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The dataset generation failed because of a cast error
Error code: DatasetGenerationCastError
Exception: DatasetGenerationCastError
Message: An error occurred while generating the dataset
All the data files must have the same columns, but at some point there are 1 new columns ({'year'})
This happened while the json dataset builder was generating data using
hf://datasets/jonaskg/imo-problems-complete/all_imo_problems.jsonl (at revision 4f40a3b416450594be05a46d08b472eb857388e7)
Please either edit the data files to have matching columns, or separate them into different configurations (see docs at https://hf.co/docs/hub/datasets-manual-configuration#multiple-configurations)
Traceback: Traceback (most recent call last):
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/builder.py", line 1831, in _prepare_split_single
writer.write_table(table)
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/arrow_writer.py", line 644, in write_table
pa_table = table_cast(pa_table, self._schema)
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/table.py", line 2272, in table_cast
return cast_table_to_schema(table, schema)
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/table.py", line 2218, in cast_table_to_schema
raise CastError(
datasets.table.CastError: Couldn't cast
year: int64
problem_number: int64
problem: string
solution: string
to
{'problem_number': Value('int64'), 'problem': Value('string'), 'solution': Value('string')}
because column names don't match
During handling of the above exception, another exception occurred:
Traceback (most recent call last):
File "/src/services/worker/src/worker/job_runners/config/parquet_and_info.py", line 1456, in compute_config_parquet_and_info_response
parquet_operations = convert_to_parquet(builder)
File "/src/services/worker/src/worker/job_runners/config/parquet_and_info.py", line 1055, in convert_to_parquet
builder.download_and_prepare(
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/builder.py", line 894, in download_and_prepare
self._download_and_prepare(
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/builder.py", line 970, in _download_and_prepare
self._prepare_split(split_generator, **prepare_split_kwargs)
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/builder.py", line 1702, in _prepare_split
for job_id, done, content in self._prepare_split_single(
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/builder.py", line 1833, in _prepare_split_single
raise DatasetGenerationCastError.from_cast_error(
datasets.exceptions.DatasetGenerationCastError: An error occurred while generating the dataset
All the data files must have the same columns, but at some point there are 1 new columns ({'year'})
This happened while the json dataset builder was generating data using
hf://datasets/jonaskg/imo-problems-complete/all_imo_problems.jsonl (at revision 4f40a3b416450594be05a46d08b472eb857388e7)
Please either edit the data files to have matching columns, or separate them into different configurations (see docs at https://hf.co/docs/hub/datasets-manual-configuration#multiple-configurations)Need help to make the dataset viewer work? Make sure to review how to configure the dataset viewer, and open a discussion for direct support.
problem_number int64 | problem string | solution string |
|---|---|---|
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 ... | 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 $1/2$ respecti... |
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 $W$, resp... | Let $\ol{BTV}$ meet the circle again at $U_1$,
so that $AU_1 UB$ is an isosceles trapezoid.
Define $U_2$ similarly.
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);
fil... |
3 | Let $x_1$, $x_2$, \dots, $x_n$ be real numbers satisfying the conditions:
|x_1 + x_2 + \dots + x_n| &= 1 \\
|x_i| &\le \frac{n+1}{2} \qquad \text{for } i= 1,2, \dots, n
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 ... | WLOG $\sum x_i = 1$ (by negating $x_i$) and $x_1 \le x_2 \le \dots \le x_n$.
Notice that
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
\[ B = nx_1 + (n-1)x_2 + \dots + x_n. \]
Meanwhile, the \emph{average} value across all permutations is... |
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:
[(a)]
there is no silver matrix for $n = 1997$;
silver matrices exist for inf... | \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 elemen... |
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 $1$.
Every integer greater than $1$
is uniquely of the form $t^n$ for some $t \in T$, $n \in \NN$.
Clear.
Let $a = s^m$, $b = t^n$.
\[ s^{m \cdot (t^n)^2} = t^{n \cdot s^m}. \... |
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 th... | 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 ... |
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 ... | In general, we say the 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.
T... |
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 $W$, resp... | Let $BTV$ meet the circle again at $U_1$,
so that $AU_1 UB$ is an isosceles trapezoid.
Define $U_2$ similarly.
[Figure omitted]
Now from the isosceles trapezoids we get
$ AU = BU_1 = BT + TU_1 = BT + TC $
as desired. |
3 | Let $x_1$, $x_2$, \dots, $x_n$ be real numbers satisfying the conditions:
|x_1 + x_2 + \dots + x_n| &= 1 \\
|x_i| &\le \frac{n+1}{2} \qquad \text{for } i= 1,2, \dots, n
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... | WLOG $\sum x_i = 1$ (by negating $x_i$) and $x_1 \le x_2 \le \dots \le x_n$.
Notice that
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. $
Meanwhile, the average value across all permutations is
$ 1 \cdot \f... |
4 | An $n \times n$ matrix whose entries come
from the set $S = \{1, 2, \dots , 2n - 1\}$
is called a 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:
[(a)]
there is no silver matrix for $n = 1997$;
silver matrices exist for infinitely m... | Solution to (a).:
Define a 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 ... |
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 $1$.
Every integer greater than $1$
is uniquely of the form $t^n$ for some $t \in T$, $n \in \NN$.
Clear.
Let $a = s^m$, $b = t^n$.
$ s^{m \cdot (t^n)^2} = t^{n \cdot s^m}. $
He... |
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 th... | 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\lfloor \frac N2 \right\rfloor )
+ f( \left\lfloor \frac N2 \right\rfloor -1 )
+ f( \left\lfloor \frac N2 \right\rfloor -2 )
+ \dots
+ f(1) + f(0). \quad (\bigstar)
$
Upper bound.:
We now prove the ... |
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.
[Figure omitted]
Let $M$ and $N$ be the midpoints of $AB$ and $CD$.
Unconditionally, we have $∠ NEM = ∠ MPN$.
Note that $EN$ is the median of right triangle $\triangle ECD$, and similarly for $E... |
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... |
3 | For any positive integer $n$,
let $\tau(n)$ denote the number of its positive divisors (including $1$ and itself).
Determine all positive integers $m$ for which
there exists a positive integer $n$ such that
$ \frac{\tau(n^{2})}{\tau(n)}=m. $ | The answer is odd integers $m$ only.
If we write $n = p_1^{e_1} \dots p_k^{e_k}$ we get
$ \prod \frac{2e_i+1}{e_i+1} = m. $
It's clear now that $m$ must be odd,
since every fraction has odd numerator.
We now endeavor to construct odd numbers.
The proof is by induction, in which we are curating sets of
fractions of the... |
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... |
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.
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}... |
6 | Classify all functions $f \colon \NN \to \NN$
satisfying the identity
$ f(n^2 f(m)) = m f(n)^2. $ | 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... |
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 fi... | Let $G$ be the centroid of $S$.
All points of $S$ lie on a sphere $\Gamma$ centered at $G$.
Each perpendicular bisector plane passes through $G$.
So if $A,B \in S$ it follows $GA = GB$.
Consider any plane passing through three or more points of $S$.
The points of $S$ in the plane form a regular polygon.
The cross s... |
2 | Find the least constant $C$ such that for any integer $n > 1$ the inequality
$\sum_{1 \le i < j \le n} x_i x_j (x_i^2 + x_j^2)
\le C ( \sum_{1 \le i \le n} x_i )^4$
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.
