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: 3585

Let $ABCD$ be a rectangle. Points $E$ and $F$ are on diagonal $AC$ such that $F$ lies between $A$ and $E$; and $E$ lies between $C$ and $F$. The circumcircle of triangle $BEF$ intersects $AB$ and $BC$ at $G$ and $H$ respectively, and the circumcircle of triangle $DEF$ intersects $AD$ and $CD$ at $I$ and $J$ respectively. Prove that the lines $GJ, IH$ and $AC$ concur at a point.
Let $\omega$ be the circumcircle of an actue triangle $ABC$ and let $H$ be the feet of aliitude from $A$ to $BC$. Let $M$ and $N$ be the midpoints of the sides $AC$ and $AB$. The lines $BM$ and $CN$ intersect each other at $G$ and intersect $\omega$ at $P$ and $Q$ respectively. The circles $(HMG)$ and $(HNG)$ intersect the segments $HP$ and $HQ$ again at $R$ and $S$ respectively. Prove that $PQ\parallel RS$.
Let $P$ be a point inside triangle $ABC$. Let $AP$ meet $BC$ at $A_1$, let $BP$ meet $CA$ at $B_1$, and let $CP$ meet $AB$ at $C_1$. Let $A_2$ be the point such that $A_1$ is the midpoint of $PA_2$, let $B_2$ be the point such that $B_1$ is the midpoint of $PB_2$, and let $C_2$ be the point such that $C_1$ is the midpoint of $PC_2$. Prove that points $A_2, B_2$, and $C_2$ cannot all lie strictly inside the circumcircle of triangle $ABC$. (Australia)
Let $ABC$ be an acute scalene triangle with orthocenter $H$. Let $M$ be the midpoint of $BC$. $N$ is the point on line $AM$ such that $(BMN)$ is tangent to $AB$. Finally, let $H'$ be the reflection of $H$ in $B$. Prove that $\angle ANH'=90^{\circ}$. [i]Proposed by Malay Mahajan and Siddharth Choppara[/i]
Let $ABCD$ be a parallelogram and $P$ be a point on diagonal $BD$ such that $\angle PCB=\angle ACD$. Circumcircle of triangle $ABD$ intersects line $AC$ at points $A$ and $E$. Prove that \[\angle AED=\angle PEB.\]
In triangle $ABC$, $AB=AC$ and $B\hat{A}C=100^\circ$. Let $D$ be on $AC$ such that $A\hat{B}D=C\hat{B}D$. Prove that $AD+DB=BC$.
Let the area and the perimeter of a cyclic quadrilateral $C$ be $A_C$ and $P_C$, respectively. If the area and the perimeter of the quadrilateral which is tangent to the circumcircle of $C$ at the vertices of $C$ are $A_T$ and $P_T$ , respectively, prove that $\frac{A_C}{A_T} \geq \left (\frac{P_C}{P_T}\right )^2$.
Let $ABC$ be a nondegenerate acute triangle with circumcircle $\omega$ and let its incircle $\gamma$ touch $AB, AC, BC$ at $X, Y, Z$ respectively. Let $XY$ hit arcs $AB, AC$ of $\omega$ at $M, N$ respectively, and let $P \neq X, Q \neq Y$ be the points on $\gamma$ such that $MP=MX, NQ=NY$. If $I$ is the center of $\gamma$, prove that $P, I, Q$ are collinear if and only if $\angle BAC=90^\circ$. [i]Proposed by David Stoner[/i]
Let $ABC$ be a triangle, and let $D$, $A$, $B$, $E$ be points on line $AB$, in that order, such that $AC=AD$ and $BE=BC$. Let $\omega_1, \omega_2$ be the circumcircles of $\triangle ABC$ and $\triangle CDE$, respectively, which meet at a point $F \neq C$. If the tangent to $\omega_2$ at $F$ cuts $\omega_1$ again at $G$, and the foot of the altitude from $G$ to $FC$ is $H$, prove that $\angle AGH=\angle BGH$. [i]Proposed by David Stoner[/i]
Let $ABCD$ be a convex quadrilateral and let $P$ and $Q$ be variable points inside this quadrilateral so that $\angle APB=\angle CPD=\angle AQB=\angle CQD$. Prove that the lines $PQ$ obtained in this way all pass through a fixed point , or they are all parallel.
