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: 509

On the side $AB$ of the non-isosceles triangle $ABC$, let the points $P$ and $Q$ be so that $AC = AP$ and $BC = BQ$. The perpendicular bisector of the segment $PQ$ intersects the angle bisector of the $\angle C$ at the point $R$ (inside the triangle). Prove that $\angle ACB + \angle PRQ = 180^o$.
The circle $\omega$ passes through the vertices $B$ and $C$ of triangle $ABC$ and intersects its sides $AC,AB$ at points $A,E$, respectively. On the ray $BD$, a point $K$ such that $BK = AC$ is chosen , and on the ray $CE$, a point $L$ such that $CL = AB$ is chosen. Prove that the center $O$ of the circumscribed circle of the triangle $AKL$ lies on the circle $\omega$.
Point $D$ is chosen on the side $AC$ of an acute-angled triangle $ABC$. The median $AM$ intersects the altitude $CH$ and the segment $BD$ at points $N$ and $K$ respectively. Prove that if $AK = BK$, then $AN = 2KM$.
Inside the equilateral triangle $ABC$, points $P$ and $Q$ are chosen so that the quadrilateral $APQC$ is convex, $AP = PQ = QC$ and $\angle PBQ = 30^o$. Prove that $AQ = BP$.
The points $C$ and $D$ lie on a halfline from the midpoint $M$ of a segment $AB$, so that $|AC| = |BD|$. Prove that the angles $u = \angle ACM$ and $v = \angle BDM$ are equal. [img]https://1.bp.blogspot.com/-tQEJ1VBCa8U/XzT7IhwlZHI/AAAAAAAAMVI/xpRdlj5Rl64VUt_tCRsQ1UxIsv_SGrMlACLcBGAsYHQ/s0/2014%2BMohr%2Bp3.png[/img]
Let $A$ and $B$ be points on the circle $k$ with center $O$, so that $AB> AO$. Let $C$ be the intersection of the bisectors of $\angle OAB$ and $k$, different from $A$. Let $D$ be the intersection of the straight line $AB$ with the circumcircle of the triangle $OBC$, different from $B$. Show that $AD = AO$ .
Given a convex pentagon $ABCDE$, with $\angle BAC = \angle ABE = \angle DEA - 90^o$, $\angle BCA = \angle ADE$ and also $BC = ED$. Prove that $BCDE$ is parallelogram.
$ABC$ is an acute triangle with $\angle ACB = 45^o$. Let $D$ and $E$ be points on the sides $BC$ and $AC$, respectively, such that $AB = AD = BE$. Let $M,N$ and $X$ be the midpoints of $BD, AE$ and $AB$, respectively. Let lines $AM$ and $BN$ intersect at point $P$. Show that lines $XP$ and $DE$ are perpendicular.
An isosceles right-angled triangle $ABC$ is given. On the extensions of sides $AB$ and $AC$, behind vertices $B$ and $C$ equal segments $BK$ and $CL$ were laid. $E$ and F are the points of intersection of the segment $KL$ and the lines perpendicular to the $KC$ , passing through the points $B$ and $A$, respectively. Prove that $EF = FL$.
A semicircle is erected over the segment $AB$ with center $M$. Let $P$ be one point different from $A$ and $B$ on the semicircle and $Q$ the midpoint of the arc of the circle $AP$. The point of intersection of the straight line $BP$ with the parallel to $P Q$ through $M$ is $S$. Prove that $PM = PS$ holds. [i](Karl Czakler)[/i]
The triangle $ABC$ is inscribed in a circle. At points $A$ and $B$ are tangents to this circle, which intersect at point $T$. A line drawn through the point $T$ parallel to the side $AC$ intersects the side $BC$ at the point $D$. Prove that $AD = CD$.
The points $A, B, C, D$ lie in this order on a circle with center $O$. Furthermore, the straight lines $AC$ and $BD$ should be perpendicular to each other. The base of the perpendicular from $O$ on $AB$ is $F$. Prove $CD = 2 OF$. [i](Karl Czakler)[/i]
Let $ABC$ be a triangle with centroid $T$. Denote by $M$ the midpoint of $BC$. Let $D$ be a point on the ray opposite to the ray $BA$ such that $AB = BD$. Similarly, let $E$ be a point on the ray opposite to the ray $CA$ such that $AC = CE$. The segments $T D$ and $T E$ intersect the side $BC$ in $P$ and $Q$, respectively. Show that the points $P, Q$ and $M$ split the segment $BC$ into four parts of equal length.
