1998 USAMO #3

Problem : Let $a_0,a_1,\cdots ,a_n$ be numbers from the interval $(0,\pi/2)$ such that\[ \tan (a_0-\frac{\pi}{4})+ \tan (a_1-\frac{\pi}{4})+\cdots +\tan (a_n-\frac{\pi}{4})\geq n-1.  \]Prove that\[ \tan a_0\tan a_1 \cdots \tan a_n\geq n^{n+1}.  \]

2022 USEMO #3

Point $P$ lies in the interior of a triangle $ABC$. Lines $AP$, $BP$, and $CP$ meet the opposite sides of triangle $ABC$ at $A$', $B'$, and $C'$ respectively. Let $P_A$ the midpoint of the segment joining the incenters of triangles $BPC'$ and $CPB'$, and define points $P_B$ and $P_C$ analogously. Show that if \[ AB'+BC'+CA'=AC'+BA'+CB' \]then points $P,P_A,P_B,$ and $P_C$ are concyclic. 

2019 IMO SL #A5

Problem : Let $x_1, x_2, \dots, x_n$ be different real numbers. Prove that \[\sum_{1 \leqslant i \leqslant n} \prod_{j \neq i} \frac{1-x_{i} x_{j}}{x_{i}-x_{j}}=\left\{\begin{array}{ll} 0, & \text { if } n \text { is even; } \\ 1, & \text { if } n \text { is odd. } \end{array}\right.\] Solution 1 (induction) : We induct on $n$, and the base case is trivial. Now for any $n > 2,$ define the multivariable rational function$$\sum_{1 \leqslant i \leqslant n} \prod_{j \neq i} \frac{1-x_{i} x_{j}}{x_{i}-x_{j}} = A_n(x_1, \dots x_n)$$and the multivariable polynomial$$B_n(x_1, \dots x_n) = \prod_{i<j}(x_i-x_j)(A_n - (\text{n's remainder mod 2})).$$This polynomial has at most degree $\binom{n}{2} + (n-1).$ But for every $x_i,$ we know $x_i - 1$ and $x_i+1$ divide $B_n$ by the inductive hypothesis, as well as all the terms of the form $x_i-x_j.$ That yields $\binom{n}{2} + 2n$ terms dividing $B_n,$ more than its degree. So $B_n$ is a zero polynomial as desired. $\square$ Solutio

2019 ISL #C7

 There are 60 empty boxes $B_1,\ldots,B_{60}$ in a row on a table and an unlimited supply of pebbles. Given a positive integer $n$, Alice and Bob play the following game. In the first round, Alice takes $n$ pebbles and distributes them into the 60 boxes as she wishes. Each subsequent round consists of two steps: (a) Bob chooses an integer $k$ with $1\leq k\leq 59$ and splits the boxes into the two groups $B_1,\ldots,B_k$ and $B_{k+1},\ldots,B_{60}$. (b) Alice picks one of these two groups, adds one pebble to each box in that group, and removes one pebble from each box in the other group. Bob wins if, at the end of any round, some box contains no pebbles. Find the smallest $n$ such that Alice can prevent Bob from winning.

2019 IMC #3

Problem : Let $f:(-1,1)\to \mathbb{R}$ be a twice differentiable function such that $$2f’(x)+xf''(x)\geqslant 1 \quad \text{ for } x\in (-1,1).$$Prove that $$\int_{-1}^{1}xf(x)dx\geqslant \frac{1}{3}.$$ Solution 1 : The LHS of the original inequality looks like the result of an application of the product rule. Indeed, the derivative of $f'g$ is $g'f'+gf''$. We cannot make $g=x$ and $g'=2$, but we can multiply by $x$ to get $g=x^2, g'=2x$. Now to do this we need to be careful about the sign of $x$. So, suppose that $x > 0$. Then $2xf'(x)+x^2f''(x) \ge x$ and hence $(x^2f')' \ge x$ so that $x^2f'(x) \ge \frac{1}{2} x^2$ and hence $f'(x) \ge \frac{1}{2}$. Similarly, we know that $f'(x)\ge \frac{1}{2}$ for $x < 0$, so we have proved this for all $x$ (for $x=0$ by continuity or by just plugging in $x=0$ into the original inequality). The rest is easy. WLOG, assume that $f(0)=0$ since shifting $f$ by a constant changes

