Given six real numbers $a$, $b$, $c$, $x$, $y$, $z$ such that $0 < b-c < a < b+c$ and $ax + by + cz = 0$. What is the sign of the sum $ayz + bzx + cxy$ ?

Let $ABC$ be a triangle with, $AB$ ≠ $AC$, and let $K$ is your incenter. The points $P$ and $Q$ are the points of the intersections of the circumcicle($BCK$) with the line(s) $AB$ and $AC$, respectively. Let $D$ be intersection of $AK$ and $BC$. Show that $P, Q, D$ are collinears.

Point $P$ is taken in the interior of the triangle $ABC$, so that \[\angle PAB = \angle PCB = \dfrac {1} {4} (\angle A + \angle C).\] Let $L$ be the foot of the angle bisector of $\angle B$. The line $PL$ meets the circumcircle of $\triangle APC$ at point $Q$. Prove that $QB$ is the angle bisector of $\angle AQC$. [i]Proposed by S. Berlov[/i]

Let $ABC$ be an acute-angled triangle with $AB<AC$, incentre $I$, and let $M$ be the midpoint of major arc $BAC$. Suppose the perpendicular line from $A$ to segment $BC$ meets lines $BI$, $CI$, and $MI$ at points $P$, $Q$, and $K$ respectively. Prove that the $A-$median line in $\triangle AIK$ passes through the circumcentre of $\triangle PIQ$. [i]Proposed by Pranjal Srivastava and Rohan Goyal[/i]

Let $ ABCD$ be a cyclic quadrilateral and $ r$ and $ s$ the lines obtained reflecting $ AB$ with respect to the internal bisectors of $ \angle CAD$ and $ \angle CBD$, respectively. If $ P$ is the intersection of $ r$ and $ s$ and $ O$ is the center of the circumscribed circle of $ ABCD$, prove that $ OP$ is perpendicular to $ CD$.

For a scalene triangle $ABC$ with an incenter $I$, let its incircle meets the sides $BC, CA, AB$ at $D, E, F$, respectively. Denote by $P$ the intersection of the lines $AI$ and $DF$, and $Q$ the intersection of the lines $BI$ and $EF$. Prove that $\overline{PQ}=\overline{CD}$.

Let $ABC$ be a triangle with $AB < BC$, and let $E$ and $F$ be points in $AC$ and $AB$ such that $BF = BC = CE$, both on the same halfplane as $A$ with respect to $BC$. Let $G$ be the intersection of $BE$ and $CF$. Let $H$ be a point in the parallel through $G$ to $AC$ such that $HG = AF$ (with $H$ and $C$ in opposite halfplanes with respect to $BG$). Show that $\angle EHG = \frac{\angle BAC}{2}$.

In triangle $ABC$, let $I$ be the incentre. Let $H$ be the orthocentre of triangle $BIC$. Show that $AH$ is parallel to $BC$ if and only if $H$ lies on the circle with diameter $AI$.

In right $ \triangle ABC$ with legs $ 5$ and $ 12$, arcs of circles are drawn, one with center $ A$ and radius $ 12$, the other with center $ B$ and radius $ 5$. They intersect the hypotenuse at $ M$ and $ N$. Then, $ MN$ has length: [asy]size(200); defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, B=(12,7), C=(12,0), M=12*dir(A--B), N=B+B.y*dir(B--A); real r=degrees(B); draw(A--B--C--cycle^^Arc(A,12,0,r)^^Arc(B,B.y,180+r,270)); pair point=incenter(A,B,C); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$M$", M, dir(point--M)); label("$N$", N, dir(point--N)); label("$12$", (6,0), S); label("$5$", (12,3.5), E);[/asy] $ \textbf{(A)}\ 2 \qquad \textbf{(B)}\ \frac {13}{5} \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ \frac {24}{5}$

Squares $S_1$ and $S_2$ are inscribed in right triangle $ABC$, as shown in the figures below. Find $AC + CB$ if area$(S_1) = 441$ and area$(S_2) = 440$. [asy] size(250); real a=15, b=5; real x=a*b/(a+b), y=a/((a^2+b^2)/(a*b)+1); pair A=(0,b), B=(a,0), C=origin, X=(y,0), Y=(0, y*b/a), Z=foot(Y, A, B), W=foot(X, A, B); draw(A--B--C--cycle); draw(W--X--Y--Z); draw(shift(-(a+b), 0)*(A--B--C--cycle^^(x,0)--(x,x)--(0,x))); pair point=incenter(A,B,C); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$A$", (A.x-a-b,A.y), dir(point--A)); label("$B$", (B.x-a-b,B.y), dir(point--B)); label("$C$", (C.x-a-b,C.y), dir(point--C)); label("$S_1$", (x/2-a-b, x/2)); label("$S_2$", intersectionpoint(W--Y, X--Z)); dot(A^^B^^C^^(-a-b,0)^^(-b,0)^^(-a-b,b));[/asy]

A triangle and two points inside it are marked. It is known that one of the triangle’s angles is equal to $58^{\circ}$, one of two remaining angles is equal to $59^{\circ}$, one of two given points is the incenter of the triangle and the second one is its circumcenter. Using only the ruler without partitions determine where is each of the angles and where is each of the centers.