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 ... |
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 domino of two cells.
For each domino, its 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... |
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 \p... |
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... | 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 $PTQ \perp AB$, it follows $L... |
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 $x,y \in \RR$. | The answer is $f(x) = -1/2 x^2+1$
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) = -1/2 s^2 + (c + 1). $
Thus, the given functional equa... |
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 ... | First, we have $∠ EAB = ∠ ACM = ∠ BAM$
and similarly $∠ EBA = ∠ BDM = ∠ ABM$.
Consequently, $AB$ bisects $\angle EAM$ and $\angle EBM$,
and hence $\triangle EAB \cong \triangle MAB$.
[Figure omitted]
Now it is well-known that $MN$ bisects $AB$
and since $AB \parallel PQ$
we deduce that $M$ is the midpoint of $PQ$.
As... |
2 | Let $a$, $b$, $c$ be positive real numbers with $abc = 1$.
Show that
$
( a - 1 + \frac 1b )
( b - 1 + \frac 1c )
( c - 1 + \frac 1a )
\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:
At most one term on the left-hand side is negative... |
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 ... | 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:
Select two indices $i < j$.
Then ball $B_i$ moves to $B_{i+1}$'s lo... |
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 magici... | There are $2 \cdot 3! = 12$ ways, which amount to:
Partitioning the cards modulo $3$, or
Placing $1$ alone in a box,
$100$ alone in a second box,
and all remaining cards in the third box.
These are easily checked to work so we prove they are the only ones.
First solution.:
We proceed by induction on $n \ge 3$ with ... |
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 Korean if $n \mid 2^n+1$.
First, observe that $n=9$ is Korean.
Now, the problem is solved upon the following claim:
If $n > 3$ is Korean,
there exists a prime $p$ not dividing $n$
such that $np$ is Korean too.
I claim that one can take any primitive prime divisor $p$ of $2^{2n}-1$,
wh... |
6 | Let $AH_1$, $BH_2$, and $CH_3$ 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_3... | 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 = 1/2 ( \frac{2bc}{b+c} + a + a - a^... |
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
\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.
Now
$ \cos C \sin B
= 1/2 ( \sin(C+B)-\sin(C-B) )
\le 1/2 ( 1 - 1/2 ) = \frac 14 ... |
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
$
( \sum_{\text{cyc}} \frac{a}{\sqrt{a^2+8bc}} )^2
( \sum_{\text{cyc}} a(a^2+8bc) )
\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. |
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... | We will show the contrapositive.
That is, assume that
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.
Assume every problem is either solved
mostly by girls (at most two boys)
or mostly by boys (at most two girls).
Then we will prove that then some contes... |
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
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 ( (n-1)! \cdot (1+\dots+n) ) \\
&= (n-1)! \cdot \frac{n(n+1)}{2} \sum... |
5 | Let $ABC$ be a triangle.
Let $AP$ bisect $\angle BAC$ and let $BQ$ bisect $\angle ABC$,
with $P$ on $BC$ and $Q$ on $AC$.
If $AB + BP = AQ + QB$ and $\angle BAC = 60\dg$,
what are the angles of the triangle? | The answer is $\angle B = 80\dg$ and $\angle C = 40\dg$.
Set $x = \angle ABQ = \angle QBC$, so that $\angle QCB = 120\dg - 2x$.
We observe $\angle AQB = 120\dg - x$ and $\angle APB = 150\dg - 2x$.
[Figure omitted]
Now by the law of sines, we may compute
BP &= AB \cdot \frac{\sin 30\dg}{\sin (150\dg- 2x)} \\
AQ &= AB... |
6 | Let $a > b > c > d > 0$ be integers satisfying
$ ac + bd = (b+d+a-c)(b+d-a+c). $
Prove that $ab + cd$ is not prime. | The problem condition is equivalent to
$ ac + bd = (b+d)^2 - (a-c)^2 $
or
$ a^2-ac+c^2 = b^2+bd+d^2. $
Let us construct a quadrilateral $WXYZ$ such that
$WX = a$, $XY = c$, $YZ = b$, $ZW = d$,
and $ WY = \sqrt{a^2-ac+c^2} = \sqrt{b^2+bd+d^2}.$
Then by the law of cosines, we obtain $\angle WXY = 60\dg$
and $\angle WZY ... |
1 | Let $n$ be a positive integer.
Let $T$ be the set of points $(x,y)$ in the plane
where $x$ and $y$ are non-negative integers with $x+y<n$.
Each point of $T$ is coloured red or blue, subject to the following condition:
if a point $(x,y)$ is red,
then so are all points $(x',y')$ of $T$ with $x'\leq x$ and $y'\leq y$.
Let... | Let $a_x$ denote the number of blue points
with a given $x$-coordinate.
Define $b_y$ to be the number of blue points
with a given $y$-coordinate.
We actually claim that
The multisets $\mathcal A \coloneq \{ a_x \mid x \}$
and $\mathcal B \coloneq \{ b_y \mid y \}$ are equal.
By induction on the number of red points.... |
2 | Let $BC$ be a diameter of circle $\omega$ with center $O$. Let $A$ be a point
of circle $\omega$ such that $0^\circ < \angle AOB < 120^\circ$. Let $D$ be the midpoint
of arc $AB$ not containing $C$. Line $\ell$ passes through $O$ and is
parallel to line $AD$. Line $\ell$ intersects line $AC$ at $J$.
The perpendicular b... | By construction, $AEOF$ is a rhombus with $60\dg$-$120\dg$ angles.
Consequently, we may set $s = AO = AE = AF = EO = EF$.
[Figure omitted]
We have $AJ = s$ too.
It suffices to show $AJ = AO$ which is angle chasing.
Let $\theta = \angle BOD = \angle DOA$,
so $\angle BOA = 2\theta$.
Thus $\angle CAO = 1/2 \angle BOA = ... |
3 | Find all pairs of positive integers $m, n \ge 3$ for which there exist infinitely many
positive integers $a$ such that
$ \frac{a^m+a-1}{a^n+a^2-1} $
is itself an integer. | The condition is equivalent to $a^n+a^2-1$
dividing $a^m+a-1$ as polynomials.
The big step is the following analytic one.
We must have $m \le 2n$.
Assume on contrary $m > 2n$
and let $0 < r < 1$ be the unique real number
with $r^n+r^2 = 1$, hence $r^m+r = 1$.
But now
0 &= r^m + r - 1 \le r^{2n+1} + r - 1 \\
&= r( (1... |
4 | Let $n \ge 2$ be a positive integer
with divisors $1 = d_1 < d_2 < \dots < d_k = n$.
Prove that $d_1d_2 + d_2d_3 + \dots + d_{k-1} d_k$ is always less than $n^2$,
and determine when it is a divisor of $n^2$. | We always have
d_k d_{k-1} + d_{k-1} d_{k-2} + \dots + d_2 d_1
&< n \cdot \frac n2 + \frac n2 \cdot \frac n3 + \dots \\
&= ( \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots ) n^2 = n^2.
This proves the first part.
For the second, we claim that this only happens
when $n$ is prime (in which case we get $d_1 d_2 = n$... |
5 | Find all functions $f \colon \RR \to \RR$ such that
$ (f(x)+f(z))(f(y)+f(t))
= f(xy-zt)+f(xt+yz) $
for all real numbers $x$, $y$, $z$, $t$. | The answer is $f(x) \equiv 0$, $f(x) \equiv 1/2$
and $f(x) \equiv x^2$ which are easily seen to work.
