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: 3585

Point $D$ is selected inside acute $\triangle ABC$ so that $\angle DAC = \angle ACB$ and $\angle BDC = 90^{\circ} + \angle BAC$. Point $E$ is chosen on ray $BD$ so that $AE = EC$. Let $M$ be the midpoint of $BC$. Show that line $AB$ is tangent to the circumcircle of triangle $BEM$. [i]Proposed by Anton Trygub[/i]
Let $ABC$ be a triangle and $M$ be the midpoint of side $BC$. The perpendicular bisector of $BC$ intersects the circumcircle of $\triangle ABC$ at points $K$ and $L$ ($K$ and $A$ lie on the opposite sides of $BC$). A circle passing through $L$ and $M$ intersects $AK$ at points $P$ and $Q$ ($P$ lies on the line segment $AQ$). $LQ$ intersects the circumcircle of $\triangle KMQ$ again at $R$. Prove that $BPCR$ is a cyclic quadrilateral.
In isosceles $\triangle ABC$, $AB=AC$, points $D,E,F$ lie on segments $BC,AC,AB$ such that $DE\parallel AB$, $DF\parallel AC$. The circumcircle of $\triangle ABC$ $\omega_1$ and the circumcircle of $\triangle AEF$ $\omega_2$ intersect at $A,G$. Let $DE$ meet $\omega_2$ at $K\neq E$. Points $L,M$ lie on $\omega_1,\omega_2$ respectively such that $LG\perp KG, MG\perp CG$. Let $P,Q$ be the circumcenters of $\triangle DGL$ and $\triangle DGM$ respectively. Prove that $A,G,P,Q$ are concyclic.
In a triangle $ABC$, with $AB \ne BC$, $E$ is a point on the line $AC$ such that $BE$ is perpendicular to $AC$. A circle passing through $A$ and touching the line $BE$ at a point $P \ne B$ intersects the line $AB$ for the second time at $X$. Let $Q$ be a point on the line $PB$ different from $P$ such that $BQ = BP$. Let $Y$ be the point of intersection of the lines $CP$ and $AQ$. Prove that the points $C, X, Y, A$ are concyclic if and only if $CX$ is perpendicular to $AB$.
The incircle of a non-isosceles triangle $ABC$ touches $AB$ at point $C'$. The circle with diameter $BC'$ meets the incircle and the bisector of angle $B$ again at points $A_1$ and $A_2$ respectively. The circle with diameter $AC'$ meets the incircle and the bisector of angle $A$ again at points $B_1$ and $B_2$ respectively. Prove that lines $AB, A_1B_1, A_2B_2$ concur. (E. H. Garsia)
In a triangle $ABC$ with $AB\not = AC$, the incircle touches the sides $BC, CA, AB$ at $D, E, F$ , respectively. Line $AD$ meets the incircle again at $P$ . The line $EF$ and the line through $P$ perpendicular to $AD$ meet at $Q$. Line $AQ$ intersects $DE$ at $X$ and $DF$ at $Y$ . Prove that $A$ is the midpoint of $XY$.
Given a nonequilateral triangle $ABC$, the vertices listed counterclockwise, find the locus of the centroids of the equilateral triangles $A'B'C'$ (the vertices listed counterclockwise) for which the triples of points $A,B', C'; A',B, C';$ and $A',B', C$ are collinear. [i]Proposed by Poland.[/i]
Given a trapezoid $ABCD$ ($AD \parallel BC$) with $\angle ABC > 90^\circ$ . Point $M$ is chosen on the lateral side $AB$. Let $O_1$ and $O_2$ be the circumcenters of the triangles $MAD$ and $MBC$, respectively. The circumcircles of the triangles $MO_1D$ and $MO_2C$ meet again at the point $N$. Prove that the line $O_1O_2$ passes through the point $N$.
In triangle $ABC$, the incircle touches sides $BC,CA,AB$ at $D,E,F$ respectively. Assume there exists a point $X$ on the line $EF$ such that \[\angle{XBC} = \angle{XCB} = 45^{\circ}.\] Let $M$ be the midpoint of the arc $BC$ on the circumcircle of $ABC$ not containing $A$. Prove that the line $MD$ passes through $E$ or $F$. United Kingdom
Let $ ABC$ be a tringle and let $ P$ be an interior point such that $ \angle BPC \equal{} 90 ,\angle BAP \equal{} \angle BCP$.Let $ M,N$ be the mid points of $ AC,BC$ respectively.Suppose $ BP \equal{} 2PM$.Prove that $ A,P,N$ are collinear.
Let $S$ be the set of positive integers $k$ such that the two parabolas$$y=x^2-k~~\text{and}~~x=2(y-20)^2-k$$intersect in four distinct points, and these four points lie on a circle with radius at most $21$. Find the sum of the least element of $S$ and the greatest element of $S$.
