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: 433

Let $\triangle{ABC}$ be a triangle, $D$ the midpoint of $BC$, and $M$ be the midpoint of $AD$. The line $BM$ intersects the side $AC$ on the point $N$. Show that $AB$ is tangent to the circuncircle to the triangle $\triangle{NBC}$ if and only if the following equality is true: \[\frac{{BM}}{{MN}} =\frac{({BC})^2}{({BN})^2}.\]
The quadrilateral $ ABCD$ inscribed in a circle wich has diameter $ BD$. Let $ A',B'$ are symmetric to $ A,B$ with respect to the line $ BD$ and $ AC$ respectively. If $ A'C \cap BD \equal{} P$ and $ AC\cap B'D \equal{} Q$ then prove that $ PQ \perp AC$
Convex circumscribed quadrilateral $ABCD$ with its incenter $I$ is given such that its incircle is tangent to $\overline{AD},\overline{DC},\overline{CB},$ and $\overline{BA}$ at $K,L,M,$ and $N$. Lines $\overline{AD}$ and $\overline{BC}$ meet at $E$ and lines $\overline{AB}$ and $\overline{CD}$ meet at $F$. Let $\overline{KM}$ intersects $\overline{AB}$ and $\overline{CD}$ at $X,Y$, respectively. Let $\overline{LN}$ intersects $\overline{AD}$ and $\overline{BC}$ at $Z,T$, respectively. Prove that the circumcircle of triangle $\triangle XFY$ and the circle with diameter $EI$ are tangent if and only if the circumcircle of triangle $\triangle TEZ$ and the circle with diameter $FI$ are tangent. [i]Proposed by Mahdi Etesamifard[/i]
Let $ABC$ be an acute triangle with altitude $\overline{AH}$, and let $P$ be a variable point such that the angle bisectors $k$ and $\ell$ of $\angle PBC$ and $\angle PCB$, respectively, meet on $\overline{AH}$. Let $k$ meet $\overline{AC}$ at $E$, $\ell$ meet $\overline{AB}$ at $F$, and $\overline{EF}$ meet $\overline{AH}$ at $Q$. Prove that as $P$ varies, line $PQ$ passes through a fixed point.
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 a triangle. Points $D$ and $E$ lie on sides $BC$ and $CA$, respectively, such that $BD=AE$. Segments $AD$ and $BE$ meet at $P$. The bisector of angle $BCA$ meet segments $AD$ and $BE$ at $Q$ and $R$, respectively. Prove that $\frac{PQ}{AD}=\frac{PR}{BE}$.
Let $ ABCD$ be a convex quadrilateral. Points $ P,Q$ and $ R$ are the feets of the perpendiculars from point $ D$ to lines $ BC, CA$ and $ AB$, respectively. Prove that $ PQ\equal{}QR$ if and only if the bisectors of the angles $ ABC$ and $ ADC$ meet on segment $ AC$.
The circle $\omega$ with center $O$ has given. From an arbitrary point $T$ outside of $\omega$ draw tangents $TB$ and $TC$ to it. $K$ and $H$ are on $TB$ and $TC$ respectively. [b]a)[/b] $B'$ and $C'$ are the second intersection point of $OB$ and $OC$ with $\omega$ respectively. $K'$ and $H'$ are on angle bisectors of $\angle BCO$ and $\angle CBO$ respectively such that $KK' \bot BC$ and $HH'\bot BC$. Prove that $K,H',B'$ are collinear if and only if $H,K',C'$ are collinear. [b]b)[/b] Consider there exist two circle in $TBC$ such that they are tangent two each other at $J$ and both of them are tangent to $\omega$.and one of them is tangent to $TB$ at $K$ and other one is tangent to $TC$ at $H$. Prove that two quadrilateral $BKJI$ and $CHJI$ are cyclic ($I$ is incenter of triangle $OBC$).