Let's prove they are the only ones;
we show two solutions.
First solution (multiplicativity).:
Let $P(x,y,z,t)$ denote the given statement.
By comparing $P(x,1,0,0)$ and $P(0,0,1,x)$
we get $\boxed{f \text{ even}}$.... |
6 | Let $n \ge 3$ be a positive integer.
Let $C_1$, $C_2$, \dots, $C_n$ be unit circles in the plane,
with centers $O_1$, $O_2$, \dots, $O_n$ respectively.
If no line meets more than two of the circles, prove that
$ \sum_{1 \le i < j \le n } \frac{1}{O_i O_j}
\le \frac{(n-1)\pi}{4}. $ | For brevity, let $d_{ij}$ be the length of $O_{ij}$
and let $\angle(ijk)$ be shorthand for $\angle O_i O_j O_k$
(or its measure in radians).
First, we eliminate the circles completely
and reduce the problem to angles using the following fact
(which is in part motivated by the mysterious
presence of $\pi$ on right-hand... |
1 | Let $A$ be a $101$-element subset of $S=\{1,2,\dots,10^6\}$.
Prove that there exist numbers $t_1$, $t_2, \dots, t_{100}$ in $S$ such that the sets
$ A_j=\{x+t_j\mid x\in A\},\qquad j=1,2,\dots,100 $
are pairwise disjoint. | A greedy algorithm works: suppose we have picked
$ T = \left\{ t_1, \dots, t_n \right\} $
as large as possible, meaning it's impossible to add any more elements to $T$.
That means, for each $t \in \left\{ 1, \dots, 10^6 \right\}$ either $t \in T$ already
or there exists two distinct elements $a, b \in A$ and $t_i \in T... |
2 | Determine all pairs of positive integers $(a,b)$ such that
$ \frac{a^2}{2ab^2-b^3+1} $
is a positive integer. | The answer is $(a,b) = (2\ell, 1)$, $(a,b) = (\ell, 2\ell)$
and $(a,b) = (8\ell^4-\ell, 2\ell)$, for any $\ell$.
Check these work.
In the sequel, assume $b > 1$,
and integers $a$, $b$, $k$ obey $k = \frac{a^2}{2ab^2-b^3+1}$.
Expanding, we have the polynomial
$ X^2 - 2kb^2 \cdot X + k(b^3-1) = 0 $
has two integer roots... |
3 | Each pair of opposite sides of convex hexagon has the property that
the distance between their midpoints is $\frac{\sqrt3}{2}$
times the sum of their lengths.
Prove that the hexagon is equiangular. | Unsurprisingly, this is a geometric inequality.
Denote the hexagon by $ABCDEF$.
Then we have that
$
\left|
\frac{\vec D + \vec E}{2} - \frac{\vec A + \vec B}{2}
\right|
= \sqrt3 \cdot \frac{\left| \vec B - \vec A \right|
+ \left| \vec E - \vec D \right|}{2}
\ge \sqrt 3 \cdot
\left| \frac{(\vec B - \vec A) - (\vec E - \... |
4 | Let $ABCD$ be a cyclic quadrilateral.
Let $P, Q$ and $R$ be the feet of perpendiculars
from $D$ to lines $BC$, $CA$ and $AB$, respectively.
Show that $PQ = QR$ if and only if the
bisectors of angles $ABC$ and $ADC$ meet on segment $AC$. | Let $\gamma$ denote the circumcircle of $ABCD$.
The condition on bisectors is equivalent to $(AC;BD)_\gamma = -1$.
Meanwhile if $\infty$ denotes the point at infinity along Simson line $PQR$
then $PQ = QR$ if and only if $(PR;Q\infty) = -1$.
Let rays $BQ$ and $DQ$ meet the circumcircle again at $F$ and $E$.
[Figure o... |
5 | Let $n$ be a positive integer and
let $x_1 \le x_2 \le \dots \le x_n$ be real numbers.
Prove that
$ (\sum_{i=1}^{n}\sum_{j=1}^{n} |x_i - x_j|)^2
\le \frac{2(n^2-1)}{3}\sum_{i=1}^{n}\sum_{j=1}^{n} (x_i - x_j)^2 $
with equality if and only if $x_1$, $x_2$, \dots, $x_n$
form an arithmetic sequence. | Let $d_1 = x_2 - x_1$, \dots, $d_{n-1} = x_n - x_{n-1}$.
The inequality in question becomes:
$
( \sum_i i(n-i) d_i )^2
\le
\frac{n^2-1}{3} \cdot
( \sum_i i(n-i) d_i^2 + 2\sum_{i<j} i(n-j) d_i d_j ).
$
Clearing the square on the right-hand side we want to show
$
\sum_{i<j} ( 3ij(n-i)(n-j) - (n^2-1)i(n-j) ) \cdot 2d_i d_... |
6 | Let $p$ be a prime number.
Prove that there exists a prime number $q$ such that for every integer $n$,
the number $n^p-p$ is not divisible by $q$. | By orders, we must have $q=pk+1$ for this to be possible
(since if $q \not \equiv 1 \pmod p$, then $n^p$ can be any residue modulo $q$).
Since $p \equiv n^p \pmod q \implies p^k \equiv 1 \pmod q$,
it suffices to prevent the latter situation from happening.
So we need a prime $q \equiv 1 \pmod p$ such that $p^k \not\eq... |
1 | Let $ABC$ be an acute-angled triangle with $AB\neq AC$.
The circle with diameter $BC$ intersects the sides $AB$ and $AC$
at $M$ and $N$ respectively.
Denote by $O$ the midpoint of the side $BC$.
The bisectors of the angles $\angle BAC$ and $\angle MON$ intersect at $R$.
Prove that the circumcircles of the triangles $BM... | By Miquel's theorem it's enough to show $AMRN$ is cyclic.
[Figure omitted]
In fact, since the bisector of $\angle MON$
is just the perpendicular bisector of $MN$,
the point $R$ is actually just the arc midpoint
of $\widehat{MN}$ of $(AMN)$ as desired. |
2 | Find all polynomials $P$ with real coefficients such that
for all reals $a$, $b$, $c$ such that $ab+bc+ca = 0$, we have
$ P(a-b) + P(b-c) + P(c-a) = 2P(a+b+c). $ | The answer is $ P(x) = \alpha x^4 + \beta x^2 $
which can be checked to work, for any real numbers $\alpha$ and $\beta$.
It is easy to obtain that $P$ is even and $P(0) = 0$.
The trick is now to choose $(a,b,c) = (6x,3x,-2x)$
and then compare the leading coefficients to get
$ 3^n + 5^n + 8^n = 2 \cdot 7^n $
for $n = \... |
3 | Define a ``hook'' to be a figure made up of six unit squares
as shown below in the picture,
or any of the figures obtained by applying rotations
and reflections to this figure.
[Figure omitted]
Which $m \times n$ rectangles can be tiled by hooks? | The answer is that one requires:
$\{1,2,5\} \notin \{m,n\}$,
$3 \mid m$ or $3 \mid n$,
$4 \mid m$ or $4 \mid n$.