Let $\Delta ABC$ be an acute-angled triangle with $AB < AC$, $H$ be a point on $BC$ such that $AH\ \bot BC$ and $G$ be the centroid of $\Delta ABC$. Let $P,Q$ be the point of tangency of the inscribed circle of $\Delta ABC$ with $AB,AC$, correspondingly. Define $M,N$ to be the midpoint of $PB,QC$, correspondingly. Let $D,E$ be points on the inscribed circle of $\Delta ABC$ such that $\angle BDH + \angle ABC = 180^{\circ}$, $\angle CEH + \angle ACB = 180^{\circ}$. Prove that lines $MD,NE,HG$ share a common point.
In acute triangle $ABC, \angle A = 45^o$. Points $O,H$ are the circumcenter and the orthocenter of $ABC$, respectively. $D$ is the foot of altitude from $B$. Point $X$ is the midpoint of arc $AH$ of the circumcircle of triangle $ADH$ that contains $D$. Prove that $DX = DO$. Proposed by Fatemeh Sajadi
Let $ABCD$ be a convex quadrilateral with $\angle ABC = \angle ADC < 90^{\circ}$. The internal angle bisectors of $\angle ABC$ and $\angle ADC$ meet $AC$ at $E$ and $F$ respectively, and meet each other at point $P$. Let $M$ be the midpoint of $AC$ and let $\omega$ be the circumcircle of triangle $BPD$. Segments $BM$ and $DM$ intersect $\omega$ again at $X$ and $Y$ respectively. Denote by $Q$ the intersection point of lines $XE$ and $YF$. Prove that $PQ \perp AC$.
$ABC$ is a triangle and $R,Q,P$ are midpoints of $AB,AC,BC$. Line $AP$ intersects $RQ$ in $E$ and circumcircle of $ABC$ in $F$. $T,S$ are on $RP,PQ$ such that $ES\perp PQ,ET\perp RP$. $F'$ is on circumcircle of $ABC$ that $FF'$ is diameter. The point of intersection of $AF'$ and $BC$ is $E'$. $S',T'$ are on $AB,AC$ that $E'S'\perp AB,E'T'\perp AC$. Prove that $TS$ and $T'S'$ are perpendicular.
Suppose $\triangle ABC$ has angles $\angle BAC = 84^\circ, \angle ABC=60^\circ,$ and $\angle ACB = 36^\circ.$ Let $D, E,$ and $F$ be the midpoints of sides $\overline{BC}, \overline{AC},$ and $\overline{AB},$ respectively. The circumcircle of $\triangle DEF$ intersects $\overline{BD}, \overline{AE},$ and $\overline{AF}$ at points $G, H,$ and $J,$ respectively. The points $G, D, E, H, J,$ and $F$ divide the circumcircle of $\triangle DEF$ into six minor arcs, as shown. Find $\overarc{DE}+2\cdot \overarc{HJ} + 3\cdot \overarc{FG},$ where the arcs are measured in degrees. [asy] import olympiad; size(6cm); defaultpen(fontsize(10pt)); pair B = (0, 0), A = (Cos(60), Sin(60)), C = (Cos(60)+Sin(60)/Tan(36), 0), D = midpoint(B--C), E = midpoint(A--C), F = midpoint(A--B); guide circ = circumcircle(D, E, F); pair G = intersectionpoint(B--D, circ), J = intersectionpoints(A--F, circ)[0], H = intersectionpoints(A--E, circ)[0]; draw(B--A--C--cycle); draw(D--E--F--cycle); draw(circ); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); dot(G); dot(H); dot(J); label("$A$", A, (0, .8)); label("$B$", B, (-.8, -.8)); label("$C$", C, (.8, -.8)); label("$D$", D, (0, -.8)); label("$E$", E, (.8, .2)); label("$F$", F, (-.8, .2)); label("$G$", G, (0, .8)); label("$H$", H, (-.2, -1)); label("$J$", J, (.2, -.8)); [/asy]
Triangle $ \triangle ABC$ is given. Points $ D$ i $ E$ are on line $ AB$ such that $ D \minus{} A \minus{} B \minus{} E, AD \equal{} AC$ and $ BE \equal{} BC$. Bisector of internal angles at $ A$ and $ B$ intersect $ BC,AC$ at $ P$ and $ Q$, and circumcircle of $ ABC$ at $ M$ and $ N$. Line which connects $ A$ with center of circumcircle of $ BME$ and line which connects $ B$ and center of circumcircle of $ AND$ intersect at $ X$. Prove that $ CX \perp PQ$.
