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: 433

Let $ABCD$ be a cyclic quadrilateral inscribed in circle $\omega$ whose diagonals meet at $F$. Lines $AB$ and $CD$ meet at $E$. Segment $EF$ intersects $\omega$ at $X$. Lines $BX$ and $CD$ meet at $M$, and lines $CX$ and $AB$ meet at $N$. Prove that $MN$ and $BC$ concur with the tangent to $\omega$ at $X$. [i]Proposed by Allen Liu[/i]
Let $ABC$ be an acute angled triangle with circumcircle $O$. Line $AO$ intersects the circumcircle of triangle $ABC$ again at point $D$. Let $P$ be a point on the side $BC$. Line passing through $P$ perpendicular to $AP$ intersects lines $DB$ and $DC$ at $E$ and $F$ respectively . Line passing through $D$ perpendicular to $BC$ intersects $EF$ at point $Q$. Prove that $EQ = FQ$ if and only if $BP = CP$.
On the extensions of sides $CA$ and $AB$ of triangle $ABC$ beyond points $A$ and $B$, respectively, the segments $AE = BC$ and $BF = AC$ are drawn. A circle is tangent to segment $BF$ at point $N$, side $BC$ and the extension of side $AC$ beyond point $C$. Point $M$ is the midpoint of segment $EF$. Prove that the line $MN$ is parallel to the bisector of angle $A$.
Point $ P$ lies on side $ AB$ of a convex quadrilateral $ ABCD$. Let $ \omega$ be the incircle of triangle $ CPD$, and let $ I$ be its incenter. Suppose that $ \omega$ is tangent to the incircles of triangles $ APD$ and $ BPC$ at points $ K$ and $ L$, respectively. Let lines $ AC$ and $ BD$ meet at $ E$, and let lines $ AK$ and $ BL$ meet at $ F$. Prove that points $ E$, $ I$, and $ F$ are collinear. [i]Author: Waldemar Pompe, Poland[/i]
Let $O$ be the circumcentre, and $\Omega$ be the circumcircle of an acute-angled triangle $ABC$. Let $P$ be an arbitrary point on $\Omega$, distinct from $A$, $B$, $C$, and their antipodes in $\Omega$. Denote the circumcentres of the triangles $AOP$, $BOP$, and $COP$ by $O_A$, $O_B$, and $O_C$, respectively. The lines $\ell_A$, $\ell_B$, $\ell_C$ perpendicular to $BC$, $CA$, and $AB$ pass through $O_A$, $O_B$, and $O_C$, respectively. Prove that the circumcircle of triangle formed by $\ell_A$, $\ell_B$, and $\ell_C$ is tangent to the line $OP$.
Let $ABCD$ be a convex quadrilateral, $M$ the midpoint of $AB$, $N$ the midpoint of $BC$, $E$ the intersection of the segments $AN$ and $BD$, $F$ the intersection of the segments $DM$ and $AC$. Prove that if $BE = \frac 13 BD$ and $AF = \frac 13 AC$, then $ABCD$ is a parallelogram.
In acute triangle ABC, which is not equilateral, let $P$ denote the foot of the altitude from $C$ to side $AB$; let $H$ denote the orthocenter; let $O$ denote the circumcenter; let $D$ denote the intersection of line $CO$ with $AB$; and let $E$ denote the midpoint of $CD$. Prove that line $EP$ passes through the midpoint of $OH$.
Let $ABC$ be a triangle with $CA=CB$ and $\angle{ACB}=120^\circ$, and let $M$ be the midpoint of $AB$. Let $P$ be a variable point of the circumcircle of $ABC$, and let $Q$ be the point on the segment $CP$ such that $QP = 2QC$. It is given that the line through $P$ and perpendicular to $AB$ intersects the line $MQ$ at a unique point $N$. Prove that there exists a fixed circle such that $N$ lies on this circle for all possible positions of $P$.
