IMO Math

  Problem : We have $2^m$ sheets of paper, with the number $1$ written on each of them. We perform the following operation. In every step we choose two distinct sheets; if the numbers on the two sheets are $a$ and $b$, then we erase these numbers and write the number $a + b$ on both sheets. Prove that after $m2^{m -1}$ steps, the sum of the numbers on all the sheets is at least $4^m$ .

Problem : Consider $2009$ cards, each having one gold side and one black side, lying on parallel on a long table. Initially all cards show their gold sides. Two player, standing by the same long side of the table, play a game with alternating moves. Each move consists of choosing a block of $50$ consecutive cards, the leftmost of which is showing gold, and turning them all over, so those which showed gold now show black and vice versa. The last player who can make a legal move wins. (a) Does the game necessarily end? (b) Does there exist a winning strategy for the starting player?

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\[ \tan a_0\tan a_1 \cdots \tan a_n\geq n^{n+1}.  \]

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. 

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

 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.

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

  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

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_

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

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

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.

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 $CD$.

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 $P(x)$.

 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 $n\geq2$.

 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 $a_i=N$).

Hello,  I have enabled ads on this blog as a fund for my future education. Please cooperate with this and if they annoy you, click the down arrow button next to the ad to hide it. Thank you!

 Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc = 1$. Prove that \[ \frac {1}{a^{3}\left(b + c\right)} + \frac {1}{b^{3}\left(c + a\right)} + \frac {1}{c^{3}\left(a + b\right)}\geq \frac {3}{2}. \]

 Let $n > 1$ be an integer and let $f(x) = x^n + 5 \cdot x^{n-1} + 3.$ Prove that there do not exist polynomials $g(x),h(x),$ each having integer coefficients and degree at least one, such that $f(x) = g(x) \cdot h(x).$ Solution 1 (overkill):  Since $5>1+3$, from Perron's Criterion this polynomial is irreducible over the integers as desired. Solution 2 (normal solution): Assume FTSOC that there do exist $g(x), h(x)$ such that $f(x)=g(x)\cdot h(x)$. Observe that there are no integer roots of $f(x)$ from the Rational Root Theorem. Thus, $g(x), h(x)$ cannot be linear, and their degree is greater than or equal to $2$. We have $g(0)h(0)=3$. WLOG $g(0)=1$. Let $r_1, r_2, \dots r_j \in\mathbb{C}$ be the roots of $g(x)$. We have $r_1r_2\dots r_n=\pm 1$. Multiplying the equalities $r_i^{n-1}(r_i+5)=-3$ for all $1\leq i\leq j$, we obtain\[\vert g(-5)\vert = \vert (r_1+5)(r_2+5)\dots (r_m+5)\vert = 3^m.\]But $g(-5)f(-5)=3$, contradiction.

 Prove for all positive reals $a,b,c$, $$\sum_{cyc} \frac{a}{\sqrt{3ab+bc}} \ge \frac 32.$$ I'll wait for comments and will post the solution in 2 days! EDIT - Well now that it's been 3 days (oops i forgot to post), I'll show my solution now.  By Holder,$$\left( \sum_{cyc} \frac{a}{\sqrt{3ab+bc}} \right)^2 \left( \sum_{cyc} a(3ab+bc) \right) \ge (a+b+c)^3$$ So it suffices to show$$(a+b+c)^3 \ge \frac 94 \sum_{cyc} a(3ab+bc)$$ Expanding and cancelling terms, we wish to show$$f(a,b,c)=4\sum_{cyc} a^3 + 12\sum_{cyc} b^2a-15\sum_{cyc} a^2b-3abc\ge 0$$ Claim: $f(a,b,c) \le f(a+d, b+d, c+d)$ if $d>0$ Proof:$$f(a+d,b+d,c+d)-f(a,b,c) =d (4\sum_{cyc} 3a^2 + 12\sum_{cyc} (b^2+2ba) - 15\sum_{cyc} (a^2+2ba) - 3(ab+bc+ca)) $$ $$+d^2 (4\sum_{cyc} 3a + 12 \sum_{cyc} (2b+a) - 15\sum_{cyc} (2a+b) - 3(a+b+c)$$ The $d^2, d^3$ stuff cancel. So $f(a+d,b+d,c+d) - f(a,b,c) = \frac{9d}{2} \sum_{cyc} (a-b)^2 > 0$. Let $a=\min\{a,b,c\}$, then $f(a,a+b,a+c) \ge f(0,b,c)$. Let $x=\frac bc$, then

Last one. I promise.   Prove that for all positive real numbers $a,b,c$,\[ \frac{a}{\sqrt{a^2 + 8bc}} + \frac{b}{\sqrt{b^2 + 8ca}} + \frac{c}{\sqrt{c^2 + 8ab}} \geq 1.  \]

Since the last one was super easy, I have to do another one... right?  We have $2^m$ sheets of paper, with the number $1$ written on each of them. We perform the following operation. In every step we choose two distinct sheets; if the numbers on the two sheets are $a$ and $b$, then we erase these numbers and write the number $a + b$ on both sheets. Prove that after $m2^{m -1}$ steps, the sum of the numbers on all the sheets is at least $4^m$ .

