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

A point $D$ is chosen on the side $AC$ of a triangle $ABC$ with $\angle C < \angle A < 90^\circ$ in such a way that $BD=BA$. The incircle of $ABC$ is tangent to $AB$ and $AC$ at points $K$ and $L$, respectively. Let $J$ be the incenter of triangle $BCD$. Prove that the line $KL$ intersects the line segment $AJ$ at its midpoint.
Let $M$ be the incenter of the tangential quadrilateral $A_1A_2A_3A_4$. Let line $g_1$ through $A_1$ be perpendicular to $A_1M$; define $g_2,g_3$ and $g_4$ similarly. The lines $g_1,g_2,g_3$ and $g_4$ define another quadrilateral $B_1B_2B_3B_4$ having $B_1$ be the intersection of $g_1$ and $g_2$; similarly $B_2,B_3$ and $B_4$ are intersections of $g_2$ and $g_3$, $g_3$ and $g_4$, resp. $g_4$ and $g_1$. Prove that the diagonals of quadrilateral $B_1B_2B_3B_4$ intersect in point $M$. [asy] import graph; size(15cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-9.773972777861085,xmax=12.231603726660566,ymin=-3.9255487671791487,ymax=7.37238601960895; pair M=(2.,2.), A_4=(-1.6391623316400197,1.2875505916864178), A_1=(3.068893183992864,-0.5728665455336459), A_2=(4.30385937824148,2.2922812065339455), A_3=(2.221541124684679,4.978916319940133), B_4=(-0.9482172571022687,-2.24176848577888), B_1=(4.5873184669543345,0.057960746374459436), B_2=(3.9796042717514277,4.848169684238838), B_3=(-2.4295496490492385,5.324816563638236); draw(circle(M,2.),linewidth(0.8)); draw(A_4--A_1,linewidth(0.8)); draw(A_1--A_2,linewidth(0.8)); draw(A_2--A_3,linewidth(0.8)); draw(A_3--A_4,linewidth(0.8)); draw(M--A_3,linewidth(0.8)+dotted); draw(M--A_2,linewidth(0.8)+dotted); draw(M--A_1,linewidth(0.8)+dotted); draw(M--A_4,linewidth(0.8)+dotted); draw((xmin,-0.07436970390935019*xmin+5.144131675605378)--(xmax,-0.07436970390935019*xmax+5.144131675605378),linewidth(0.8)); draw((xmin,-7.882338401302275*xmin+36.2167572574517)--(xmax,-7.882338401302275*xmax+36.2167572574517),linewidth(0.8)); draw((xmin,0.4154483588930812*xmin-1.847833182441644)--(xmax,0.4154483588930812*xmax-1.847833182441644),linewidth(0.8)); draw((xmin,-5.107958950031516*xmin-7.085223310768749)--(xmax,-5.107958950031516*xmax-7.085223310768749),linewidth(0.8)); dot(M,linewidth(3.pt)+ds); label("$M$",(2.0593440948136896,2.0872038897020024),NE*lsf); dot(A_4,linewidth(3.pt)+ds); label("$A_4$",(-2.6355449660387147,1.085078446888477),NE*lsf); dot(A_1,linewidth(3.pt)+ds); label("$A_1$",(3.1575637581709772,-1.2486383377457595),NE*lsf); dot(A_2,linewidth(3.pt)+ds); label("$A_2$",(4.502882845783654,2.30684782237346),NE*lsf); dot(A_3,linewidth(3.pt)+ds); label("$A_3$",(2.169166061149418,5.203402184478307),NE*lsf); label("$g_3$",(-9.691606303109287,5.354407388189934),NE*lsf); label("$g_2$",(3.0889250292111465,6.727181967386543),NE*lsf); label("$g_1$",(-4.763345563793459,-3.4725331560442676),NE*lsf); label("$g_4$",(-2.663000457622647,6.878187171098171),NE*lsf); dot(B_4,linewidth(3.pt)+ds); label("$B_4$",(-1.5647807942653595,-3.0332452907013523),NE*lsf); dot(B_1,linewidth(3.pt)+ds); label("$B_1$",(4.955898456918535,-0.6583452686912173),NE*lsf); dot(B_2,linewidth(3.pt)+ds); label("$B_2$",(4.104778217816637,5.0661247265586455),NE*lsf); dot(B_3,linewidth(3.pt)+ds); label("$B_3$",(-3.4454819677647146,5.656417795613188),NE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]
A bicentric quadrilateral is one that is both inscribable in and circumscribable about a circle, i.e. both the incircle and circumcircle exists. Show that for such a quadrilateral, the centers of the two associated circles are collinear with the point of intersection of the diagonals.
