IMO Math

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

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

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_

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

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 $n\geq2$.

 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 $a,b,c$,\[ \frac{a}{\sqrt{a^2 + 8bc}} + \frac{b}{\sqrt{b^2 + 8ca}} + \frac{c}{\sqrt{c^2 + 8ab}} \geq 1.  \]

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$.

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 $i + j + k \geq n$.

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 $\sum \limits_{i=1}^{n} ( x_{i} -\ y_{i} )^{2} \leq \sum \limits_{i=1}^{n}$ $( x_{i} - z_{i})^{2}.$

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 $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 \[0 \leq (n + 1)(a_n - a_{n+1}) \leq 2 \qquad \text{ for } n = 1, 2, 3, \ldots\]

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

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 $a$ 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.

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\[\sum^{2011}_{j=1} j  x^n_j = a^{n+1} + 1\]

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.