\ol{PTQ} \perp \ol{AB}$, it follows $\ol{LW}$ is a diameter of $\Omega$.
Let $O$ be its center.
\begin{center}
\begin{asy}
size(10cm);
pen xfqqff = rgb(0.49803,0.,1.); pen qqffff = rgb(0.,1.,1.); pen cqcqcq = rgb(0.75294,0.75294,0.75294);
pair O = (0.,0.), M = (-0.54799,0.83648), P = (-0.16957,0.25884), Q = (-0.16861,-0.43170), A = (-0.97449,-0.22439), B = (0.97512,-0.22166), C = (-0.84251,0.10388), D = (0.50379,0.10577), T = (-0.16936,0.10483), L = (-0.00139,0.99999);
draw(circle(O, 1.), linewidth(0.6));
draw(circle(P, 0.69055), linewidth(0.6) + xfqqff);
draw(circle(Q, 0.53653), linewidth(0.6) + xfqqff);
draw(O--M, linewidth(0.6) + red);
draw(M--A, linewidth(0.6) + qqffff);
draw(M--B, linewidth(0.6) + qqffff);
draw(O--(-0.36380,-0.93147), linewidth(0.6) + red);
draw(P--Q, linewidth(0.6));
draw(M--(0.00139,-0.99999), linewidth(0.6) + green);
draw((0.00139,-0.99999)--L, linewidth(0.6));
draw(L--(-0.36380,-0.93147), linewidth(0.6) + green);
draw(M--(-0.36380,-0.93147), linewidth(0.6));
draw(A--B, linewidth(0.6));
dot("$O$", O, dir((1.868, -2.897)));
dot("$N$", (-0.36380,-0.93147), dir((-5.700, -12.058)));
dot("$M$", M, dir((-8.656, 10.478)));
dot("$P$", P, dir((-2.789, 6.521)));
dot("$Q$", Q, dir((3.600, -6.558)));
dot("$A$", A, dir((-15.478, -5.915)));
dot("$B$", B, dir((4.982, -4.507)));
dot("$C$", C, dir((-8.023, -3.199)));
dot("$D$", D, dir((1.442, 0.454)));
dot("$T$", T, dir((-7.133, 1.989)));
dot("$W$", (0.00139,-0.99999), dir((-2.113, -15.052)));
dot("$L$", L, dir((1.046, 1.812)));
dot("$E$", midpoint(A--B), dir(-45));
\end{asy}
\end{center}
\begin{claim*}
$MNTQ$ is cyclic.
\end{claim*}
\begin{proof}
By Reim: $\dang TQM = \dang LWM = \dang LNM = \dang TNM$.
\end{proof}
Let $E$ be the midpoint of $\ol{AB}$.
\begin{claim*}
$OEMN$ is cyclic.
\end{claim*}
\begin{proof}
By radical axis, the lines $MM$, $NN$, $AEB$ meet at a point $R$.
Then $OEMN$ is on the circle with diameter $\ol{OR}$.
\end{proof}
\begin{claim*}
$MTE$ are collinear.
\end{claim*}
\begin{proof}
$\dang NMT = \dang TQN = \dang LON = \dang NOE = \dang NME$.
\end{proof}
Now consider the homothety mapping $\triangle WAB$ to $\triangle QCD$.
It should map $E$ to a point on line $ME$
which is also on the line through $Q$ perpendicular to $\ol{AB}$; that is, to point $T$.
Hence $TCD$ are collinear, and it's immediate that $T$ is the desired tangency point.
|
IMO-1999-notes_6
|
Find all the functions $f \colon \RR \to \RR$ such that
\[f(x-f(y))=f(f(y))+xf(y)+f(x)-1\]
for all $x,y \in \RR$.
\end{enumerate}
|
The answer is $f(x) = -\half x^2+1$
which obviously works.
For the other direction, first note that
\[ P(f(y),y) \implies 2f(f(y)) + f(y)^2 - 1 = f(0). \]
We introduce the notation $c = \frac{f(0)-1}{2}$,
and $S = \opname{img} f$.