 Let $ABC$ be a triangle. Circle $\Gamma$ passes through $A$, meets segments $AB$ and $AC$ again at points $D$ and $E$ respectively, and intersects segment $BC$ at $F$ and $G$ such that $F$ lies between $B$ and $G$. The tangent to circle $BDF$ at $F$ and the tangent to circle $CEG$ at $G$ meet at point $T$. Suppose that points $A$ and $T$ are distinct. Prove that line $AT$ is parallel to $BC$.

Let $ABC$ be an acute triangle with orthocenter $H$, and let $W$ be a point on the side $BC$, lying strictly between $B$ and $C$. The points $M$ and $N$ are the feet of the altitudes from $B$ and $C$, respectively. Denote by $\omega_1$ is the circumcircle of $BWN$, and let $X$ be the point on $\omega_1$ such that $WX$ is a diameter of $\omega_1$. Analogously, denote by $\omega_2$ the circumcircle of triangle $CWM$, and let $Y$ be the point such that $WY$ is a diameter of $\omega_2$. Prove that $X,Y$ and $H$ are collinear.

 Find the smallest positive integer $n$ or show no such $n$ exists, with the following property: there are infinitely many distinct $n$-tuples of positive rational numbers $(a_1, a_2, \ldots, a_n)$ such that both $$a_1+a_2+\dots +a_n \quad \text{and} \quad \frac{1}{a_1} + \frac{1}{a_2} + \dots + \frac{1}{a_n}$$are integers.

Let $ABC$ be a triangle with circumcircle $\Omega$ and incentre $I$. Let the line passing through $I$ and perpendicular to $CI$ intersect the segment $BC$ and the arc $BC$ (not containing $A$) of $\Omega$ at points $U$ and $V$ , respectively. Let the line passing through $U$ and parallel to $AI$ intersect $AV$ at $X$, and let the line passing through $V$ and parallel to $AI$ intersect $AB$ at $Y$ . Let $W$ and $Z$ be the midpoints of $AX$ and $BC$, respectively. Prove that if the points $I, X,$ and $Y$ are collinear, then the points $I, W ,$ and $Z$ are also collinear.

A site is any point $(x, y)$ in the plane such that $x$ and $y$ are both positive integers less than or equal to 20. Initially, each of the 400 sites is unoccupied. Amy and Ben take turns placing stones with Amy going first. On her turn, Amy places a new red stone on an unoccupied site such that the distance between any two sites occupied by red stones is not equal to $\sqrt{5}$. On his turn, Ben places a new blue stone on any unoccupied site. (A site occupied by a blue stone is allowed to be at any distance from any other occupied site.) They stop as soon as a player cannot place a stone. Find the greatest $K$ such that Amy can ensure that she places at least $K$ red stones, no matter how Ben places his blue stones.

Two squirrels, Bushy and Jumpy, have collected 2021 walnuts for the winter. Jumpy numbers the walnuts from 1 through 2021, and digs 2021 little holes in a circular pattern in the ground around their favorite tree. The next morning Jumpy notices that Bushy had placed one walnut into each hole, but had paid no attention to the numbering. Unhappy, Jumpy decides to reorder the walnuts by performing a sequence of 2021 moves. In the $k$-th move, Jumpy swaps the positions of the two walnuts adjacent to walnut $k$. Prove that there exists a value of $k$ such that, on the $k$-th move, Jumpy swaps some walnuts $a$ and $b$ such that $a<k<b$.

 Hello,  I just noticed that the blog was private for a while. I fixed the problem now. Sorry for the inconvenience.

Four real numbers $ p$, $ q$, $ r$, $ s$ satisfy $ p+q+r+s = 9$ and $ p^{2}+q^{2}+r^{2}+s^{2}= 21$. Prove that there exists a permutation $ \left(a,b,c,d\right)$ of $ \left(p,q,r,s\right)$ such that $ ab-cd \geq 2$.

As I said in an earlier post, I will also post problems that are not from the IMO but I believe that they are of the same level. Let $ABC$ be a triangle with $AB<AC$ and $I$ be your incenter. Let $M$ and $N$ be the midpoints of the sides $BC$ and $AC$, respectively. If the lines $AI$ and $IN$ are perpendicular, prove that the line $AI$ is tangent to the circumcircle of $\triangle IMC$.

 From now on, I will only stream on Twitch if someone wants one-on-one detailed help on a solution they don't understand on my blog. 

This was the huge surprise I was talking about in the other post! Now, we have a site like this for Physics as well, which you can find at  thinkphysics1.blogspot.com .  I have combined these two to form my organization named Thonks, and its website can be found at  thonks123123.github.io/Thonks .  I hope you like it!

