Found problems: 3585
Let $ABC$ be an acute triangle with orthocenter $H$. Let $D$ and $E$ be feet of the altitudes from $B$ and $C$ respectively. Let $M$ be the midpoint of segment $AH$ and $F$ be the intersection point of $AH$ and $DE$. Furthermore, let $P$ and $Q$ be the points
inside triangle $ADE$ so that $P$ is an intersection of $CM$ and the circumcircle of $DFH$, and $Q$ is an intersection of $BM$ and the circumcircle of $EFH$. Prove that the intersection of lines $DQ$ and $EP$ lies on segment $AH$.
Let $K$ be a point inside an acute triangle $ ABC $, such that $ BC $ is a common tangent of the circumcircles of $ AKB $ and $ AKC$. Let $ D $ be the intersection of the lines $ CK $ and $ AB $, and let $ E $ be the intersection of the lines $ BK $ and $ AC $ . Let $ F $ be the intersection of the line $BC$ and the perpendicular bisector of the segment $DE$. The circumcircle of $ABC$ and the circle $k$ with centre $ F$ and radius $FD$ intersect at points $P$ and $Q$.
Prove that the segment $PQ$ is a diameter of $k$.
Point $P$ lies inside of scalene triangle $ABC$ with incenter $I$ such that $:$
$$ 2\angle ABP = \angle BCA , 2\angle ACP = \angle CBA $$
Lines $PB$ and $PC$ intersect line $AI$ respectively at $B'$ and $C'$. Line through $B'$ parallel to $AB$ intersects $BI$ at $X$ and line through $C'$ parallel to $AC$ intersects $CI$ at $Y$. Prove that triangles $PXY$ and $ABC$ are similar.
Let $ABC$ be a triangle with $AB = AC \neq BC$ and let $I$ be its incentre. The line $BI$ meets $AC$ at $D$, and the line through $D$ perpendicular to $AC$ meets $AI$ at $E$. Prove that the reflection of $I$ in $AC$ lies on the circumcircle of triangle $BDE$.
Given is the acute triangle $ABC$ with the intersection of altitudes $H$. The angle bisector of angle $BHC$ intersects side $BC$ at point $D$. Mark $E$ and $F$ the symmetrics of the point $D$ wrt lines $AB$ and $AC$. Prove that the circle circumscribed around the triangle $AEF$ passes through the midpoint of the arc $BAC$
The quadrilateral $ABCD$ is inscribed in circle $\Omega$ with center $O$, not lying on either of the diagonals. Suppose that the circumcircle of triangle $AOC$ passes through the midpoint of the diagonal $BD$. Prove that the circumcircle of triangle $BOD$ passes through the midpoint of diagonal $AC$.
[i](A. Zaslavsky)[/i]
(Translated from [url=http://sasja.shap.homedns.org/Turniry/TG/index.html]here.[/url])
The triangle $ABC$ has $CA = CB$. $P$ is a point on the circumcircle between $A$ and $B$ (and on the opposite side of the line $AB$ to $C$). $D$ is the foot of the perpendicular from $C$ to $PB$. Show that $PA + PB = 2 \cdot PD$.
$ABCDEF$ is a cyclic hexagon with $AB=BC=CD=DE$. $K$ is a point on segment $AE$ satisfying $\angle BKC=\angle KFE, \angle CKD = \angle KFA$. Prove that $KC=KF$.
Let $ABC$ be an acute-angled triangle with circumcircle $\omega$. A circle $\Gamma$ is internally tangent to $\omega$ at $A$ and also tangent to $BC$ at $D$. Let $AB$ and $AC$ intersect $\Gamma$ at $P$ and $Q$ respectively. Let $M$ and $N$ be points on line $BC$ such that $B$ is the midpoint of $DM$ and $C$ is the midpoint of $DN$. Lines $MP$ and $NQ$ meet at $K$ and intersect $\Gamma$ again at $I$ and $J$ respectively. The ray $KA$ meets the circumcircle of triangle $IJK$ again at $X\neq K$.
Prove that $\angle BXP = \angle CXQ$.