In $\triangle ABC$, consider points $D$ and $E$ on $AC$ and $AB$, respectively, and assume that they do not coincide with any of the vertices $A$, $B$, $C$. If the segments $BD$ and $CE$ intersect at $F$, consider areas $w$, $x$, $y$, $z$ of the quadrilateral $AEFD$ and the triangles $BEF$, $BFC$, $CDF$, respectively. [list=a] [*] Prove that $y^2 > xz$. [*] Determine $w$ in terms of $x$, $y$, $z$. [/list] [asy] import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(12); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -2.8465032978885407, xmax = 9.445649196374966, ymin = -1.7618066305534972, ymax = 4.389732795464592; /* image dimensions */ draw((3.8295013012181283,2.816337276198864)--(-0.7368327629589799,-0.5920813291311117)--(5.672613975760373,-0.636902634996282)--cycle, linewidth(0.5)); /* draw figures */ draw((3.8295013012181283,2.816337276198864)--(-0.7368327629589799,-0.5920813291311117), linewidth(0.5)); draw((-0.7368327629589799,-0.5920813291311117)--(5.672613975760373,-0.636902634996282), linewidth(0.5)); draw((5.672613975760373,-0.636902634996282)--(3.8295013012181283,2.816337276198864), linewidth(0.5)); draw((-0.7368327629589799,-0.5920813291311117)--(4.569287648059735,1.430279997142299), linewidth(0.5)); draw((5.672613975760373,-0.636902634996282)--(1.8844000180622977,1.3644681598392678), linewidth(0.5)); label("$y$",(2.74779188172294,0.23771684184669772),SE*labelscalefactor); label("$w$",(3.2941097703568736,1.8657441499758196),SE*labelscalefactor); label("$x$",(1.6660824622277512,1.0025618859342047),SE*labelscalefactor); label("$z$",(4.288408327670633,0.8168138037986672),SE*labelscalefactor); /* dots and labels */ dot((3.8295013012181283,2.816337276198864),dotstyle); label("$A$", (3.8732067323088435,2.925600853925651), NE * labelscalefactor); dot((-0.7368327629589799,-0.5920813291311117),dotstyle); label("$B$", (-1.1,-0.7565817154670613), NE * labelscalefactor); dot((5.672613975760373,-0.636902634996282),dotstyle); label("$C$", (5.763466626982254,-0.7784344310124186), NE * labelscalefactor); dot((4.569287648059735,1.430279997142299),dotstyle); label("$D$", (4.692683565259744,1.5051743434774234), NE * labelscalefactor); dot((1.8844000180622977,1.3644681598392678),dotstyle); label("$E$", (1.775346039954538,1.4942479857047448), NE * labelscalefactor); dot((2.937230516274804,0.8082418657164665),linewidth(4.pt) + dotstyle); label("$F$", (2.889834532767763,0.954), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]
In acute triangle $ABC$, $\angle{A}<\angle{B}$ and $\angle{A}<\angle{C}$. Let $P$ be a variable point on side $BC$. Points $D$ and $E$ lie on sides $AB$ and $AC$, respectively, such that $BP=PD$ and $CP=PE$. Prove that as $P$ moves along side $BC$, the circumcircle of triangle $ADE$ passes through a fixed point other than $A$.
Let $ABCD$ be a convex quadrilateral, and let $P$, $Q$, $R$, and $S$ be points on the sides $AB$, $BC$, $CD$, and $DA$, respectively. Let the line segment $PR$ and $QS$ meet at $O$. Suppose that each of the quadrilaterals $APOS$, $BQOP$, $CROQ$, and $DSOR$ has an incircle. Prove that the lines $AC$, $PQ$, and $RS$ are either concurrent or parallel to each other.
Given an acute scalene triangle $ABC$ inscribed in circle $(O)$. Let $H$ be its orthocenter and $M$ be the midpoint of $BC$. Let $D$ lie on the opposite rays of $HA$ so that $BC=2DM$. Let $D'$ be the reflection of $D$ through line $BC$ and $X$ be the intersection of $AO$ and $MD$. a) Show that $AM$ bisects $D'X$. b) Similarly, we define the points $E,F$ like $D$ and $Y,Z$ like $X$. Let $S$ be the intersection of tangent lines from $B,C$ with respect to $(O)$. Let $G$ be the projection of the midpoint of $AS$ to the line $AO$. Show that there exists a point with the same power to all the circles $(BEY),(CFZ),(SGO)$ and $(O)$.
Let $\triangle ABC$ be a triangle. Let $M$ be the midpoint of $BC$ and let $D$ be a point on the interior of side $AB$. The intersection of $AM$ and $CD$ is called $E$. Suppose that $|AD|=|DE|$. Prove that $|AB|=|CE|$.
The incircle of a triangle $ABC$ touches the side $AB$ and $AC$ at respectively at $X$ and $Y$. Let $K$ be the midpoint of the arc $\widehat{AB}$ on the circumcircle of $ABC$. Assume that $XY$ bisects the segment $AK$. What are the possible measures of angle $BAC$?
Let $ABC$ be a triangle with $CA \neq CB$. Let $D$, $F$, and $G$ be the midpoints of the sides $AB$, $AC$, and $BC$ respectively. A circle $\Gamma$ passing through $C$ and tangent to $AB$ at $D$ meets the segments $AF$ and $BG$ at $H$ and $I$, respectively. The points $H'$ and $I'$ are symmetric to $H$ and $I$ about $F$ and $G$, respectively. The line $H'I'$ meets $CD$ and $FG$ at $Q$ and $M$, respectively. The line $CM$ meets $\Gamma$ again at $P$. Prove that $CQ = QP$. [i]Proposed by El Salvador[/i]
Triangle $ ABC$ has $ AC \equal{} 450$ and $ BC \equal{} 300$. Points $ K$ and $ L$ are located on $ \overline{AC}$ and $ \overline{AB}$ respectively so that $ AK \equal{} CK$, and $ \overline{CL}$ is the angle bisector of angle $ C$. Let $ P$ be the point of intersection of $ \overline{BK}$ and $ \overline{CL}$, and let $ M$ be the point on line $ BK$ for which $ K$ is the midpoint of $ \overline{PM}$. If $ AM \equal{} 180$, find $ LP$.