Let $n$ be a positive integer. Define a chameleon to be any sequence of $3n$ letters, with exactly $n$ occurrences of each of the letters $a, b,$ and $c$. Define a swap to be the transposition of two adjacent letters in a chameleon. Prove that for any chameleon $X$ , there exists a chameleon $Y$ such that $X$ cannot be changed to $Y$ using fewer than $3n^2/2$ swaps.

 In an acute-angled triangle $ABC$ on the sides $AB$, $BC$, $AC$ the points $H$, $L$, $K$ so that $CH \perp AB$, $HL \parallel AC$, $HK \parallel BC$. Let $P$ and $Q$ feet of altitudes of a triangle $HBL$, drawn from the vertices $H$ and $B$ respectively. Prove that the feet of the altitudes of the triangle $AKH$, drawn from the vertices $A$ and $H$ lie on the line $PQ$.

 Hi,  I'm sorry but I won't be able to stream today.  I have a little pain in neck and it hurts a bit when I talk. I will stream again next week (Friday, April 15th) at 7:00 PM

Now, we are officially spread all throughout the world! We have views from every continent (well other than Antarctica, of course).  We already had it spread in all continents other than Africa but now we have many views from Africa as well. Before, it was technically "spread" in Africa, but there were minimal views from there.  I'm really happy about this :) Thank you so much to everyone who views my blog :D I hope it continues to help you.

Hi,  I have decided that sometimes, I will pick problems that are not from the IMO shortlists but I believe that it is an IMO-level problem (in other words, I think that the problem could have appeared in the IMO as well). Because after all, there are a limited number of problems.  

 A permutation $ \{x_1, x_2, \ldots, x_{2n}\}$ of the set $ \{1,2, \ldots, 2n\}$ where $ n$ is a positive integer, is said to have property $ T$ if $ |x_i - x_{i + 1}| = n$ for at least one $ i$ in $ \{1,2, \ldots, 2n - 1\}.$ Show that, for each $ n$, there are more permutations with property $ T$ than without.

  Find all triplets of positive integers $ (a,m,n)$ such that $ a^m + 1 \mid (a + 1)^n$.

Let $\mathcal{A}$ denote the set of all polynomials in three variables $x, y, z$ with integer coefficients. Let $\mathcal{B}$ denote the subset of $\mathcal{A}$ formed by all polynomials which can be expressed as \begin{align*} (x + y + z)P(x, y, z) + (xy + yz + zx)Q(x, y, z) + xyzR(x, y, z) \end{align*}with $P, Q, R \in \mathcal{A}$. Find the smallest non-negative integer $n$ such that $x^i y^j z^k \in \mathcal{B}$ for all non-negative integers $i, j, k$ satisfying $i + j + k \geq n$.

In  this  post, Anonymous suggested that I should start a Twitch stream. Thanks for the suggestion! I will stream every Friday at 7:00 PM EST, starting March 25th. You can find my Twitch account here .  You can submit problems that you want me to solve during the stream at my email, imomath@imomath.xyz, or at my discord handle, math4l#4750.  If I get no problem requests, then I will pick a random problem to solve and explain. 

Today (March 18th) is my birthday! I was born on March 18th, 2009.  Thanks to my parents for supporting me in my journey so much :) I probably would've given up a long time back if it wasn't for them.  Now it's time for 2009 IMO SL #18, which is 2009 IMO SL #G3.  Let $ABC$ be a triangle. The incircle of $ABC$ touches the sides $AB$ and $AC$ at the points $Z$ and $Y$, respectively. Let $G$ be the point where the lines $BY$ and $CZ$ meet, and let $R$ and $S$ be points such that the two quadrilaterals $BCYR$ and $BCSZ$ are parallelogram. Prove that $GR=GS$. Let the incircle touch $\overline{BC}$ at $X$ and let the $A$-excircle $\omega_A$ touch $\overline{BC}$ at $X'$ and $\overline{AC}$ at $Y'$. Denote $\omega_R$ and $\omega_S$ as the circles centered at $R$ and $S$ respectively, both with radius $0$. Notice that $BR=CY=CX=BX'$ and $YR=BC=AY'-AY=YY'$, so we have that $\overline{BY}$ is the radical axis of $\omega_A$ and $\omega_R$. Similarly, $\overline{CZ}

We consider two sequences of real numbers $x_{1} \geq x_{2} \geq \ldots \geq x_{n}$ and $\ y_{1} \geq y_{2} \geq \ldots \geq y_{n}.$ Let $z_{1}, z_{2}, .\ldots, z_{n}$ be a permutation of the numbers $y_{1}, y_{2}, \ldots, y_{n}.$ Prove that $\sum \limits_{i=1}^{n} ( x_{i} -\ y_{i} )^{2} \leq \sum \limits_{i=1}^{n}$ $( x_{i} - z_{i})^{2}.$

I got several emails about how I haven't posted in a week, while I usually post during the first few days of the week. So don't worry, I'm still alive :D  Now let's get to the math.