First, we check all of these work, in fact we claim:
Any rectangle satisfying these conditions
can be tiled by $3 \times 4$ rectangles (and hence by hooks).
If $3 \mid m$ and $4 \mid n$, this is clear.
E... |
4 | Let $n \ge 3$ be an integer
and $t_1$, $t_2$, \dots, $t_n$ positive real numbers such that
$ n^2+1 > (t_1 + t_2 + \dots + t_n)
( \frac{1}{t_1} + \frac{1}{t_2} + \dots + \frac{1}{t_n} ). $
Show that $t_i$, $t_j$, $t_k$ are the sides of a triangle
for all $i$, $j$, $k$ with $1 \le i < j < k \le n$. | Let $a = t_1$, $b = t_2$, $c = t_3$.
Expand:
n^2+1 &> (t_1 + t_2 + \dots + t_n)
( \frac{1}{t_1} + \dots + \frac{1}{t_n} ) \\
&= n + \sum_{1 \le i < j \le n}
( \frac{t_i}{t_j} + \frac{t_j}{t_i} ) \\
&= n + \sum_{1 \le i < j \le n}
( \frac{t_i}{t_j} + \frac{t_j}{t_i} ) \\
&\ge n + \sum_{1 \le i < j \le 3}
( \frac{t_i}{t... |
5 | In a convex quadrilateral $ABCD$,
the diagonal $BD$ bisects neither the angle $ABC$ nor the angle $CDA$.
The point $P$ lies inside $ABCD$ and satisfies
$\angle PBC=\angle DBA \quad\text{and}\quad \angle PDC=\angle BDA. $
Prove that $ABCD$ is a cyclic quadrilateral if and only if $AP=CP$. | We show two solutions.
We note that the first hypothesis cannot be dropped, because $ABCD$ being a kite
with $BA=BC$ and $DA=DC$ with center $P$ is a counterexample
(where $AP=CP$ but $ABCD$ is not cyclic).
However, the condition that $BD$ bisects neither angle $\angle B$ nor $\angle D$
is equivalent to requiring that ... |
6 | We call a positive integer alternating if every two consecutive digits
in its decimal representation are of different parity.
Find all positive integers $n$ which have an alternating multiple. | If $20 \mid n$, then clearly $n$ has no alternating
multiple since the last two digits are both even.
We will show the other values of $n$ all work.
The construction is just rush-down do-it.
The meat of the solution is the two following steps.
[Tail construction]
For every even integer $w \ge 2$,
there exists an eve... |
1 | Six points are chosen on the sides of an equilateral triangle $ABC$:
$A_1$, $A_2$ on $BC$, $B_1$, $B_2$ on $CA$ and $C_1$, $C_2$ on $AB$,
such that they are the vertices of a
convex hexagon $A_1A_2B_1B_2C_1C_2$ with equal side lengths.
Prove that the lines $A_1B_2$, $B_1C_2$ and $C_1A_2$ are concurrent. | The six sides of the hexagon, when oriented, comprise
six vectors with vanishing sum.
However note that $ \overrightarrow{A_1A_2}
+ \overrightarrow{B_1B_2}
+ \overrightarrow{C_1C_2} = 0. $
Thus
$ \overrightarrow{A_2B_1} + \overrightarrow{B_2C_1} +
\overrightarrow{C_2A_1} = 0 $
and since three unit vectors with vanishin... |
2 | Let $a_1$, $a_2$, \dots\ be a sequence of integers
with infinitely many positive and negative terms.
Suppose that for every positive integer $n$
the numbers $a_1$, $a_2$, \dots, $a_n$
leave $n$ different remainders upon division by $n$.
Prove that every integer occurs exactly once in the sequence. | Obviously every integer appears at most once
(otherwise take $n$ much larger).
So we will prove every integer appears at least once.
For any $i < j$ we have $\left| a_i-a_j \right| < j$.
Otherwise, let $n = \left| a_i-a_j \right| \neq 0$.
Then $i,j \in [1,n]$ and $a_i \equiv a_j \pmod n$,
contradiction.
For any $n$,... |
3 | Let $x,y,z > 0$ satisfy $xyz\geq 1$. Prove that
$ \frac { x^5-x^2 }{x^5+y^2+z^2}
+ \frac {y^5-y^2}{x^2+y^5+z^2}
+ \frac {z^5-z^2}{x^2+y^2+z^5} \geq 0. $ | Negating both sides and adding $3$ eliminates the minus signs:
$ \sum_{\text{cyc}} \frac{1}{x^5+y^2+z^2}
\le \frac{3}{x^2+y^2+z^2}. $
Thus we only need to consider the case $xyz = 1$.
Direct expansion and Muirhead works now!
As advertised, once we show it suffices to analyze if $xyz=1$
the inequality becomes more econ... |
4 | Determine all positive integers relatively
prime to all the terms of the infinite sequence
$ a_n = 2^n+3^n+6^n-1, \quad n \ge 1. $ | The answer is $1$ only (which works).
It suffices to show there are no primes.
For the primes $p=2$ and $p=3$, take $a_2=48$.
For any prime $p \ge 5$ notice that
a_{p-2} &= 2^{p-2} + 3^{p-2} + 6^{p-2} - 1 \\
&\equiv \frac 12 + \frac 13 + \frac 16 - 1 \pmod p \\
&\equiv 0 \pmod p
so no other larger prime works. |
5 | Let $ABCD$ be a fixed convex quadrilateral
with $BC=DA$ and $BC \nparallel DA$.
Let two variable points $E$ and $F$ lie on the
sides $BC$ and $DA$, respectively, and satisfy $BE=DF$.
The lines $AC$ and $BD$ meet at $P$,
the lines $BD$ and $EF$ meet at $Q$,
the lines $EF$ and $AC$ meet at $R$.
Prove that the circumcircl... | Let $M$ be the Miquel point of complete quadrilateral $ADBC$;
in other words, let $M$ be the second intersection point
of the circumcircles of $\triangle APD$ and $\triangle BPC$.
(A good diagram should betray this secret;
all the points are given in the picture.)
This makes lots of sense since we know $E$ and $F$
will... |
6 | In a mathematical competition $6$ problems were posed to the contestants.
Each pair of problems was solved by more than $\frac{2}{5}$ of the contestants.
Nobody solved all 6 problems.
Show that there were at least $2$ contestants
who each solved exactly $5$ problems. | Assume not and at most one contestant solved five problems.
By adding in solves,
we can assume WLOG that one contestant solved problems one through five,
and every other contestant solved four of the six problems.
We split the remaining contestants based on whether they solved P6.
Let $a_i$ denote the number of contes... |
1 | Let $ABC$ be a triangle with incenter $I$.
A point $P$ in the interior of the triangle satisfies
$ \angle PBA + \angle PCA = \angle PBC + \angle PCB. $
Show that $AP \ge AI$ and that equality holds if and only if $P=I$. | The condition rewrites as
$
\angle PBC + \angle PCB
= (\angle B - \angle PBC)
+ (\angle C - \angle PCB)
\implies
\angle PBC + \angle PCB = \frac{\angle B + \angle C}{2}
$
which means that
$ \angle BPC = 180\dg - \frac{\angle B + \angle C}{2}
= 90\dg + \frac{\angle A}{2}
= \angle BIC.
$
Since $P$ and $I$ are both inside... |
2 | Let $P$ be a regular $2006$-gon.
