A scalene triangle $ABC$ is inscribed within circle $\omega$. The tangent to the circle at point $C$ intersects line $AB$ at point $D$. Let $I$ be the center of the circle inscribed within $\triangle ABC$. Lines $AI$ and $BI$ intersect the bisector of $\angle CDB$ in points $Q$ and $P$, respectively. Let $M$ be the midpoint of $QP$. Prove that $MI$ passes through the middle of arc $ACB$ of circle $\omega$.

A triangle $ABC$, its circumcircle, and its incenter $I$ are drawn on the plane. Construct the circumcenter of $ABC$ using only a ruler.

Let the line $\ell$ intersects the sides $AC,BC$ of the triangle $ABC$ respectively at the points $E$ and $F$. Prove that the line $\ell$ is passing through the incenter of the triangle $ABC$ if and only if the following equality is true: $$BC\cdot\frac{AE}{CE}+AC\cdot\frac{BF}{CF}=AB.$$ [i]H. Lesov[/i]

In acute-angled $\bigtriangleup ABC$, $H$ is the orthocenter, $O$ is the circumcenter and $I$ is the incenter. Given that $\angle C > \angle B > \angle A$, prove that $I$ lies within $\bigtriangleup BOH$.

The diagonals of a convex quadrilateral divide it into four triangles. Restore the quadrilateral by the circumcenters of two adjacent triangles and the incenters of two mutually opposite triangles

Let $I$ and $I_a$ be the incenter and excenter (opposite vertex $A$) of a triangle $ABC$, respectively. Let $A'$ be the point on its circumcircle opposite to $A$, and $A_1$ be the foot of the altitude from $A$. Prove that $\angle IA_1I_a=\angle IA'I_a$. [i](Proposed by Pavel Kozhevnikov)[/i]

The bisectors of the angles $A$ and $B$ of the triangle $ABC$ meet the sides $BC$ and $CA$ at the points $D$ and $E$, respectively. Assuming that $AE+BD=AB$, determine the angle $C$.

We have $\vartriangle ABC$ with $I$ as its incenter. Let $D$ be the intersection of $AI$ and $BC$ and define $E, F$ in a similar way. Furthermore, let $Y = CI \cap DE, Z = BI \cap DF$. Prove that if $\angle BAC = 120^o$, then $E, F, Y,Z$ are concyclic. [img]https://1.bp.blogspot.com/-5IFojUbPE3o/XnSKTlTISqI/AAAAAAAALd0/0OwKMl02KJgqPs-SDOlujdcWXM0cWJiegCK4BGAYYCw/s1600/imoc2017%2Bg5.png[/img]

Let $ \Gamma_1,\Gamma_2$ and $ \Gamma_3$ be three non-intersecting circles,which are tangent to the circle $ \Gamma$ at points $ A_1,B_1,C_1$,respectively.Suppose that common tangent lines to $ (\Gamma_2,\Gamma_3)$,$ (\Gamma_1,\Gamma_3)$,$ (\Gamma_2,\Gamma_1)$ intersect in points $ A,B,C$. Prove that lines $ AA_1,BB_1,CC_1$ are concurrent.

Let $ABC$ be an isosceles triangle with $AC=BC$, whose incentre is $I$. Let $P$ be a point on the circumcircle of the triangle $AIB$ lying inside the triangle $ABC$. The lines through $P$ parallel to $CA$ and $CB$ meet $AB$ at $D$ and $E$, respectively. The line through $P$ parallel to $AB$ meets $CA$ and $CB$ at $F$ and $G$, respectively. Prove that the lines $DF$ and $EG$ intersect on the circumcircle of the triangle $ABC$. [i]Proposed by Hojoo Lee, Korea[/i]

[b]i.)[/b] Four balls of radius 1 are mutually tangent, three resting on the floor and the fourth resting on the others. A tedrahedron, each of whose edges has length $ s,$ is circumscribed around the balls. Find the value of $ s.$ [b]ii.)[/b] Suppose that $ ABCD$ and $ EFGH$ are opposite faces of a retangular solid, with $ \angle DHC \equal{} 45^{\circ}$ and $ \angle FHB \equal{} 60^{\circ}.$ Find the cosine of $ \angle BHD.$

Let $ M$ be a point on $ BC$ and $ N$ be a point on $ AB$ such that $ AM$ and $ CN$ are angle bisectors of the triangle $ ABC$. Given that $ \frac {\angle BNM}{\angle MNC} \equal{} \frac {\angle BMN}{\angle NMA}$, prove that the triangle $ ABC$ is isosceles.

