Found problems: 3585
In a cyclic quadrilateral $ABCD$ with circumcenter $O$, the diagonals $AC$ and $BD$ intersect at $X$. Let the circumcircles of triangles $AXD$ and $BXC$ intersect at $Y$. Let the circumcircles of triangles $AXB$ and $CXD$ intersect at $Z$. If $O$ lies inside $ABCD$ and if the points $O,X,Y,Z$ are all distinct, prove that $O,X,Y,Z$ lie on a circle.
Let $\omega$ be the circumcircle of triangle $ABC$ with center $O$, and the $A$ inmixtilinear circle is tangent to $AB, AC, \omega$ at $D,E,T$ respectively. $P$ is the intersection of $TO$ and $DE$ and $X$ is the intersection of $AP$ and $\omega$. Prove that the isogonal conjugate of $P$ lies on the line passing through the midpoint of $BC$ and $X$.
The diagonals $ AC$ and $ BD$ of a convex quadrilateral $ ABCD$ intersect at point $ M$. The bisector of $ \angle ACD$ meets the ray $ BA$ at $ K$. Given that $ MA \cdot MC \plus{}MA \cdot CD \equal{} MB \cdot MD$, prove that $ \angle BKC \equal{} \angle CDB$.
Points $Y,X$ lies on $AB,BC$ of $\triangle ABC$ and $X,Y,A,C$ are concyclic. $AX$ and $CY$ intersect in $O$. Points $M,N$ are midpoints of $AC$ and $XY$. Prove, that $BO$ is tangent to circumcircle of $\triangle MON$
Let $ABCDE$ be a convex pentagon such that $BC \parallel AE,$ $AB = BC + AE,$ and $\angle ABC = \angle CDE.$ Let $M$ be the midpoint of $CE,$ and let $O$ be the circumcenter of triangle $BCD.$ Given that $\angle DMO = 90^{\circ},$ prove that $2 \angle BDA = \angle CDE.$
[i]Proposed by Nazar Serdyuk, Ukraine[/i]
The chords $AC$ and $BD$ of the circle intersect at point $P$. The perpendiculars to $AC$ and $BD$ at points $C$ and $D$, respectively, intersect at point $Q$. Prove that the lines $AB$ and $PQ$ are perpendicular.
The points $A, B, C, D$ lie in this order on a circle $o$. The point $S$ lies inside $o$ and has properties $\angle SAD=\angle SCB$ and $\angle SDA= \angle SBC$. Line which in which angle bisector of $\angle ASB$ in included cut the circle in points $P$ and $Q$. Prove $PS =QS$.
The diagonals of a parallelogram $ABCD$ meet at point $O$. The tangent to the circumcircle of triangle $BOC$ at $O$ meets ray $CB$ at point $F$. The circumcircle of triangle $FOD$ meets $BC$ for the second time at point $G$. Prove that $AG=AB$.
In isoceles $\triangle ABC$, $AB=AC$, $I$ is its incenter, $D$ is a point inside $\triangle ABC$ such that $I,B,C,D$ are concyclic. The line through $C$ parallel to $BD$ meets $AD$ at $E$. Prove that $CD^2=BD\cdot CE$.
Consider $\omega$ is circumcircle of an acute triangle $ABC$. $D$ is midpoint of arc $BAC$ and $I$ is incenter of triangle $ABC$. Let $DI$ intersect $BC$ in $E$ and $\omega$ for second time in $F$. Let $P$ be a point on line $AF$ such that $PE$ is parallel to $AI$. Prove that $PE$ is bisector of angle $BPC$.
[i]Proposed by Mr.Etesami[/i]
Angle bisector at $A$, altitude from $B$ to $CA$ and altitude of $C$ to $AB$ on a scalene triangle $ABC$ forms a triangle $\triangle$. Let $P$ and $Q$ points on lines $AB$ and $AC$, respectively, such that the midpoint of segment $PQ$ is the orthocenter of the triangle $\triangle$. Prove that the points $B, C, P$ and $Q$ lie on a circle.
Two real numbers $a$ and $b$ are chosen independently and uniformly at random from the interval $(0, 75)$. Let $O$ and $P$ be two points on the plane with $OP = 200$. Let $Q$ and $R$ be on the same side of line $OP$ such that the degree measures of $\angle POQ$ and $\angle POR$ are $a$ and $b$ respectively, and $\angle OQP$ and $\angle ORP$ are both right angles. The probability that $QR \leq 100$ is equal to $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.
Consider an acute-angled triangle $\triangle ABC$ ($AC>AB$) with its orthocenter $H$ and circumcircle $\Gamma$.Points $M$,$P$ are midpoints of $BC$ and $AH$ respectively.The line $\overline{AM}$ meets $\Gamma$ again at $X$ and point $N$ lies on the line $\overline{BC}$ so that $\overline{NX}$ is tangent to $\Gamma$.
Points $J$ and $K$ lie on the circle with diameter $MP$ such that $\angle AJP=\angle HNM$ ($B$ and $J$ lie one the same side of $\overline{AH}$) and circle $\omega_1$, passing through $K,H$, and $J$, and circle $\omega_2$ passing through $K,M$, and $N$, are externally tangent to each other. Prove that the common external tangents of $\omega_1$ and $\omega_2$ meet on the line $\overline{NH}$.