A diagonal is called good if its endpoints
divide the boundary of $P$ into two parts,
each composed of an odd number of sides of $P$.
The sides of $P$ are also called good.
Suppose $P$ has been dissected into triangles by $2003$ diagonals,
no two of which have a common point in the inte... | Call a triangle with the desired property special.
We prove the maximum number of special triangles is $1003$,
achieved by paring up the sides of the polygon.
We present two solutions for the upper bound.
Both of them rely first on two geometric notes:
In a special triangle, the good sides are congruent
(and not cong... |
3 | Determine the least real number $M$ such that the inequality
$ \left| ab(a^2-b^2)+bc(b^2-c^2)+ca(c^2-a^2) \right|
\leq M( a^2+b^2+c^2 )^2 $
holds for all real numbers $a$, $b$ and $c$. | It's the same as
$ \left| (a-b)(b-c)(c-a)(a+b+c) \right|
\le M ( a^2+b^2+c^2 )^2. $
Let $x=a-b$, $y=b-c$, $z=c-a$, $s=a+b+c$.
Then we want to have
$
\left| xyzs \right|
\le \frac{M}{9} (x^2+y^2+z^2+s^2)^2.
$
Here $x+y+z=0$.
Now if $x$ and $y$ have the same sign,
we can replace them with the average
(this increases the... |
4 | Determine all pairs $(x,y)$ of integers such that
$ 1 + 2^x + 2^{2x+1} = y^2. $ | Answers: $(0, \pm 2)$, $(4, \pm 23)$, which work.
Assume $x \ge 4$.
$ 2^x ( 1 + 2^{x+1} )
= 2^x + 2^{2x+1} = y^2 - 1 = (y-1)(y+1). $
So either:
$y = 2^{x-1} m + 1$ for some odd $m$, so
$ 1 + 2^{x+1} = m( 2^{x-2}m+1 )
\implies 2^x = \frac{4(1-m)}{m^2-8}. $
$y = 2^{x-1} m - 1$ for some odd $m$, so
$ 1 + 2^{x+1} = m( 2^... |
5 | Let $P(x)$ be a polynomial of degree $n > 1$
with integer coefficients and let $k$ be a positive integer.
Consider the polynomial
$ Q(x) = P(P(\dots P(P(x)) \dots )) $ where $P$ occurs $k$ times.
Prove that there are at most $n$ integers $t$ such that $Q(t) = t$. | First, we prove that:
[Putnam 2000 et al]
If a number is periodic under $P$
then in fact it's fixed by $P \circ P$.
Let $x_1$, $x_2$, \dots, $x_n$ be a minimal orbit.
Then
$ x_i - x_{i+1} \mid P(x_i) - P(x_{i+1})
= x_{i+1} - x_{i+2} $
and so on cyclically.
If any of the quantities are zero we are done.
Else, we must... |
6 | Assign to each side $b$ of a convex polygon $P$
the maximum area of a triangle that has $b$ as a side and is contained in $P$.
Show that the sum of the areas assigned to the sides of $P$ is at least twice the area of $P$. | We say a polygon in almost convex
if all its angles are at most $180\dg$.
Note that given any convex or almost convex polygon,
we can take any side $b$ and add another vertex on it, and the sum of the labels doesn't change
(since the label of a side is the length of the side times the distance of the farthest point).
... |
1 | Real numbers $a_1$, $a_2$, \dots, $a_n$ are fixed.
For each $1 \le i \le n$ we let
$d_i = \max\{a_j : 1 \le j \le i\} - \min\{a_j : i \le j \le n\}$
and let $d = \max \{d_i : 1 \le i \le n\}$.
[(a)]
Prove that for any real numbers $x_1 \le \dots \le x_n$ we have
$
\max \left\{ \left| x_i - a_i \right| :
1 \le i \le n ... | Note that we can dispense of $d_i$ immediately
by realizing that the definition of $d$ just says
$ d = \max_{1 \le i \le j \le n} ( a_i - a_j ). $
If $a_1 \le \dots \le a_n$ are already nondecreasing
then $d = 0$ and there is nothing to prove
(for the equality case, just let $x_i = a_i$),
so we will no longer consider... |
2 | Consider five points $A$, $B$, $C$, $D$ and $E$
such that $ABCD$ is a parallelogram and $BCED$ is a cyclic quadrilateral.
Let $\ell$ be a line passing through $A$.
Suppose that $\ell$ intersects the interior of the segment $DC$ at $F$
and intersects line $BC$ at $G$.
Suppose also that $EF = EG = EC$.
Prove that $\ell$ ... | Let $M$, $N$, $P$ denote the midpoints of $CF$, $CG$, $AC$
(noting $P$ is also the midpoint of $BD$).
By a homothety at $C$ with ratio $1/2$,
we find $MNP$ is the image of line $\ell \equiv AGF$.
[Figure omitted]
However, since we also have $EM \perp CF$
and $EN \perp CG$ (from $EF=EG=EC$)
we conclude $PMN$ is the S... |
3 | In a mathematical competition some competitors are (mutual) friends.
Call a group of competitors a clique if each two of them are friends.
Given that the largest size of a clique is even,
prove that the competitors can be arranged into two rooms
such that the largest size of a clique contained in one room
is the same a... | Take the obvious graph interpretation $G$.
We paint red any vertices in one of the maximal cliques $K$,
which we assume has $2r$ vertices, and paint the remaining vertices green.
We let $\alpha(\bullet)$ denote the clique number.
Initially, let the two rooms $A = K$, $B = G-K$.
We can move at most $r$ vertices of $A$... |
4 | In triangle $ABC$ the bisector of $\angle BCA$
meets the circumcircle again at $R$,
the perpendicular bisector of $BC$ at $P$,
and the perpendicular bisector of $AC$ at $Q$.
The midpoint of $BC$ is $K$ and the midpoint of $AC$ is $L$.
Prove that the triangles $RPK$ and $RQL$ have the same area. | We first begin by proving the following claim.
We have $CQ = PR$ (equivalently, $CP = QR$).
Let $O = LQ \cap KP$ be the circumcenter.
Then
$ ∠ OPQ = ∠ KPC = 90\dg - ∠ PCK
= 90\dg - ∠ LCQ = ∠ ∠ CQL = ∠ PQO. $
Thus $OP = OQ$.
Since $OC = OR$ as well, we get the conclusion.
Denote by $X$ and $Y$ the feet from $R$ to $C... |
5 | Let $a$ and $b$ be positive integers.
Show that if $4ab - 1$ divides $(4a^{2} - 1)^{2}$, then $a = b$. | As usual,
$ 4ab-1 \mid (4a^2-1)^2 \iff 4ab-1 \mid (4ab \cdot a-b)^2
\iff 4ab-1 \mid (a-b)^2. $
Then we use a typical Vieta jumping argument.
Define $ k = \frac{(a-b)^2}{4ab-1}. $
Note that $k = 0 \iff a = b$.
So we will prove that $k > 0$ leads to a contradiction.
Indeed, suppose $(a, b)$ is a minimal solution with $a... |
6 | Let $n$ be a positive integer.
Consider
$ S = \left\{ (x,y,z) \mid
x,y,z \in \{ 0, 1, \dots, n\}, \;
x+y+z > 0 \right\} $
as a set of $(n+1)^3-1$ points in the three-dimensional space.