A cyclic quadrilateral $ABCD$ has diameter $AC$ with circumcircle $\omega$. Let $K$ be the foot of the perpendicular from $C$ to $BD$, and the tangent to $\omega$ at $A$ meets $BD$ at $T$. Let the line $AK$ meets $\omega$ at $X$ and choose a point $Y$ on line $AK$ such that $\angle TYA=90^{\circ}$. Prove that $AY=KX$. [i]Proposed by Anzo Teh Zhao Yang[/i]
A chord $CD$ of a circle with center $O$ is perpendicular to a diameter $AB$. A chord $AE$ bisects the radius $OC$. Show that the line $DE$ bisects the chord $BC$ [i]V. Gordon[/i]
Let $ABCD$ be a cyclic quadrilateral, and suppose that $BC = CD = 2$. Let $I$ be the incenter of triangle $ABD$. If $AI = 2$ as well, find the minimum value of the length of diagonal $BD$.
$ABC$ is an isosceles triangle. The base $BC$ is $1$ cm long, and legs $AB$ and $AC$ are $2$ cm long. Let the midpoint of $AB$ be $F$, and the midpoint of $AC$ be $G$. Additionally, $k$ is a circle, that is tangent to $AB$ and A$C$, and it’s points of tangency are $F$ and $G$ accordingly. Prove, that the intersection of $CF$ and $BG$ falls on the circle $k$.
Let $O$ be the circumcenter of acute $\triangle ABC$($AB<AC$), the angle bisector of $\angle BAC$ meets $BC$ at $T$ and $M$ is the midpoint of $AT$. Point $P$ lies inside $\triangle ABC$ such that $PB\perp PC$. $D,E$ distinct from $P$ lies on the perpendicular to $AP$ through $P$ such that $BD=BP, CE=CP$. If $AO$ bisects segment $DE$, prove that $AO$ is tangent to the circumcircle of $\triangle AMP$.
Let $ABCD$ be a cyclic quadrilateral. Let $ P$ be the intersection of the lines $BC$ and $AD$. Line $AC$ cuts the circumscribed circle of the triangle $BDP$ in $S$ and $T$, with $S$ between $ A$ and $C$. The line $BD$ intersects the circumscribed circle of the triangle $ACP$ in $U$ and $V$, with $U$ between $ B$ and $D$. Prove that $PS = PT = PU = PV$.
Let $n\ge 4$ be a positive integer and let $M$ be a set of $n$ points in the plane, where no three points are collinear and not all of the $n$ points being concyclic. Find all real functions $f:M\to\mathbb{R}$ such that for any circle $\mathcal{C}$ containing at least three points from $M$, the following equality holds: \[\sum_{P\in\mathcal{C}\cap M} f(P)=0\] [i]Dorel Mihet[/i]
Given a triangle $ABC$, let $I$ be the incenter, and $J$ be the $A$-excenter. A line $\ell$ through $A$ perpendicular to $BC$ intersect the lines $BI$, $CI$, $BJ$, $CJ$ at $P$, $Q$, $R$, $S$ respectively. Suppose the angle bisector of $\angle BAC$ meet $BC$ at $K$, and $L$ is a point such that $AL$ is a diameter in $(ABC)$. Prove that the line $KL$, $\ell$, and the line through the centers of circles $(IPQ)$ and $(JRS)$, are concurrent. [i]Proposed by Chuah Jia Herng & Ivan Chan Kai Chin[/i]
Let $B$ be a point on a circle $S_1$, and let $A$ be a point distinct from $B$ on the tangent at $B$ to $S_1$. Let $C$ be a point not on $S_1$ such that the line segment $AC$ meets $S_1$ at two distinct points. Let $S_2$ be the circle touching $AC$ at $C$ and touching $S_1$ at a point $D$ on the opposite side of $AC$ from $B$. Prove that the circumcentre of triangle $BCD$ lies on the circumcircle of triangle $ABC$.
Let $ABC$ be an isosceles triangle with $BC=CA$, and let $D$ be a point inside side $AB$ such that $AD< DB$. Let $P$ and $Q$ be two points inside sides $BC$ and $CA$, respectively, such that $\angle DPB = \angle DQA = 90^{\circ}$. Let the perpendicular bisector of $PQ$ meet line segment $CQ$ at $E$, and let the circumcircles of triangles $ABC$ and $CPQ$ meet again at point $F$, different from $C$. Suppose that $P$, $E$, $F$ are collinear. Prove that $\angle ACB = 90^{\circ}$.