[i]Proposed by Alireza Dadgarnia[/i]
Let $ABC$ be a triangle with $\angle ABC \neq 90^\circ$ and $AB$ its shortest side. Denote by $H$ the intersection of the altitudes of triangle $ABC$. Let $K$ be the circle through $A$ with centre $B$. Let $D$ be the other intersection of $K$ and $AC$. Let $K$ intersect the circumcircle of $BCD$ again at $E$. If $F$ is the intersection of $DE$ and $BH$, show that $BD$ is tangent to the circle through $D$, $F$, and $H$.
Convex quadrilateral $ABCD$ has $\angle ABC = \angle CDA = 90^{\circ}$. Point $H$ is the foot of the perpendicular from $A$ to $BD$. Points $S$ and $T$ lie on sides $AB$ and $AD$, respectively, such that $H$ lies inside triangle $SCT$ and \[
\angle CHS - \angle CSB = 90^{\circ}, \quad \angle THC - \angle DTC = 90^{\circ}. \] Prove that line $BD$ is tangent to the circumcircle of triangle $TSH$.
Let $ABC$ be a triangle with $AB \neq AC$ and circumcenter $O$. The bisector of $\angle BAC$ intersects $BC$ at $D$. Let $E$ be the reflection of $D$ with respect to the midpoint of $BC$. The lines through $D$ and $E$ perpendicular to $BC$ intersect the lines $AO$ and $AD$ at $X$ and $Y$ respectively. Prove that the quadrilateral $BXCY$ is cyclic.
Let $ABCD$ be a convex quadrilateral such that the circle with diameter $AB$ is tangent to the line $CD$, and the circle with diameter $CD$ is tangent to the line $AB$. Prove that the two intersection points of these circles and the point $AC \cap BD$ are collinear.
Let $ABC$ be an equilateral triangle. Let $A_1,B_1,C_1$ be interior points of $ABC$ such that $BA_1=A_1C$, $CB_1=B_1A$, $AC_1=C_1B$, and
$$\angle BA_1C+\angle CB_1A+\angle AC_1B=480^\circ$$
Let $BC_1$ and $CB_1$ meet at $A_2,$ let $CA_1$ and $AC_1$ meet at $B_2,$ and let $AB_1$ and $BA_1$ meet at $C_2.$
Prove that if triangle $A_1B_1C_1$ is scalene, then the three circumcircles of triangles $AA_1A_2, BB_1B_2$
and $CC_1C_2$ all pass through two common points.
(Note: a scalene triangle is one where no two sides have equal length.)
[i]Proposed by Ankan Bhattacharya, USA[/i]
Let $H, K, L$ be the feet from the altitudes from vertices $A, B, C$ of the triangle $ABC$, respectively. Prove that
$| AK | \cdot | BL | \cdot| CH | = | HK | \cdot | KL | \cdot | LH | = | AL | \cdot | BH | \cdot | CK | $.
Let $ABC$ be a triangle with $\angle BAC = 60^{\circ}$. Let $AP$ bisect $\angle BAC$ and let $BQ$ bisect $\angle ABC$, with $P$ on $BC$ and $Q$ on $AC$. If $AB + BP = AQ + QB$, what are the angles of the triangle?
In a trapezoid $ABCD$ such that the internal angles are not equal to $90^{\circ}$, the diagonals $AC$ and $BD$ intersect at the point $E$. Let $P$ and $Q$ be the feet of the altitudes from $A$ and $B$ to the sides $BC$ and $AD$ respectively. Circumscribed circles of the triangles $CEQ$ and $DEP$ intersect at the point $F\neq E$. Prove that the lines $AP$, $BQ$ and $EF$ are either parallel to each other, or they meet at exactly one point.
Janabel has a device that, when given two distinct points $U$ and $V$ in the plane, draws the perpendicular bisector of $UV$. Show that if three lines forming a triangle are drawn, Janabel can mark the orthocenter of the triangle using this device, a pencil, and no other tools.
[i]Proposed by Fedir Yudin.[/i]
Let triangle$ABC(AB<AC)$ with incenter $I$ circumscribed in $\odot O$. Let $M,N$ be midpoint of arc $\widehat{BAC}$ and $\widehat{BC}$, respectively. $D$ lies on $\odot O$ so that $AD//BC$, and $E$ is tangency point of $A$-excircle of $\bigtriangleup ABC$. Point $F$ is in $\bigtriangleup ABC$ so that $FI//BC$ and $\angle BAF=\angle EAC$. Extend $NF$ to meet $\odot O$ at $G$, and extend $AG$ to meet line $IF$ at L. Let line $AF$ and $DI$ meet at $K$. Proof that $ML\bot NK$.
Let $K, L, M$ denote three points on the sides $BC$, $AB$ and $BC$ of $\triangle{ABC}$, so that $ALKM$ is a parallelogram. Points $S$ and $T$ are chosen on lines $KL$ and $KM$ respectively, so that the quadrilaterals $AKBS$ and $AKCT$ are both cyclic. Prove that $MLST$ is cyclic if and only if $K$ is the midpoint of $BC$.
There are $6$ points on the plane such that no three of them are collinear. It's known that between every $4$ points of them, there exists a point that it's power with respect to the circle passing through the other three points is a constant value $k$.(Power of a point in the interior of a circle has a negative value.)
Prove that $k=0$ and all $6$ points lie on a circle.
[i]Proposed by Morteza Saghafian[/I]