Suppose that $ a_1$, $ a_2$, $ \ldots$, $ a_n$ are integers such that $ n\mid a_1 + a_2 + \ldots + a_n$. Prove that there exist two permutations $ \left(b_1,b_2,\ldots,b_n\right)$ and $ \left(c_1,c_2,\ldots,c_n\right)$ of $ \left(1,2,\ldots,n\right)$ such that for each integer $ i$ with $ 1\leq i\leq n$, we have \[ n\mid a_i - b_i - c_i \]

Let $ a$, $ b$, $ c$, $ d$ be positive real numbers such that $ abcd = 1$ and $ a + b + c + d > \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{d} + \dfrac{d}{a}$. Prove that \[ a + b + c + d < \dfrac{b}{a} + \dfrac{c}{b} + \dfrac{d}{c} + \dfrac{a}{d}\]

Let $S$ be a convex quadrilateral $ABCD$ and $O$ a point inside it. The feet of the perpendiculars from $O$ to $AB, BC, CD, DA$ are $A_1, B_1, C_1, D_1$ respectively. The feet of the perpendiculars from $O$ to the sides of $S_i$, the quadrilateral $A_iB_iC_iD_i$, are $A_{i+1}B_{i+1}C_{i+1}D_{i+1}$, where $i = 1, 2, 3.$ Prove that $S_4$ is similar to S.

Let $a_1, a_2, \ldots , a_n, \ldots  $ be a sequence of real numbers such that $0 \leq a_n \leq 1$ and $a_n - 2a_{n+1} + a_{n+2} \geq  0$ for $n = 1, 2, 3, \ldots$. Prove that \[0 \leq (n + 1)(a_n - a_{n+1}) \leq 2 \qquad \text{ for } n = 1, 2, 3, \ldots\]

Due to several emails requesting to close QOTD (people found them distracting to the main purpose of the blog), I will not be posting any more quotes. I have deleted the two quotes I posted as well. By the way, if you didn't know, you can contact me at imomath@imomath.xyz or at math4l#4750 on Discord for any questions or suggestions. 

For each positive integer $n$, the Bank of Cape Town issues coins of denomination $\frac1n$. Given a finite collection of such coins (of not necessarily different denominations) with total value at most most $99+\frac12$, prove that it is possible to split this collection into $100$ or fewer groups, such that each group has total value at most $1$.

Let $\omega$ be the circumcircle of a triangle $ABC$. Denote by $M$ and $N$ the midpoints of the sides $AB$ and $AC$, respectively, and denote by $T$ the midpoint of the arc $BC$ of $\omega$ not containing $A$. The circumcircles of the triangles $AMT$ and $ANT$ intersect the perpendicular bisectors of $AC$ and $AB$ at points $X$ and $Y$, respectively; assume that $X$ and $Y$ lie inside the triangle $ABC$. The lines $MN$ and $XY$ intersect at $K$. Prove that $KA=KT$.

Let $ p \geq 2$ be a prime number. Eduardo and Fernando play the following game making moves alternately: in each move, the current player chooses an index $i$ in the set $\{0,1,2,\ldots, p-1 \}$ that was not chosen before by either of the two players and then chooses an element $a_i$ from the set $\{0,1,2,3,4,5,6,7,8,9\}$. Eduardo has the first move. The game ends after all the indices have been chosen .Then the following number is computed: $$M=a_0+a_110+a_210^2+\cdots+a_{p-1}10^{p-1}= \sum_{i=0}^{p-1}a_i \cdot 10^i$$. The goal of Eduardo is to make $M$ divisible by $p$, and the goal of Fernando is to prevent this. Prove that Eduardo has a winning strategy. 

This blog just reached 1 million views... I'm glad to know that my solutions are helping people! Keep visiting this blog, more interesting solutions are to come.  As of now, I have no plans of closing this blog, and there will usually be 2-3 posts a week.   Also, if you have any questions or suggestions for this blog, please contact me at imomath@imomath.xyz  

Let $\,{\mathbb{R}}\,$ denote the set of all real numbers. Find all functions $\,f: {\mathbb{R}}\rightarrow {\mathbb{R}}\,$ such that\[ f\left( x^{2}+f(y)\right) =y+\left( f(x)\right) ^{2}\hspace{0.2in}\text{for all}\,x,y\in \mathbb{R}. \]

Let $ABC$ be an acute-angled triangle and let $D, E$, and $F$ be the feet of altitudes from $A, B$, and $C$ to sides $BC, CA$, and $AB$, respectively. Denote by $\omega_B$ and $\omega_C$ the incircles of triangles $BDF$ and $CDE$, and let these circles be tangent to segments $DF$ and $DE$ at $M$ and $N$, respectively. Let line $MN$ meet circles $\omega_B$ and $\omega_C$ again at $P \ne M$ and $Q \ne N$, respectively. Prove that $MP = NQ$.

