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: 1049

1998 Romania Team Selection Test, 1

We are given an isosceles triangle $ABC$ such that $BC=a$ and $AB=BC=b$. The variable points $M\in (AC)$ and $N\in (AB)$ satisfy $a^2\cdot AM \cdot AN = b^2 \cdot BN \cdot CM$. The straight lines $BM$ and $CN$ intersect in $P$. Find the locus of the variable point $P$. [i]Dan Branzei[/i]

2024 Yasinsky Geometry Olympiad, 5

Let \( AL \) be the bisector of triangle \( ABC \), \( O \) the center of its circumcircle, and \( D \) and \( E \) the midpoints of \( BL \) and \( CL \), respectively. Points \( P \) and \( Q \) are chosen on segments \( AD \) and \( AE \) such that \( APLQ \) is a parallelogram. Prove that \( PQ \perp AO \). [i]Proposed by Mykhailo Plotnikov[/i]

Ukraine Correspondence MO - geometry, 2008.11

Let $ABCD$ be a parallelogram. A circle with diameter $AC$ intersects line $BD$ at points $P$ and $Q$. The perpendicular on $AC$ passing through point $C$, intersects lines $AB$ and $AD$ at points $X$ and $Y$, respectively. Prove that the points $P, Q, X$ and $Y$ lie on the same circle.

1990 AMC 8, 24

Three $ \Delta $'s and a $ \diamondsuit $ will balance nine $ \bullet $'s. One $ \Delta $ will balance a $ \diamondsuit $ and a $ \bullet $. [asy] unitsize(5.5); fill((0,0)--(-4,-2)--(4,-2)--cycle,black); draw((-12,2)--(-12,0)--(12,0)--(12,2)); draw(ellipse((-12,5),8,3)); draw(ellipse((12,5),8,3)); label("$\Delta \hspace{2 mm}\Delta \hspace{2 mm}\Delta \hspace{2 mm}\diamondsuit $",(-12,6.5),S); label("$\bullet \hspace{2 mm}\bullet \hspace{2 mm}\bullet \hspace{2 mm} \bullet $",(12,5.2),N); label("$\bullet \hspace{2 mm}\bullet \hspace{2 mm}\bullet \hspace{2 mm}\bullet \hspace{2 mm}\bullet $",(12,5.2),S); fill((44,0)--(40,-2)--(48,-2)--cycle,black); draw((34,2)--(34,0)--(54,0)--(54,2)); draw(ellipse((34,5),6,3)); draw(ellipse((54,5),6,3)); label("$\Delta $",(34,6.5),S); label("$\bullet \hspace{2 mm}\diamondsuit $",(54,6.5),S);[/asy] How many $ \bullet $'s will balance the two $ \diamondsuit $'s in this balance? [asy] unitsize(5.5); fill((0,0)--(-4,-2)--(4,-2)--cycle,black); draw((-12,4)--(-12,2)--(12,-2)--(12,0)); draw(ellipse((-12,7),6.5,3)); draw(ellipse((12,3),6.5,3)); label("$?$",(-12,8.5),S); label("$\diamondsuit \hspace{2 mm}\diamondsuit $",(12,4.5),S);[/asy] $ \text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 3\qquad\text{(D)}\ 4\qquad\text{(E)}\ 5 $

1963 All Russian Mathematical Olympiad, 029

a) Each diagonal of the quadrangle halves its area. Prove that it is a parallelogram. b) Three main diagonals of the hexagon halve its area. Prove that they intersect in one point.

2004 Turkey MO (2nd round), 1

In a triangle $\triangle ABC$ with$\angle B>\angle C$, the altitude, the angle bisector, and the median from $A$ intersect $BC$ at $H, L$ and $D$, respectively. Show that $\angle HAL=\angle DAL$ if and only if $\angle BAC=90^{\circ}$.

2017 Romania National Olympiad, 3

In the square $ABCD$ denote by $M$ the midpoint of the side $[AB]$, with $P$ the projection of point $B$ on the line $CM$ and with $N$ the midpoint of the segment $[CP]$, Bisector of the angle $DAN$ intersects the line $DP$ at point $Q$. Show that the quadrilateral $BMQN$ is a parallelogram.

