In a triangle $A_1A_2A_3$, the excribed circles corresponding to sides $A_2A_3$, $A_3A_1$, $A_1A_2$ touch these sides at $T_1$, $T_2$, $T_3$, respectively. If $H_1$, $H_2$, $H_3$ are the orthocenters of triangles $A_1T_2T_3$, $A_2T_3T_1$, $A_3T_1T_2$, respectively, prove that lines $H_1T_1$, $H_2T_2$, $H_3T_3$ are concurrent.

The circles $ C_{1}$ and $ C_{2}$ touch externally at $ M$ and the radius of $ C_{2}$ is larger than that of $ C_{1}$. $ A$ is any point on $ C_{2}$ which does not lie on the line joining the centers of the circles. $ B$ and $ C$ are points on $ C_{1}$ such that $ AB$ and $ AC$ are tangent to $ C_{1}$. The lines $ BM$, $ CM$ intersect $ C_{2}$ again at $ E$, $ F$ respectively. $ D$ is the intersection of the tangent at $ A$ and the line $ EF$. Show that the locus of $ D$ as $ A$ varies is a straight line.

Triangle $ABC$ is inscribed in circle $\omega$. Circle $\gamma$ is tangent to $AB$ and $AC$ at points $P$ and $Q$ respectively. Also circle $\gamma$ is tangent to circle $\omega$ at point $S$. Let the intesection of $AS$ and $PQ$ be $T$. Prove that $\angle{BTP}=\angle{CTQ}$.

Let $ABC$ be an acute angled triangle satisfying the conditions $AB>BC$ and $AC>BC$. Denote by $O$ and $H$ the circumcentre and orthocentre, respectively, of the triangle $ABC.$ Suppose that the circumcircle of the triangle $AHC$ intersects the line $AB$ at $M$ different from $A$, and the circumcircle of the triangle $AHB$ intersects the line $AC$ at $N$ different from $A.$ Prove that the circumcentre of the triangle $MNH$ lies on the line $OH$.

Let $k$ be a circle centered at $M$ and let $t$ be a tangentline to $k$ through some point $T\in k$. Let $P$ be a point on $t$ and let $g\neq t$ be a line through $P$ intersecting $k$ at $U$ and $V$. Let $S$ be the point on $k$ bisecting the arc $UV$ not containing $T$ and let $Q$ be the the image of $P$ under a reflection over $ST$. Prove that $Q$, $T$, $U$ and $V$ are vertices of a trapezoid.

$A,B,C$ are on circle $\mathcal C$. $I$ is incenter of $ABC$ , $D$ is midpoint of arc $BAC$. $W$ is a circle that is tangent to $AB$ and $AC$ and tangent to $\mathcal C$ at $P$. ($W$ is in $\mathcal C$) Prove that $P$ and $I$ and $D$ are on a line.

Let $ k$ be the inscribed circle of non-isosceles triangle $ \triangle ABC$, which center is $ S$. Circle $ k$ touches sides $ BC,CA,AB$ in points $ P,Q,R$ respectively. Line $ QR$ intersects $ BC$ in point $ M$. Let a circle which contains points $ B$ and $ C$ touch $ k$ in point $ N$. Circumscribed circle of $ \triangle MNP$ intersects line $ AP$ in point $ L$, different from $ P$. Prove that points $ S,L$ and $ M$ are collinear.

The triangle $ABC$ lies on the coordinate plane. The midpoint of $\overline{AB}$ has coordinates $(-16, -63)$, the midpoint of $\overline{AC}$ has coordinates $(13, 50)$, and the midpoint of $\overline{BC}$ has coordinates $(6, -85)$. What are the coordinates of point $A$?

