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: 1065

Given a square $ABCD$ whose side length is $1$, $P$ and $Q$ are points on the sides $AB$ and $AD$. If the perimeter of $APQ$ is $2$ find the angle $PCQ$.
Let $l,m$ be two parallel lines in the plane. Let $P$ be a fixed point between them. Let $E,F$ be variable points on $l,m$ such that the angle $EPF$ is fixed to a number like $\alpha$ where $0<\alpha<\frac{\pi}2$. (By angle $EPF$ we mean the directed angle) Show that there is another point (not $P$) such that it sees the segment $EF$ with a fixed angle too.
The angle bisector of $\angle ABC$ of triangle $ABC$ ($AB>BC$) cuts the circumcircle of that triangle in $K$. The foot of the perpendicular from $K$ to $AB$ is $N$, and $P$ is the midpoint of $BN$. The line through $P$ parallel to $BC$ cuts line $BK$ in $T$. Prove that the line $NT$ passes through the midpoint of $AC$.
Given is a triangle $ABC$ with its circumscribed circle and $| AC | <| AB |$. On the short arc $AC$, there is a variable point $D\ne A$. Let $E$ be the reflection of $A$ wrt the inner bisector of $\angle BDC$. Prove that the line $DE$ passes through a fixed point, regardless of point $D$.
The angles of an acute triangle $ABC$ at $\alpha , \beta, \gamma$. Let $AD$ be a height, $CF$ a median, and $BE$ the bisector of $\angle B$. Show that $AD,CF$ and $BE$ are concurrent if and only if $\cos \gamma \tan\beta = \sin \alpha$ .
In triangle $ABC$, the point $H$ is the orthocenter. A circle centered at point $H$ and with radius $AH$ intersects the lines $AB$ and $AC$ at points $E$ and $D$, respectively. The point $X$ is the symmetric of the point $A$ with respect to the line $BC$ . Prove that $XH$ is the bisector of the angle $DXE$. (Matthew of Kursk)
In the diagram below, angle $ABC$ is a right angle. Point $D$ is on $\overline{BC}$, and $\overline{AD}$ bisects angle $CAB$. Points $E$ and $F$ are on $\overline{AB}$ and $\overline{AC}$, respectively, so that $AE=3$ and $AF=10.$ Given that $EB=9$ and $FC=27$, find the integer closest to the area of quadrilateral $DCFG.$ [asy] size(250); pair A=(0,12), E=(0,8), B=origin, C=(24*sqrt(2),0), D=(6*sqrt(2),0), F=A+10*dir(A--C), G=intersectionpoint(E--F, A--D); draw(A--B--C--A--D^^E--F); pair point=G+1*dir(250); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$E$", E, dir(point--E)); label("$F$", F, dir(point--F)); label("$G$", G, dir(point--G)); markscalefactor=0.1; draw(rightanglemark(A,B,C)); label("10", A--F, dir(90)*dir(A--F)); label("27", F--C, dir(90)*dir(F--C)); label("3", (0,10), W); label("9", (0,4), W);[/asy]
Let $ABCD$ be an isosceles trapezoid with bases $AD> BC$. Let $X$ be the intersection of the bisectors of $\angle BAC$ and $BC$. Let $E$ be the intersection of$ DB$ with the parallel to the bisector of $\angle CBD$ through $X$ and let $F$ be the intersection of $DC$ with the parallel to the bisector of $\angle DCB$ through $X$. Show that quadrilateral $AEFD$ is cyclic.
Let ABC be an acute isosceles triangle ($AB=BC$) inscribed in a circle with center $O$ . The line through the midpoint of the chord $AB$ and point $O$ intersects the line $AC$ at $L$ and the circle at the point $P$. Let the bisector of angle $BAC$ intersects the circle at point $K$. Lines $AB$ and $PK$ intersect at point $D$. Prove that the points $L,B,D$ and $P$ lie on the same circle.
Suppose $ABCD$ is a trapezoid with $AB\parallel CD$ and $AB\perp BC$. Let $X$ be a point on segment $\overline{AD}$ such that $AD$ bisects $\angle BXC$ externally, and denote $Y$ as the intersection of $AC$ and $BD$. If $AB=10$ and $CD=15$, compute the maximum possible value of $XY$.
Circle $k$ and its diameter $AB$ are given. Find the locus of the centers of circles inscribed in the triangles having one vertex on $AB$ and two other vertices on $k.$
Let $AD$ be the diameter of a circle $\omega$ and $BC$ is a chord of $\omega$ which is perpendicular to $AD$. Let $M,N,P$ be points on the segments $AB,AC,BC$ respectively, such that $MP\parallel AC$ and $PN\parallel AB$. The line $MN$ cuts the line $PD$ in the point $Q$ and the angle bisector of $\angle MPN$ in the point $R$. Prove that the points $B,R,Q,C$ are concyclic.
The incircle $\omega$ of a triangle $ABC$ is tangent to the sides $AB$ and $BC$ at $P$ and $Q$ respectively. The angle bisector at $A$ meets $PQ$ at point $S$. Prove $\angle ASC = 90^o$ .