2012 IMO SL #G6

  Problem:  Let $ABC$ be a triangle with circumcenter $O$ and incenter $I$. The points $D,E$ and $F$ on the sides $BC,CA$ and $AB$ respectively are such that $BD+BF=CA$ and $CD+CE=AB$. The circumcircles of the triangles $BFD$ and $CDE$ intersect at $P \neq D$. Prove that $OP=OI$. Solution 1:  Let circle with radius $OI$ intersect $BI, CI$ at $S, T \ne I$ respectively. Let $H$ be the foot of $S$ to $AB$ and let $BI$ meet $(ABC)$ at $M \ne A.$ It's known that if any circle passing through $S,B$ intersects $BC,BA$ at $D',F'$ respectively then $BD'+BF' = 2BH.$ But$$BH = BS \cdot \cos \left(\frac{1}{2} \angle ABC \right) = IM \cdot \cos \left(\frac{1}{2} \angle ABC \right) = MA \cdot \cos \left(\frac{1}{2} \angle ABC \right) = \frac{1}{2}AC$$so in fact $BFSD$ cyclic, similarly $CDTE$ cyclic. Then$$\angle TPS = 360^\circ - \angle TPD - \angle SPD = \frac{1}{2} \angle ABC  + \frac{1}{2} \angle ACB = \angle TIS$$so $P$ lies on $(TIS)$ centered at $O,$ done. $\square$ Soluti

2021 IMO SL #A1

Problem:  Let $n$ be a positive integer. Given is a subset $A$ of $\{0,1,...,5^n\}$ with $4n+2$ elements. Prove that there exist three elements $a<b<c$ from $A$ such that $c+2a>3b$. Solution:  Suppose for the sake of contradiction that $c+2a\le 3b$ for all $a,b,c\in A$ with $a<b<c$.  We claim that if the elements of $A$ are $s_1<s_2<\dots<s_{4n+2}$, then $$s_i\ge s_{4n+2}(1-(\tfrac{2}{3})^{i-1})$$ for all $i$. We use induction to prove this -- the base case is trivially true because $s_1\geq 0$. As for the inductive step, consider $s_k.$ Take $(a,b,c)=(s_{k-1},s_k,s_{4n+2})$; by our assumption, we have $$s_{4n+2}+2s_{k-1}\ge 3s_k.$$ Therefore, $$s_k\ge\frac{s_{4n+2}+2s_{k-1}}{3}\ge\frac{s_{4n+2}+2(s_{4n+2}(1-(\tfrac{2}{3})^{k-2}))}{3}=\frac{s_{4n+2}(3-2(\tfrac{2}{3})^{k-2})}{3}=s_{4n+2}(1-(\tfrac{2}{3})^{k-1}),$$ which completes the induction.  Now, using our claim, we have $$s_{4n+1}\ge s_{4n+2}(1-(\tfrac{2}{3})^{4n})=s_{4n+2}-(\tfrac{16}{81})^{n}s_{4n+2}>s_