Determine the smallest possible number of planes,
the union of which contains $S$ but does not include $(0,0,0)$. | The answer is $3n$.
Here are two examples of constructions with $3n$ planes:
$x+y+z=i$ for $i=1,\dots,3n$.
$x=i$, $y=i$, $z=i$ for $i=1,\dots,n$.
Suppose for contradiction we have $N < 3n$ planes.
Let them be $a_i x + b_i y + c_i z + 1 = 0$, for $i = 1, \dots, N$.
Define the polynomials
A(x,y,z) &= \prod_{i=1}^n (x-... |
1 | Let $H$ be the orthocenter of an acute-angled triangle $ABC$.
The circle $\Gamma_{A}$ centered at the midpoint of $BC$ and passing
through $H$ intersects the sideline $BC$ at points $A_1$ and $A_2$.
Similarly, define the points $B_1$, $B_2$, $C_1$, and $C_2$.
Prove that six points $A_1$, $A_2$, $B_1$, $B_2$, $C_1$, $C_... | We show two solutions.
First solution using power of a point.:
Let $D$, $E$, $F$ be the centers of $\Gamma_A$, $\Gamma_B$, $\Gamma_C$
(in other words, the midpoints of the sides).
We first show that $B_1$, $B_2$, $C_1$, $C_2$ are concyclic.
It suffices to prove that $A$
lies on the radical axis of the circles $\Gamm... |
2 | Let $x$, $y$, $z$ be real numbers with $xyz = 1$, all different from $1$.
Prove that
$ \frac{x^2}{(x-1)^2} + \frac{y^2}{(y-1)^2} + \frac{z^2}{(z-1)^2} \ge 1 $
and show that equality holds for infinitely many choices
of rational numbers $x$, $y$, $z$. | Let $x=a/b$, $y=b/c$, $z=c/a$, so we want to show
$ (\frac{a}{a-b})^2+(\frac{b}{b-c})^2
+(\frac{c}{c-a})^2\ge 1.$
A boring computation shows this is equivalent to
$ \frac{(a^2b+b^2c+c^2a-3abc)^2}{(a-b)^2(b-c)^2(c-a)^2} \ge 0 $
which proves the inequality
(and it is unsurprising we are in such a situation,
given that th... |
3 | Prove that there are infinitely many positive integers $n$
such that $n^2+1$ has a prime factor greater than $2n + \sqrt{2n}$. | The idea is to pick the prime $p$ first!
Select any large prime $p \ge 2013$,
and let $h = \left\lceil \sqrt p \right\rceil$.
We will try to find an $n$ such that
$ n \le \frac 12 (p-h) \quad \text{and} \quad p \mid n^2+1. $
This implies $p \ge 2n+\sqrt{p}$
which is enough to ensure $p \ge 2n + \sqrt{2n}$.
Assume $p ... |
4 | Find all functions $f$ from the positive reals to the positive reals such that
$ \frac{f(w)^2 + f(x)^2}{f(y^2)+f(z^2)} = \frac{w^2+x^2}{y^2+z^2} $
for all positive real numbers $w$, $x$, $y$, $z$ satisfying $wx=yz$. | The answers are $f(x) \equiv x$ and $f(x) \equiv 1/x$.
These work, so we show they are the only ones.
First, setting $(t,t,t,t)$ gives $f(t^2) = f(t)^2$.
In particular, $f(1) = 1$.
Next, setting $(t, 1, \sqrt t, \sqrt t)$ gives
$ \frac{f(t)^2 + 1}{2f(t)} = \frac{t^2 + 1}{2t} $
which as a quadratic implies $f(t) \in \{... |
5 | Let $n$ and $k$ be positive integers with $k \geq n$ and $k - n$ an even number.
There are $2n$ lamps labelled $1$, $2$, \dots, $2n$ each of which can be either on or off.
Initially all the lamps are off.
We consider sequences of steps: at each step one of the lamps is switched
(from on to off or from off to on).
Let $... | The answer is $2^{k-n}$.
Consider the following map $\Psi$ from $N$-sequences to $M$-sequences:
change every instance of $n+1$ to $1$;
change every instance of $n+2$ to $2$;
[$\vdots$]
change every instance of $2n$ to $n$.
(For example, suppose $k=9$, $n=3$;
then $144225253 \mapsto 111222223$.)
Clearly this is map ... |
6 | Let $ABCD$ be a convex quadrilateral with $BA \neq BC$.
Denote the incircles of triangles $ABC$ and $ADC$
by $\omega_1$ and $\omega_2$ respectively.
Suppose that there exists a circle $\omega$ tangent
to ray $BA$ beyond $A$ and to the ray $BC$ beyond $C$,
which is also tangent to the lines $AD$ and $CD$.
Prove that the... | By the external version of Pitot theorem, the existence
of $\omega$ implies that
$ BA + AD = CB + CD. $
Let $PQ$ and $ST$ be diameters of $\omega_1$ and $\omega_2$
with $P, T \in AC$.
Then the length relation on $ABCD$ implies that $P$ and $T$
are reflections about the midpoint of $AC$.
Now orient $AC$ horizontally an... |
1 | Let $n, k \ge 2$ be positive integers and let $a_1$, $a_2$, $a_3$, \dots, $a_k$
be distinct integers in the set $\left\{ 1,2,\dots,n \right\}$
such that $n$ divides $a_i(a_{i+1} - 1)$ for $i = 1,2,\dots,k-1$.
Prove that $n$ does not divide $a_k(a_1 - 1)$. | We proceed indirectly and assume that
$ a_i (a_{i+1}-1) \equiv 0 \pmod n $
for $i = 1, \dots, k$ (indices taken modulo $k$).
We claim that this implies all the $a_i$ are equal modulo $n$.
Let $q = p^e$ be any prime power dividing $n$.
Then, $a_1 (a_2 - 1) \equiv 0 \pmod q$, so $p$ divides either $a_1$ or $a_2-1$.
If ... |
2 | Let $ABC$ be a triangle with circumcenter $O$.
The points $P$ and $Q$ are interior points of the sides $CA$ and $AB$ respectively.
Let $K$, $L$, $M$ be the midpoints of $BP$, $CQ$, $PQ$.
Suppose that $PQ$ is tangent to the circumcircle of $\triangle KLM$.
Prove that $OP = OQ$. | By power of a point, we have $-AQ \cdot QB = OQ^2 - R^2$
and $-AP \cdot PC = OP^2 - R^2$.
Therefore, it suffices to show $AQ \cdot QB = AP \cdot PC$.
[Figure omitted]
As $ML \parallel AC$ and $MK \parallel AB$ we have that
∠ APQ &= ∠ LMP = ∠ LKM \\
∠ PQA &= ∠ KMQ = ∠ MLK
and consequently we have the (opposite orien... |
3 | Suppose that $s_1,s_2,s_3, \dotsc$ is a strictly increasing sequence of
positive integers such that the sub-sequences
$s_{s_1}$, $s_{s_2}$, $s_{s_3}$, \dots
and $s_{s_1 + 1}$, $s_{s_2 + 1}$, $s_{s_3 + 1}$, \dots
are both arithmetic progressions.
Prove that the sequence $s_1$, $s_2$, $s_3$, \dots\ is itself an arithmeti... | We present two solutions.