Let $ABC$ be acute, non-isosceles triangle, inscribed in the circle $(O)$. Let $D$ be perpendicular projection of $A$ onto $BC$, and $E, F$ be perpendicular projections of $D$ onto $CA,AB$ respectively. (a) Prove that $AO \perp EF$. (b) The line $AO$ intersects $DE,DF$ at $I,J$ respectively. Prove that $\vartriangle DIJ$ and $\vartriangle ABC$ are similar. (c) Prove that circumcenter of $\vartriangle DIJ$ is equidistant from $B$ and $C$
Let $ABCD$ be a cyclic quadrilateral whose diagonals $AC$ and $BD$ are perpendicular. Let $O$ be the circumcenter of $ABCD$, $K$ the intersection of the diagonals, $ L\neq O $ the intersection of the circles circumscribed to $OAC$ and $OBD$, and $G$ the intersection of the diagonals of the quadrilateral whose vertices are the midpoints of the sides of $ABCD$. Prove that $O, K, L$ and $G$ are collinear
Let $ABCD$ be a quadrilateral inscribed in the circle $(O)$. Let $(K)$ be an arbitrary circle passing through $B,C$. Circle $(O_1)$ tangent to $AB,AC$ and is internally tangent to $(K)$. Circle $(O_2)$ touches $DB,DC$ and is internally tangent to $(K)$. Prove that one of the two external common tangents of $(O_1)$ and $(O_2)$ is parallel to $AD$. Trần Quang Hùng
Triangle $ABC$ is given with its centroid $G$ and cicumcentre $O$ is such that $GO$ is perpendicular to $AG$. Let $A'$ be the second intersection of $AG$ with circumcircle of triangle $ABC$. Let $D$ be the intersection of lines $CA'$ and $AB$ and $E$ the intersection of lines $BA'$ and $AC$. Prove that the circumcentre of triangle $ADE$ is on the circumcircle of triangle $ABC$.
Let $ABCD$ be a cyclic quadrilateral with circumradius $100\sqrt{3}$ and $AC=300$. If $\angle DBC = 15^{\circ}$, then find $AD^2$. [i]Proposed by Anand Iyer[/i]
Show that the vertices of a regular pentagon are concyclic. If the length of each side of the pentagon is $x$, show that the radius of the circumcircle is $\frac{x}{2\sin 36^\circ}$.
Let $ABC$ be an acute triangle and let $M$ be the midpoint of $AC$. A circle $\omega$ passing through $B$ and $M$ meets the sides $AB$ and $BC$ at points $P$ and $Q$ respectively. Let $T$ be the point such that $BPTQ$ is a parallelogram. Suppose that $T$ lies on the circumcircle of $ABC$. Determine all possible values of $\frac{BT}{BM}$.
Given trapezoid $ ABCD$ with parallel sides $ AB$ and $ CD$, assume that there exist points $ E$ on line $ BC$ outside segment $ BC$, and $ F$ inside segment $ AD$ such that $ \angle DAE \equal{} \angle CBF$. Denote by $ I$ the point of intersection of $ CD$ and $ EF$, and by $ J$ the point of intersection of $ AB$ and $ EF$. Let $ K$ be the midpoint of segment $ EF$, assume it does not lie on line $ AB$. Prove that $ I$ belongs to the circumcircle of $ ABK$ if and only if $ K$ belongs to the circumcircle of $ CDJ$. [i]Proposed by Charles Leytem, Luxembourg[/i]
Given an acute triangle whose circumcenter is $O$.let $K$ be a point on $BC$,different from its midpoint.$D$ is on the extension of segment $AK,BD$ and $AC$,$CD$and$AB$intersect at $N,M$ respectively.prove that $A,B,D,C$ are concyclic.
Let $ABC$ be a triangle such that $\angle A=90^{\circ }$ and $\angle B<\angle C$. The tangent at $A$ to the circumcircle $\omega$ of triangle $ABC$ meets the line $BC$ at $D$. Let $E$ be the reflection of $A$ in the line $BC$, let $X$ be the foot of the perpendicular from $A$ to $BE$, and let $Y$ be the midpoint of the segment $AX$. Let the line $BY$ intersect the circle $\omega$ again at $Z$. Prove that the line $BD$ is tangent to the circumcircle of triangle $ADZ$. [hide="comment"] [i]Edited by Orl.[/i] [/hide]
Let $ABC$ be a triangle with $\angle{C} = 90^{\circ}$, and let $H$ be the foot of the altitude from $C$. A point $D$ is chosen inside the triangle $CBH$ so that $CH$ bisects $AD$. Let $P$ be the intersection point of the lines $BD$ and $CH$. Let $\omega$ be the semicircle with diameter $BD$ that meets the segment $CB$ at an interior point. A line through $P$ is tangent to $\omega$ at $Q$. Prove that the lines $CQ$ and $AD$ meet on $\omega$.
Let $ ABC$ be an acute triangle with $ \angle BAC\equal{}60^{\circ}$. Let $ P$ be a point in triangle $ ABC$ with $ \angle APB\equal{}\angle BPC\equal{}\angle CPA\equal{}120^{\circ}$. The foots of perpendicular from $ P$ to $ BC,CA,AB$ are $ X,Y,Z$, respectively. Let $ M$ be the midpoint of $ YZ$. a) Prove that $ \angle YXZ\equal{}60^{\circ}$ b) Prove that $ X,P,M$ are collinear.
Given an acute triangle $ABC, H$ is the foot of the altitude drawn from the point $A$ on the line $BC, P$ and $K \ne H$ are arbitrary points on the segments $AH$ and$ BC$ respectively. Segments $AC$ and $BP$ intersect at point $B_1$, lines $AB$ and $CP$ at point $C_1$. Let $X$ and $Y$ be the projections of point $H$ on the lines $KB_1$ and $KC_1$, respectively. Prove that points $A, P, X$ and $Y$ lie on one circle.
Let $ABC$ be a triangle where $\angle BAC = 30^\circ$. Construct $D$ in $\triangle ABC$ such that $\angle ABD = \angle ACD = 30^\circ$. Let the circumcircle of $\triangle ABD$ intersect $AC$ at $X$. Let the circumcircle of $\triangle ACD$ intersect $AB$ at $Y$. Given that $DB - DC = 10$ and $BC = 20$, find $AX \cdot AY$.