[i]Kian Moshiri, United Kingdom[/i]
In a triangle $ABC ~(\overline{AB} < \overline{AC})$, points $D (\neq A, B)$ and $E (\neq A, C)$ lies on side $AB$ and $AC$ respectively. Point $P$ satisfies $\overline{PB}=\overline{PD}, \overline{PC}=\overline{PE}$. $X (\neq A, C)$ is on the arc $AC$ of the circumcircle of triangle $ABC$ not including $B$. Let $Y (\neq A)$ be the intersection of circumcircle of triangle $ADE$ and line $XA$. Prove that $\overline{PX} = \overline{PY}$.
Let be given a semicircle with diameter $AB$ and center $O$, and a line intersecting the semicircle at $C$ and $D$ and the line $AB$ at $M$ ($MB < MA$, $MD < MC$). The circumcircles of the triangles $AOC$ and $DOB$ meet again at $L$. Prove that $\angle MKO$ is right.
[i]L. Kuptsov[/i]
As shown in the figure, in $\vartriangle ABC$, $AB>AC$, the inscribed circle $I$ is tangent to the sides $BC$, $CA$, $AB$ at points $D$, $E$, $F$ respectively, and the straight lines $BC$ and $EF$ intersect at point $K$, $DG \perp EF$ at point $G$, ray $IG$ intersects the circumscribed circle of $\vartriangle ABC$ at point $H$. Prove that points $H$, $G$, $D$, $K$ lie on a circle.
[img]https://cdn.artofproblemsolving.com/attachments/5/e/804fb919e9c2f9cf612099e44bad9c75699b2e.png[/img]
An acute triangle $ABC$ with $AB<AC<BC$ is inscribed in a circle $c(O,R)$. The circle $c_1(A,AC)$ intersects the circle $c$ at point $D$ and intersects $CB$ at $E$. If the line $AE$ intersects $c$ at $F$ and $G$ lies in $BC$ such that $EB=BG$, prove that $F,E,D,G$ are concyclic.
Let $ABC$ be an isosceles triangle with $BC=CA$, and let $D$ be a point inside side $AB$ such that $AD< DB$. Let $P$ and $Q$ be two points inside sides $BC$ and $CA$, respectively, such that $\angle DPB = \angle DQA = 90^{\circ}$. Let the perpendicular bisector of $PQ$ meet line segment $CQ$ at $E$, and let the circumcircles of triangles $ABC$ and $CPQ$ meet again at point $F$, different from $C$.
Suppose that $P$, $E$, $F$ are collinear. Prove that $\angle ACB = 90^{\circ}$.
Let $ABCD$ be a square. $E$ and $F$ lie on sides $AB$ and $BC$, respectively, such that $BE = BF$. The line perpendicular to $CE$, which passes through $B$, intersects $CE$ and $AD$ at points $G$ and $H$, respectively. The lines $FH$ and $CE$ intersect at point $P$ and the lines $GF$ and $CD$ intersect at point $Q$. Prove that the line $DP$ is perpendicular to the line $BQ$.
We are given a triangle $ABC$. Let $M$ be the mid-point of its side $AB$.
Let $P$ be an interior point of the triangle. We let $Q$ denote the point symmetric to $P$ with respect to $M$.
Furthermore, let $D$ and $E$ be the common points of $AP$ and $BP$ with sides $BC$ and $AC$, respectively.
Prove that points $A$, $B$, $D$, and $E$ lie on a common circle if and only if $\angle ACP = \angle QCB$ holds.
(Karl Czakler)
Let $\triangle ABC$ be a right-angled triangle with $\angle BAC = 90^{\circ}$ and let $E$ be the foot of the perpendicular from $A$ to $BC$. Let $Z \ne A$ be a point on the line $AB$ with $AB = BZ$. Let $(c)$ be the circumcircle of the triangle $\triangle AEZ$. Let $D$ be the second point of intersection of $(c)$ with $ZC$ and let $F$ be the antidiametric point of $D$ with respect to $(c)$. Let $P$ be the point of intersection of the lines $FE$ and $CZ$. If the tangent to $(c)$ at $Z$ meets $PA$ at $T$, prove that the points $T$, $E$, $B$, $Z$ are concyclic.