2006 QEDMO 2nd, 7

Let $H$ be the orthocenter of a triangle $ABC$, and let $D$ be the midpoint of the segment $AH$. The altitude $BH$ of triangle $ABC$ intersects the perpendicular to the line $AB$ through the point $A$ at the point $M$. The altitude $CH$ of triangle $ABC$ intersects the perpendicular to the line $CA$ through the point $A$ at the point $N$. The perpendicular bisector of the segment $AB$ intersects the perpendicular to the line $BC$ through the point $B$ at the point $U$. The perpendicular bisector of the segment $CA$ intersects the perpendicular to the line $BC$ through the point $C$ at the point $V$. Finally, let $E$ be the midpoint of the side $BC$ of triangle $ABC$. Prove that the points $D$, $M$, $N$, $U$, $V$ all lie on one and the same perpendicular to the line $AE$. [i]Extensions.[/i] In other words, we have to show that the points $M$, $N$, $U$, $V$ lie on the perpendicular to the line $AE$ through the point $D$. Additionally, one can find two more points on this perpendicular: [b](a)[/b] The nine-point circle of triangle $ABC$ is known to pass through the midpoint $E$ of its side $BC$. Let $D^{\prime}$ be the point where this nine-point circle intersects the line $AE$ apart from $E$. Then, the point $D^{\prime}$ lies on the perpendicular to the line $AE$ through the point $D$. [b](b)[/b] Let the tangent to the circumcircle of triangle $ABC$ at the point $A$ intersect the line $BC$ at a point $X$. Then, the point $X$ lies on the perpendicular to the line $AE$ through the point $D$. [i]Comment.[/i] The actual problem was created by Victor Thébault around 1950 (cf. Hyacinthos messages #1102 and #1551). The extension [b](a)[/b] initially was a (pretty trivial) lemma in Thébault's solution of the problem. Extension [b](b)[/b] is rather new; in the form "prove that $X\in UV$", it was [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=3659]proposed by Valentin Vornicu for the Balkan MO 2003[/url], however it circulated in the Hyacinthos newsgroup before (Hyacinthos messages #7240 and #7242), where different solutions of the problem were discussed as well. Hereby, "Hyacinthos" always refers to the triangle geometry newsgroup "Hyacinthos", which can be found at http://groups.yahoo.com/group/Hyacinthos . I proposed the problem for the QEDMO math fight wishing to draw some attention to it. It has a rather short and elementary solution, by the way (without using radical axes or inversion like the standard solutions). Darij

2006 Team Selection Test For CSMO, 2

Let $AA_1$ and $BB_1$ be the altitudes of an acute-angled, non-isosceles triangle $ABC$. Also, let $A_0$ and $B_0$ be the midpoints of its sides $BC$ and $CA$, respectively. The line $A_1B_1$ intersects the line $A_0B_0$ at a point $C'$. Prove that the line $CC'$ is perpendicular to the Euler line of the triangle $ABC$ (this is the line that joins the orthocenter and the circumcenter of the triangle $ABC$).

2002 China Team Selection Test, 2

Circles $ \omega_{1}$ and $ \omega_{2}$ intersect at points $ A$ and $ B.$ Points $ C$ and $ D$ are on circles $ \omega_{1}$ and $ \omega_{2},$ respectively, such that lines $ AC$ and $ AD$ are tangent to circles $ \omega_{2}$ and $ \omega_{1},$ respectively. Let $ I_{1}$ and $ I_{2}$ be the incenters of triangles $ ABC$ and $ ABD,$ respectively. Segments $ I_{1}I_{2}$ and $ AB$ intersect at $ E$. Prove that: $ \frac {1}{AE} \equal{} \frac {1}{AC} \plus{} \frac {1}{AD}$

2000 Baltic Way, 1

Let $ K$ be a point inside the triangle $ ABC$. Let $ M$ and $ N$ be points such that $ M$ and $ K$ are on opposite sides of the line $ AB$, and $ N$ and $ K$ are on opposite sides of the line $ BC$. Assume that $ \angle MAB \equal{} \angle MBA \equal{} \angle NBC \equal{} \angle NCB \equal{} \angle KAC \equal{} \angle KCA$. Show that $ MBNK$ is a parallelogram.

2006 Iran MO (3rd Round), 2

$ABC$ is a triangle and $R,Q,P$ are midpoints of $AB,AC,BC$. Line $AP$ intersects $RQ$ in $E$ and circumcircle of $ABC$ in $F$. $T,S$ are on $RP,PQ$ such that $ES\perp PQ,ET\perp RP$. $F'$ is on circumcircle of $ABC$ that $FF'$ is diameter. The point of intersection of $AF'$ and $BC$ is $E'$. $S',T'$ are on $AB,AC$ that $E'S'\perp AB,E'T'\perp AC$. Prove that $TS$ and $T'S'$ are perpendicular.

2003 Romania National Olympiad, 4

In triangle $ ABC$, $ P$ is the midpoint of side $ BC$. Let $ M\in(AB)$, $ N\in(AC)$ be such that $ MN\parallel BC$ and $ \{Q\}$ be the common point of $ MP$ and $ BN$. The perpendicular from $ Q$ on $ AC$ intersects $ AC$ in $ R$ and the parallel from $ B$ to $ AC$ in $ T$. Prove that: (a) $ TP\parallel MR$; (b) $ \angle MRQ\equal{}\angle PRQ$. [i]Mircea Fianu[/i]

2006 JBMO ShortLists, 12

Let $ ABC$ be an equilateral triangle of center $ O$, and $ M\in BC$. Let $ K,L$ be projections of $ M$ onto the sides $ AB$ and $ AC$ respectively. Prove that line $ OM$ passes through the midpoint of the segment $ KL$.