Let $ABCD$ be a cyclic quadrilateral which is not a trapezoid and whose diagonals meet at $E$. The midpoints of $AB$ and $CD$ are $F$ and $G$ respectively, and $\ell$ is the line through $G$ parallel to $AB$. The feet of the perpendiculars from E onto the lines $\ell$ and $CD$ are $H$ and $K$, respectively. Prove that the lines $EF$ and $HK$ are perpendicular.
Let $ABCD$ be a convex quadrilateral with $\angle ABC>90$, $CDA>90$ and $\angle DAB=\angle BCD$. Denote by $E$ and $F$ the reflections of $A$ in lines $BC$ and $CD$, respectively. Suppose that the segments $AE$ and $AF$ meet the line $BD$ at $K$ and $L$, respectively. Prove that the circumcircles of triangles $BEK$ and $DFL$ are tangent to each other. $\emph{Slovakia}$
Let $ABC$ be a scalene triangle with incenter $I$ and circumcircle $\omega$. The internal and external bisectors of angle $\angle BAC$ intersect $BC$ at $D$ and $E$, respectively. Let $M$ be the point on segment $AC$ such that $MC = MB$. The tangent to $\omega$ at $B$ meets $MD$ at $S$. The circumcircles of triangles $ADE$ and $BIC$ intersect each other at $P$ and $Q$. If $AS$ meets $\omega$ at a point $K$ other than $A$, prove that $K$ lies on $PQ$. [i]Proposed by Alexandru Lopotenco (Moldova)[/i]
In triangle $ABC$, $AB = AC = 3$ and $\angle A = 90^o$. Let $M$ be the midpoint of side $BC$. Points $D$ and $E$ lie on sides $AC$ and $AB$ respectively such that $AD > AE$ and $ADME$ is a cyclic quadrilateral. Given that triangle $EMD$ has area $2$, find the length of segment $CD$.
Let $\triangle ABC$ be a right-angled triangle with $\angle BAC = 90^{\circ}$, and let $E$ be the foot of the perpendicular from $A$ to $BC$. Let $Z \neq A$ be a point on the line $AB$ with $AB = BZ$. Let $(c)$ and $(c_1)$ be the circumcircles of the triangles $\triangle AEZ$ and $\triangle BEZ$, respectively. Let $(c_2)$ be an arbitrary circle passing through the points $A$ and $E$. Suppose $(c_1)$ meets the line $CZ$ again at the point $F$, and meets $(c_2)$ again at the point $N$. If $P$ is the other point of intersection of $(c_2)$ with $AF$, prove that the points $N$, $B$, $P$ are collinear.
Let $\triangle ABC$ be a scalene triangle, and let $D$ and $E$ be points on sides $AB$ and $AC$ respectively such that the circumcircles of triangles $\triangle ACD$ and $\triangle ABE$ are tangent to $BC$. Let $F$ be the intersection point of $BC$ and $DE$. Prove that $AF$ is perpendicular to the Euler line of $\triangle ABC$.
Cyclic quadrilateral $ABCD$ has circumcircle $(O)$. Points $M$ and $N$ are the midpoints of $BC$ and $CD$, and $E$ and $F$ lie on $AB$ and $AD$ respectively such that $EF$ passes through $O$ and $EO=OF$. Let $EN$ meet $FM$ at $P$. Denote $S$ as the circumcenter of $\triangle PEF$. Line $PO$ intersects $AD$ and $BA$ at $Q$ and $R$ respectively. Suppose $OSPC$ is a parallelogram. Prove that $AQ=AR$.
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]
Through the midpoint $M$ of the side $BC$ of the triangle $ABC$ passes a line which intersects the rays $AB$ and $AC$ at $D$ and $E$, respectively, such that $AD=AE$. Let $F$ be the foot of the perpendicular from $A$ onto $BC$ and $P{}$ the circumcenter of triangle $ADE$. Prove that $PF=PM$.