2017 USAJMO #3

Let $ABC$ be an equilateral triangle, and point $P$ on its circumcircle. Let $PA$ and $BC$ intersect at $D$, $PB$ and $AC$ intersect at $E$, and $PC$ and $AB$ intersect at $F$. Prove that the area of $\triangle DEF$ is twice the area of $\triangle ABC$. Solution 1:  Note that $$\angle{DPF}=\angle{FPE}=\angle{EPD}=120^{\circ}.$$ Now, using the sin area formula, we get $$[DEF]=[DPF]+[FPE]+[EPD]=\frac{\sqrt{3}}{4}\left(DP \cdot FP+FP \cdot EP+EP \cdot DP\right).$$We also have that $$[ABC]=BC^2 \cdot \frac{\sqrt{3}}{4}.$$ So, it suffices to prove $$\frac{\sqrt{3}}{4}\left(DP \cdot FP+FP \cdot EP+EP \cdot DP\right)=2 \cdot \frac{BC^2\sqrt{3}}{4} \implies DP \cdot FP+FP \cdot EP+EP \cdot DP=2BC^2.$$By the Law of Cosines on $\triangle{BPC}$, we obtain $$BC^2=b^2+c^2-2 \cdot b \cdot c \cdot \cos{120^{\circ}}=b^2+c^2+bc,$$ and Ptolemy's Theorem on quadrilateral $ABPC$ gives$$AP \cdot BC=BP \cdot AC+CP \cdot AB \implies AP=b+c.$$From some angle chasing, we know that $\triangle{ACD} \sim \tri

2000 Putnam A1

Problem: Let $A$ be a positive real number. What are the possible values of $$\sum_{j=0}^{\infty} x_j^2,$$ given that $x_0,x_1,\dots$ are positive numbers for which $$\sum_{j=0}^{\infty}=A?$$ Solution:  The answer is $\boxed{(0,A^2)}$. \\ There are two things we need to prove: first, that the possible values of $\sum_{j=0}^{\infty} x_j^2$ must lie in the interval $(0,A^2)$, and second, all values in this interval can be achieved. \\ For our first problem, the values are obviously positive. Also, note that $$\sum_{j=0}^{\infty} \frac{x_j}{A}=1,$$ and because $x_j>0$, we must have $$0 <  \frac{x_j}A   < 1,$$ which implies that $$\left(\frac{x_j}A\right)^2   <  \frac{x_j}A  ,$$$$\implies \sum_{j=0}^{\infty}\left(\frac{x_j}A\right)^2   <  \sum_{j=0}^{\infty}\left(\frac{x_j}A\right)   = 1.$$Multiplying by $A^2$, we get $$\sum_{j=0}^{\infty}x_j^2 < A^2.$$ \\ Now, for the second part, let the sequence of $x_j$'s be a geometric sequence with ratio $r$. Then, we have $$\su

Occasional Putnam problems

Starting now, I will occasionally post Putnam problems on this blog. However, the main content will still be IMO Problems.  These problems will be under the tag #putnam.

2006 IMO #G3

Let $ ABCDE$ be a convex pentagon such that \[ \angle BAC = \angle CAD = \angle DAE\qquad \text{and}\qquad \angle ABC = \angle ACD = \angle ADE. \]The diagonals $BD$ and $CE$ meet at $P$. Prove that the line $AP$ bisects the side $CD$.

Problem request: 2022 AIME II #13

Hi everyone,  I hope you had a great summer!  I was not posting for about a month or so because not many people were viewing the blog in the summer.  I'm going to start to post frequently again, and for now I have a problem request I got a couple days ago.  There is a polynomial $P(x)$ with integer coefficients such that\[P(x)=\frac{(x^{2310}-1)^6}{(x^{105}-1)(x^{70}-1)(x^{42}-1)(x^{30}-1)}\]holds for every $0<x<1.$ Find the coefficient of $x^{2022}$ in $P(x)$.

2015 IMO SL #A1

 Suppose that a sequence $a_1,a_2,\ldots$ of positive real numbers satisfies\[a_{k+1}\geq\frac{ka_k}{a_k^2+(k-1)}\]for every positive integer $k$. Prove that $a_1+a_2+\ldots+a_n\geq n$ for every $n\geq2$.

2019 IMO SL #C1

 The infinite sequence $a_0,a _1, a_2, \dots$ of (not necessarily distinct) integers has the following properties: $0\le a_i \le i$ for all integers $i\ge 0$, and\[\binom{k}{a_0} + \binom{k}{a_1} + \dots + \binom{k}{a_k} = 2^k\]for all integers $k\ge 0$. Prove that all integers $N\ge 0$ occur in the sequence (that is, for all $N\ge 0$, there exists $i\ge 0$ with $a_i=N$).

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