Let $\omega,\omega_1,\omega_2$ be three mutually tangent circles such that $\omega_1,\omega_2$ are externally tangent at $P$, $\omega_1,\omega$ are internally tangent at $A$, and $\omega,\omega_2$ are internally tangent at $B$. Let $O,O_1,O_2$ be the centers of $\omega,\omega_1,\omega_2$, respectively. Given that $X$ is the foot of the perpendicular from $P$ to $AB$, prove that $\angle{O_1XP}=\angle{O_2XP}$. [i]David Yang.[/i]
Let $ABCD$ be a quadrilateral with $\measuredangle DAB=\measuredangle ABC=90^{\circ}$. Denote by $M$ the midpoint of the side $AB$, and assume that $\measuredangle CMD=90^{\circ}$. Let $K$ be the foot of the perpendicular from the point $M$ to the line $CD$. The line $AK$ meets $BD$ at $P$, and the line $BK$ meets $AC$ at $Q$. Show that $\angle{AKB}=90^{\circ}$ and $\frac{KP}{PA}+\frac{KQ}{QB}=1$. [color=red][Moderator edit: The proposed solution can be found at http://erdos.fciencias.unam.mx/mexproblem3.pdf .][/color]
Let $AA_0$ be the altitude of the isosceles triangle $ABC~(AB = AC)$. A circle $\gamma$ centered at the midpoint of $AA_0$ touches $AB$ and $AC$. Let $X$ be an arbitrary point of line $BC$. Prove that the tangents from $X$ to $\gamma$ cut congruent segments on lines $AB$ and $AC$
Let $P$ be a point in the interior of a triangle $ABC$ and let the rays $\overrightarrow{AP}, \overrightarrow{BP}$ and $\overrightarrow{CP}$ intersect the sides $BC, CA$ and $AB$ in $A_1,B_1$ and $C_1$, respectively. Let $D$ be the foot of the perpendicular from $A_1$ to $B_1C_1$. Show that \[\frac{CD}{BD}=\frac{B_1C}{BC_1} \cdot \frac{C_1A}{AB_1}.\]
Consider an acute-angled triangle $ABC$ with $AB<AC$ and let $\omega$ be its circumscribed circle. Let $t_B$ and $t_C$ be the tangents to the circle $\omega$ at points $B$ and $C$, respectively, and let $L$ be their intersection. The straight line passing through the point $B$ and parallel to $AC$ intersects $t_C$ in point $D$. The straight line passing through the point $C$ and parallel to $AB$ intersects $t_B$ in point $E$. The circumcircle of the triangle $BDC$ intersects $AC$ in $T$, where $T$ is located between $A$ and $C$. The circumcircle of the triangle $BEC$ intersects the line $AB$ (or its extension) in $S$, where $B$ is located between $S$ and $A$. Prove that $ST$, $AL$, and $BC$ are concurrent. $\text{Vangelis Psychas and Silouanos Brazitikos}$
Let $ABCD$ be a convex quadrilateral, and let $P$, $Q$, $R$, and $S$ be points on the sides $AB$, $BC$, $CD$, and $DA$, respectively. Let the line segment $PR$ and $QS$ meet at $O$. Suppose that each of the quadrilaterals $APOS$, $BQOP$, $CROQ$, and $DSOR$ has an incircle. Prove that the lines $AC$, $PQ$, and $RS$ are either concurrent or parallel to each other.
Let $ABCD$ be a quadrilateral inscribed in a circle $\omega$, and let $P$ be a point on the extension of $AC$ such that $PB$ and $PD$ are tangent to $\omega$. The tangent at $C$ intersects $PD$ at $Q$ and the line $AD$ at $R$. Let $E$ be the second point of intersection between $AQ$ and $\omega$. Prove that $B$, $E$, $R$ are collinear.
The diagonals $AC$ and $CE$ of the regular hexagon $ABCDEF$ are divided by inner points $M$ and $N$ respectively, so that \[ {AM\over AC}={CN\over CE}=r. \] Determine $r$ if $B,M$ and $N$ are collinear.
In a convex hexagon $AC'BA'CB'$, every two opposite sides are equal. Let $A_1$ denote the point of intersection of $BC$ with the perpendicular bisector of $AA'$. Define $B_1$ and $C_1$ similarly. Prove that $A_1$, $B_1$, and $C_1$ are collinear.