Let $ABC$ be a triangle and let $\omega$ be its incircle. Denote by $D_1$ and $E_1$ the points where $\omega$ is tangent to sides $BC$ and $AC$, respectively. Denote by $D_2$ and $E_2$ the points on sides $BC$ and $AC$, respectively, such that $CD_2=BD_1$ and $CE_2=AE_1$, and denote by $P$ the point of intersection of segments $AD_2$ and $BE_2$. Circle $\omega$ intersects segment $AD_2$ at two points, the closer of which to the vertex $A$ is denoted by $Q$. Prove that $AQ=D_2P$.

Consider the two triangles $ ABC$ and $ PQR$ shown below. In triangle $ ABC, \angle ADB \equal{} \angle BDC \equal{} \angle CDA \equal{} 120^\circ$. Prove that $ x\equal{}u\plus{}v\plus{}w$. [asy]unitsize(7mm); defaultpen(linewidth(.7pt)+fontsize(10pt)); pair C=(0,0), B=4*dir(5); pair A=intersectionpoints(Circle(C,5), Circle(B,6))[0]; pair Oc=scale(sqrt(3)/3)*rotate(30)*(B-A)+A; pair Ob=scale(sqrt(3)/3)*rotate(30)*(A-C)+C; pair D=intersectionpoints(Circle(Ob,length(Ob-C)), Circle(Oc,length(Oc-B)))[1]; real s=length(A-D)+length(B-D)+length(C-D); pair P=(6,0), Q=P+(s,0), R=rotate(60)*(s,0)+P; pair M=intersectionpoints(Circle(P,length(B-C)), Circle(Q,length(A-C)))[0]; draw(A--B--C--A--D--B); draw(D--C); label("$B$",B,SE); label("$C$",C,SW); label("$A$",A,N); label("$D$",D,NE); label("$a$",midpoint(B--C),S); label("$b$",midpoint(A--C),WNW); label("$c$",midpoint(A--B),NE); label("$u$",midpoint(A--D),E); label("$v$",midpoint(B--D),N); label("$w$",midpoint(C--D),NNW); draw(P--Q--R--P--M--Q); draw(M--R); label("$P$",P,SW); label("$Q$",Q,SE); label("$R$",R,N); label("$M$",M,NW); label("$x$",midpoint(P--R),NW); label("$x$",midpoint(P--Q),S); label("$x$",midpoint(Q--R),NE); label("$c$",midpoint(R--M),ESE); label("$a$",midpoint(P--M),NW); label("$b$",midpoint(Q--M),NE);[/asy]

Given an equilateral triangle $ABC$ and a point $M$ in the plane ($ABC$). Let $A', B', C'$ be respectively the symmetric through $M$ of $A, B, C$. [b]I.[/b] Prove that there exists a unique point $P$ equidistant from $A$ and $B'$, from $B$ and $C'$ and from $C$ and $A'$. [b]II.[/b] Let $D$ be the midpoint of the side $AB$. When $M$ varies ($M$ does not coincide with $D$), prove that the circumcircle of triangle $MNP$ ($N$ is the intersection of the line $DM$ and $AP$) pass through a fixed point.

Let $ABC$ be an acute triangle with $AB<AC$, and let $D$ and $E$ be points on side $BC$ such that $BD=CE$ and $D$ lies between $B$ and $E$. Suppose there exists a point $P$ inside $ABC$ such that $PD\parallel AE$ and $\angle PAB=\angle EAC$. Prove that $\angle PBA=\angle PCA$. [i]Calvin Deng.[/i]

If $x$ and $y$ are real numbers with $(x+y)^4=x-y$, what is the maximum possible value of $y$?

The median to a 10 cm side of a triangle has length 9 cm and is perpendicular to a second median of the triangle. Find the exact value in centimeters of the length of the third median.

Given points $A,B,C$ and $D$ lie a circle. $AC\cap BD=K$. $I_1, I_2,I_3$ and $I_4$ incenters of $ABK,BCK,CDK,DKA$. $M_1,M_2,M_3,M_4$ midpoints of arcs $AB,BC,CA,DA$ . Then prove that $M_1I_1,M_2I_2,M_3I_3,M_4I_4$ are concurrent.