Then the above assertion says
\[ f(s) = -\half s^2 + (c + 1). \]
Thus, the given functional equation can be rewritten as
\[ Q(x,s) : f(x-s)=-\half s^2 + sx + f(x) - c. \]
\begin{claim*}
[Main claim]
We can find a function $g \colon \RR \to \RR$ such that
\[ f(x-z) = zx + f(x) + g(z). \qquad (\spadesuit). \]
\end{claim*}
\begin{proof}
If $z \neq 0$,
the idea is to fix a nonzero value $s_0 \in S$ (it exists)
and then choose $x_0$ such that $- \half s_0^2 + s_0 x_0 - c = z$.
Then, $Q(x_0, s)$ gives an pair $(u,v)$ with $u-v = z$.
But now for any $x$, using $Q(x+v,u)$ and $Q(x,-v)$ gives
\begin{align*}
f(x-z)-f(x) &= f(x-u+v)-f(x)
= f(x+v)-f(x) + u(x+v) - \half u^2 + c \\
&= -vx-\half s^2-c + u(x+v) - \half u^2 + c \\
&= -vx-\half v^2 + u(x+v) - \half u^2 = zx + g(z)
\end{align*}
where $g(z) = -\half(u^2+v^2)$ depends only on $z$.
\end{proof}
Now, let
\[ h(x) \coloneq \half x^2 + f(x) - (2c+1), \]
so $h(0) = 0$.
\begin{claim*}
The function $h$ is additive.
\end{claim*}
\begin{proof}
We just need to rewrite $(\spadesuit)$.
Letting $x=z$ in $(\spadesuit)$,
we find that actually $g(x)=f(0)-x^2-f(x)$.
Using the definition of $h$ now gives
\[ h(x-z) = h(x) + h(z). \qedhere \]
\end{proof}
To finish, we need to remember that $f$, hence $h$, is known
on the image
\[ S = \left\{ f(x) \mid x \in \RR \right\}
= \left\{ h(x) - \half x^2 + (2c+1) \mid x \in \RR \right\}. \]
Thus, we derive
\[ h\left( h(x)-\half x^2+(2c+1) \right) = -c
\qquad \forall x \in \RR. \qquad(\heartsuit) \]
We can take the following two instances of $\heartsuit$:
\begin{align*}
h\left( h(2x)-2x^2+(2c+1) \right) &= -c \\
h\left( 2h(x)-x^2+2(2c+1) \right) &= -2c.
\end{align*}
Now subtracting these and using $2h(x)=h(2x)$ gives
\[ c = h\left( -x^2 - (2c+1) \right). \]
Together with $h$ additive, this implies readily $h$ is constant.
That means $c=0$ and the problem is solved.
|
IMO-2000-notes_1
|
Two circles $G_1$ and $G_2$ intersect at two points $M$ and $N$.
Let $AB$ be the line tangent to these circles at $A$ and $B$,
respectively, so that $M$ lies closer to $AB$ than $N$.
Let $CD$ be the line parallel to $AB$
and passing through the point $M$,
with $C$ on $G_1$ and $D$ on $G_2$.
Lines $AC$ and $BD$ meet at $E$; lines $AN$ and $CD$ meet at $P$;
lines $BN$ and $CD$ meet at $Q$.
Show that $EP = EQ$.
|
First, we have $\dang EAB = \dang ACM = \dang BAM$
and similarly $\dang EBA = \dang BDM = \dang ABM$.
Consequently, $\ol{AB}$ bisects $\angle EAM$ and $\angle EBM$,
and hence $\triangle EAB \cong \triangle MAB$.