Given triangle ABC with incenter I. Let P,Q be point on circumcircle such that $\angle API=\angle CPI$ and $\angle BQI=\angle CQI$.Prove that $BP,AQ$ and $OI$ are concurrent.

On the circle $K$ there are given three distinct points $A,B,C$. Construct (using only a straightedge and a compass) a fourth point $D$ on $K$ such that a circle can be inscribed in the quadrilateral thus obtained.

Given triangle $ ABC$ with $ C$ a right angle, show that the diameter of the incenter is $ a\plus{}b\minus{}c$, where $ a\equal{}BC$, $ b\equal{}CA$, and $ c\equal{}AB$.

Consider an acute triangle $ABC$ with circumcircle $\mathcal{C}$. Let $H$ be the orthocenter of $ABC$ and $M$ the midpoint of $BC$. Lines $AH$, $BH$ and $CH$ cut $\mathcal{C}$ again at points $D$, $E$, and $F$ respectively; line $MH$ cuts $\mathcal{C}$ at $J$ such that $H$ lies between $J$ and $M$. Let $K$ and $L$ be the incenters of triangles $DEJ$ and $DFJ$ respectively. Prove $KL$ is parallel to $BC$.

Let $ABC$ be a triangle with orthocenter $H$, incenter $I$ and centroid $S$, and let $d$ be the diameter of the circumcircle of triangle $ABC$. Prove the inequality \[9\cdot HS^2+4\left(AH\cdot AI+BH\cdot BI+CH\cdot CI\right)\geq 3d^2,\] and determine when equality holds.

Let $ABC$ be a triangle, $I$ its incenter, and $\omega$ a circle of center $I$. Points $A',B', C'$ are on $\omega$ such that rays $IA', IB', IC',$ starting from $I$ intersect perpendicularly sides $BC, CA, AB$, respectively. Prove that lines $AA', BB', CC'$ are concurrent.

Given a triangle $ABC$ with circumcircle $\Gamma$. Points $E$ and $F$ are the foot of angle bisectors of $B$ and $C$, $I$ is incenter and $K$ is the intersection of $AI$ and $EF$. Suppose that $T$ be the midpoint of arc $BAC$. Circle $\Gamma$ intersects the $A$-median and circumcircle of $AEF$ for the second time at $X$ and $S$. Let $S'$ be the reflection of $S$ across $AI$ and $J$ be the second intersection of circumcircle of $AS'K$ and $AX$. Prove that quadrilateral $TJIX$ is cyclic. [i]Proposed by Alireza Dadgarnia and Amir Parsa Hosseini[/i]

In the acute triangle $ ABC $, the point $ I $ is the center of the inscribed the circle, the point $ O $ is the center of the circumscribed circle and the point $ I_a $ is the center the excircle tangent to the side $ BC $ and the extensions of the sides $ AB $ and $ AC $. Point $ A'$ is symmetric to vertex $ A $ with respect to the line $ BC $. Prove that $ \angle IOI_a = \angle IA'I_a $.

Let $ABC$ a triangle, $I$ the incenter, $D$ the contact point of the incircle with the side $BC$ and $E$ the foot of the bisector of the angle $A$. If $M$ is the midpoint of the arc $BC$ which contains the point $A$ of the circumcircle of the triangle $ABC$ and $\{F\} = DI \cap AM$, prove that $MI$ passes through the midpoint of $[EF]$.

Let $ABC$ be a triangle, and $I$ its incenter. Consider a circle which lies inside the circumcircle of triangle $ABC$ and touches it, and which also touches the sides $CA$ and $BC$ of triangle $ABC$ at the points $D$ and $E$, respectively. Show that the point $I$ is the midpoint of the segment $DE$.

Let $ABC$ be an acute triangle with $AT, AS$ respectively are the internal, external angle bisector of $ABC$ and $T, S \in BC$. On the circle with diameter $TS$, take an arbitrary point $P$ that lies inside the triangle ABC. Denote $D, E, F, I$ as the incenter of triangle $PBC, PCA, PAB, ABC$. Prove that four lines $AD, BE, CF$ and $IP$ are concurrent.

Let $I$ be the incentre of a triangle $ABC$ and $\Gamma_a$ be the excircle opposite $A$ touching $BC$ at $D$. If $ID$ meets $\Gamma_a$ again at $S$, prove that $DS$ bisects $\angle BSC$.

In $\triangle ABC$, $AB>AC$. In the circumcircle $(O)$ of $\triangle ABC$, $M$ is the midpoint of arc $BAC$. The incircle $(I)$ of $\triangle ABC$ touches $BC$ at $D$, the line through $D$ parallel to $AI$ intersects $(I)$ again at $P$. Prove that $AP$ and $IM$ intersect at a point on $(O)$.

In triangle $ABC$ it holds $|BC|= \frac{1}{2}(|AB|+|AC|)$. Let $M$ and $N$ be midpoints of $AB$ and $AC$, and let $I$ be the incenter of $ABC$. Prove that $A, M, I, N$ are concyclic.