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]

Let $ABCD$ and $BCFG$ be two faces of a cube with $AB=12.$ A beam of light emanates from vertex $A$ and reflects off face $BCFG$ at point $P,$ which is 7 units from $\overline{BG}$ and 5 units from $\overline{BC}.$ The beam continues to be reflected off the faces of the cube. The length of the light path from the time it leaves point $A$ until it next reaches a vertex of the cube is given by $m\sqrt{n},$ where $m$ and $n$ are integers and $n$ is not divisible by the square of any prime. Find $m+n.$

Consider the following transformation of the Cartesian plane: choose a lattice point and rotate the plane $90^\circ$ counterclockwise about that lattice point. Is it possible, through a sequence of such transformations, to take the triangle with vertices $(0,0)$, $(1,0)$ and $(0,1)$ to the triangle with vertices $(0,0)$, $(1,0)$ and $(1,1)$?

Let $ ABC$ be a triangle. Point $ D$ lies on its sideline $ BC$ such that $ \angle CAD \equal{} \angle CBA.$ Circle $ (O)$ passing through $ B,D$ intersects $ AB,AD$ at $ E,F$, respectively. $ BF$ meets $ DE$ at $ G$.Denote by$ M$ the midpoint of $ AG.$ Show that $ CM\perp AO.$

Let $ABC$ be a triangle and $D$ be a point on $BC$ such that $AB+BD=AC+CD$. The line $AD$ intersects the incircle of triangle $ABC$ at $X$ and $Y$ where $X$ is closer to $A$ than $Y$ i. Suppose $BC$ is tangent to the incircle at $E$, prove that: 1) $EY$ is perpendicular to $AD$; 2) $XD=2IM$ where $I$ is the incentre and $M$ is the midpoint of $BC$.

Let $ABCDA'B'C'D'$ be a cube (where $ABCD$ is a square and $AA'\parallel BB'\parallel CC'\parallel DD'$). Furthermore, let $\mathcal R$ be a rotation (with respect some line) that maps vertex $A$ to $B.$ Find the set of all images $X=\mathcal R(C)$ such that $X$ lies on the surface of the cube for some rotation $\mathcal R(A)=B.$

Points $A$, $B$, $C$ and $D$ are chosen on a line in that order, with $AB$ and $CD$ greater than $BC$. Equilateral triangles $APB$, $BCQ$ and $CDR$ are constructed so that $P$, $Q$ and $R$ are on the same side with respect to $AD$. If $\angle PQR=120^\circ$, show that \[\frac{1}{AB}+\frac{1}{CD}=\frac{1}{BC}.\]

In an acute scalene triangle $ABC$ the bisector of the acute angle between the altitudes $AA_1$ and $CC_1$ meets the sides $AB$ and $BC$ at $P$ and $Q$ respectively. The bisector of the angle $B$ intersects the segment joining the orthocenter of $ABC$ and the midpoint of $AC$ at point $R$. Prove that $P$, $B$, $Q$, $R$ lie on a circle.

1-Let $ \triangle ABC$ be a triangle and $ (O)$ its circumcircle. $ D$ is the midpoint of arc $ BC$ which doesn't contain $ A$. We draw a circle $ W$ that is tangent internally to $ (O)$ at $ D$ and tangent to $ BC$.We draw the tangent $ AT$ from $ A$ to circle $ W$.$ P$ is taken on $ AB$ such that $ AP \equal{} AT$.$ P$ and $ T$ are at the same side wrt $ A$.PROVE $ \angle APD \equal{} 90^\circ$.

Two circles are tangent to each other internally at a point $\ T $. Let the chord $\ AB $ of the larger circle be tangent to the smaller circle at a point $\ P $. Prove that the line $\ TP $ bisects $\ \angle ATB $.