This website contains problems from math contests. Problems and corresponding tags were obtained from the Art of Problem Solving website.

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Found problems: 114

Let $O(0,0)$, $A(\tfrac{1}{2},0)$, and $B(0, \tfrac{\sqrt{3}}{2})$ be points in the coordinate plane. Let $\mathcal{F}$ be the family of segments $\overline{PQ}$ of unit length lying in the first quadrant with $P$ on the $x$-axis and $Q$ on the $y$-axis. There is a unique point $C$ on $\overline{AB}$, distinct from $A$ and $B$, that does not belong to any segment from $\mathcal{F}$ other than $\overline{AB}$. Then $OC^2 = \tfrac{p}{q}$ where $p$ and $q$ are relatively prime positive integers. Find $p + q$.
Let $a, b, c$ be positive real numbers satisfying $a^3+b^3+c^3=a^4+b^4+c^4$. Show that \[ \frac{a}{a^2+b^3+c^3}+\frac{b}{a^3+b^2+c^3}+\frac{c}{a^3+b^3+c^2} \geq 1 \]
If $a$, $b$ ,$c$ are three positive real numbers such that $ab+bc+ca = 1$, prove that \[ \sqrt[3]{ \frac{1}{a} + 6b} + \sqrt[3]{\frac{1}{b} + 6c} + \sqrt[3]{\frac{1}{c} + 6a } \leq \frac{1}{abc}. \]
If $a$, $b$ ,$c$ are three positive real numbers such that $ab+bc+ca = 1$, prove that \[ \sqrt[3]{ \frac{1}{a} + 6b} + \sqrt[3]{\frac{1}{b} + 6c} + \sqrt[3]{\frac{1}{c} + 6a } \leq \frac{1}{abc}. \]
Let $a,b,c$ be positive real numbers. Prove that $\frac{a^3+3b^3}{5a+b}+\frac{b^3+3c^3}{5b+c}+\frac{c^3+3a^3}{5c+a} \geq \frac{2}{3}(a^2+b^2+c^2)$.
Let $a$, $b$, $c$ be positive real numbers such that $ab+bc+ca\leq 3abc$. Prove that \[\sqrt{\frac{a^2+b^2}{a+b}}+\sqrt{\frac{b^2+c^2}{b+c}}+\sqrt{\frac{c^2+a^2}{c+a}}+3\leq \sqrt{2}\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)\] [i]Proposed by Dzianis Pirshtuk, Belarus[/i]
If $a$, $b$ ,$c$ are three positive real numbers such that $ab+bc+ca = 1$, prove that \[ \sqrt[3]{ \frac{1}{a} + 6b} + \sqrt[3]{\frac{1}{b} + 6c} + \sqrt[3]{\frac{1}{c} + 6a } \leq \frac{1}{abc}. \]
Let $a, b, c$ be the sides of a triangle with $abc = 1$. Prove that $$\frac{\sqrt{b + c -a}}{a}+\frac{\sqrt{c + a - b}}{b}+\frac{\sqrt{a + b - c}}{c} \ge a + b + c$$
Find the maximal value of \[S = \sqrt[3]{\frac{a}{b+7}} + \sqrt[3]{\frac{b}{c+7}} + \sqrt[3]{\frac{c}{d+7}} + \sqrt[3]{\frac{d}{a+7}},\] where $a$, $b$, $c$, $d$ are nonnegative real numbers which satisfy $a+b+c+d = 100$. [i]Proposed by Evan Chen, Taiwan[/i]
Let all possible $2023$-degree real polynomials: $P(x)=x^{2023}+a_1x^{2022}+a_2x^{2021}+\cdots+a_{2022}x+a_{2023}$, where $P(0)+P(1)=0$, and the polynomial has 2023 real roots $r_1, r_2,\cdots r_{2023}$ [not necessarily distinct] so that $0\leq r_1,r_2,\cdots r_{2023}\leq1$. What is the maximum value of $r_1 \cdot r_2 \cdots r_{2023}?$
Let $\,S\,$ be a finite set of points in three-dimensional space. Let $\,S_{x},\,S_{y},\,S_{z}\,$ be the sets consisting of the orthogonal projections of the points of $\,S\,$ onto the $yz$-plane, $zx$-plane, $xy$-plane, respectively. Prove that \[ \vert S\vert^{2}\leq \vert S_{x} \vert \cdot \vert S_{y} \vert \cdot \vert S_{z} \vert, \] where $\vert A \vert$ denotes the number of elements in the finite set $A$. [hide="Note"] Note: The orthogonal projection of a point onto a plane is the foot of the perpendicular from that point to the plane. [/hide]
Let $x, y, z$ be positive real numbers whose sum is $2012$. Find the maximum value of $$ \frac{(x^2 + y^2 + z^2)(x^3 + y^3 + z^3)}{(x^4 + y^4 + z^4)}$$
Let $a, b, c$ be positive reals such that $abc=1$. Show that \[\frac{1}{a^5(b+2c)^2} + \frac{1}{b^5(c+2a)^2} + \frac{1}{c^5(a+2b)^2} \ge \frac{1}{3}.\]
If $a$, $b$ ,$c$ are three positive real numbers such that $ab+bc+ca = 1$, prove that \[ \sqrt[3]{ \frac{1}{a} + 6b} + \sqrt[3]{\frac{1}{b} + 6c} + \sqrt[3]{\frac{1}{c} + 6a } \leq \frac{1}{abc}. \]