\begin{center}
\begin{asy}
pair M = dir(120);
pair U = dir(170);
pair V = dir(10);
pair N = 0.3*V+0.7*U;
pair O_1 = circumcenter(M, U, N);
pair O_2 = circumcenter(M, V, N);
real r1 = abs(U-O_1);
real r2 = abs(V-O_2);
pair Se = (r1*O_2-r2*O_1)/(r1-r2);
path w1 = CP(O_1,N);
path w2 = CP(O_2,N);
filldraw(w1, opacity(0.1)+lightcyan, blue);
filldraw(w2, opacity(0.1)+lightcyan, blue);
pair A = IP(w1, CP(midpoint(Se--O_1), Se));
pair B = IP(w2, CP(midpoint(Se--O_2), Se));
draw(A--B, red);
pair C = -M+2*foot(O_1, M, M+B-A);
pair D = -M+2*foot(O_2, M, M+B-A);
draw(C--D, red);
pair E = extension(A, C, B, D);
pair P = extension(A, N, C, D);
pair Q = extension(B, N, C, D);
draw(C--E--D, blue);
draw(A--N--B, blue);
draw(P--E--Q, heavygreen);
pair T = extension(M, N, A, B);
draw(T--N, heavycyan);
draw(A--M--B, purple);
draw(E--M, red);
dot("$M$", M, dir(350));
dot("$N$", N, dir(260));
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
dot("$P$", P, dir(260));
dot("$Q$", Q, dir(330));
dot("$T$", T, dir(160));
/* TSQ Source:
M = dir 120 R350
U := dir 170
V := dir 10
N = 0.3*V+0.7*U R260
O_1 := circumcenter M U N R-90
O_2 := circumcenter M V N R-90
!real r1 = abs(U-O_1);
!real r2 = abs(V-O_2);
Se := (r1*O_2-r2*O_1)/(r1-r2)
!path w1 = CP(O_1,N);
!path w2 = CP(O_2,N);
w1 0.1 lightcyan / blue
w2 0.1 lightcyan / blue
A = IP w1 CP midpoint Se--O_1 Se
B = IP w2 CP midpoint Se--O_2 Se
A--B red
C = -M+2*foot O_1 M M+B-A
D = -M+2*foot O_2 M M+B-A
C--D red
E = extension A C B D
P = extension A N C D R260
Q = extension B N C D R330
C--E--D blue
A--N--B blue
P--E--Q heavygreen
T = extension M N A B R160
T--N heavycyan
A--M--B purple
E--M red
*/
\end{asy}
\end{center}
Now it is well-known that $\ol{MN}$ bisects $\ol{AB}$
and since $\ol{AB} \parallel \ol{PQ}$
we deduce that $M$ is the midpoint of $\ol{PQ}$.
As $\ol{AB}$ is the perpendicular bisector of $\ol{EM}$,
it follows that $EP = EQ$ as well.
|
IMO-2000-notes_2
|
Let $a$, $b$, $c$ be positive real numbers with $abc = 1$.
Show that
\[
\left( a - 1 + \frac 1b \right)
\left( b - 1 + \frac 1c \right)
\left( c - 1 + \frac 1a \right)
\le 1.
\]
|
Let $a = x/y$, $b = y/z$, $c = z/x$ for $x,y,z > 0$.
Then the inequality rewrites as
\[ (-x+y+z)(x-y+z)(x+y-z) \le xyz \]
which when expanded is equivalent to Schur's inequality.
Alternatively, if one wants to avoid appealing to Schur,
then the following argument works:
\begin{itemize}
\ii At most one term on the left-hand side is negative;
if that occurs we are done from $xyz > 0 > (-x+y+z)(x-y+z)(x+y-z)$.
\ii If all terms in the left-hand side are nonnegative,
let us denote $m = -x+y+z \ge 0$, $n = x-y+z \ge 0$, $p = x+y-z \ge 0$.
Then it becomes
\[ mnp \le \frac{(m+n)(n+p)(p+m)}{8} \]
which follows by AM-GM.
\end{itemize}
|
IMO-2000-notes_3
|
Let $n \ge 2$ be a positive integer
and $\lambda$ a positive real number.
Initially there are $n$ fleas on a horizontal line,
not all at the same point.
We define a move as choosing two fleas at some points $A$ and $B$,
with $A$ to the left of $B$,
and letting the flea from $A$ jump over the flea from $B$ to the point $C$
so that $\frac{BC}{AB} = \lambda$.
Determine all values of $ \lambda$ such that,
for any point $M$ on the line
and for any initial position of the $n$ fleas,
there exists a sequence of moves that will take
them all to the position right of $M$.
|
The answer is $\lambda \ge \frac{1}{n-1}$.
We change the problem by replacing the fleas
with \textbf{bowling balls} $B_1$, $B_2$, \dots, $B_n$ in that order.
Bowling balls aren't exactly great at jumping,
so each move can now be described as follows:
\begin{itemize}
\ii Select two indices $i < j$.
Then ball $B_i$ moves to $B_{i+1}$'s location,
$B_{i+1}$ moves to $B_{i+2}$'s location, and so on;
until $B_{j-1}$ moves to $B_j$'s location,
\ii Finally, $B_j$ moves some distance forward;
the distance is at most $\lambda \cdot |B_j B_i|$
and $B_j$ may not pass $B_{j+1}$.
