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

Tags were heavily modified to better represent problems.

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Found problems: 85335

2005 IMO, 3

Let $x,y,z$ be three positive reals such that $xyz\geq 1$. Prove that \[ \frac { x^5-x^2 }{x^5+y^2+z^2} + \frac {y^5-y^2}{x^2+y^5+z^2} + \frac {z^5-z^2}{x^2+y^2+z^5} \geq 0 . \] [i]Hojoo Lee, Korea[/i]

1991 Cono Sur Olympiad, 1

Let $A, B$ and $C$ be three non-collinear points and $E$ ($\ne B$) an arbitrary point not in the straight line $AC$. Construct the parallelograms $ABCD$ and $AECF$. Prove that $BE \parallel DF$.

2018 Austria Beginners' Competition, 2

Let $ABC$ be an acute-angled triangle, $M$ the midpoint of the side $AC$ and $F$ the foot on $AB$ of the altitude through the vertex $C$. Prove that $AM = AF$ holds if and only if $\angle BAC = 60^o$. (Karl Czakler)

Kvant 2022, M2700

What is the maximal possible number of roots on the interval (0,1) for a polynomial of degree 2022 with integer coefficients and with the leading coefficient equal to 1?

2002 India IMO Training Camp, 3

Tags: quadratic , algebra
Let $X=\{2^m3^n|0 \le m, \ n \le 9 \}$. How many quadratics are there of the form $ax^2+2bx+c$, with equal roots, and such that $a,b,c$ are distinct elements of $X$?

1970 All Soviet Union Mathematical Olympiad, 132

The digits of the $17$-digit number are rearranged in the reverse order. Prove that at list one digit of the sum of the new and the initial number is even.

2019 AMC 12/AHSME, 12

Tags: geometry
Right triangle $ACD$ with right angle at $C$ is constructed outwards on the hypotenuse $\overline{AC}$ of isosceles right triangle $ABC$ with leg length $1$, as shown, so that the two triangles have equal perimeters. What is $\sin(2\angle BAD)$? [asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8.016233639805293cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.001920114613276, xmax = 4.014313525192017, ymin = -2.552570341575814, ymax = 5.6249093771911145; /* image dimensions */ draw((-1.6742337260757447,-1.)--(-1.6742337260757445,-0.6742337260757447)--(-2.,-0.6742337260757447)--(-2.,-1.)--cycle, linewidth(2.)); draw((-1.7696484586262846,2.7696484586262846)--(-1.5392969172525692,3.)--(-1.7696484586262846,3.2303515413737154)--(-2.,3.)--cycle, linewidth(2.)); /* draw figures */ draw((-2.,3.)--(-2.,-1.), linewidth(2.)); draw((-2.,-1.)--(2.,-1.), linewidth(2.)); draw((2.,-1.)--(-2.,3.), linewidth(2.)); draw((-0.6404058554606791,4.3595941445393205)--(-2.,3.), linewidth(2.)); draw((-0.6404058554606791,4.3595941445393205)--(2.,-1.), linewidth(2.)); label("$D$",(-0.9382446143428628,4.887784444795223),SE*labelscalefactor,fontsize(14)); label("$A$",(1.9411496528285788,-1.0783204767840298),SE*labelscalefactor,fontsize(14)); label("$B$",(-2.5046350956841272,-0.9861798602345433),SE*labelscalefactor,fontsize(14)); label("$C$",(-2.5737405580962416,3.5747806589650395),SE*labelscalefactor,fontsize(14)); label("$1$",(-2.665881174645728,1.2712652452278765),SE*labelscalefactor,fontsize(14)); label("$1$",(-0.3393306067712029,-1.3547423264324894),SE*labelscalefactor,fontsize(14)); /* dots and labels */ dot((-2.,3.),linewidth(4.pt) + dotstyle); dot((-2.,-1.),linewidth(4.pt) + dotstyle); dot((2.,-1.),linewidth(4.pt) + dotstyle); dot((-0.6404058554606791,4.3595941445393205),linewidth(4.pt) + dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy] $\textbf{(A) } \dfrac{1}{3} \qquad\textbf{(B) } \dfrac{\sqrt{2}}{2} \qquad\textbf{(C) } \dfrac{3}{4} \qquad\textbf{(D) } \dfrac{7}{9} \qquad\textbf{(E) } \dfrac{\sqrt{3}}{2}$

2013 Stanford Mathematics Tournament, 4

Tags:
Given the digits $1$ through $7$, one can form $7!=5040$ numbers by forming different permutations of the $7$ digits (for example, $1234567$ and $6321475$ are two such permutations). If the $5040$ numbers are then placed in ascending order, what is the $2013^{\text{th}}$ number?