The real numbers $a_0,a_1,a_2,\ldots$ satisfy $1=a_0\le a_1\le a_2\le\ldots. b_1,b_2,b_3,\ldots$ are defined by $b_n=\sum_{k=1}^n{1-{a_{k-1}\over a_k}\over\sqrt a_k}$.  a) Prove that $0\le b_n<2$.  b) Given $c$ satisfying $0\le c<2$, prove that we can find $a_n$ so that $b_n>c$ for all sufficiently large $n$.

 Let $n$ be a positive integer. Find the number of permutations $a_1$, $a_2$, $\dots a_n$ of the sequence $1$, $2$, $\dots$ , $n$ satisfying $$a_1 \le 2a_2\le 3a_3 \le \dots \le na_n.$$

Hello,  I am making changes to this site today, so please visit after 6pm EST. The content might look messed up and weird before that.  Thank you!

Suppose medians $m_a$ and $m_b$ of a triangle are orthogonal. Prove that  a) Using medians of that triangle it is possible to construct a rectangular triangle.  b) The following inequality:\[5(a^2+b^2-c^2) \geq 8ab,\]is valid, where $a,b$ and $c$ are side length of the given triangle.

Let $a$ be a postive integer and let ${a_n}$ be defined by $a_0=0$ and \[a_{n+1}=(a_n+1)a+(a+1)a_n+2\sqrt{a(a+1)a_n(a_n+1)}\]Show that for each positive integer $n,a_n$ is a positive integer.

Let $ ABCD$ be a convex quadrilateral, and let $ R_A, R_B, R_C, R_D$ denote the circumradii of the triangles $ DAB, ABC, BCD, CDA,$ respectively. Prove that $R_A + R_C > R_B + R_D$ if and only if $ \angle A + \angle C > \angle B + \angle D.$

Prove that from $x + y = 1 \  (x, y \in \mathbb R)$ it follows that \[x^{m+1} \sum_{j=0}^n \binom{m+j}{j} y^j + y^{n+1} \sum_{i=0}^m \binom{n+i}{i} x^i = 1 \qquad (m, n = 0, 1, 2, \ldots ).\] 

This problem was requested to be solved by a user. If you would like to request a problem too, please use the form on the left menu of the blog.  Misha rolls a standard, fair six-sided die until she rolls $1-2-3$ in that order on three consecutive rolls. The probability that she will roll the die an odd number of times is $\dfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

This problem was requested to be solved by a user. If you would like to request a problem too, please use the form on the left menu of the blog.  Triangle $ABC$ has side lengths $AB=4$, $BC=5$, and $CA=6$. Points $D$ and $E$ are on ray $AB$ with $AB<AD<AE$. The point $F \neq C$ is a point of intersection of the circumcircles of $\triangle ACD$ and $\triangle EBC$ satisfying $DF=2$ and $EF=7$. Then $BE$ can be expressed as $\tfrac{a+b\sqrt{c}}{d}$, where $a$, $b$, $c$, and $d$ are positive integers such that $a$ and $d$ are relatively prime, and $c$ is not divisible by the square of any prime. Find $a+b+c+d$. 

Let $a, b$ be integers, and let $P(x) = ax^3+bx.$ For any positive integer $n$ we say that the pair $(a,b)$ is $n$-good if $n | P(m)-P(k)$ implies $n | m - k$ for all integers $m, k.$ We say that $(a,b)$ is $very \ good$ if $(a,b)$ is $n$-good for infinitely many positive integers $n.$  (a) Find a pair $(a,b)$ which is 51-good, but not very good.  (b) Show that all 2010-good pairs are very good.

$$ne^n = \sum_{k=0}^{\infty}\frac{n^k(n - k)^2}{k!}.$$   I dreamt of this identity yesterday night but I don't even know how to prove it lol

Prove that there exists a positive constant $c$ such that the following statement is true: Consider an integer $n > 1$, and a set $\mathcal S$ of $n$ points in the plane such that the distance between any two different points in $\mathcal S$ is at least 1. It follows that there is a line $\ell$ separating $\mathcal S$ such that the distance from any point of $\mathcal S$ to $\ell$ is at least $cn^{-1/3}$. (A line $\ell$ separates a set of points S if some segment joining two points in $\mathcal S$ crosses $\ell$.) Suppose that among all projections of points in $\mathcal{S}$ onto some line $m$, the maximum possible distance between two consecutive projections is $\delta$. We need to prove $\delta \ge \Omega(n^{-1/3})$.  At this point, define $A,B$ as the two farthest points in $\mathcal{S}$. So, all points lie in the intersections of the circles centered at $A,B$ with radius $AB \geq 1$.  Choose chord $XY$ in circle $B$ where $XY \perp AB$ and $d(A, XY)=\frac{1}{2}$. Also, let $\mat

Due to the request of a few users, I got a custom domain at  imomath.xyz . The  math4l.blogspot.com  link will still work, but you can just go to  imomath.xyz  now. 