Proposed by [i]Theoklitos Parayiou, Cyprus[/i]
A point $T$ is chosen inside a triangle $ABC$. Let $A_1$, $B_1$, and $C_1$ be the reflections of $T$ in $BC$, $CA$, and $AB$, respectively. Let $\Omega$ be the circumcircle of the triangle $A_1B_1C_1$. The lines $A_1T$, $B_1T$, and $C_1T$ meet $\Omega$ again at $A_2$, $B_2$, and $C_2$, respectively. Prove that the lines $AA_2$, $BB_2$, and $CC_2$ are concurrent on $\Omega$.
[i]Proposed by Mongolia[/i]
Let $\omega$ be a circle with center $A$ and radius $R$. On the circumference of $\omega$ four distinct points $B, C, G, H$ are taken in that order in such a way that $G$ lies on the extended $B$-median of the triangle $ABC$, and H lies on the extension of altitude of $ABC$ from $B$. Let $X$ be the intersection of the straight lines $AC$ and $GH$. Show that the segment $AX$ has length $2R$.
The points $A, B, C, D$ lie in this order on a circle with center $O$. Furthermore, the straight lines $AC$ and $BD$ should be perpendicular to each other. The base of the perpendicular from $O$ on $AB$ is $F$. Prove $CD = 2 OF$.
[i](Karl Czakler)[/i]
Let $ABC$ be an isosceles triangle with $AB=AC$ and let $\Gamma$ denote its circumcircle. A point $D$ is on arc $AB$ of $\Gamma$ not containing $C$. A point $E$ is on arc $AC$ of $\Gamma$ not containing $B$. If $AD=CE$ prove that $BE$ is parallel to $AD$.
A triangular pyramid $ABCD$ is given. A sphere $\omega_A$ is tangent to the face $BCD$ and to the planes of other faces in points don't lying on faces. Similarly, sphere $\omega_B$ is tangent to the face $ACD$ and to the planes of other faces in points don't lying on faces. Let $K$ be the point where $\omega_A$ is tangent to $ACD$, and let $L$ be the point where $\omega_B$ is tangent to $BCD$. The points $X$ and $Y$ are chosen on the prolongations of $AK$ and $BL$ over $K$ and $L$ such that $\angle CKD = \angle CXD + \angle CBD$ and $\angle CLD = \angle CYD +\angle CAD$. Prove that the distances from the points $X$, $Y$ to the midpoint of $CD$ are the same.
[hide=thanks ]Thanks to the user Vlados021 for translating the problem.[/hide]
In triangle $ABC$, where $AB<AC$, let $X$, $Y$, $Z$ denote the points where the incircle is tangent to $BC$, $CA$, $AB$, respectively. On the circumcircle of $ABC$, let $U$ denote the midpoint of the arc $BC$ that contains the point $A$. The line $UX$ meets the circumcircle again at the point $K$. Let $T$ denote the point of intersection of $AK$ and $YZ$. Prove that $XT$ is perpendicular to $YZ$.
Let $ABC$ be an acute scalene triangle with orthocenter $H$ and circumcenter $O$. Let the line through $A$ tangent to the circumcircle of triangle $AHO$ intersect the circumcircle of triangle $ABC$ at $A$ and $P \neq A$. Let the circumcircles of triangles $AOP$ and $BHP$ intersect at $P$ and $Q \neq P$. Let line $PQ$ intersect segment $BO$ at $X$. Suppose that $BX=2$, $OX=1$, and $BC=5$. Then $AB \cdot AC = \sqrt{k}+m\sqrt{n}$ for positive integers $k$, $m$, and $n$, where neither $k$ nor $n$ is divisible by the square of any integer greater than $1$. Compute $100k+10m+n$.
[i]Proposed by Luke Robitaille[/i]
Let $ABC$ be a triangle with $AB=AC$, and let $M$ be the midpoint of $BC$. Let $P$ be a point such that $PB<PC$ and $PA$ is parallel to $BC$. Let $X$ and $Y$ be points on the lines $PB$ and $PC$, respectively, so that $B$ lies on the segment $PX$, $C$ lies on the segment $PY$, and $\angle PXM=\angle PYM$. Prove that the quadrilateral $APXY$ is cyclic.