DMM Individual Rounds, 1999 Tie

[b]p1A.[/b] Compute $$1 + \frac{1}{2^3} + \frac{1}{3^3} + \frac{1}{4^3} + \frac{1}{5^3} + ...$$ $$1 - \frac{1}{2^3} + \frac{1}{3^3} - \frac{1}{4^3} + \frac{1}{5^3} - ...$$ [b]p1B.[/b] Real values $a$ and $b$ satisfy $ab = 1$, and both numbers have decimal expansions which repeat every five digits: $$ a = 0.(a_1)(a_2)(a_3)(a_4)(a_5)(a_1)(a_2)(a_3)(a_4)(a_5)...$$ and $$ b = 1.(b_1)(b_2)(b_3)(b_4)(b_5)(b_1)(b_2)(b_3)(b_4)(b_5)...$$ If $a_5 = 1$, find $b_5$. [b]p2.[/b] $P(x) = x^4 - 3x^3 + 4x^2 - 9x + 5$. $Q(x)$ is a $3$rd-degree polynomial whose graph intersects the graph of $P(x)$ at $x = 1$, $2$, $5$, and $10$. Compute $Q(0)$. [b]p3.[/b] Distinct real values $x_1$, $x_2$, $x_3$, $x_4 $all satisfy $ ||x - 3| - 5| = 1.34953$. Find $x_1 + x_2 + x_3 + x_4$. [b]p4.[/b] Triangle $ABC$ has sides $AB = 8$, $BC = 10$, and $CA = 11$. Let $L$ be the locus of points in the interior of triangle $ABC$ which are within one unit of either $A$, $B$, or $C$. Find the area of $L$. [b]p5.[/b] Triangles $ABC$ and $ADE$ are equilateral, and $AD$ is an altitude of $ABC$. The area of the intersection of these triangles is $3$. Find the area of the larger triangle $ABC$. PS. You had better use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].

2017 Mediterranean Mathematics Olympiad, Problem 1

Tags: geometry
Let $ABC$ be an equilateral triangle, and let $P$ be some point in its circumcircle. Determine all positive integers $n$, for which the value of the sum $S_n (P) = |PA|^n + |PB|^n + |PC|^n$ is independent of the choice of point $P$.

2023 Macedonian Balkan MO TST, Problem 4

Let $f$ be a non-zero function from the set of positive integers to the set of non-negative integers such that for all positive integers $a$ and $b$ we have $$2f(ab)=(b+1)f(a)+(a+1)f(b).$$ Prove that for every prime number $p$ there exists a prime $q$ and positive integers $x_{1}$, ..., $x_{n}$ and $m \geq 0$ so that $$\frac{f(q^{p})}{f(q)} = (px_{1}+1) \cdot ... \cdot (px_{n}+1) \cdot p^{m},$$ where the integers $px_{1}+1$,..., $px_{n}+1$ are all prime. [i]Authored by Nikola Velov[/i]

2015 Sharygin Geometry Olympiad, 8

Does there exist a rectangle which can be divided into a regular hexagon with sidelength $1$ and several congruent right-angled triangles with legs $1$ and $\sqrt{3}$?

2008 China Team Selection Test, 2

Prove that for arbitary integer $ n > 16$, there exists the set $ S$ that contains $ n$ positive integers and has the following property:if the subset $ A$ of $ S$ satisfies for arbitary $ a,a'\in A, a\neq a', a \plus{} a'\notin S$ holds, then $ |A|\leq4\sqrt n.$

2017 ELMO Shortlist, 1

Tags: algebra
Let $0<k<\frac{1}{2}$ be a real number and let $a_0, b_0$ be arbitrary real numbers in $(0,1)$. The sequences $(a_n)_{n\ge 0}$ and $(b_n)_{n\ge 0}$ are then defined recursively by $$a_{n+1} = \dfrac{a_n+1}{2} \text{ and } b_{n+1} = b_n^k$$ for $n\ge 0$. Prove that $a_n<b_n$ for all sufficiently large $n$. [i]Proposed by Michael Ma