In the acute triangle $ABC$ the orthocenter $H$ and the center of the circumscribed circle $O$ were noted. The line $AO$ intersects the side $BC$ at the point $D$. A perpendicular drawn to the side $BC$ at the point $D$ intersects the heights from the vertices $B$ and $C$ of the triangle $ABC$ at the points $X$ and $Y$ respectively. Prove that the center of the circumscribed circle $\Delta HXY$ is equidistant from the points $B$ and $C$. (Danilo Hilko)
Let $H$ be the orthocenter of triangle $ABC$. The tangents to the circumcircles of triangles $CHB$ and $AHB$ at point $H$ meet $AC$ at points $A_1$ and $C_1$ respectively. Prove that $A_1H = C_1H$.
Prove that in a triangle one of the sides is twice as large as the other if and only if a median and an angle bisector of this triangle are perpendicular
Let $ABC$ be an isosceles triangle with $AB = AC$. Let $D, E$ and $F$ be points on line segments $BC, CA$ and $AB$, respectively, such that $BF = BE$ and such that $ED$ is the angle bisector of $\angle BEC$. Prove that $BD = EF$ if and only if $AF = EC$.
In a triangle $ABC$, $\angle ABC= 60^o$, point $I$ is the incenter. Let the points $P$ and $T$ on the sides $AB$ and $BC$ respectively such that $PI \parallel BC$ and $TI \parallel AB$ , and points $P_1$ and $T_1$ on the sides $AB$ and $BC$ respectively such that $AP_1 = BP$ and $CT_1 = BT$. Prove that point $I$ lies on segment $P_1T_1$. (Anton Trygub)
Let $ABCD$ be a parallelogram with an acute angle at $A$. Let $G$ be the point on the line $AB$, distinct from $B$, such that $CG = CB$. Let $H$ be the point on the line $BC$, distinct from $B$, such that $AB = AH$. Prove that triangle $DGH$ is isosceles.
Let $ABCD$ is a trapezoid with $\angle A = \angle B = 90^o$ and let $E$ is a point lying on side $CD$. Let the circle $\omega$ is inscribed to triangle $ABE$ and tangents sides $AB, AE$ and $BE$ at points $P, F$ and $K$ respectively. Let $KF$ intersects segments $BC$ and $AD$ at points $M$ and $N$ respectively, as well as $PM$ and $PN$ intersect $\omega$ at points $H$ and $T$ respectively. Prove that $PH = PT$.
In triangle $ABC$, let $I$ be the incentre, $O$ be the circumcentre, and $H$ be the orthocentre. It is given that $IO = IH$. Show that one of the angles of triangle $ABC$ must be equal to $60$ degrees.
Let $D$ the touchpoint of the inscribed circle of triangle $ABC$ be with side $AB$ . From $A$ the perpendicular lines on the angle bisectors of vertices $B$ and $C$ intersect them at points $A_1$ and $A_2$ respectively . Prove that $A_1A_2 = AD$.
Let $\vartriangle ABC$be a triangle and let $\Gamma$ its circumscribed circle. Let $M$ be the midpoint of the side $BC$ and let $D$ be the point of intersection of the line $AM$ with $\Gamma$. By $D$ a straight line is drawn parallel to $BC$, which intersects $\Gamma$ at a point $E$. Let $N$ be the midpoint of the segment $AE$ and let $P$ be the point of intersection of $CN$ with $AM$. Show that $AP = PC$.
The inscribed circle $\omega$ of the triangle $ABC$ touches its sides $BC, CA$, and $AB$ at the points $D, E$, and $F$, respectively. Let the points $X, Y$, and $Z$ of the circle $\omega$ be diametrically opposite to the points $D, E$, and $F$, respectively. Line $AX, BY$ and $CZ$ intersect the sides $BC, CA$ and $AB$ at the points $D', E'$ and $F'$, respectively. On the segments $AD', BE'$ and $CF'$ noted the points $X', Y'$ and $Z'$, respectively, so that $D'X'= AX$, $E'Y' = BY$, $F'Z' = CZ$. Prove that the points $X', Y'$ and $Z'$ coincide.
Determine the locus of points, from which the tangent segments to two given circles are equal.