If $a,b,c$ are the sides of a triangle, prove that \[\frac{\sqrt{b+c-a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}+\frac{\sqrt{c+a-b}}{\sqrt{c}+\sqrt{a}-\sqrt{b}}+\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}\leq 3 \]

Let $ABC$ be an acute triangle and let $M$ be the midpoint of $AC$. A circle $\omega$ passing through $B$ and $M$ meets the sides $AB$ and $BC$ at points $P$ and $Q$ respectively. Let $T$ be the point such that $BPTQ$ is a parallelogram. Suppose that $T$ lies on the circumcircle of $ABC$. Determine all possible values of $\frac{BT}{BM}$.

Let $ \alpha < \frac {3 - \sqrt {5}}{2}$ be a positive real number. Prove that there exist positive integers $ n$ and $ p > \alpha \cdot 2^n$ for which one can select $ 2 \cdot p$ pairwise distinct subsets $ S_1, \ldots, S_p, T_1, \ldots, T_p$ of the set $ \{1,2, \ldots, n\}$ such that $ S_i \cap T_j \neq \emptyset$ for all $ 1 \leq i,j \leq p$.

Determine all sequences $(x_1,x_2,\ldots,x_{2011})$ of positive integers, such that for every positive integer $n$ there exists an integer $a$ with\[\sum^{2011}_{j=1} j  x^n_j = a^{n+1} + 1\]

Let $\Gamma_1$, $\Gamma_2$, $\Gamma_3$, $\Gamma_4$ be distinct circles such that $\Gamma_1$, $\Gamma_3$ are externally tangent at $P$, and $\Gamma_2$, $\Gamma_4$ are externally tangent at the same point $P$. Suppose that $\Gamma_1$ and $\Gamma_2$; $\Gamma_2$ and $\Gamma_3$; $\Gamma_3$ and $\Gamma_4$; $\Gamma_4$ and $\Gamma_1$ meet at $A$, $B$, $C$, $D$, respectively, and that all these points are different from $P$. Prove that \[ \frac{AB\cdot BC}{AD\cdot DC}=\frac{PB^2}{PD^2}. \]

I.  Am.  Completely.  Shocked.  I couldn't even think, in my dreams, for this blog to blow up so much. I'm glad to know that my solutions are helping people! Keep visiting this blog, more solutions are to come! As of now, I have no plans of closing this blog, and I plan to continue posting daily.  Btw, I won't be posting much of these "X views" anymore, since they're getting so frequent lol. I'll only post at some major number of views, like 100K, 150K, 200K, etc.

An isosceles trapezoid with bases $a$ and $c$ and altitude $h$ is given.  a) On the axis of symmetry of this trapezoid, find all points $P$ such that both legs of the trapezoid subtend right angles at $P$;  b) Calculate the distance of $p$ from either base;  c) Determine under what conditions such points $P$ actually exist. Discuss various cases that might arise.

Let $m$ be a positive integer, and consider a $m\times m$ checkerboard consisting of unit squares. At the centre of some of these unit squares there is an ant. At time $0$, each ant starts moving with speed $1$ parallel to some edge of the checkerboard. When two ants moving in the opposite directions meet, they both turn $90^{\circ}$ clockwise and continue moving with speed $1$. When more than $2$ ants meet, or when two ants moving in perpendicular directions meet, the ants continue moving in the same direction as before they met. When an ant reaches one of the edges of the checkerboard, it falls off and will not re-appear.  Considering all possible starting positions, determine the latest possible moment at which the last ant falls off the checkerboard, or prove that such a moment does not necessarily exist.

Find a method by which one can compute the coefficients of $P(x) = x^6 + a_1x^5 + \cdots+  a_6$ from the roots of $P(x) = 0$ by performing not more than $15$ additions and $15$ multiplications.

Consider a prism with pentagons $A_1A_2A_3A_4A_5$ and $B_1B_2B_3B_4B_5$ as the top and bottom faces is given. Each side of the two pentagons and each of the 25 segments $A_iB_j$ is colored red or green. Every triangle whose vertices are vertices of the prism and whose sides have all been colored has two sides of a different color. Prove that all 10 sides of the top and bottom faces have the same color.

Let $A$ be a $101$-element subset of the set $S=\{1,2,\ldots,10^6\}$. Prove that there exist numbers $t_1$, $t_2, \ldots, t_{100}$ in $S$ such that the sets\[ A_j=\{x+t_j\mid x\in A\},\qquad j=1,2,\ldots,100  \]are pairwise disjoint.

