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: 1065

Point $X$ is chosen inside the non trapezoid quadrilateral $ABCD$ such that $\angle AXD +\angle BXC=180$. Suppose the angle bisector of $\angle ABX$ meets the $D$-altitude of triangle $ADX$ in $K$, and the angle bisector of $\angle DCX$ meets the $A$-altitude of triangle $ADX$ in $L$.We know $BK \perp CX$ and $CL \perp BX$. If the circumcenter of $ADX$ is on the line $KL$ prove that $KL \perp AD$. Proposed by [i]Alireza Dadgarnia[/i]
We say that a circle is [i]half-inscribed[/i] in a triangle, if its center lies on one side of the triangle, and it is tangent to the other two sides. Show that a triangle that has two half-inscribed circles of equal radii, is isosceles. (Recall that a triangle is said to be [i]isosceles[/i], if it has two sides of equal length.)
Let $ABC$ be a triangle and the points $M$ and $N$ on the sides $AB$ respectively $BC$, such that $2 \cdot \frac{CN}{BC} = \frac{AM}{AB}$. Let $P$ be a point on the line $AC$. Prove that the lines $MN$ and $NP$ are perpendicular if and only if $PN$ is the interior angle bisector of $\angle MPC$.
Point $D$ is on side $CB$ of triangle $ABC$. If \[ \angle{CAD} = \angle{DAB} = 60^\circ,\quad AC = 3\quad\mbox{ and }\quad AB = 6, \] then the length of $AD$ is $\text{(A)} \ 2 \qquad \text{(B)} \ 2.5 \qquad \text{(C)} \ 3 \qquad \text{(D)} \ 3.5 \qquad \text{(E)} \ 4$
In triangle $ABC$, $\angle B=2\angle C$ and $\angle A>90^\circ$. Let $D$ be the point on the line $AB$ such that $CD$ is perpendicular to $AC$, and let $M$ be the midpoint of $BC$. Prove that $\angle AMB=\angle DMC$.
Point $A$ lies on one ray and points $B,C$ lie on the other ray of an angle with the vertex at $O$ such that $B$ lies between $O$ and $C$. Let $O_1$ be the incenter of $\triangle OAB$ and $O_2$ be the center of the excircle of $\triangle OAC$ touching side $AC$. Prove that if $O_1A = O_2A$, then the triangle $ABC$ is isosceles.
Let $ABC$ be a triangle, having no right angles, and let $D$ be a point on the side $BC$. Let $E$ and $F$ be the feet of the perpendiculars drawn from the point $D$ to the lines $AB$ and $AC$ respectively. Let $P$ be the point of intersection of the lines $BF$ and $CE$. Prove that the line $AP$ is the altitude of the triangle $ABC$ from the vertex $A$ if and only if the line $AD$ is the angle bisector of the angle $CAB$.
Let $ABC$ be a triangle with $|AB|> |BC|$. Let $D$ be the midpoint of $AC$. Let $E$ be the intersection of the angular bisector of $\angle ABC$ and the line $AC$. Let $F$ be the point on $BE$ such that $CF$ is perpendicular to $BE$. Finally, let $G$ be the intersection of $CF$ and $BD$. Prove that $DF$ divides the line segment $EG$ into two equal parts.
In an acute-angled triangle $ABC$, point $I$ is the incenter, $H$ is the orthocenter, $O$ is the center of the circumscribed circle, $T$ and $K$ are the touchpoints of the $A$-excircle and incircle with side $BC$ respectively. It turned out that the segment $TI$ is passing through the point $O$. Prove that $HK$ is the angle bisector of $\angle BHC$. (Matvii Kurskyi)
Let $M$ be the midpoint of the side $AC$ of a triangle $ABC$, and let $K$ be a point on the ray $BA$ beyond $A$. The line $KM$ intersects the side $BC$ at the point $L$. $P$is the point on the segment $BM$ such that $PM$ is the bisector of the angle $LPK$. The line $\ell$ passes through $A$ and is parallel to $BM$. Prove that the projection of the point $M$ onto the line $\ell$ belongs to the line $PK$.
Points $B,C$ vary on two fixed rays emanating from point $A$ such that $AB+AC$ is constant. Show that there is a point $D$, other than $A$, such that the circumcircle of triangle $ABC$ passes through $D$ for all possible choices of $B, C$.
A parallelogram of $ABCD$ is given. Line parallel to $AB$ intersects the bisectors of angles $A$ and $C$ at points $P$ and $Q$, respectively. Prove that the angles $ADP$ and $ABQ$ are equal. (A. Hakobyan)
In triangle $ ABC$, angle $ C$ is a right angle and $ CB > CA$. Point $ D$ is located on $ \overline{BC}$ so that angle $ CAD$ is twice angle $ DAB$. If $ AC/AD \equal{} 2/3$, then $ CD/BD \equal{} m/n$, where $ m$ and $ n$ are relatively prime positive integers. Find $ m \plus{} n$. $ \textbf{(A)}\ 10\qquad \textbf{(B)}\ 14\qquad \textbf{(C)}\ 18\qquad \textbf{(D)}\ 22\qquad \textbf{(E)}\ 26$
In an acute triangle $ABC$ the angle $\angle C$ is greater than $\angle A$. Let $E$ be such that $AE$ is a diameter of the circumscribed circle $\Gamma$ of \vartriangle ABC. Let $K$ be the intersection of $AC$ and the tangent line at $B$ to $\Gamma$. Let $L$ be the orthogonal projection of $K$ on $AE$ and let $D$ be the intersection of $KL$ and $AB$. Prove that $CE$ is the bisector of $\angle BCD$.
Points $A_1$, $B_1$, $C_1$ are midpoints of sides $BC$, $AC$, $AB$ of triangle $ABC$. On midlines $C_1B_1$ and $A_1B_1$ points $E$ and $F$ are chosen such that $BE$ is the angle bisector of $AEB_1$ and $BF$ is the angle bisector of $CFB_1$. Prove that bisectors of angles $ABC$ and $FBE$ coincide. [I]Proposed by F. Baharev[/i]