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: 87

Let $ABC$ be a triangle with $\angle B > \angle C$. Let $P$ and $Q$ be two different points on line $AC$ such that $\angle PBA = \angle QBA = \angle ACB $ and $A$ is located between $P$ and $C$. Suppose that there exists an interior point $D$ of segment $BQ$ for which $PD=PB$. Let the ray $AD$ intersect the circle $ABC$ at $R \neq A$. Prove that $QB = QR$.
Let $ABC$ be a triangle such that $|AB|=|AC|$ and let $D,E$ be points on the minor arcs $\overarc{AB}$ and $\overarc{AC}$ respectively. The lines $AD$ and $BC$ intersect at $F$ and the line $AE$ intersects the circumcircle of $\triangle FDE$ a second time at $G$. Prove that the line $AC$ is tangent to the circumcircle of $\triangle ECG$.
Let $\omega$ be the semicircle with diameter $PQ$. A circle $k$ is tangent internally to $\omega$ and to segment $PQ$ at $C$. Let $AB$ be the tangent to $K$ perpendicular to $PQ$, with $A$ on $\omega$ and $B$ on the segment $CQ$. Show that $AC$ bisects angle $\angle PAB$
$\Delta ABC$ is isosceles with a circumscribed circle $\omega (O)$. Let $H$ be the foot of the altitude from $C$ to $AB$ and let $M$ be the middle point of $AB$. We define a point $X$ as the second intersection point of the circle with diameter $CM$ and $\omega$ and let $XH$ intersect $\omega$ for a second time in $Y$. If $CO\cap AB=D$, then prove that the circumscribed circle of $\Delta YHD$ is tangent to $\omega$.
Let $ABC$ be a triangle with $\angle B > \angle C$. Let $P$ and $Q$ be two different points on line $AC$ such that $\angle PBA = \angle QBA = \angle ACB $ and $A$ is located between $P$ and $C$. Suppose that there exists an interior point $D$ of segment $BQ$ for which $PD=PB$. Let the ray $AD$ intersect the circle $ABC$ at $R \neq A$. Prove that $QB = QR$.
Let $I$ and $I_A$ denote the incentre and excentre opposite to $A$ of scalene $\triangle ABC$ respectively. Let $A'$ be the antipode of $A$ in $\odot (ABC)$ and $L$ be the midpoint of arc $(BAC)$. Let $LB$ and $LC$ intersect $AI$ at points $Y$ and $Z$ respectively. Prove that $\odot (LYZ)$ is tangent to $\odot (A'II_A)$. [i]~Mahavir Gandhi[/i]
Let $ABC$ be a triangle with $AB = AC \neq BC$ and let $I$ be its incentre. The line $BI$ meets $AC$ at $D$, and the line through $D$ perpendicular to $AC$ meets $AI$ at $E$. Prove that the reflection of $I$ in $AC$ lies on the circumcircle of triangle $BDE$.
Let $ABC$ be a triangle with $AB = AC \neq BC$ and let $I$ be its incentre. The line $BI$ meets $AC$ at $D$, and the line through $D$ perpendicular to $AC$ meets $AI$ at $E$. Prove that the reflection of $I$ in $AC$ lies on the circumcircle of triangle $BDE$.
Let triangle \( ABC \) be an acute-angled triangle. Square \( AEFB \) and \( ADGC \) lie outside triangle \( ABC \). \( BD \) intersects \( CE \) at point \( H \), and \( BG \) intersects \( CF \) at point \( I \). The circumcircle of triangle \( BFI \) intersects the circumcircle of triangle \( CGI \) again at point \( K \). Prove that line segment \( HK \) bisects \( BC \).
Let $\Gamma_1$, $\Gamma_2$, $\Gamma_3$, $\Gamma_4$ be distinct circles such that $\Gamma_1$, $\Gamma_3$ are externally tangent at $P$, and $\Gamma_2$, $\Gamma_4$ are externally tangent at the same point $P$. Suppose that $\Gamma_1$ and $\Gamma_2$; $\Gamma_2$ and $\Gamma_3$; $\Gamma_3$ and $\Gamma_4$; $\Gamma_4$ and $\Gamma_1$ meet at $A$, $B$, $C$, $D$, respectively, and that all these points are different from $P$. Prove that \[ \frac{AB\cdot BC}{AD\cdot DC}=\frac{PB^2}{PD^2}. \]
Let $ ABCD$ be a convex quadrilateral and let $ P$ and $ Q$ be points in $ ABCD$ such that $ PQDA$ and $ QPBC$ are cyclic quadrilaterals. Suppose that there exists a point $ E$ on the line segment $ PQ$ such that $ \angle PAE \equal{} \angle QDE$ and $ \angle PBE \equal{} \angle QCE$. Show that the quadrilateral $ ABCD$ is cyclic. [i]Proposed by John Cuya, Peru[/i]
Given six points $ A, B, C, D, E, F $ such that $ \triangle BCD \stackrel{+}{\sim} \triangle ECA \stackrel{+}{\sim} \triangle BFA $ and let $ I $ be the incenter of $ \triangle ABC. $ Prove that the circumcenter of $ \triangle AID, \triangle BIE, \triangle CIF $ are collinear. [i]Proposed by Telv Cohl[/i]