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

2007 Indonesia TST, 1

Call an $n$-gon to be [i]lattice[/i] if its vertices are lattice points. Prove that inside every lattice convex pentagon there exists a lattice point.

2002 Baltic Way, 13

Let $ABC$ be an acute triangle with $\angle BAC>\angle BCA$, and let $D$ be a point on side $AC$ such that $|AB|=|BD|$. Furthermore, let $F$ be a point on the circumcircle of triangle $ABC$ such that line $FD$ is perpendicular to side $BC$ and points $F,B$ lie on different sides of line $AC$. Prove that line $FB$ is perpendicular to side $AC$ .

1967 IMO, 1

The parallelogram $ABCD$ has $AB=a,AD=1,$ $\angle BAD=A$, and the triangle $ABD$ has all angles acute. Prove that circles radius $1$ and center $A,B,C,D$ cover the parallelogram if and only \[a\le\cos A+\sqrt3\sin A.\]

2005 All-Russian Olympiad, 2

We have an acute-angled triangle $ABC$, and $AA',BB'$ are its altitudes. A point $D$ is chosen on the arc $ACB$ of the circumcircle of $ABC$. If $P=AA'\cap BD,Q=BB'\cap AD$, show that the midpoint of $PQ$ lies on $A'B'$.

2010 Romania Team Selection Test, 1

Let $P$ be a point in the plane and let $\gamma$ be a circle which does not contain $P$. Two distinct variable lines $\ell$ and $\ell'$ through $P$ meet the circle $\gamma$ at points $X$ and $Y$, and $X'$ and $Y'$, respectively. Let $M$ and $N$ be the antipodes of $P$ in the circles $PXX'$ and $PYY'$, respectively. Prove that the line $MN$ passes through a fixed point. [i]Mihai Chis[/i]

2001 Paraguay Mathematical Olympiad, 4

In a parallelogram $ABCD$ of surface area $60$ cm$^2$ , a line is drawn by $D$ that intersects $BC$ at $P$ and the extension of $AB$ at $Q$. If the area of the quadrilateral $ABPD$ is $46$ cm$^2$ , find the area of triangle $CPQ$.

2021 Kyiv City MO Round 1, 10.3

Circles $\omega_1$ and $\omega_2$ with centers at points $O_1$ and $O_2$ intersect at points $A$ and $B$. Let point $C$ be such that $AO_2CO_1$ is a parallelogram. An arbitrary line is drawn through point $A$, which intersects the circles $\omega_1$ and $\omega_2$ at points $X$ and $Y$, respectively. Prove that $CX = CY$. [i]Proposed by Oleksii Masalitin[/i]

2004 Bosnia and Herzegovina Junior BMO TST, 4

Let $ABCD$ be a parallelogram. On the ray $(DB$ a point $E$ is given such that the ray $(AB$ is the angle bisector of $\angle CAE$. Let $F$ be the intersection of $CE$ and $AB$. Prove that $\frac{AB}{BF} - \frac{AC}{AE} = 1$

2008 IMS, 2

Let $ f$ be an entire function on $ \mathbb C$ and $ \omega_1,\omega_2$ are complex numbers such that $ \frac {\omega_1}{\omega_2}\in{\mathbb C}\backslash{\mathbb Q}$. Prove that if for each $ z\in \mathbb C$, $ f(z) \equal{} f(z \plus{} \omega_1) \equal{} f(z \plus{} \omega_2)$ then $ f$ is constant.

2006 China Team Selection Test, 1

Let the intersections of $\odot O_1$ and $\odot O_2$ be $A$ and $B$. Point $R$ is on arc $AB$ of $\odot O_1$ and $T$ is on arc $AB$ on $\odot O_2$. $AR$ and $BR$ meet $\odot O_2$ at $C$ and $D$; $AT$ and $BT$ meet $\odot O_1$ at $Q$ and $P$. If $PR$ and $TD$ meet at $E$ and $QR$ and $TC$ meet at $F$, then prove: $AE \cdot BT \cdot BR = BF \cdot AT \cdot AR$.

1953 AMC 12/AHSME, 32

Each angle of a rectangle is trisected. The intersections of the pairs of trisectors adjacent to the same side always form: $ \textbf{(A)}\ \text{a square} \qquad\textbf{(B)}\ \text{a rectangle} \qquad\textbf{(C)}\ \text{a parallelogram with unequal sides} \\ \textbf{(D)}\ \text{a rhombus} \qquad\textbf{(E)}\ \text{a quadrilateral with no special properties}$

2015 Bosnia And Herzegovina - Regional Olympiad, 3

In parallelogram $ABCD$ holds $AB=BD$. Let $K$ be a point on $AB$, different from $A$, such that $KD=AD$. Let $M$ be a point symmetric to $C$ with respect to $K$, and $N$ be a point symmetric to point $B$ with respect to $A$. Prove that $DM=DN$

2000 National High School Mathematics League, 15

$C_0:x^2+y^2=1,C_1:\frac{x^2}{a^2}+\frac{y^2}{b^2}(a>b>0)$. Find all $(a,b)$ such that for any point $P$ on $C_1$, we can find a parallelogram with an apex $P$, and it is externally tangent to $C_0$, inscribed to $C_1$.