In a trapezoid $ABCD$ with $AB<CD$ and $AB \parallel CD$, the diagonals intersect each other at $E$. Let $F$ be the midpoint of the arc $BC$ (not containing the point $E$) of the circumcircle of the triangle $EBC$. The lines $EF$ and $BC$ intersect at $G$. The circumcircle of the triangle $BFD$ intersects the ray $[DA$ at $H$ such that $A \in [HD]$. The circumcircle of the triangle $AHB$ intersects the lines $AC$ and $BD$ at $M$ and $N$, respectively. $BM$ intersects $GH$ at $P$, $GN$ intersects $AC$ at $Q$. Prove that the points $P, Q, D$ are collinear.
In an acute scalene triangle $ABC$, points $D,E,F$ lie on sides $BC, CA, AB$, respectively, such that $AD \perp BC, BE \perp CA, CF \perp AB$. Altitudes $AD, BE, CF$ meet at orthocenter $H$. Points $P$ and $Q$ lie on segment $EF$ such that $AP \perp EF$ and $HQ \perp EF$. Lines $DP$ and $QH$ intersect at point $R$. Compute $HQ/HR$. [i]Proposed by Zuming Feng[/i]
Triangle $ABC$ circumscribed $(O)$ has $A$-excircle $(J)$ that touches $AB,\ BC,\ AC$ at $F,\ D,\ E$, resp. a. $L$ is the midpoint of $BC$. Circle with diameter $LJ$ cuts $DE,\ DF$ at $K,\ H$. Prove that $(BDK),\ (CDH)$ has an intersecting point on $(J)$. b. Let $EF\cap BC =\{G\}$ and $GJ$ cuts $AB,\ AC$ at $M,\ N$, resp. $P\in JB$ and $Q\in JC$ such that $$\angle PAB=\angle QAC=90{}^\circ .$$ $PM\cap QN=\{T\}$ and $S$ is the midpoint of the larger $BC$-arc of $(O)$. $(I)$ is the incircle of $ABC$. Prove that $SI\cap AT\in (O)$.
Let triangle $ABC$ with $AB<AC$ has orthocenter $H$, and let the midpoint of $BC$ be $M$. The internal angle bisector of $\angle BAC$ meet $CH$ at $X$, and the external angle bisector of $\angle BAC$ meet $BH$ at $Y$. The circles $(BHX)$ and $(CHY)$ meet again at $Z$. Prove that $\angle HZM=90^{\circ}$. [i]Proposed by Ivan Chan Kai Chin[/i]
In $\triangle ABC$, $AB = 360$, $BC = 507$, and $CA = 780$. Let $M$ be the midpoint of $\overline{CA}$, and let $D$ be the point on $\overline{CA}$ such that $\overline{BD}$ bisects angle $ABC$. Let $F$ be the point on $\overline{BC}$ such that $\overline{DF} \perp \overline{BD}$. Suppose that $\overline{DF}$ meets $\overline{BM}$ at $E$. The ratio $DE: EF$ can be written in the form $m/n$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.
In triangle $ABC$, with orthocenter $H$ and circumcircle $\Gamma$, the bisector of angle $BAC$ meets $\overline{BC}$ at $K$. Point $Q$ lies on $\Gamma$ such that $\overline{AQ} \perp \overline{QK}$. Circumcircle of $\triangle AQH$ meets $\overline{AC}$ at $Y$ and $\overline{AB}$ at $Z$. Let $\overline{BY}$ and $\overline{CZ}$ meet at $T$. Prove that $\overline{TH} \perp \overline{KA}$
In triangle $ABC$ with $AB>BC$, $BM$ is a median and $BL$ is an angle bisector. The line through $M$ and parallel to $AB$ intersects $BL$ at point $D$, and the line through $L$ and parallel to $BC$ intersects $BM$ at point $E$. Prove that $ED$ is perpendicular to $BL$.
Let a triangle $ABC$ satisfy $AC = BC$; in other words, let $ABC$ be an isosceles triangle with base $AB$. Let $P$ be a point inside the triangle $ABC$ such that $\angle PAB = \angle PBC$. Denote by $M$ the midpoint of the segment $AB$. Show that $\angle APM + \angle BPC = 180^{\circ}$.
In convex quadrilateral $ ABCD$, the rays $ BA,CD$ meet at $ P$, and the rays $ BC,AD$ meet at $ Q$. $ H$ is the projection of $ D$ on $ PQ$. Prove that there is a circle inscribed in $ ABCD$ if and only if the incircles of triangles $ ADP,CDQ$ are visible from $ H$ under the same angle.