Let $ABC$ be a triangle. Let $T$ be the point of tangency of the circumcircle of triangle $ABC$ and the $A$-mixtilinear incircle. The incircle of triangle $ABC$ has center $I$ and touches sides $BC,CA$ and $AB$ at points $D,E$ and $F,$ respectively. Let $N$ be the midpoint of line segment $DF.$ Prove that the circumcircle of triangle $BTN,$ line $TI$ and the perpendicular from $D$ to $EF$ are concurrent. [i]Proposed by Diaconescu Tashi, Romania[/i]

Let $\omega$ be the circumscribed circle of the triangle $ABC$, in which $AC< AB$, $K$ is the center of the arc $BAC$, $KW$ is the diameter of the circle $\omega$. The circle $\gamma$ is inscribed in the curvilinear triangle formed by the segments $BC$, $AB$ and the arc $AC$ of the circle $\omega$. It turned out that circle $\gamma$ also touches $KW$ at point $F$. Let $I$ be the center of the triangle $ABC$, $M$ is the midpoint of the smaller arc $AK$, and $T$ is the second intersection point of $MI$ with the circle $\omega$. Prove that lines $FI$, $TW$ and $BC$ intersect at one point. (Mykhailo Sydorenko)

In triangle $ABC$ the incenter is $I$. The center of the excircle opposite $A$ is $I_A$, and it is tangent to $BC$ at $D$. The midpoint of arc $BAC$ is $N$, and $NI$ intersects $(ABC)$ again at $T$. The center of $(AID)$ is $K$. Prove that $TI_A\perp KI$.

In an acute triangle $ABC$ with $|AB| \not= |AC|$, let $I$ be the incenter and $O$ the circumcenter. The incircle is tangent to $\overline{BC}, \overline{CA}$ and $\overline{AB}$ in $D,E$ and $F$ respectively. Prove that if the line parallel to $EF$ passing through $I$, the line parallel to $AO$ passing through $D$ and the altitude from $A$ are concurrent, then the point of concurrence is the orthocenter of the triangle $ABC$. [i]Proposed by Petar Nizié-Nikolac[/i]

Denote by $ M$ midpoint of side $ BC$ in an isosceles triangle $ \triangle ABC$ with $ AC = AB$. Take a point $ X$ on a smaller arc $ \overarc{MA}$ of circumcircle of triangle $ \triangle ABM$. Denote by $ T$ point inside of angle $ BMA$ such that $ \angle TMX = 90$ and $ TX = BX$. Prove that $ \angle MTB - \angle CTM$ does not depend on choice of $ X$. [i]Author: Farzan Barekat, Canada[/i]

Let $\Delta ABC$ be a triangle with circumcenter $O$ and circumcircle $w$. Let $S$ be the center of the circle which is tangent with $AB$, $AC$, and $w$ (in the inside), and let the circle meet $w$ at point $K$. Let the circle with diameter $AS$ meet $w$ at $T$. If $M$ is the midpoint of $BC$, show that $K,T,M,O$ are concyclic.

Let $ABC$ be a triangle, let $I$ be its incenter, let $\Omega$ be its circumcircle, and let $\omega$ be the $A$- mixtilinear incircle. Let $D,E$ and $T$ be the intersections of $\omega$ and $AB,AC$ and $\Omega$, respectively, let the line $IT$ cross $\omega$ again at $P$, and let lines $PD$ and $PE$ cross the line $BC$ at $M$ and $N$ respectively. Prove that points $D,E,M,N$ are concyclic. What is the center of this circle?

Let $ABC$ be a triangle with circumcircle $\Gamma$ and incenter $I$ and let $M$ be the midpoint of $\overline{BC}$. The points $D$, $E$, $F$ are selected on sides $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ such that $\overline{ID} \perp \overline{BC}$, $\overline{IE}\perp \overline{AI}$, and $\overline{IF}\perp \overline{AI}$. Suppose that the circumcircle of $\triangle AEF$ intersects $\Gamma$ at a point $X$ other than $A$. Prove that lines $XD$ and $AM$ meet on $\Gamma$. [i]Proposed by Evan Chen, Taiwan[/i]