A circle $\omega$ and a point $T$ outside the circle is given. Let a tangent from $T$ to $\omega$ touch $\omega$ at $A$, and take points $B,C$ lying on $\omega$ such that $T,B,C$ are colinear. The bisector of $\angle ATC$ intersects $AB$ and $AC$ at $P$ and $Q$,respectively. Prove that $PA=\sqrt{PB\cdot QC}$
In a triangle $ABC$, the angle bisector at $A,B,C$ meet the opposite sides at $A_1,B_1,C_1$, respectively. Prove that if the quadrilateral $BA_1B_1C_1$ is cyclic, then $$\frac{AC}{AB+BC}=\frac{AB}{AC+CB}+\frac{BC}{BA+AC}.$$
Triangle $ABC$ has $AC = 3$, $BC = 5$, $AB = 7$. A circle is drawn internally tangent to the circumcircle of $ABC$ at $C$, and tangent to $AB$. Let $D$ be its point of tangency with $AB$. Find $BD - DA$. [asy] /* File unicodetex not found. */ /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(6cm); real labelscalefactor = 2.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.5, xmax = 7.01, ymin = -3, ymax = 8.02; /* image dimensions */ /* draw figures */ draw(circle((1.37,2.54), 5.17)); draw((-2.62,-0.76)--(-3.53,4.2)); draw((-3.53,4.2)--(5.6,-0.44)); draw((5.6,-0.44)--(-2.62,-0.76)); draw(circle((-0.9,0.48), 2.12)); /* dots and labels */ dot((-2.62,-0.76),dotstyle); label("$C$", (-2.46,-0.51), SW * labelscalefactor); dot((-3.53,4.2),dotstyle); label("$A$", (-3.36,4.46), NW * labelscalefactor); dot((5.6,-0.44),dotstyle); label("$B$", (5.77,-0.17), SE * labelscalefactor); dot((0.08,2.37),dotstyle); label("$D$", (0.24,2.61), SW * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); label("$7$",(-3.36,4.46)--(5.77,-0.17), NE * labelscalefactor); label("$3$",(-3.36,4.46)--(-2.46,-0.51),SW * labelscalefactor); label("$5$",(-2.46,-0.51)--(5.77,-0.17), SE * labelscalefactor); /* end of picture */ [/asy]
On an acute triangle $ABC$ we draw the internal bisector of $<ABC$, $BE$, and the altitude $AD$, ($D$ on $BC$), show that $<CDE$ it's bigger than 45 degrees.
The incircle of the triangle $\triangle{ABC}$ has center at $O$ and it is tangent to the sides $BC$, $AC$ and $AB$ at the points $X$, $Y$ and $Z$, respectively. The lines $BO$ and $CO$ intersect the line $YZ$ at the points $P$ and $Q$, respectively. Show that if the segments $XP$ and $XQ$ has the same length, then the triangle $\triangle ABC$ is isosceles.
$ ABC$ is a triangle and $ AA'$ , $ BB'$ and $ CC'$ are three altitudes of this triangle . Let $ P$ be the feet of perpendicular from $ C'$ to $ A'B'$ , and $ Q$ is a point on $ A'B'$ such that $ QA \equal{} QB$ . Prove that : $ \angle PBQ \equal{} \angle PAQ \equal{} \angle PC'C$
In the triangle $ABC, \angle BAC$ is acute, the angle bisector of $\angle BAC$ meets $BC$ at $D, K$ is the foot of the perpendicular from $B$ to $AC$, and $\angle ADB = 45^o$. Point $P$ lies between $K$ and $C$ such that $\angle KDP = 30^o$. Point $Q$ lies on the ray $DP$ such that $DQ = DK$. The perpendicular at $P$ to $AC$ meets $KD$ at $L$. Prove that $PL^2 = DQ \cdot PQ$.
Let $ABC$ be a triangle with $AB < AC$ and circumcenter $O$. The angle bisector of $\angle BAC$ meets the side $BC$ at $D$. The line through $D$ perpendicular to $BC$ meets the segment $AO$ at $X$. Furthermore, let $Y$ be the midpoint of segment $AD$. Prove that points $B, C, X, Y$ are concyclic.
In a trapezoid $ABCD$ with bases $AD$ and $BC$, the bisector of the angle $\angle DAB$ intersects the bisectors of the angles $\angle ABC$ and $\angle CDA$ at the points $P$ and $S$, respectively, and the bisector of the angle $\angle BCD$ intersects the bisectors of the angles $\angle ABC$ and $\angle CDA$ at the points $Q$ and $R$, respectively. Prove that if $PS\parallel RQ$, then $AB = CD$.
In a triangle $ABC$ the bisectors $BB'$ and $CC'$ are drawn. After that, the whole picture except the points $A, B'$, and $C'$ is erased. Restore the triangle using a compass and a ruler. (A.Karlyuchenko)
An acute triangle $ABC$ is given. Let us denote by $D$ and $E$ the orthogonal projections, respectively of points $ B$ and $C$ on the bisector of the external angle $BAC$. Let $F$ be the point of intersection of the lines $BE$ and $CD$. Show that the lines $AF$ and $DE$ are perpendicular.
In a non-equilateral acute-angled triangle $ABC$ with $\angle C = 60^\circ$, $U$ is the circumcenter, $H$ the orthocenter and $D$ the intersection of $AH$ and $BC$. Prove that the Euler line $HU$ bisects the angle $BHD$.