\end{itemize}
\begin{claim*}
If $\lambda < \frac{1}{n-1}$
the bowling balls have bounded movement.
\end{claim*}
\begin{proof}
Let $a_i \ge 0$ denote the initial distance
between $B_i$ and $B_{i+1}$,
and let $\Delta_i$ denote the distance travelled by ball $i$.
Of course we have
$\Delta_1 \le a_1 + \Delta_2$,
$\Delta_2 \le a_2 + \Delta_3$,
\dots,
$\Delta_{n-1} \le a_{n-1} + \Delta_n$
by the relative ordering of the bowling balls.
Finally, distance covered by $B_n$ is always
$\lambda$ times distance travelled by other bowling balls, so
\begin{align*}
\Delta_n &\le \lambda \sum_{i=1}^{n-1} \Delta_i
\le \lambda \sum_{i=1}^{n-1}
\left( \left( a_i + a_{i+1} + \dots + a_{n-1} \right)
+ \Delta_n \right) \\
&= (n-1)\lambda \cdot \Delta_n + \sum_{i=1}^{n-1} i a_i
\end{align*}
and since $(n-1)\lambda > 1$, this gives an upper bound.
\end{proof}
\begin{remark*}
Equivalently, you can phrase the proof without
bowling balls as follows:
if $x_1 < \dots < x_n$ are the positions of the fleas,
the quantity
\[ L = x_n - \lambda(x_1 + \dots + x_{n-1}) \]
is a monovariant which never increases;
i.e.\ $L$ is bounded above.
Since $L > (1-(n-1)\lambda) x_n$, it follows
$\lambda < \frac{1}{n-1}$ is enough to stop the fleas.
\end{remark*}
\begin{claim*}
When $\lambda \ge \frac{1}{n-1}$,
it suffices to always jump the leftmost flea
over the rightmost flea.
\end{claim*}
\begin{proof}
If we let $x_i$ denote the distance travelled by $B_1$
in the $i$th step,
then $x_i = a_i$ for $1 \le i \le n-1$
and $x_i = \lambda(x_{i-1} + x_{i-2} + \dots + x_{i-(n-1)})$.
In particular, if $\lambda \ge \frac{1}{n-1}$
then each $x_i$ is at least the average of the previous $n-1$ terms.
So if the $a_i$ are not all zero,
then $\{x_{n}, \dots, x_{2n-2}\}$ are all positive
and thereafter $x_i \ge \min \left\{ x_n, \dots, x_{2n-2} \right\} > 0$
for every $i \ge 2n-1$.
So the partial sums of $x_i$ are unbounded, as desired.
\end{proof}
\begin{remark*}
Other inductive constructions are possible.
Here is the idea of one of them,
although the details are more complicated.
We claim in general that given $n-1$ fleas at $0$
and one flea at $1$,
we can get all the fleas arbitrarily close to
$\frac{1}{1-(n-1)\lambda}$
(or as far as we want if $\lambda > \frac{1}{n-1}$.).
The proof is induction by $n \ge 2$;
for $n=2$ we get a geometric series.
For $n \ge 3$, we leave one flea at zero
and move the remainder close to $\frac{1}{1-(n-2)\lambda}$,
then jump the last flea to
$\frac{1+\lambda}{1-(n-2)\lambda}$.
Now we're in the same situation,
except we shifted $\frac{1}{1-(n-2)\lambda}$ right
and have then scaled everything by
$r = \frac{\lambda}{1-(n-2)\lambda}$.
If we repeat this process again and check the geometric series,
we see the fleas converge to
\[ \frac{1}{1-(n-2)\lambda}
\left( 1 + r + r^2 + r^3 + \dots \right)
= \frac{1}{1-(n-2)\lambda} \cdot \frac{1}{1-r}
= \frac{1}{1-(n-1)\lambda}. \]
\end{remark*}
|
IMO-2000-notes_4
|
A magician has one hundred cards numbered $1$ to $100$.
He puts them into three boxes, a red one, a white one and a blue one,
so that each box contains at least one card.
A member of the audience draws two cards from two different boxes
and announces the sum of numbers on those cards.
Given this information,
the magician locates the box from which no card has been drawn.
How many ways are there to put the cards
in the three boxes so that the trick works?
|
There are $2 \cdot 3! = 12$ ways, which amount to:
\begin{itemize}
\ii Partitioning the cards modulo $3$, or
\ii Placing $1$ alone in a box,
$100$ alone in a second box,
and all remaining cards in the third box.