2018 CCA Math Bonanza, L5.2

Tags:
Two circles of equal radii are drawn to intersect at $X$ and $Y$. Suppose that the two circles bisect each other's areas. If the measure of minor arc $\widehat{XY}$ is $\theta$ degrees, estimate $\left\lfloor1000\theta\right\rfloor$. An estimate of $E$ earns $2e^{-\frac{\left|A-E\right|}{50000}}$ points, where $A$ is the actual answer. [i]2018 CCA Math Bonanza Lightning Round #5.2[/i]

2020 MBMT, 9

Tags:
Consider a regular pentagon $ABCDE$, and let the intersection of diagonals $\overline{CA}$ and $\overline{EB}$ be $F$. Find $\angle AFB$. [i]Proposed by Justin Chen[/i]

2011 Czech and Slovak Olympiad III A, 1

Tags: geometry
In a certain triangle $ABC$, there are points $K$ and $M$ on sides $AB$ and $AC$, respectively, such that if $L$ is the intersection of $MB$ and $KC$, then both $AKLM$ and $KBCM$ are cyclic quadrilaterals with the same size circumcircles. Find the measures of the interior angles of triangle $ABC$.

2019 Durer Math Competition Finals, 15

The positive integer $m$ and non-negative integers $x_0, x_1,..., x_{1001}$ satisfy the following equation: $$m^{x_0} =\sum_{i=1}^{1001}m^{x_i}.$$ How many possibilities are there for the value of $m$?

2006 Thailand Mathematical Olympiad, 6

Let $a, b, c$ be positive reals. Show that $$1 +\frac{3}{ab + bc + ca}\ge \frac{6}{a + b + c}$$

2008 Swedish Mathematical Competition, 1

A rhombus is inscribed in a convex quadrilateral. The sides of the rhombus are parallel with the diagonals of the quadrilateral, which have the lengths $d_1$ and $d_2$. Calculate the length of side of the rhombus , expressed in terms of $d_1$ and $d_2$.

2019 AMC 10, 3

Tags: percent
In a high school with $500$ students, $40\%$ of the seniors play a musical instrument, while $30\%$ of the non-seniors do not play a musical instrument. In all, $46.8\%$ of the students do not play a musical instrument. How many non-seniors play a musical instrument? $\textbf{(A) } 66 \qquad\textbf{(B) } 154 \qquad\textbf{(C) } 186 \qquad\textbf{(D) } 220 \qquad\textbf{(E) } 266$

1999 USAMTS Problems, 2

Tags:
The Fibonacci numbers are defined by $F_1=F_2=1$ and $F_n=F_{n-1}+F_{n-2}$ for $n>2$. It is well-known that the sum of any $10$ consecutive Fibonacci numbers is divisible by $11$. Determine the smallest integer $N$ so that the sum of any $N$ consecutive Fibonacci numbers is divisible by $12$.

Ukraine Correspondence MO - geometry, 2007.11

Denote by $B_1$ and $C_1$, the midpoints of the sides $AB$ and $AC$ of the triangle $ABC$. Let the circles circumscribed around the triangles $ABC_1$ and $AB_1C$ intersect at points $A$ and $P$, and let the line $AP$ intersect the circle circumscribed around the triangle $ABC$ at points $A$ and $Q$. Find the ratio $\frac{AQ}{AP}$.

2016 Brazil Team Selection Test, 2

Determine all functions $f:\mathbb{Z}\rightarrow\mathbb{Z}$ with the property that \[f(x-f(y))=f(f(x))-f(y)-1\] holds for all $x,y\in\mathbb{Z}$.

2024 Korea National Olympiad, 1

Tags: geometry
Let there be a circle with center $O$, and three distinct points $A, B, X$ on the circle, where $A, B, O$ are not collinear. Let $\Omega$ be the circumcircle of triangle $ABO$. Segments $AX, BX$ intersect $\Omega$ at points $C(\neq A), D(\neq B)$, respectively. Prove that $O$ is the orthocenter of triangle $CXD$.