The incircles of the faces $ABC$ and $ABD$ of a tetrahedron $ABCD$ are tangent to the edge $AB$ in the same point. Prove that the points of tangency of these incircles to the edges $AC,BC,AD,BD$ are concyclic.
Given a triangle $ABC$. The points $A$, $B$, $C$ divide the circumcircle $\Omega$ of the triangle $ABC$ into three arcs $BC$, $CA$, $AB$. Let $X$ be a variable point on the arc $AB$, and let $O_{1}$ and $O_{2}$ be the incenters of the triangles $CAX$ and $CBX$. Prove that the circumcircle of the triangle $XO_{1}O_{2}$ intersects the circle $\Omega$ in a fixed point.
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 $PQ \perp AC$.
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]
Let $\triangle ABC$ be an acute triangle with circumcenter $O$, and let $Q\neq A$ denote the point on $\odot (ABC)$ for which $AQ\perp BC$. The circumcircle of $\triangle BOC$ intersects lines $AC$ and $AB$ for the second time at $D$ and $E$ respectively. Suppose that $AQ$, $BC$, and $DE$ are concurrent. If $OD=3$ and $OE=7$, compute $AQ$.
Let $k$ be the circle and $A$ and $B$ points on circle which are not diametrically opposite. On minor arc $AB$ lies point arbitrary point $C$. Let $D$, $E$ and $F$ be foots of perpendiculars from $C$ on chord $AB$ and tangents of circle $k$ in points $A$ and $B$. Prove that $CD= \sqrt {CE \cdot CF}$
Let $\Gamma$ be the circumcircle of $\triangle ABC$. Let $D$ be a point on the side $BC$. The tangent to $\Gamma$ at $A$ intersects the parallel line to $BA$ through $D$ at point $E$. The segment $CE$ intersects $\Gamma$ again at $F$. Suppose $B$, $D$, $F$, $E$ are concyclic. Prove that $AC$, $BF$, $DE$ are concurrent.
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 $Z \ne A$ be a point on the line $AB$ with $AB = BZ$. Let $(c)$ be the circumcircle of the triangle $\triangle AEZ$. Let $D$ be the second point of intersection of $(c)$ with $ZC$ and let $F$ be the antidiametric point of $D$ with respect to $(c)$. Let $P$ be the point of intersection of the lines $FE$ and $CZ$. If the tangent to $(c)$ at $Z$ meets $PA$ at $T$, prove that the points $T$, $E$, $B$, $Z$ are concyclic. Proposed by [i]Theoklitos Parayiou, Cyprus[/i]
$E$ is the intersection point of the diagonals of the cyclic quadrilateral, $ABCD, F$ is the intersection point of the lines $AB$ and $CD, M$ is the midpoint of the side $AB$, and $N$ is the midpoint of the side $CD$. The circles circumscribed around the triangles $ABE$ and $ACN$ intersect for the second time at point $K$. Prove that the points $F, K, M$ and $N$ lie on one circle.
Let $\triangle{ABC}$ be an acute triangle with orthocenter $H.$ Points $E$ and $F$ are on segments $\overline{AC}$ and $\overline{AB},$ respectively, such that $\angle{EHF}=90^\circ.$ Let $X$ be the foot of the perpendicular from $H$ to $\overline{EF}.$ Prove that $\angle{BXC}=90^\circ.$
Let $ w $ be the incircle of triangle $ ABC $. Segments $ BC, CA $ meet with $ w $ at points $ D, E$. A line passing through $ B $ and parallel to $ DE $ meets $ w $ at $ F $ and $ G $. ($ F $ is nearer to $ B $ than $ G $.) Line $ CG $ meets $ w $ at $ H ( \ne G ) $. A line passing through $ G $ and parallel to $ EH $ meets with line $ AC $ at $ I $. Line $ IF $ meets with circle $ w $ at $ J (\ne F ) $. Lines $ CJ $ and $ EG $ meets at $ K $. Let $ l $ be the line passing through $ K $ and parallel to $ JD $. Prove that $ l, IF, ED $ meet at one point.
Let $ABC$ be an isosceles triangle with $BC=CA$, and let $D$ be a point inside side $AB$ such that $AD< DB$. Let $P$ and $Q$ be two points inside sides $BC$ and $CA$, respectively, such that $\angle DPB = \angle DQA = 90^{\circ}$. Let the perpendicular bisector of $PQ$ meet line segment $CQ$ at $E$, and let the circumcircles of triangles $ABC$ and $CPQ$ meet again at point $F$, different from $C$. Suppose that $P$, $E$, $F$ are collinear. Prove that $\angle ACB = 90^{\circ}$.