\end{itemize}
These are easily checked to work so we prove they are the only ones.
\paragraph{First solution.}
We proceed by induction on $n \ge 3$ with the base case being immediate.
For the inductive step,
consider a working partition of $\{1, 2, \dots, n\}$.
Then either $n$ is in its own box; or
deleting $n$ gives a working partition of $\{1, 2, \dots, n-1\}$.
Similarly, either $1$ is in its own box; or
deleting $1$ gives a working partition of $\{2, 3, \dots, n\}$,
and we can reduce all numbers by $1$ to get
a working partition of $\{1, 2, \dots, n-1\}$.
Therefore, we only need to consider there cases.
\begin{itemize}
\ii If $1$ and $n$ are both in their own box,
this yields one type of solution we already found.
\ii If $n$ is not in a box by itself,
then by induction hypothesis the cards $1$ through $n-1$
are either arranged mod $3$,
or as $\{1\} \cup \{2,3,\dots,n-2\} \cup \{n-1\}$.
\begin{itemize}
\ii In the former mod $3$ situation,
since $n + (n-3) = (n-1) + (n-2)$,
so $n$ must be in the same box as $n-3$.
\ii In the latter case and for $n > 4$,
since $n + 1 = 2 + (n-1)$,
$n$ must be in the same box as $1$.
But now $n + 2 = (n-1) + 3$ for $n > 4$, contradiction.
\end{itemize}
\ii The case where $1$ is in a box by itself is analogous.
\end{itemize}
This exhausts all cases, completing the proof.
\paragraph{Second solution.}
Let $A$, $B$, $C$ be the sets of cards in the three boxes.
Then $A+B$, $B+C$, $C+A$ should be disjoint,
and contained in $\{3, 4, \dots, 199\}$.
On the other hand, we have the following famous fact.
\begin{lemma*}
Let $X$ and $Y$ be finite nonempty sets of real numbers.
We have $|X+Y| \ge |X|+|Y|-1$,
with equality if and only if $X$ and $Y$ are arithmetic
progressions with the same common difference,
or one of $X$ and $Y$ is a singleton set.
\end{lemma*}
Putting these two together gives the estimates
\[ 197 \ge |A+B| + |B+C| + |C+A|
\ge 2\left( |A|+|B|+|C| \right)-3 = 197. \]
So all the inequalities must be sharp.
Consequently we conclude that:
\begin{claim*}
Either the sets $A$, $B$, $C$ are disjoint arithmetic progressions
with the same common difference
$d = \min_{x \neq y \text{ in same set}} |x-y|$,
or two of the sets are two singleton.
Moreover, $\{3, 4, \dots, 199\}
= (A+B) \sqcup (B+C) \sqcup (C+A)$.
\end{claim*}
From here it is not hard to deduce the layouts above are the only ones,
but there are some details.
First, we make the preliminary observation that
$3=1+2$, $4=1+3$, $198=98+100$, $199=99+100$
and these numbers can't be decomposed in other ways;
thus from the remark about the disjoint union:
\begin{claim*}
[Convenient corollary]
The pairs $(1,2)$, $(1,3)$, $(98,100)$, $(99,100)$
are all in different sets.
\end{claim*}
We now consider the four cases.
\begin{itemize}
\ii If two of the boxes are singletons,
it follows from the corollary that we should have $A = \{1\}$,
$B = \{100\}$ and $C = \{2, \dots, 99\}$, up to permutation.
\ii Otherwise $A$, $B$, $C$ are disjoint arithmetic
progressions with the same common difference $d$.
As two of $\{1,2,3,4\}$ are in the same box
(by pigeonhole), we must have $d \le 3$.
\begin{itemize}
\ii If $d=3$, then no two elements of different residues
modulo $3$ can be in the same box,
so we must be in the first construction claimed earlier.
\ii If $d=2$, then the convenient corollary
tells us we may assume WLOG that $1 \in A$ and $2 \in B$,
hence $3 \in C$
(since $3 \notin A$ by convenient corollary,
and $3 \notin B$ because common difference $2$).
Thus we must have $A = \{1\}$, $B = \{2, 4, \dots, 100\}$
and $C = \{3, 5, \dots 99\}$
which does not work since $1+4 = 2+3$.
Therefore there are no solutions in this case.