Let $ABC$ be a triangle, $I$ its incenter, and $D$ a point on the arc $BC$ of the circumcircle of $ABC$ not containing $A$. The bisector of the angle $\angle ADB$ intesects the segment $AB$ at $E$. The bisector of the angle $\angle CDA$ intesects the segment $AC$ at $F$. Prove that the points $E, F,I$ are collinear. (Malik Talbi)

Let $ABC$ be a triangle with all angles $\leq 120^{\circ}$. Let $F$ be the Fermat point of triangle $ABC$, that is, the interior point of $ABC$ such that $\angle AFB = \angle BFC = \angle CFA = 120^\circ$. For each one of the three triangles $BFC$, $CFA$ and $AFB$, draw its Euler line - that is, the line connecting its circumcenter and its centroid. Prove that these three Euler lines pass through one common point. [i]Remark.[/i] The Fermat point $F$ is also known as the [b]first Fermat point[/b] or the [b]first Toricelli point[/b] of triangle $ABC$. [i]Floor van Lamoen[/i]

Given six points $ A, B, C, D, E, F $ such that $ \triangle BCD \stackrel{+}{\sim} \triangle ECA \stackrel{+}{\sim} \triangle BFA $ and let $ I $ be the incenter of $ \triangle ABC. $ Prove that the circumcenter of $ \triangle AID, \triangle BIE, \triangle CIF $ are collinear. [i]Proposed by Telv Cohl[/i]

Let $ABC$ be a triangle with $AB<AC$ and incenter $I.$ A point $D$ lies on segment $AC$ such that $AB=AD,$ and the line $BI$ intersects $AC$ at $E.$ Suppose the line $CI$ intersects $BD$ at $F,$ and $G$ lies on segment $DI$ such that $FD=FG.$ Prove that the lines $AG$ and $EF$ intersect on the circumcircle of triangle $CEI.$ \\ Proposed by Avan Lim Zenn Ee, Malaysia

Given that circle $ \omega$ is tangent internally to circle $ \Gamma$ at $ S.$ $ \omega$ touches the chord $ AB$ of $ \Gamma$ at $ T$. Let $ O$ be the center of $ \omega.$ Point $ P$ lies on the line $ AO.$ Show that $ PB\perp AB$ if and only if $ PS\perp TS.$

Let $\triangle ABC$ be acute. The feet of altitudes from the corners $A, B, C$ are $ D, E, F$, respectively. If $|DF|=3, |FE|=4,$ and $|DE|=5$, then what is the radius of the circle with center $C$ and tangent to $DE$? $\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 6 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ 3$

