Found problems: 3585
In the triangle $\bigtriangleup ABC$ let $D$ be the foot of the altitude from $A$ to the side $BC$ and $I$, $I_A$, $I_C$ be the incenter, $A$-excenter, and $C$-excenter, respectively. Denote by $P\neq B$ and $Q\neq D$ the other intersection points of the circle $\bigtriangleup BDI_C$ with the lines $BI$ and $DI_A$, respectively. Prove that $AP=AQ$.
[i]Proposed Michal Jan'ik - Czech Republic[/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]
In square $ABCD$, points $E$ and $H$ lie on $\overline{AB}$ and $\overline{DA}$, respectively, so that $AE=AH.$ Points $F$ and $G$ lie on $\overline{BC}$ and $\overline{CD}$, respectively, and points $I$ and $J$ lie on $\overline{EH}$ so that $\overline{FI} \perp \overline{EH}$ and $\overline{GJ} \perp \overline{EH}$. See the figure below. Triangle $AEH$, quadrilateral $BFIE$, quadrilateral $DHJG$, and pentagon $FCGJI$ each has area $1.$ What is $FI^2$?
[asy]
real x=2sqrt(2);
real y=2sqrt(16-8sqrt(2))-4+2sqrt(2);
real z=2sqrt(8-4sqrt(2));
pair A, B, C, D, E, F, G, H, I, J;
A = (0,0);
B = (4,0);
C = (4,4);
D = (0,4);
E = (x,0);
F = (4,y);
G = (y,4);
H = (0,x);
I = F + z * dir(225);
J = G + z * dir(225);
draw(A--B--C--D--A);
draw(H--E);
draw(J--G^^F--I);
draw(rightanglemark(G, J, I), linewidth(.5));
draw(rightanglemark(F, I, E), linewidth(.5));
dot("$A$", A, S);
dot("$B$", B, S);
dot("$C$", C, dir(90));
dot("$D$", D, dir(90));
dot("$E$", E, S);
dot("$F$", F, dir(0));
dot("$G$", G, N);
dot("$H$", H, W);
dot("$I$", I, SW);
dot("$J$", J, SW);
[/asy]
$\textbf{(A) } \frac{7}{3} \qquad \textbf{(B) } 8-4\sqrt2 \qquad \textbf{(C) } 1+\sqrt2 \qquad \textbf{(D) } \frac{7}{4}\sqrt2 \qquad \textbf{(E) } 2\sqrt2$
On the sides $AB$ and $AC$ of the triangle $ABC$, the points $P$ and $Q$ are chosen, respectively, so that $PQ\parallel BC$. Segments $BQ$ and $CP$ intersect at point $O$. Point $A'$ is symmetric to point $A$ relative to line $BC$. The segment $A'O$ intersects the circumcircle $w$ of the triangle $APQ$ at the point $S$. Prove that circumcircle of $BSC$ is tangent to the circle $w$.
The incircle of a triangle $ABC$ is tangent to the sides $AB,AC$ at $M,N$ respectively. Suppose $P$ is the intersection between $MN$ and the bisector of $\angle ABC$. Prove that $BP$ and $CP$ are perpendicular.
Does there exist a set of $n > 2, n < \infty$ points in the plane such that no three are collinear and the circumcenter of any three points of the set is also in the set?
Let $ABC$ be a triangle with $AB = AC$. The incircle touches $BC$, $AC$ and $AB$ at $D$, $E$ and $F$ respectively. Let $P$ be a point on the arc $\overarc{EF}$ that does not contain $D$. Let $Q$ be the second point of intersection of $BP$ and the incircle of $ABC$. The lines $EP$ and $EQ$ meet the line $BC$ at $M$ and $N$, respectively.
Prove that the four points $P, F, B, M$ lie on a circle and $\frac{EM}{EN} = \frac{BF}{BP}$.
Let $ABC$ be a triangle with $AB < AC$, incenter $I$, and $A$ excenter $I_{A}$. The incircle meets $BC$ at $D$. Define $E = AD\cap BI_{A}$, $F = AD\cap CI_{A}$. Show that the circumcircle of $\triangle AID$ and $\triangle I_{A}EF$ are tangent to each other
Let $n \geq 3$ be a positive integer. Find the maximum number of diagonals in a regular $n$-gon one can select, so that any two of them do not intersect in the interior or they are perpendicular to each other.
Let $ ABCD$ be a trapezoid with parallel sides $ AB > CD$. Points $ K$ and $ L$ lie on the line segments $ AB$ and $ CD$, respectively, so that $AK/KB=DL/LC$. Suppose that there are points $ P$ and $ Q$ on the line segment $ KL$ satisfying \[\angle{APB} \equal{} \angle{BCD}\qquad\text{and}\qquad \angle{CQD} \equal{} \angle{ABC}.\] Prove that the points $ P$, $ Q$, $ B$ and $ C$ are concyclic.
[i]Proposed by Vyacheslev Yasinskiy, Ukraine[/i]
Let $I, C, \omega$ and $\Omega$ be the incenter, circumcenter, incircle and circumcircle, respectively, of the scalene triangle $XYZ$ with $XZ > YZ > XY$. The incircle $\omega$ is tangent to the sides $YZ, XZ$ and $XY$ at the points $D, E$ and $F$. Let $S$ be the point on $\Omega$ such that $XS, CI$ and $YZ$ are concurrent. Let $(XEF) \cap \Omega = R$, $(RSD) \cap (XEF) = U$, $SU \cap CI = N$, $EF \cap YZ = A$, $EF \cap CI = T$ and $XU \cap YZ = O$.