\ii If $d=1$, then by convenient corollary
the numbers $1$ and $2$ are in different sets,
as are $99$ and $100$.
So we must have $A = \{1\}$, $B = \{2, \dots, 99\}$, $C = \{100\}$
which we have already seen is valid.
\end{itemize}
\end{itemize}
|
IMO-2000-notes_5
|
Does there exist a positive integer $n$
such that $n$ has exactly 2000 distinct prime divisors
and $n$ divides $2^n + 1$?
|
Answer: Yes.
We say that $n$ is \emph{Korean} if $n \mid 2^n+1$.
First, observe that $n=9$ is Korean.
Now, the problem is solved upon the following claim:
\begin{claim*}
If $n > 3$ is Korean,
there exists a prime $p$ not dividing $n$
such that $np$ is Korean too.
\end{claim*}
\begin{proof}
I claim that one can take any primitive prime divisor $p$ of $2^{2n}-1$,
which exists by Zsigmondy theorem.
Obviously $p \neq 2$.
Then:
\begin{itemize}
\ii Since $p \nmid 2^{\varphi(n)}-1$ it follows then that $p \nmid n$.
\ii Moreover, $p \mid 2^n+1$ since $p \nmid 2^n-1$.
\end{itemize}
Hence $np \mid 2^n+1 \mid 2^{np} + 1$ by Chinese Theorem,
since $\gcd(n,p) = 1$.
% Thanks Wanlin for catching this
\end{proof}
|
IMO-2000-notes_6
|
Let $\ol{AH_1}$, $\ol{BH_2}$, and $\ol{CH_3}$ be the
altitudes of an acute triangle $ABC$.
The incircle $\omega$ of triangle $ABC$ touches the sides
$BC$, $CA$ and $AB$ at $T_1$, $T_2$ and $T_3$, respectively.
Consider the reflections of the lines $H_1H_2$, $H_2H_3$, and
$H_3H_1$ with respect to the lines $T_1T_2$, $T_2T_3$, and $T_3T_1$.
Prove that these images form a triangle whose vertices lie on $\omega$.
\end{enumerate}
|
We use complex numbers with $\omega$ the unit circle.
Let $T_1 = a$, $T_2 = b$, $T_3 = c$.
The main content of the problem is to show that
the triangle in question has vertices
$ab/c$, $bc/a$, $ca/b$
(which is evident from a good diagram).
Since $A = \frac{2bc}{b+c}$, we have
\[ H_1 = \half \left( \frac{2bc}{b+c} + a + a - a^2 \cdot
\frac{2}{b+c} \right)
= \frac{ab+bc+ca-a^2}{b+c}. \]
The reflection of $H_1$ over $\ol{T_1 T_2}$ is
\begin{align*}
H_1^C &= a + b - ab \ol{H_1}
= a + b - b \cdot \frac{ac+ab+a^2-bc}{a(b+c)} \\
&= \frac{a(a+b)(b+c) - b(a^2+ab+ac-bc)}{a(b+c)}
= \frac{c(a^2+b^2)}{a(b+c)}.
\end{align*}
Now, we claim that $H_1^C$ lies on the chord joining
$\frac{ca}{b}$ and $\frac{cb}{a}$;
by symmetry so will $H_2^C$
and this will imply the problem
(it means that the desired triangle has vertices
$ab/c$, $bc/a$, $ca/b$).
A cartoon of this is shown below.
\begin{center}
\begin{asy}
pair A = dir(110);
pair B = dir(210);
pair C = dir(330);
dot("$\frac{bc}{a}$", A, dir(10), blue);
dot("$\frac{ca}{b}$", B, dir(140), blue);
dot("$\frac{ab}{c}$", C, dir(50), blue);
dot("$H_1^C$", 2*A-B, dir(A-B));
dot("$H_2^C$", 2*B-A, dir(B-A));
dot("$H_2^A$", 2*B-C, dir(B-C));
dot("$H_3^A$", 2*C-B, dir(C-B));
dot("$H_3^B$", 2*C-A, dir(C-A));
dot("$H_1^B$", 2*A-C, dir(A-C));
draw( (2*A-B)--(2*B-A) );
draw( (2*B-C)--(2*C-B) );
draw( (2*C-A)--(2*A-C) );
\end{asy}
\end{center}
To see this, it suffices to compute
\begin{align*}
H_1^C + \left( \frac{ca}{b} \right)\left( \frac{cb}{a} \right) \ol{H_1^C}
&= \frac{c(a^2+b^2)}{a(b+c)}
+ c^2 \frac{\frac 1c \cdot \frac{a^2+b^2}{a^2b^2}}%
{\frac1a\left( \frac{b+c}{bc} \right)} \\
&= \frac{c(a^2+b^2)}{a(b+c)}
+ \frac{c(a^2+b^2)}{abc\inv(b+c)} \\
&= \frac{c(a^2+b^2)}{a(b+c)} \left( \frac{b+c}{b} \right) \\
&= \frac{c(a^2+b^2)}{ab}= \frac{ca}{b} + \frac{cb}{a}
\end{align*}
as desired.