2004 239 Open Mathematical Olympiad, 8

Given a triangle $ABC$. A point $X$ is chosen on a side $AC$. Some circle passes through $X$, touches the side $AC$ and intersects the circumcircle of triangle $ABC$ in points $M$ and $N$ such that the segment $MN$ bisects $BX$ and intersects sides $AB$ and $BC$ in points $P$ and $Q$. Prove that the circumcircle of triangle $PBQ$ passes through a fixed point different from $B$. [b]proposed by Sergej Berlov[/b]

2005 Tuymaada Olympiad, 7

Let $I$ be the incentre of triangle $ABC$. A circle containing the points $B$ and $C$ meets the segments $BI$ and $CI$ at points $P$ and $Q$ respectively. It is known that $BP\cdot CQ=PI\cdot QI$. Prove that the circumcircle of the triangle $PQI$ is tangent to the circumcircle of $ABC$. [i]Proposed by S. Berlov[/i]

2006 Germany Team Selection Test, 3

Let $ABCD$ be a parallelogram. A variable line $g$ through the vertex $A$ intersects the rays $BC$ and $DC$ at the points $X$ and $Y$, respectively. Let $K$ and $L$ be the $A$-excenters of the triangles $ABX$ and $ADY$. Show that the angle $\measuredangle KCL$ is independent of the line $g$. [i]Proposed by Vyacheslev Yasinskiy, Ukraine[/i]

2015 All-Russian Olympiad, 2

Given is a parallelogram $ABCD$, with $AB <AC <BC$. Points $E$ and $F$ are selected on the circumcircle $\omega$ of $ABC$ so that the tangenst to $\omega$ at these points pass through point $D$ and the segments $AD$ and $CE$ intersect. It turned out that $\angle ABF = \angle DCE$. Find the angle $\angle{ABC}$. A. Yakubov, S. Berlov

1998 All-Russian Olympiad, 6

In triangle $ABC$ with $AB>BC$, $BM$ is a median and $BL$ is an angle bisector. The line through $M$ and parallel to $AB$ intersects $BL$ at point $D$, and the line through $L$ and parallel to $BC$ intersects $BM$ at point $E$. Prove that $ED$ is perpendicular to $BL$.

2025 EGMO, 4

Let $ABC$ be an acute triangle with incentre $I$ and $AB \neq AC$. Let lines $BI$ and $CI$ intersect the circumcircle of $ABC$ at $P \neq B$ and $Q \neq C$, respectively. Consider points $R$ and $S$ such that $AQRB$ and $ACSP$ are parallelograms (with $AQ \parallel RB, AB \parallel QR, AC \parallel SP$, and $AP \parallel CS$). Let $T$ be the point of intersection of lines $RB$ and $SC$. Prove that points $R, S, T$, and $I$ are concyclic.

2022 USA TSTST, 7

Let $ABCD$ be a parallelogram. Point $E$ lies on segment $CD$ such that \[2\angle AEB=\angle ADB+\angle ACB,\] and point $F$ lies on segment $BC$ such that \[2\angle DFA=\angle DCA+\angle DBA.\] Let $K$ be the circumcenter of triangle $ABD$. Prove that $KE=KF$. [i]Merlijn Staps[/i]

2024 Singapore Junior Maths Olympiad, Q2

Let $ABCD$ be a parallelogram and points $E,F$ be on its exterior. If triangles $BCF$ and $DEC$ are similar, i.e. $\triangle BCF \sim \triangle DEC$, prove that triangle $AEF$ is similar to these two triangles.

1988 Austrian-Polish Competition, 3

In a ABCD cyclic quadrilateral 4 points K, L ,M, N are taken on AB , BC , CD and DA , respectively such that KLMN is a parallelogram. Lines AD, BC and KM have a common point. And also lines AB, DC and NL have a common point. Prove that KLMN is rhombus.

2004 Tuymaada Olympiad, 3

An acute triangle $ABC$ is inscribed in a circle of radius 1 with centre $O;$ all the angles of $ABC$ are greater than $45^\circ.$ $B_{1}$ is the foot of perpendicular from $B$ to $CO,$ $B_{2}$ is the foot of perpendicular from $B_{1}$ to $AC.$ Similarly, $C_{1}$ is the foot of perpendicular from $C$ to $BO,$ $C_{2}$ is the foot of perpendicular from $C_{1}$ to $AB.$ The lines $B_{1}B_{2}$ and $C_{1}C_{2}$ intersect at $A_{3}.$ The points $B_{3}$ and $C_{3}$ are defined in the same way. Find the circumradius of triangle $A_{3}B_{3}C_{3}.$ [i]Proposed by F.Bakharev, F.Petrov[/i]

1994 Vietnam Team Selection Test, 1

Given a parallelogram $ABCD$. Let $E$ be a point on the side $BC$ and $F$ be a point on the side $CD$ such that the triangles $ABE$ and $BCF$ have the same area. The diaogonal $BD$ intersects $AE$ at $M$ and intersects $AF$ at $N$. Prove that: [b]I. [/b] There exists a triangle, three sides of which are equal to $BM, MN, ND$. [b]II.[/b] When $E, F$ vary such that the length of $MN$ decreases, the radius of the circumcircle of the above mentioned triangle also decreases.

1998 Hungary-Israel Binational, 2

On the sides of a convex hexagon $ ABCDEF$ , equilateral triangles are constructd in its exterior. Prove that the third vertices of these six triangles are vertices of a regular hexagon if and only if the initial hexagon is [i]affine regular[/i]. (A hexagon is called affine regular if it is centrally symmetric and any two opposite sides are parallel to the diagonal determine by the remaining two vertices.)