Suppose circles $\mathit{W}_1$ and $\mathit{W}2$, with centres $\mathit{O}_1$ and $\mathit{O}_2$ respectively, intersect at points $\mathit{M}$ and $\mathit{N}$. Let the tangent on $\mathit{W}_2$ at point $\mathit{N}$ intersect $\mathit{W}_1$ for the second time at $\mathit{B}_1$. Similarly, let the tangent on $\mathit{W}_1$ at point $\mathit{N}$ intersect $\mathit{W}_2$ for the second time at $\mathit{B}_2$. Let $\mathit{A}_1$ be a point on $\mathit{W}_1$ which is on arc $\mathit{B}_1\mathit{N}$ not containing $\mathit{M}$ and suppose line $\mathit{A}_1\mathit{N}$ intersects $\mathit{W}_2$ at point $\mathit{A}_2$. Denote the incentres of triangles $\mathit{B}_1\mathit{A}_1\mathit{N}$ and $\mathit{B}_2\mathit{A}_2\mathit{N}$ by $\mathit{I}_1$ and $\mathit{I}_2$, respectively.* [asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(10.1cm); 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 = -0.9748626324969808, xmax = 13.38440254515721, ymin = 0.5680051903627492, ymax = 10.99430986899034; /* image dimensions */ pair O_2 = (7.682929606970993,6.084708172218866), O_1 = (2.180000000000002,6.760000000000007), M = (4.560858774883258,8.585242858926296), B_2 = (10.07334553576748,9.291873850408265), A_2 = (11.49301008867042,4.866805580476367), B_1 = (2.113311869970955,9.759258690628950), A_1 = (0.2203184186713625,4.488514120712773); /* draw figures */ draw(circle(O_2, 4.000000000000000)); draw(circle(O_1, 3.000000000000000)); draw((4.048892687647541,4.413249028538064)--B_2); draw(B_2--A_2); draw(A_2--(4.048892687647541,4.413249028538064)); draw((4.048892687647541,4.413249028538064)--B_1); draw(B_1--A_1); draw(A_1--(4.048892687647541,4.413249028538064)); /* dots and labels */ dot(O_2,dotstyle); label("$O_2$", (7.788512439159622,6.243082420501817), NE * labelscalefactor); dot(O_1,dotstyle); label("$O_1$", (2.298205165350667,6.929370829727937), NE * labelscalefactor); dot(M,dotstyle); label("$M$", (4.383466101076183,8.935444641311980), NE * labelscalefactor); dot((4.048892687647541,4.413249028538064),dotstyle); label("$N$", (3.855551940133015,3.761885864068922), NE * labelscalefactor); dot(B_2,dotstyle); label("$B_2$", (10.19052187145104,9.463358802255147), NE * labelscalefactor); dot(A_2,dotstyle); label("$A_2$", (11.80066006232771,4.659339937672310), NE * labelscalefactor); dot(B_1,dotstyle); label("$B_1$", (1.981456668784765,10.09685579538695), NE * labelscalefactor); dot(A_1,dotstyle); label("$A_1$", (0.08096568938935705,3.973051528446190), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy] Show that \[\angle\mathit{I}_1\mathit{MI}_2=\angle\mathit{O}_1\mathit{MO}_2.\] *[size=80]Given a triangle ABC, the incentre of the triangle is defined to be the intersection of the angle bisectors of A, B, and C. To avoid cluttering, the incentre is omitted in the provided diagram. Note also that the diagram serves only as an aid and is not necessarily drawn to scale.[/size]

The incircle of the quadrilateral $ABCD$ touches $AB,BC, CD$ and $DA$ at $E, F,G$ and $H$ respectively. Prove that the line joining the incentres of triangles $HAE$ and $FCG$ is perpendicular to the line joining the incentres of triangles $EBF$ and $GDH$. (4)

In an acute angled triangle $ABC$ with $AB < AC$, let $I$ denote the incenter and $M$ the midpoint of side $BC$. The line through $A$ perpendicular to $AI$ intersects the tangent from $M$ to the incircle (different from line $BC$) at a point $P$> Show that $AI$ is tangent to the circumcircle of triangle $MIP$. [i]Proposed by Tejaswi Navilarekallu[/i]

Let $ABC$ be a triangle such that $AB<AC$. The perpendicular bisector of the side $BC$ meets the side $AC$ at the point $D$, and the (interior) bisectrix of the angle $ADB$ meets the circumcircle $ABC$ at the point $E$. Prove that the (interior) bisectrix of the angle $AEB$ and the line through the incentres of the triangles $ADE$ and $BDE$ are perpendicular.

Let $ABC$ be a triangle. Suppose that $X,Y$ are points in the plane such that $BX,CY$ are tangent to the circumcircle of $ABC$, $AB=BX,AC=CY$ and $X,Y,A$ are in the same side of $BC$. If $I$ be the incenter of $ABC$ prove that $\angle BAC+\angle XIY=180$.

Let $\triangle ABC$ have incenter $I$ and centroid $G$. Suppose that $P_A$ is the foot of the perpendicular from $C$ to the exterior angle bisector of $B$, and $Q_A$ is the foot of the perpendicular from $B$ to the exterior angle bisector of $C$. Define $P_B$, $P_C$, $Q_B$, and $Q_C$ similarly. Show that $P_A, P_B, P_C, Q_A, Q_B,$ and $Q_C$ lie on a circle whose center is on line $IG$.

Let $ABC$ be an acute triangle with incenter $I$ and circumcircle $\Omega$. The line passing $I$ and perpendicular to $AI$ meets $AB, AC$ at $D, E$, respectively. $A$-excircle of $\triangle{ABC}$ meets $BC$ at $T$. $AT$ meets $\Omega$ at $P$. The line passing $P$ and parallel to $BC$ meets $\Omega$ at $Q$. The intersection of $QI$ and $AT$ is $K$. Prove that $Q,D,K,E$ are concyclic.