|
IMO-2001-notes_1
|
Let $ABC$ be an acute-angled triangle with $O$ as its circumcenter.
Let $P$ on line $BC$ be the foot of the altitude from $A$.
Assume that $\angle BCA \ge \angle ABC + 30\dg$.
Prove that $\angle CAB + \angle COP < 90\dg$.
|
The conclusion rewrites as
\begin{align*}
\angle COP &< 90\dg - \angle A = \angle OCP \\
\iff PC &< PO \\
\iff PC^2 &< PO^2 \\
\iff PC^2 &< R^2 - PB \cdot PC \\
\iff PC \cdot BC &< R^2 \\
\iff ab \cos C &< R^2 \\
\iff \sin A \sin B \cos C & < \frac14.
\end{align*}
Now
\[ \cos C \sin B
= \half \left( \sin(C+B)-\sin(C-B) \right)
\le \half \left( 1 - \half \right) = \frac 14 \]
which finishes when combined with $\sin A < 1$.
\begin{remark*}
If we allow $ABC$ to be right then
equality holds when $\angle A = 90\dg$,
$\angle C = 60\dg$, $\angle B = 30\dg$.
This motivates the choice of estimates
after reducing to a trig inequality.
\end{remark*}
|
IMO-2001-notes_2
|
Let $a$, $b$, $c$ be positive reals. Prove that
\[ \frac{a}{\sqrt{a^2+8bc}} + \frac{b}{\sqrt{b^2+8ca}} + \frac{c}{\sqrt{c^2+8ab}} \ge 1. \]
|
By Holder, we have
\[
\left( \sum_{\text{cyc}} \frac{a}{\sqrt{a^2+8bc}} \right)^2
\left( \sum_{\text{cyc}} a(a^2+8bc) \right)
\ge (a+b+c)^3.
\]
So it suffices to show $(a+b+c)^3 \ge a^3+b^3+c^3+24abc$ which is clear by expanding.
|
IMO-2001-notes_3
|
Twenty-one girls and twenty-one boys took part in a mathematical competition.
It turned out that each contestant solved at most six problems,
and for each pair of a girl and a boy,
there was at least one problem that was solved by both the girl and the boy.
Show that there is a problem that was solved by at least three girls and at least three boys.
|
We will show the contrapositive.
That is, assume that
\begin{itemize}
\ii For each pair of a girl and a boy,
there was at least one problem that was
solved by both the girl and the boy.
\ii Assume every problem is either solved
mostly by girls (at most two boys)
or mostly by boys (at most two girls).
\end{itemize}
Then we will prove that then some contestant
solved more than six problems.
Create a $21 \times 21$ grid with boys as columns
and girls as rows, and in each cell
write the name of a problem solved by the pair.
Color the cell \textbf{green} if at most two girls solved that problem,
and color it \textbf{blue} if at most two boys solved that problem.
(G for girl, B for boy.
It's possible both colors are used for some cell.)
WLOG, there are more green cells than blue,
so (by pigeonhole) take a column (boy) with that property.
That means the boy's column has at least $11$ green squares.
By hypothesis, those corresponds to at least $6$ different problems
solved. Now there are two cases:
\begin{itemize}
\ii If there are any blue-only squares,
then that square means a seventh distinct problems.
\ii If the entire column is green,
then among the $21$ green squares
there are at least $11$ distinct problems solved
in that column.
\end{itemize}
\begin{remark*}
The number $21$ cannot be decreased.