Prove that $NARUTO$ is cyclic.
Let $ABCD$ be a convex quadrilateral, where $R$ and $S$ are points in $DC$ and $AB$, respectively, such that $AD=RC$ and $BC=SA$. Let $P$, $Q$ and $M$ be the midpoints of $RD$, $BS$ and $CA$, respectively. If $\angle MPC + \angle MQA = 90$, prove that $ABCD$ is cyclic.
Let the convex quadrilateral $ABCD$ with $AD = BC$ ¸and $\angle A + \angle B = 120^o$. Take a point $P$ in the plane so that the line $CD$ separates the points $A$ and $P$, and the $DCP$ triangle is equilateral. Show that the triangle $ABP$ is equilateral. It is the true statement for a non-convex quadrilateral?
Let $A$, $B$, $C$, $D$ be four points on a line in this order. Suppose that $AC = 25$, $BD = 40$, and $AD = 57$. Compute $AB \cdot CD + AD \cdot BC$.
[i]Proposed by Evan Chen[/i]
In the quadrilateral $ABCD$, $AB=AD$, $CB=CD$, $\angle ABC =90^\circ$. $E$, $F$ are on $AB$, $AD$ and $P$, $Q$ are on $EF$($P$ is between $E, Q$), satisfy $\frac{AE}{EP}=\frac{AF}{FQ}$. $X, Y$ are on $CP, CQ$ that satisfy $BX \perp CP, DY \perp CQ$. Prove that $X, P, Q, Y$ are concyclic.
Segments $AC$ and $BD$ intersect at point $P$ such that $PA = PD$ and $PB = PC$. Let $E$ be the foot of the perpendicular from $P$ to the line $CD$. Prove that the line $PE$ and the perpendicular bisectors of the segments $PA$ and $PB$ are concurrent.
Point $H$ is the orthocenter of the acute triangle $\Delta ABC$ and point $K$,situated on the line $(BC)$, is the foot of the perpendicular from point $A$ .The circle $\Omega$ passes through points $A$ and $K$ ,intersecting the sides $(AB)$ and $(AC)$ in points $M$ and $N$ .The line that passes through point $A$ and is parallel with $BC$ intersects again the circumcircles of triangles $\Delta AHM$ and $\Delta AHN$ in points $X$ and $Y$.Prove that $XY =BC$.
Let $ABC$ be a triangle with $AB < AC < BC$. Let the incenter and incircle of triangle $ABC$ be $I$ and $\omega$, respectively. Let $X$ be the point on line $BC$ different from $C$ such that the line through $X$ parallel to $AC$ is tangent to $\omega$. Similarly, let $Y$ be the point on line $BC$ different from $B$ such that the line through $Y$ parallel to $AB$ is tangent to $\omega$. Let $AI$ intersect the circumcircle of triangle $ABC$ at $P \ne A$. Let $K$ and $L$ be the midpoints of $AC$ and $AB$, respectively.
Prove that $\angle KIL + \angle YPX = 180^{\circ}$.
[i]Proposed by Dominik Burek, Poland[/i]
Let $ABCD$ be a cyclic quadrilateral that inscribed in the circle $\omega$.Let $I_{1},I_{2}$ and $r_{1},r_{2}$ be incenters and radii of incircles of triangles $ACD$ and $ABC$,respectively.assume that $r_{1}=r_{2}$. let $\omega'$ be a circle that touches $AB,AD$ and touches $\omega$ at $T$. tangents from $A,T$ to $\omega$ meet at the point $K$.prove that $I_{1},I_{2},K$ lie on a line.
Let $P$ be a point on the circumcircle of square $ABCD$ such that $PA \cdot PC = 56$ and $PB \cdot PD = 90.$ What is the area of square $ABCD?$
A point $E$ is located inside a parallelogram $ABCD$ such that $\angle BAE = \angle BCE$. The centers of the circumcircles of the triangles $ABE,ECB, CDE$ and $DAE$ are concyclic.
A cyclic quadrilateral $ABCD$ has circumcenter $O$, its diagonals $AC$ and $BD$ intersect at $G$. Let $P, Q, R, S$ be the circumcenters of $\vartriangle AGB, \vartriangle BGC, \vartriangle CGD, \vartriangle DGA$ respectively. Lines $P R$ and $QS$ intersect at $M$. Show that $M$ is the midpoint of $OG$.
Triangle $ABC$ is an scalene triangle. Let $I$ the incenter, $\Omega$ the circumcircle, $E$ the $A$-excenter of triangle $ABC$. Let $\Gamma$ the circle centered at $E$ and passes $A$. $\Gamma$ and $\Omega$ intersect at point $D(\neq A)$, and the perpendicular line of $BC$ which passes $A$ meets $\Gamma$ at point $K(\neq A)$. $L$ is the perpendicular foot from $I$ to $AC$. Now if $AE$ and $DK$ intersects at $F$, prove that $BE\cdot CI=2\cdot CF\cdot CL$.
Circle $I$ is the incircle of $\triangle ABC$. Circle $I$ is tangent to sides $BC$ and $AC$ at $M,N$ respectively. $E,F$ are midpoints of sides $AB$ and $AC$ respectively. Lines $EF, BI$ intersect at $D$. Show that $M,N,D$ are collinear.
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]