First solution (Alex Zhai).:
Let $s(n) \coloneq s_n$ and write
s(s(n)) &= Dn + A \\
s(s(n)+1) &= D'n + B.
In light of the bounds $s(s(n)) \le s(s(n)+1) \le s(s(n+1))$
we right away recover $D = D'$ and $A \le B$.
Let $d_n = s(n+1)-s(n)$.
Note that $\sup d_n < \infty$ since $d_n$ is bounde... |
4 | Let $ABC$ be a triangle with $AB = AC$.
The angle bisectors of $\angle CAB$ and $\angle ABC$
meet the sides $BC$ and $CA$ at $D$ and $E$, respectively.
Let $K$ be the incenter of triangle $ADC$.
Suppose that $\angle BEK = 45^\circ$.
Find all possible values of $\angle CAB$. | Here is the solution presented in my book EGMO.
Let $I$ be the incenter of $ABC$,
and set $\angle DAC = 2x$ (so that $0\dg < x < 45\dg$).
From $\angle AIE = \angle DIC$, it is easy to compute
$
\angle KIE = 90\dg - 2x, \;
\angle ECI = 45\dg -x, \;
\angle IEK = 45\dg, \;
\angle KEC = 3x. $
Having chased all the angles ... |
5 | Find all functions $f \colon \ZZ_{>0} \to \ZZ_{>0}$
such that for positive integers $a$ and $b$, the numbers
$ a, \qquad f(b), \qquad f(b+f(a)-1) $
are the sides of a non-degenerate triangle. | The only function is the identity function (which works).
We prove it is the only one.
Let $P(a,b)$ denote the given statement.
We have $f(1) = 1$, and $f(f(n)) = n$.
(In particular $f$ is a bijection.)
Note that $ P(1,b) \implies f(b) = f(b+f(1)-1). $
Otherwise, the function $f$ is periodic modulo $N = f(1)-1 \ge 1... |
6 | Let $a_1$, $a_2$, \dots, $a_n$ be distinct positive integers and
let $M$ be a set of $n-1$ positive integers not containing $s = a_1 + \dots + a_n$.
A grasshopper is to jump along the real axis, starting at the point $0$ and
making $n$ jumps to the right with lengths $a_1$, $a_2$, \dots, $a_n$ in some order.
Prove that... | The proof is by induction on $n$.
Assume $a_1 < \dots < a_n$ and call each element of $M$ a mine.
Let $x = s - a_n$.
We consider four cases, based on whether $x$ has a mine
and whether there is a mine past $x$.
If $x$ has no mine, and there is a mine past $x$,
then at most $n-2$ mines in $[0, x]$ and so we use inducti... |
1 | Find all functions $f \colon \RR \to \RR$ such that for all $x,y \in \RR$,
$ f(\left\lfloor x\right\rfloor y) = f(x)\left\lfloor f(y)\right\rfloor. $ | The only solutions are $f(x) \equiv c$,
where $c = 0$ or $1 \le c < 2$.
It's easy to see these work.
Plug in $x=0$ to get $f(0) = f(0) \left\lfloor f(y) \right\rfloor$,
so either
$ 1 \le f(y) < 2 \quad \forall y
\qquad\text{or}\qquad f(0) = 0 $
In the first situation,
plug in $y=0$ to get $f(x) \left\lfloor f(0) \rig... |
2 | Let $I$ be the incenter of a triangle $ABC$ and let $\Gamma$ be its circumcircle.
Let line $AI$ intersect $\Gamma$ again at $D$.
Let $E$ be a point on arc $\widehat{BDC}$ and $F$ a point on side $BC$ such that
$ \angle BAF = \angle CAE < \tfrac12 \angle BAC. $
Finally, let $G$ be the midpoint of $IF$.
Prove that $DG$ a... | Let $EI$ meet $\Gamma$ again at $K$.
Then it suffices to show that $KD$ bisects $IF$.
Let $AF$ meet $\Gamma$ again at $H$, so $HE \parallel BC$.
By Pascal theorem on $ AHEKDD $
we then obtain that $P = AH \cap KD$ lies on a line through $I$
parallel to $BC$.
Let $I_A$ be the $A$-excenter,
and set $Q = I_AF \cap IP$, a... |
3 | Find all functions $g \colon \ZZ_{>0} \to \ZZ_{>0}$ such that
$ ( g(m)+n )( g(n)+m ) $
is always a perfect square. | For $c \ge 0$, the function $g(n) = n+c$ works; we prove this is the only possibility.
First, the main point of the problem is that:
We have $g(n) \equiv g(n') \pmod p \implies n \equiv n' \pmod p$.
Pick a large integer $M$ such that
$ \nu_p(M+g(n)), \quad \nu_p(M+g(n')) \quad \text{are both odd}. $
(It's not hard t... |
4 | Let $P$ be a point interior to triangle $ABC$ (with $CA \neq CB$).
The lines $AP$, $BP$ and $CP$ meet again its circumcircle $\Gamma$
at $K$, $L$, $M$, respectively.
The tangent line at $C$ to $\Gamma$ meets the line $AB$ at $S$.
Show that from $SC = SP$ follows $MK = ML$. | We present two solutions using harmonic bundles.
First solution (Evan Chen).:
Let $N$ be the antipode of $M$, and let $NP$ meet $\Gamma$ again at $D$.
Focus only on $CDMN$ for now (ignoring the condition).
Then $C$ and $D$ are feet of altitudes in $\triangle MNP$;
it is well-known that the circumcircle of $\triangle ... |
5 | Each of the six boxes $B_1$, $B_2$, $B_3$, $B_4$, $B_5$, $B_6$
initially contains one coin.
The following two types of operations are allowed:
Choose a non-empty box $B_j$, $1\leq j \leq 5$,
remove one coin from $B_j$ and add two coins to $B_{j+1}$;
Choose a non-empty box $B_k$, $1\leq k \leq 4$,
remove one coin from ... | First,
(1,1,1,1,1,1) &\to (0,3,1,0,3,1) \to (0,0,7,0,0,7) \\
&\to (0,0,6,2,0,7) \to (0,0,6,1,2,7) \to (0,0,6,1,0,11) \\
&\to (0,0,6,0,11,0) \to (0,0,5,11,0,0).
and henceforth we ignore boxes $B_1$ and $B_2$,
looking at just the last four boxes;
so we write the current position as $(5,11,0,0)$.
We prove a lemma:
Let... |
6 | Let $a_1, a_2, a_3, \dots$ be a sequence of positive real numbers, and $s$ be a positive integer, such that
$
a_n =
\max \{ a_k + a_{n-k} \mid 1 \leq k \leq n-1 \}
\text{ for all $n > s$}.
$
Prove there exist positive integers $\ell \leq s$ and $N$, such that
$
a_n =
a_{\ell} + a_{n - \ell} \text{ for all $n \ge N$}.
$ | Let $ w_1 = \frac{a_1}{1}, \quad w_2 = \frac{a_2}{2},
\quad \dots, \quad w_s = \frac{a_s}{s}. $
(The choice of the letter $w$ is for ``weight''.)
We claim the right choice of $\ell$
is the one maximizing $w_\ell$.
Our plan is to view each $a_n$ as a linear combination
of the weights $w_1, \dots, w_s$ and track their c... |
1 | Given any set $A = \{a_1, a_2, a_3, a_4\}$ of four distinct
positive integers, we denote the sum $a_1+a_2+a_3+a_4$ by $s_A$.
