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: 7

2021 Oral Moscow Geometry Olympiad, 6

$ABCD$ is a square and $XYZ$ is an equilateral triangle such that $X$ lies on $AB$, $Y$ lies on $BC$ and $Z$ lies on $DA$. Line through the centers of $ABCD$ and $XYZ$ intersects $CD$ at $T$. Find angle $CTY$

2007 AMC 12/AHSME, 20

The parallelogram bounded by the lines $ y \equal{} ax \plus{} c,y \equal{} ax \plus{} d,y \equal{} bx \plus{} c$ and $ y \equal{} bx \plus{} d$ has area $ 18$. The parallelogram bounded by the lines $ y \equal{} ax \plus{} c,y \equal{} ax \minus{} d,y \equal{} bx \plus{} c,$ and $ y \equal{} bx \minus{} d$ has area $ 72.$ Given that $ a,b,c,$ and $ d$ are positive integers, what is the smallest possible value of $ a \plus{} b \plus{} c \plus{} d$? $ \textbf{(A)}\ 13 \qquad \textbf{(B)}\ 14 \qquad \textbf{(C)}\ 15 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 17$

2016 CentroAmerican, 2

Let $ABC$ be an acute-angled triangle, $\Gamma$ its circumcircle and $M$ the midpoint of $BC$. Let $N$ be a point in the arc $BC$ of $\Gamma$ not containing $A$ such that $\angle NAC= \angle BAM$. Let $R$ be the midpoint of $AM$, $S$ the midpoint of $AN$ and $T$ the foot of the altitude through $A$. Prove that $R$, $S$ and $T$ are collinear.

2025 Vietnam National Olympiad, 3

Let $ABC$ be an acute, scalene triangle with circumcenter $O$, circumcircle $(O)$, orthocenter $H$. Line $AH$ meets $(O)$ again at $D \neq A$. Let $E, F$ be the midpoint of segments $AB, AC$ respectively. The line through $H$ and perpendicular to $HF$ meets line $BC$ at $K$. a) Line $DK$ meets $(O)$ again at $Y \neq D$. Prove that the intersection of line $BY$ and the perpendicular bisector of $BK$ lies on the circumcircle of triangle $OFY$. b) The line through $H$ and perpendicular to $HE$ meets line $BC$ at $L$. Line $DL$ meets $(O)$ again at $Z \neq D$. Let $M$ be the intersection of lines $BZ, OE$; $N$ be the intersection of lines $CY, OF$; $P$ be the intersection of lines $BY, CZ$. Let $T$ be the intersection of lines $YZ, MN$ and $d$ be the line through $T$ and perpendicular to $OA$. Prove that $d$ bisects $AP$.

2023 USAJMO, 4

Tags: hard
Two players, $B$ and $R$, play the following game on an infinite grid of unit squares, all initially colored white. The players take turns starting with $B$. On $B$'s turn, $B$ selects one white unit square and colors it blue. On $R$'s turn, $R$ selects two white unit squares and colors them red. The players alternate until $B$ decides to end the game. At this point, $B$ gets a score, given by the number of unit squares in the largest (in terms of area) simple polygon containing only blue unit squares. What is the largest score $B$ can guarantee? (A [i]simple polygon[/i] is a polygon (not necessarily convex) that does not intersect itself and has no holes.) [i]Proposed by David Torres[/i]

2008 Gheorghe Vranceanu, 2

Consider the $ 4\times 4 $ integer matrices that have the property that each one of them multiplied by its transpose is $ 4I. $ [b]a)[/b] Show that the product of the elements of such a matrix is either $ 0, $ either $ 1. $ [b]b)[/b] How many such matrices have the property that the product of its elements is $ 0? $

2023 USAJMO Solutions by peace09, 4

Tags: hard
Two players, $B$ and $R$, play the following game on an infinite grid of unit squares, all initially colored white. The players take turns starting with $B$. On $B$'s turn, $B$ selects one white unit square and colors it blue. On $R$'s turn, $R$ selects two white unit squares and colors them red. The players alternate until $B$ decides to end the game. At this point, $B$ gets a score, given by the number of unit squares in the largest (in terms of area) simple polygon containing only blue unit squares. What is the largest score $B$ can guarantee? (A [i]simple polygon[/i] is a polygon (not necessarily convex) that does not intersect itself and has no holes.) [i]Proposed by David Torres[/i]