Let $O$ be the circumcentre, and $\Omega$ be the circumcircle of an acute-angled triangle $ABC$. Let $P$ be an arbitrary point on $\Omega$, distinct from $A$, $B$, $C$, and their antipodes in $\Omega$. Denote the circumcentres of the triangles $AOP$, $BOP$, and $COP$ by $O_A$, $O_B$, and $O_C$, respectively. The lines $\ell_A$, $\ell_B$, $\ell_C$ perpendicular to $BC$, $CA$, and $AB$ pass through $O_A$, $O_B$, and $O_C$, respectively. Prove that the circumcircle of triangle formed by $\ell_A$, $\ell_B$, and $\ell_C$ is tangent to the line $OP$.
Distinct points $A$ and $B$ are given on the plane. Consider all triangles $ABC$ in this plane on whose sides $BC,CA$ points $D,E$ respectively can be taken so that (i) $\frac{BD}{BC}=\frac{CE}{CA}=\frac{1}{3}$; (ii) points $A,B,D,E$ lie on a circle in this order. Find the locus of the intersection points of lines $AD$ and $BE$.
Let $ABC$ be a triangle. Let $E$ be a point on the segment $BC$ such that $BE = 2EC$. Let $F$ be the mid-point of $AC$. Let $BF$ intersect $AE$ in $Q$. Determine $BQ:QF$.
Let $I$ be the incentre of acute triangle $ABC$ with $AB\neq AC$. The incircle $\omega$ of $ABC$ is tangent to sides $BC, CA$, and $AB$ at $D, E,$ and $F$, respectively. The line through $D$ perpendicular to $EF$ meets $\omega$ at $R$. Line $AR$ meets $\omega$ again at $P$. The circumcircles of triangle $PCE$ and $PBF$ meet again at $Q$. Prove that lines $DI$ and $PQ$ meet on the line through $A$ perpendicular to $AI$. [i]Proposed by Anant Mudgal, India[/i]
In $\Delta ABC$, $AC=kAB$, with $k>1$. The internal angle bisector of $\angle BAC$ meets $BC$ at $D$. The circle with $AC$ as diameter cuts the extension of $AD$ at $E$. Express $\dfrac{AD}{AE}$ in terms of $k$.
In triangle $ABC$, $\omega$ is its circumcircle and $O$ is the center of this circle. Points $M$ and $N$ lie on sides $AB$ and $AC$ respectively. $\omega$ and the circumcircle of triangle $AMN$ intersect each other for the second time in $Q$. Let $P$ be the intersection point of $MN$ and $BC$. Prove that $PQ$ is tangent to $\omega$ iff $OM=ON$. [i]proposed by Mr.Etesami[/i]
In triangle $ABC$ shown in the adjoining figure, $M$ is the midpoint of side $BC$, $AB=12$ and $AC=16$. Points $E$ and $F$ are taken on $AC$ and $AB$, respectively, and lines $EF$ and $AM$ intersect at $G$. If $AE=2AF$ then $\frac{EG}{GF}$ equals [asy] draw((0,0)--(12,0)--(14,7.75)--(0,0)); draw((0,0)--(13,3.875)); draw((5,0)--(8.75,4.84)); label("A", (0,0), S); label("B", (12,0), S); label("C", (14,7.75), E); label("E", (8.75,4.84), N); label("F", (5,0), S); label("M", (13,3.875), E); label("G", (7,1)); [/asy] $ \textbf{(A)}\ \frac{3}{2} \qquad\textbf{(B)}\ \frac{4}{3} \qquad\textbf{(C)}\ \frac{5}{4} \qquad\textbf{(D)}\ \frac{6}{5} \\ \qquad\textbf{(E)}\ \text{not enough information to solve the problem} $
Let $ABC$ be a triangle with incenter $I$. Points $K$ and $L$ are chosen on segment $BC$ such that the incircles of $\triangle ABK$ and $\triangle ABL$ are tangent at $P$, and the incircles of $\triangle ACK$ and $\triangle ACL$ are tangent at $Q$. Prove that $IP=IQ$. [i]Ankan Bhattacharya[/i]