I am speechless. Thank you so much for 10K views! I will try to keep posting interesting content every day. Keep visiting! By the way, I have recently added a new feature in this blog. If you have a specific contest math problem you would like me to solve, please use the "Problem/Advice request" form on the left side of the homepage. It doesn't have to be an olympiad problem, but as long as it is a contest problem you can submit it through the form. You can use the same form if you want to ask for any advice. I will make a blog post regarding your problem/advice request.

Let $n$ be a positive integer and let $(x_1,\ldots,x_n)$, $(y_1,\ldots,y_n)$ be two sequences of positive real numbers. Suppose $(z_2,\ldots,z_{2n})$ is a sequence of positive real numbers such that $z_{i+j}^2 \geq x_iy_j$ for all $1\le i,j \leq n$. Let $M=\max\{z_2,\ldots,z_{2n}\}$. Prove that\[ \left( \frac{M+z_2+\dots+z_{2n}}{2n} \right)^2 \ge \left( \frac{x_1+\dots+x_n}{n} \right) \left( \frac{y_1+\dots+y_n}{n} \right). \]

Let $ABCD$ be a convex quadrilateral whose vertices do not lie on a circle. Let $A'B'C'D'$ be a quadrilateral such that $A',B', C',D'$ are the centers of the circumcircles of triangles $BCD,ACD,ABD$, and $ABC$. We write $T (ABCD) = A'B'C'D'$. Let us define $A''B''C''D'' = T (A'B'C'D') = T (T (ABCD)).$  (a) Prove that $ABCD$ and $A''B''C''D''$ are similar.  (b) The ratio of similitude depends on the size of the angles of $ABCD$. Determine this ratio.

Let $ a, b \in \mathbb{N}$ with $ 1 \leq a \leq b,$ and $ M = \left[\frac {a + b}{2} \right]$. Define a function $ f: \mathbb{Z} \mapsto \mathbb{Z}$ by \[ f(n) = \begin{cases} n + a, & \text{if } n \leq M, \\ n - b, & \text{if } n >M. \end{cases} \]Let $ f^1(n) = f(n),$ $ f_{i + 1}(n) = f(f^i(n)),$ $ i = 1, 2, \ldots$ Find the smallest natural number $ k$ such that $ f^k(0) = 0$

Let $ a_1, a_2, \ldots, a_{100}$ be nonnegative real numbers such that $ a^2_1 + a^2_2 + \ldots + a^2_{100} = 1.$ Prove that \[ a^2_1 \cdot a_2 + a^2_2 \cdot a_3 + \ldots + a^2_{100} \cdot a_1 < \frac {12}{25}. \]

Thanks everyone for 3K views on this blog! I didn't expect it to blow up this much. Keep viewing it, more interesting content is to come!

 Let $ABCDEF$ be a convex hexagon with $AB=DE$, $BC=EF$, $CD=FA$, and $\angle A-\angle D = \angle C -\angle F = \angle E -\angle B$. Prove that the diagonals $AD$, $BE$, and $CF$ are concurrent.

Let $\mathbb{Q}_{>0}$ denote the set of all positive rational numbers. Determine all functions $f:\mathbb{Q}_{>0}\to \mathbb{Q}_{>0}$ satisfying$$f(x^2f(y)^2)=f(x)^2f(y)$$for all $x,y\in\mathbb{Q}_{>0}$

At this point,  I have decided that I am only going to do IMO Shortlist problems.  Once I finish EGMO, I will keep doing this if I don't feel like the problems have started to repeat.  If I do feel that way, then I will then go towards research and most likely close this blog.  These plans might change in the future.

Let $a_1,a_2,\ldots a_n,k$, and $M$ be positive integers such that $$\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}=k\quad\text{and}\quad a_1a_2\cdots a_n=M.$$If $M>1$, prove that the polynomial $$P(x)=M(x+1)^k-(x+a_1)(x+a_2)\cdots (x+a_n)$$has no positive roots.

Since winter break is now over for me, although I will still be posting daily, I will post 3-6 problems instead of more.  However, the content will still be high quality! You won't ever run out of posts to view, and you won't ever be bored. 

Let $a,b,c,d$ be real numbers satisfying $|a|,|b|,|c|,|d|>1$ and $abc+abd+acd+bcd+a+b+c+d=0$. Prove that $\frac {1} {a-1}+\frac {1} {b-1}+ \frac {1} {c-1}+ \frac {1} {d-1} >0$

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 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. (A domino is a tile of size $2 \times 1$ or $1 \times 2$. Dominoes are placed on the board in such a way that each domino covers exactly two cells of the board, and dominoes do not overlap. Two cells are said to be adjacent if they are different and share a common side.)