Let $ABC$ be a triangle with circumcircle $\Gamma$ and incenter $I$ and let $M$ be the midpoint of $\overline{BC}$. The points $D$, $E$, $F$ are selected on sides $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ such that $\overline{ID} \perp \overline{BC}$, $\overline{IE}\perp \overline{AI}$, and $\overline{IF}\perp \overline{AI}$. Suppose that the circumcircle of $\triangle AEF$ intersects $\Gamma$ at a point $X$ other than $A$. Prove that lines $XD$ and $AM$ meet on $\Gamma$. [i]Proposed by Evan Chen, Taiwan[/i]

Denote by $ M$ midpoint of side $ BC$ in an isosceles triangle $ \triangle ABC$ with $ AC = AB$. Take a point $ X$ on a smaller arc $ \overarc{MA}$ of circumcircle of triangle $ \triangle ABM$. Denote by $ T$ point inside of angle $ BMA$ such that $ \angle TMX = 90$ and $ TX = BX$. Prove that $ \angle MTB - \angle CTM$ does not depend on choice of $ X$. [i]Author: Farzan Barekat, Canada[/i]

Let $I$ be the incentre of acute triangle $ABC$ with $AB\neq AC$. The incircle $\omega$ of $ABC$ is tangent to sides $BC, CA$, and $AB$ at $D, E,$ and $F$, respectively. The line through $D$ perpendicular to $EF$ meets $\omega$ at $R$. Line $AR$ meets $\omega$ again at $P$. The circumcircles of triangle $PCE$ and $PBF$ meet again at $Q$. Prove that lines $DI$ and $PQ$ meet on the line through $A$ perpendicular to $AI$. [i]Proposed by Anant Mudgal, India[/i]

Denote by $ M$ midpoint of side $ BC$ in an isosceles triangle $ \triangle ABC$ with $ AC = AB$. Take a point $ X$ on a smaller arc $ \overarc{MA}$ of circumcircle of triangle $ \triangle ABM$. Denote by $ T$ point inside of angle $ BMA$ such that $ \angle TMX = 90$ and $ TX = BX$. Prove that $ \angle MTB - \angle CTM$ does not depend on choice of $ X$. [i]Author: Farzan Barekat, Canada[/i]

Let $ABC$ be a triangle with circumcircle $\Gamma$ and incenter $I$ and let $M$ be the midpoint of $\overline{BC}$. The points $D$, $E$, $F$ are selected on sides $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ such that $\overline{ID} \perp \overline{BC}$, $\overline{IE}\perp \overline{AI}$, and $\overline{IF}\perp \overline{AI}$. Suppose that the circumcircle of $\triangle AEF$ intersects $\Gamma$ at a point $X$ other than $A$. Prove that lines $XD$ and $AM$ meet on $\Gamma$. [i]Proposed by Evan Chen, Taiwan[/i]

Let $I$ be the incentre of acute triangle $ABC$ with $AB\neq AC$. The incircle $\omega$ of $ABC$ is tangent to sides $BC, CA$, and $AB$ at $D, E,$ and $F$, respectively. The line through $D$ perpendicular to $EF$ meets $\omega$ at $R$. Line $AR$ meets $\omega$ again at $P$. The circumcircles of triangle $PCE$ and $PBF$ meet again at $Q$. Prove that lines $DI$ and $PQ$ meet on the line through $A$ perpendicular to $AI$. [i]Proposed by Anant Mudgal, India[/i]

Triangle $ABC$ is inscribed in the circle $\Gamma$. Let $\Gamma_a$ denote the circle internally tangent to $\Gamma$ and also tangent to sides $AB$ and $AC$. Let $A'$ denote the point of tangency of $\Gamma$ and $\Gamma_a$. Define $B'$ and $C'$ similarly. Prove that $AA'$, $BB'$ and $CC'$ are concurrent.

Given is a triangle $ABC$ with incenter $I$ and circumcircle $\omega$. The incircle is tangent to $BC$ at $D$. The perpendicular at $I$ to $AI$ meets $AB, AC$ at $E, F$ and the circle $(AEF)$ meets $\omega$ and $AI$ at $G, H$. The tangent at $G$ to $\omega$ meets $BC$ at $J$ and $AJ$ meets $\omega$ at $K$. Prove that $(DJK)$ and $(GIH)$ are tangent to each other.