Here is an example of $20$ boys and $20$ girls
who solve problems named $A$-$J$
and $0$-$9$, which motivates the solution to begin with.
\begin{center}
\scriptsize \ttfamily
{\color{green}0000000000}{\color{blue}AABBCCDDEE} \\
{\color{green}0000000000}{\color{blue}AABBCCDDEE} \\
{\color{green}1111111111}{\color{blue}AABBCCDDEE} \\
{\color{green}1111111111}{\color{blue}AABBCCDDEE} \\
{\color{green}2222222222}{\color{blue}AABBCCDDEE} \\
{\color{green}2222222222}{\color{blue}AABBCCDDEE} \\
{\color{green}3333333333}{\color{blue}AABBCCDDEE} \\
{\color{green}3333333333}{\color{blue}AABBCCDDEE} \\
{\color{green}4444444444}{\color{blue}AABBCCDDEE} \\
{\color{green}4444444444}{\color{blue}AABBCCDDEE} \\
{\color{blue}FFGGHHIIJJ}{\color{green}5555555555} \\
{\color{blue}FFGGHHIIJJ}{\color{green}5555555555} \\
{\color{blue}FFGGHHIIJJ}{\color{green}6666666666} \\
{\color{blue}FFGGHHIIJJ}{\color{green}6666666666} \\
{\color{blue}FFGGHHIIJJ}{\color{green}7777777777} \\
{\color{blue}FFGGHHIIJJ}{\color{green}7777777777} \\
{\color{blue}FFGGHHIIJJ}{\color{green}8888888888} \\
{\color{blue}FFGGHHIIJJ}{\color{green}8888888888} \\
{\color{blue}FFGGHHIIJJ}{\color{green}9999999999} \\
{\color{blue}FFGGHHIIJJ}{\color{green}9999999999}
\end{center}
\end{remark*}
\begin{remark*}
This took me embarrassingly long,
but part of the work of the problem seemed to be
finding the right ``data structure'' to get a foothold.
I think the grid is good because we want to fill each intersection,
then we consider for each cell a problem to put.
I initially wanted to capture the full data by writing
in each green cell the row index of the other girl who solved it,
and similarly for the blue cells.
(That led naturally to the colors, there were two types of cells.)
This was actually helpful for finding the equality case above,
but once I realized the equality case
I also realized that I could discard the extra information
and only remember the colors.
\end{remark*}
|
IMO-2001-notes_4
|
Let $n > 1$ be an odd integer and let $c_1$, $c_2$, \dots, $c_n$ be integers.
For each permutation $a = (a_1, a_2, \dots, a_n)$
of $\{1,2,\dots,n\}$, define $S(a) = \sum_{i=1}^n c_i a_i$.
Prove that there exist two permutations $a \neq b$
of $\{1,2,\dots,n\}$ such that $n!$ is a divisor of $S(a)-S(b)$.
|
Assume for contradiction that all the $S(a)$ are distinct modulo $n!$.
Then summing across all permutations gives
\begin{align*}
1 + 2 + \dots + n!
&\equiv \sum_a S(a) \\
&= \sum_a \sum_{i=1}^n c_i a_i \\
&= \sum_{i=1}^n c_i \sum_a a_i \\
&= \sum_{i=1}^n c_i \cdot \left( (n-1)! \cdot (1+\dots+n) \right) \\
&= (n-1)! \cdot \frac{n(n+1)}{2} \sum_{i=1}^n c_i \\
&= n! \cdot \frac{n+1}{2} \sum_{i=1}^n c_i \\
&\equiv 0
\end{align*}
since $\half(n+1)$ is an integer.
But on the other hand
$1 + 2 + \dots + n! = \frac{n!(n!+1)}{2}$
which is not divisible by $n!$ if $n > 1$,
as the quotient is the non-integer $\frac{n!+1}{2}$.
This is a contradiction.
|
End of preview. Expand
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e1-proof
e1-proof contains mathematics olympiad problems and proofs. The proofs were sourced from Evan Chen's website (https://web.evanchen.cc/problems.html) and are provided here for the use of the community with Evan's permission.
This dataset was collected in this work: Didactic to Constructive: Turning Expert Solutions into Learnable Reasoning. If you use this dataset, please cite this paper:
@article{mendes2026didactic,
title={Didactic to Constructive: Turning Expert Solutions into Learnable Reasoning},
author={Mendes, Ethan and Park, Jungsoo and Ritter, Alan},
journal={arXiv preprint arXiv:2602.02405},
year={2026}
}
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