2014 Vietnam National Olympiad, 4

Let $ABC$ be an acute triangle, $(O)$ be the circumcircle, and $AB<AC.$ Let $I$ be the midpoint of arc $BC$ (not containing $A$). $K$ lies on $AC,$ $K\ne C$ such that $IK=IC.$ $BK$ intersects $(O)$ at the second point $D,$ $D\ne B$ and intersects $AI$ at $E.$ $DI$ intersects $AC$ at $F.$ a) Prove that $EF=\frac{BC}{2}.$ b) $M$ lies on $DI$ such that $CM$ is parallel to $AD.$ $KM$ intersects $BC$ at $N.$ The circumcircle of triangle $BKN$ intersects $(O)$ at the second point $P.$ Prove that $PK$ passes through the midpoint of segment $AD.$

1947 Moscow Mathematical Olympiad, 136

Prove that no convex $13$-gon can be cut into parallelograms.

2023 Regional Olympiad of Mexico West, 5

We have a rhombus $ABCD$ with $\angle BAD=60^\circ$. We take points $F,H,G$ on the sides $AD,DC$ and the diagonal $AC$, respectively, such that $DFGH$ is a parallelogram. Prove that $BFH$ is equilateral.

2014 Dutch Mathematical Olympiad, 2 seniors

On the sides of triangle $ABC$, isosceles right-angled triangles $AUB, CVB$, and $AWC$ are placed. These three triangles have their right angles at vertices $U, V$ , and $W$, respectively. Triangle $AUB$ lies completely inside triangle $ABC$ and triangles $CVB$ and $AWC$ lie completely outside $ABC$. See the figure. Prove that quadrilateral $UVCW$ is a parallelogram. [asy] import markers; unitsize(1.5 cm); pair A, B, C, U, V, W; A = (0,0); B = (2,0); C = (1.7,2.5); U = (B + rotate(90,A)*(B))/2; V = (B + rotate(90,C)*(B))/2; W = (C + rotate(90,A)*(C))/2; draw(A--B--C--cycle); draw(A--W, StickIntervalMarker(1,1,size=2mm)); draw(C--W, StickIntervalMarker(1,1,size=2mm)); draw(B--V, StickIntervalMarker(1,2,size=2mm)); draw(C--V, StickIntervalMarker(1,2,size=2mm)); draw(A--U, StickIntervalMarker(1,3,size=2mm)); draw(B--U, StickIntervalMarker(1,3,size=2mm)); draw(rightanglemark(A,U,B,5)); draw(rightanglemark(B,V,C,5)); draw(rightanglemark(A,W,C,5)); dot("$A$", A, S); dot("$B$", B, S); dot("$C$", C, N); dot("$U$", U, NE); dot("$V$", V, NE); dot("$W$", W, NW); [/asy]

2007 Sharygin Geometry Olympiad, 4

Does a parallelogram exist such that all pairwise meets of bisectors of its angles are situated outside it?

1992 Baltic Way, 16

All faces of a convex polyhedron are parallelograms. Can the polyhedron have exactly 1992 faces?

2002 Manhattan Mathematical Olympiad, 4

A triangle has sides with lengths $a,b,c$ such that \[ a^2 + b^2 = 5c^2 \] Prove that medians to the sides of lengths $a$ and $b$ are perpendicular.

2008 National Olympiad First Round, 29

$[AB]$ and $[CD]$ are not parallel in the convex quadrilateral $ABCD$. Let $E$ and $F$ be the midpoints of $[AD]$ and $[BC]$, respectively. If $|CD|=12$, $|AB|=22$, and $|EF|=x$, what is the sum of integer values of $x$? $ \textbf{(A)}\ 110 \qquad\textbf{(B)}\ 114 \qquad\textbf{(C)}\ 118 \qquad\textbf{(D)}\ 121 \qquad\textbf{(E)}\ \text{None of the above} $

2014 USAMO, 4

Let $k$ be a positive integer. Two players $A$ and $B$ play a game on an infinite grid of regular hexagons. Initially all the grid cells are empty. Then the players alternately take turns with $A$ moving first. In his move, $A$ may choose two adjacent hexagons in the grid which are empty and place a counter in both of them. In his move, $B$ may choose any counter on the board and remove it. If at any time there are $k$ consecutive grid cells in a line all of which contain a counter, $A$ wins. Find the minimum value of $k$ for which $A$ cannot win in a finite number of moves, or prove that no such minimum value exists.

2014 Greece Junior Math Olympiad, 1

Let $ABC$ be a triangle and let $M$ be the midpoint $BC$. On the exterior of the triangle, consider the parallelogram $BCDE$ such that $BE//AM$ and $BE=AM/2$ . Prove that line $EM$ passes through the midpoint of segment $AD$.

2012 Tournament of Towns, 4

A circle touches sides $AB, BC, CD$ of a parallelogram $ABCD$ at points $K, L, M$ respectively. Prove that the line $KL$ bisects the height of the parallelogram drawn from the vertex $C$ to $AB$.