From now on, I will post one IMO shortlist, then two of these below.        1 : "EGMO" ,     2 : "Romanian Masters of Mathematics" ,     3 : "Tournament of Towns" ,     4 : "USA ELMO Shortlist" ,     5 : "USA Math Prize for Girls Olympiad" ,     7 : "USAJMO" ,     8 : "Canada National Olympiad" ,     9 : "China TST" ,     10 : "India IMO Training Camp" ,     11 : "India National Olympiad" For example, it could go like: 2021 IMO #1, 2019 EGM0 #2, China TST #3.

Thanks for 1600 views, everybody! Keep viewing this blog!

Everyone has trouble with math olympiad problems, but what makes the difference is what you do  after .  Below are steps you should do (in order) if you do not know how to solve a problem.  1. Keep thinking about the problem until you completely run out of ideas, and you have tried every single one and it hasn't worked. Don't let any idea go; e 2. Okay, now it's time to give up. Read the solution carefully and understand every step of it. Don't think about the motivation for it yet.  3. When you have finished reading the solution and have verified that it is correct, in your mind summarize what they did, yet without thinking about the motivation. 4. Now, think about what the 'key' in the solution was, or what the 'heart of the problem' was. Basically, think about the main insight in the solution after which it got easy. Make sure to remember this 'insight', because it will apply to many, many other problems than just this one. 5. After a couple d

Real numbers $ a_{1}$, $ a_{2}$, $ \ldots$, $ a_{n}$ are given. For each $ i$, $ (1 \leq i \leq n )$, define \[ d_{i} = \max \{ a_{j}\mid 1 \leq j \leq i \} - \min \{ a_{j}\mid i \leq j \leq n \} \] and let $ d = \max \{d_{i}\mid 1 \leq i \leq n \}$.  (a) Prove that, for any real numbers $ x_{1}\leq x_{2}\leq \cdots \leq x_{n}$, \[ \max \{ |x_{i} - a_{i}| \mid 1 \leq i \leq n \}\geq \frac {d}{2}. \quad \quad (*) \]  (b) Show that there are real numbers $ x_{1}\leq x_{2}\leq \cdots \leq x_{n}$ such that the equality holds in (*). 

Let $ a, b, c$ be positive real numbers so that $ abc = 1$. Prove that \[ \left( a - 1 + \frac 1b \right) \left( b - 1 + \frac 1c \right) \left( c - 1 + \frac 1a \right) \leq 1. \]

Let $\mathbb{N}$ denote the set of positive integers. Find all functions $f: \mathbb{N} \to \mathbb{N}$ such that\[ f(m+n)f(m-n) = f(m^2)  \]for $m,n \in \mathbb{N}$.

Let $ABC$ be a scalene triangle with orthocenter $H$ and circumcenter $O$. Let $P$ be the midpoint of $\overline{AH}$ and let $T$ be on line $BC$ with $\angle TAO=90^{\circ}$. Let $X$ be the foot of the altitude from $O$ onto line $PT$. Prove that the midpoint of $\overline{PX}$ lies on the nine-point circle* of $\triangle ABC$. *The nine-point circle of $\triangle ABC$ is the unique circle passing through the following nine points: the midpoint of the sides, the feet of the altitudes, and the midpoints of $\overline{AH}$, $\overline{BH}$, and $\overline{CH}$

I will be posting higher quality math olympiad problems from now on. Using the random math olympiad problem generator Python program (which I had posted before), most problems are not high quality.  So, in the Python code, I have now removed the olympiads that do not have many good problems.  My updated list from which the computer randomizes looks like this:    1 : "IMO Shortlist" ,   2 : "IMO Shortlist" ,   3 : "APMO" ,   4 : "Austrian-Polish" ,     5 : "Balkan MO Shortlist" ,     6 : "Balkan MO Shortlist" ,     7 : "Baltic Way" ,     8 : "Benelux" ,     9 : "EGMO" ,     10 : "Iranian Geometry Olympiad" ,     11 : "Romanian Masters of Mathematics" ,     12 : "Silk Road" ,     13 : "Tournament of Towns" ,     14 : "USA BAMO" ,     15 : "USA ELMO Shortlist" ,     16 : "USA ELMO Shortlist" ,     17 : "USA USEMO" ,     1

Let $\{a_n\}$ be a sequence of integers satisfying $(n-1)a_{n+1}=(n+1)a_n-2(n-1) \forall n\ge 1$. If $2000|a_{1999}$, find the smallest $n\ge 2$ such that $2000|a_n$.

 In an acute-angled triangle $ABC$ on the sides $AB$, $BC$, $AC$ the points $H$, $L$, $K$ so that $CH \perp AB$, $HL \parallel AC$, $HK \parallel BC$. Let $P$ and $Q$ feet of altitudes of a triangle $HBL$, drawn from the vertices $H$ and $B$ respectively. Prove that the feet of the altitudes of the triangle $AKH$, drawn from the vertices $A$ and $H$ lie on the line $PQ$.

 It is now officially the very first moment of 2022 (in EST).  Welcome, 2022! I hope you bring great happiness and health to everyone.  Time to eat cake.  via GIPHY