Let $n_A$ denote the number of pairs $(i,j)$ with $1 \le i < j \le 4$
for which $a_i + a_j$ divides $s_A$.
Find all sets $A$ of four distinct positive integers which achieve
the largest possible ... | There are two curves of solutions,
namely $\{x,5x,7x,11x\}$ and $\{x,11x,19x,29x\}$, for any positive integer $x$,
achieving $n_A = 4$ (easy to check).
We'll show that $n_A \le 4$ and equality holds only in one of the curves.
Let $A = \{a < b < c < d\}$.
We have $n_A \le 4$ with equality iff
$ a+b \mid c+d, \quad a+c... |
2 | Let $\mathcal{S}$ be a finite set of at least two points in the plane.
Assume that no three points of $\mathcal S$ are collinear.
A windmill is a process that starts with a
line $\ell$ going through a single point $P \in \mathcal S$.
The line rotates clockwise about the pivot $P$ until the first time
that the line meet... | Orient $\ell$ in some direction,
and color the plane such that its left half is red
and right half is blue.
The critical observation is that:
The number of points on the red side of $\ell$ does not change,
nor does the number of points on the blue side
(except at a moment when $\ell$ contains two points).
Thus, if $|... |
3 | Let $f \colon \RR \to \RR$ be a real-valued function
defined on the set of real numbers that satisfies
$ f(x+y) \leq yf(x) + f(f(x))$
for all real numbers $x$ and $y$.
Prove that $f(x) = 0$ for all $x \leq 0$. | We begin by rewriting the given as
$ f(z) \le (z-x)f(x) + f(f(x)) \quad
\forall x,z \in \RR \qquad (\heartsuit) $
(which is better anyways since control over inputs to $f$ is more valuable).
We start by eliminating the double $f$:
let $z = f(w)$ to get
$ f(f(w)) \le (f(w)-x)f(x) + f(f(x)) $
and then use the symmetry tr... |
4 | Let $n > 0$ be an integer.
We are given a balance and $n$ weights of weight $2^0$, $2^1$, \dots, $2^{n-1}$.
We are to place each of the $n$ weights on the balance, one after another,
in such a way that the right pan is never heavier than the left pan.
At each step we choose one of the weights
that has not yet been plac... | The answer is $a_n = (2n-1)!!$.
We refer to what we're counting as a valid $n$-sequence:
an order of which weights to place,
and whether to place them on the left or right pan.
We use induction, with $n=1$ being obvious.
Now consider the weight $2^0 = 1$.
If we delete it from any valid $n$-sequence,
we get a valid $... |
5 | Let $f \colon \ZZ \to \ZZ_{>0}$ be a function such that
$f(m-n) \mid f(m) - f(n)$ for $m,n \in \ZZ$.
Prove that if $m,n \in \ZZ$ satisfy $f(m) \le f(n)$
then $f(m) \mid f(n)$. | Let $P(m,n)$ denote the given assertion.
First, we claim $f$ is even.
This is straight calculation:
$P(x,0) \implies f(x) \mid f(x)-f(0)
\implies f(x) \mid M \coloneq f(0)$.
$P(0,x) \implies f(-x) \mid M-f(x) \implies
f(-x) \mid f(x)$.
Analogously, $f(x) \mid f(-x)$.
So $f(x) = f(-x)$ and $f$ is even.
Let $x$, $y$, $... |
6 | Let $ABC$ be an acute triangle with circumcircle $\Gamma$.
Let $\ell$ be a tangent line to $\Gamma$, and let $\ell_a$, $\ell_b$, $\ell_c$ be the lines obtained
by reflecting $\ell$ in the lines $BC$, $CA$, and $AB$, respectively.
Show that the circumcircle of the triangle determined by the lines $\ell_a$, $\ell_b$, and... | This is a hard problem with many beautiful solutions.
The following solution is not very beautiful but not too hard to find during an olympiad,
as the only major insight it requires is the construction of $A_2$, $B_2$, and $C_2$.
[Figure omitted]
We apply complex numbers with $\omega$ the unit circle and $p=1$. Let $... |
1 | Let $ABC$ be a triangle and $J$ the center of the $A$-excircle.
This excircle is tangent to the side $BC$ at $M$,
and to the lines $AB$ and $AC$ at $K$ and $L$, respectively.
The lines $LM$ and $BJ$ meet at $F$, and the lines $KM$ and $CJ$ meet at $G$.
Let $S$ be the point of intersection of the lines $AF$ and $BC$,
an... | We employ barycentric coordinates with reference $\triangle ABC$.
As usual $a = BC$, $b = CA$, $c = AB$, $s = 1/2(a+b+c)$.
It's obvious that $K = ( -(s-c): s : 0)$, $M = ( 0 : s-b : s-c)$.
Also, $J = (-a : b : c)$.
We then obtain
$ G = ( -a: b : \frac{-as + (s-c)b}{s-b} ). $
It follows that
$ T = ( 0 : b : \frac{-as +... |
2 | Let $a_2$, $a_3$, \dots, $a_n$ be positive reals with product $1$,
where $n \ge 3$.
Show that
$ (1+a_2)^2 (1+a_3)^3 \dots (1+a_n)^n > n^n. $ | Try the dumbest thing possible: by AM-GM,
(1 + a_2)^2 &\ge 2^2 a_2 \\
(1 + a_3)^3 = ( 1/2 + 1/2 + a_3 )^3 &\ge \frac{3^3}{2^2} a_3 \\
(1 + a_4)^4 = ( \frac13 + \frac13 + \frac13 + a_4 )^4
&\ge \frac{4^4}{3^3} a_4 \\
&\vdotswithin=
and so on.
Multiplying these all gives the result.
The inequality is strict since it's ... |
3 | The liar's guessing game is a game played between two players $A$ and $B$.
The rules of the game depend on two fixed positive integers $k$ and $n$
which are known to both players.
At the start of the game $A$
chooses integers $x$ and $N$ with $1 \le x \le N$.
Player $A$ keeps $x$ secret, and truthfully tells $N$ to pl... | Call the players Alice and Bob.
\textbf{Part (a)}: We prove the following.
If $N \ge 2^k+1$, then in $2k+1$ questions,
Bob can rule out some number in $\{1, \dots, 2^k+1\}$
form being equal to $x$.
First, Bob asks the question $S_0 = \{ 2^k+1 \}$
until Alice answers ``yes''
or until Bob has asked $k+1$ questions.
If... |
4 | Find all functions $f \colon \ZZ \to \ZZ$ such that,
for all integers $a$, $b$, $c$ that satisfy $a+b+c=0$,
the following equality holds:
$ f(a)^2+f(b)^2+f(c)^2 = 2f(a)f(b)+2f(b)f(c)+2f(c)f(a). $ | Answer: for arbitrary $k \in \ZZ$, we have
[(i)]
$f(x) = kx^2$,
$f(x) = 0$ for even $x$, and $f(x) = k$ for odd $x$, and
$f(x) = 0$ for $x \equiv 0 \pmod 4$,
$f(x) = k$ for odd $x$, and $f(x) = 4k$ for $x \equiv 2 \pmod 4$.
These can be painfully seen to work.
(It's more natural to think of these as
$f(x) = x^2$, $f(x... |
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