Given an triangle $ABC$ isosceles at the vertex $A$, let $P$ and $Q$ be the touchpoints with $AB$ and $AC$, respectively with the circle $T$, which is tangent to both and is internally tangent to the circumcircle of $ABC$. Let $R$ and $S$ be the points of the circumscribed circle of $ABC$ such that $AP = AR = AS$ . Prove that $RS$ is tangent to $T$ .

Denote by $ M$ midpoint of side $ BC$ in an isosceles triangle $ \triangle ABC$ with $ AC = AB$. Take a point $ X$ on a smaller arc $ \overarc{MA}$ of circumcircle of triangle $ \triangle ABM$. Denote by $ T$ point inside of angle $ BMA$ such that $ \angle TMX = 90$ and $ TX = BX$. Prove that $ \angle MTB - \angle CTM$ does not depend on choice of $ X$. [i]Author: Farzan Barekat, Canada[/i]

Let $\triangle ABC$ be given, it's $A-$mixtilinear incirlce, $\omega$, and it's excenter $I_A$. Let $H$ be the foot of altitude from $A$ to $BC$, $E$ midpoint of arc $\overarc{BAC}$ and denote by $M$ and $N$, midpoints of $BC$ and $AH$, respectively. Suposse that $MN\cap AE=\{ P \}$ and that line $I_AP$ meet $\omega$ at $S$ and $T$ in this order: $I_A-T-S-P$. Prove that circumcircle of $\triangle BSC$ and $\omega$ are tangent to each other. [hide=Diagram] [asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(10.48006497171429cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -14.1599662489562, xmax = 15.320098722758091, ymin = -10.77766346689139, ymax = 5.5199497728160765; /* image dimensions */ /* draw figures */ draw(circle((0.025918528679950675,-0.7846460299371317), 4.597564438637676)); draw(circle((-0.9401249699191643,-2.0521899279943225), 3.003855690249927), red); draw((-2.4211259444978057,3.107599095143759)--(-4.242260757102907,-2.493518275554076), linewidth(1.2) + blue); draw((-4.242260757102907,-2.493518275554076)--(4.286443606492271,-2.5125131627798423), linewidth(1.2) + blue); draw((4.286443606492271,-2.5125131627798423)--(-2.4211259444978057,3.107599095143759), linewidth(1.2) + blue); draw((-2.4211259444978057,3.107599095143759)--(-2.433609564871039,-2.4975464519291233)); draw((-4.381282878515476,2.5449802910435655)--(1.435215864174395,-10.327847927108488), linewidth(1.2) + dotted); draw((-4.381282878515476,2.5449802910435655)--(0.022091424694681727,-2.5030157191669593), linetype("4 4")); draw(circle((0.0212183867796688,-2.8950097429721975), 4.282341626812516), red); draw((0.03615806773666919,3.8129070061099433)--(-2.4211259444978057,3.107599095143759)); draw((-2.4211259444978057,3.107599095143759)--(-4.381282878515476,2.5449802910435655), linetype("2 2") + green); /* dots and labels */ dot((-4.242260757102907,-2.493518275554076),linewidth(4.pt) + dotstyle); label("$B$", (-4.8714663963993114,-2.647851734263423), NE * labelscalefactor); dot((4.286443606492271,-2.5125131627798423),linewidth(4.pt) + dotstyle); label("$C$", (4.474018117829703,-2.5908670725861245), NE * labelscalefactor); dot((-2.4211259444978057,3.107599095143759),linewidth(4.pt) + dotstyle); label("$A$", (-2.5350952678420575,3.278553080175656), NE * labelscalefactor); dot((0.022091424694681727,-2.5030157191669593),linewidth(3.pt) + dotstyle); label("$M$", (-0.027770154268419445,-3.0657392532302814), NE * labelscalefactor); label("$\omega$", (-1.1294736132628969,0.5812790941168441), NE * labelscalefactor,red); dot((-2.433609564871039,-2.4975464519291233),linewidth(3.pt) + dotstyle); label("$H$", (-2.9529827867709972,-3.0657392532302814), NE * labelscalefactor); dot((-2.4273677546844223,0.3050263216073179),linewidth(3.pt) + dotstyle); label("$N$", (-2.2691668467054598,0.25836601127881736), NE * labelscalefactor); dot((1.435215864174395,-10.327847927108488),linewidth(3.pt) + dotstyle); label("$I_A$", (1.5678003725511684,-10.11284241398957), NE * labelscalefactor); dot((-4.381282878515476,2.5449802910435655),linewidth(3.pt) + dotstyle); label("$P$", (-4.643527749710799,2.708706463402667), NE * labelscalefactor); dot((-3.1988410259345286,-0.0719498450384039),linewidth(3.pt) + dotstyle); label("$S$", (-3.0859469973392963,-0.17851639491380702), NE * labelscalefactor); dot((-0.9468150550874253,-5.056038168270003),linewidth(3.pt) + dotstyle); label("$T$", (-0.8255554176782134,-4.908243314129611), NE * labelscalefactor); dot((0.03615806773666919,3.8129070061099433),linewidth(3.pt) + dotstyle); label("$E$", (-0.008775267044376731,3.962369020303242), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/hide]

Denote by $ M$ midpoint of side $ BC$ in an isosceles triangle $ \triangle ABC$ with $ AC = AB$. Take a point $ X$ on a smaller arc $ \overarc{MA}$ of circumcircle of triangle $ \triangle ABM$. Denote by $ T$ point inside of angle $ BMA$ such that $ \angle TMX = 90$ and $ TX = BX$. Prove that $ \angle MTB - \angle CTM$ does not depend on choice of $ X$. [i]Author: Farzan Barekat, Canada[/i]

Let $I,\ O$ and $\Gamma$ be the incenter, circumcenter and the circumcircle of triangle $ABC$, respectively. Line $AI$ meets $\Gamma$ at $M$ $(M\neq A)$. The circumference $\omega$ is tangent internally to $\Gamma$ at $T$, and is tangent to the lines $AB$ and $AC$. The tangents through $A$ and $T$ to $\Gamma$ intersect at $P$. Lines $PI$ and $TM$ meet at $Q$. Prove that the lines $QA$ and $MO$ meet at a point on $\Gamma$.

Let $I$ be the incenter of a non-isosceles triangle $ABC$. The line $AI$ intersects the circumcircle of the triangle $ABC$ at $A$ and $D$. Let $M$ be the middle point of the arc $BAC$. The line through the point $I$ perpendicular to $AD$ intersects $BC$ at $F$. The line $MI$ intersects the circle $BIC$ at $N$. Prove that the line $FN$ is tangent to the circle $BIC$.

Given $\triangle ABC$, let $I,O,\Gamma$ denote its incenter, circumcenter and circumcircle respecitvely. Let $AI$ intersect $\Gamma$ at $M(\neq A)$. Circle $\omega$ is tangent to $AB$, $AC$ and $\Gamma$ internally at $T$ (i.e. the mixtilinear incircle opposite $A$). Let the tangents at $A$ and $T$ to $\Gamma$ meet at $P$, and let $PI$ and $TM$ intersect at $Q$. Prove that $QA$ and $MO$ intersect at a point on $\Gamma$.

Let $ABC$ be a scalene triangle with circumcircle $\Omega$, and suppose the incircle of $ABC$ touches $BC$ at $D$. The angle bisector of $\angle A$ meets $BC$ and $\Omega$ at $E$ and $F$. The circumcircle of $\triangle DEF$ intersects the $A$-excircle at $S_1$, $S_2$, and $\Omega$ at $T \neq F$. Prove that line $AT$ passes through either $S_1$ or $S_2$. [i]Proposed by Evan Chen[/i]

Let $ABC$ be a triangle with circumcircle $\Gamma$ and incenter $I$ and let $M$ be the midpoint of $\overline{BC}$. The points $D$, $E$, $F$ are selected on sides $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ such that $\overline{ID} \perp \overline{BC}$, $\overline{IE}\perp \overline{AI}$, and $\overline{IF}\perp \overline{AI}$. Suppose that the circumcircle of $\triangle AEF$ intersects $\Gamma$ at a point $X$ other than $A$. Prove that lines $XD$ and $AM$ meet on $\Gamma$. [i]Proposed by Evan Chen, Taiwan[/i]