Found problems: 3585
Let $P$ be an arbitrary point inside triangle $ABC$, denote by $A_{1}$ (different from $P$) the second intersection of line $AP$ with the circumcircle of triangle $PBC$ and define $B_{1},C_{1}$ similarly. Prove that $\left(1 \plus{} 2\cdot\frac {PA}{PA_{1}}\right)\left(1 \plus{} 2\cdot\frac {PB}{PB_{1}}\right)\left(1 \plus{} 2\cdot\frac {PC}{PC_{1}}\right)\geq 8$.
Let $\Omega$ be the circumcircle of an acute triangle $ABC$. Points $D$, $E$, $F$ are the midpoints of the inferior arcs $BC$, $CA$, $AB$, respectively, on $\Omega$. Let $G$ be the antipode of $D$ in $\Omega$. Let $X$ be the intersection of lines $GE$ and $AB$, while $Y$ the intersection of lines $FG$ and $CA$. Let the circumcenters of triangles $BEX$ and $CFY$ be points $S$ and $T$, respectively. Prove that $D$, $S$, $T$ are collinear.
[i]Proposed by kyou46 and Li4.[/i]
Let $AEBC$ be a cyclic quadrilateral. Let $D$ be a point on the ray $AE$ which is outside the circumscribed circumference of $AEBC$. Suppose that $\angle CAB=\angle BAE$. Prove that $AB=BD$ if and only if $DE=AC$.
Let \(ABCDEF\) be a convex cyclic hexagon such that quadrilateral \(ABDF\) is a square, and the incenter of \(\triangle ACE\) lines on \(\overline{BF}\). Diagonal \(CE\) intersects diagonals \(BD\) and \(DF\) at points \(P\) and \(Q\), respectively. Prove that the circumcircle of \(\triangle DPQ\) is tangent to \(\overline{BF}\).
[i]Proposed by Elliott Liu[/i]
Let $ABC$ be a triangle with orthocenter $H,$ and let $M$ be the midpoint of $\overline{BC}.$ Suppose that $P$ and $Q$ are distinct points on the circle with diameter $\overline{AH},$ different from $A,$ such that $M$ lies on line $PQ.$ Prove that the orthocenter of $\triangle APQ$ lies on the circumcircle of $\triangle ABC.$
[i]Proposed by Michael Ren[/i]
In triangle $ABC$, point $A_1$ lies on side $BC$ and point $B_1$ lies on side $AC$. Let $P$ and $Q$ be points on segments $AA_1$ and $BB_1$, respectively, such that $PQ$ is parallel to $AB$. Let $P_1$ be a point on line $PB_1$, such that $B_1$ lies strictly between $P$ and $P_1$, and $\angle PP_1C=\angle BAC$. Similarly, let $Q_1$ be the point on line $QA_1$, such that $A_1$ lies strictly between $Q$ and $Q_1$, and $\angle CQ_1Q=\angle CBA$.
Prove that points $P,Q,P_1$, and $Q_1$ are concyclic.
[i]Proposed by Anton Trygub, Ukraine[/i]
Let $AD, BE$, and $CF$ denote the altitudes of triangle $\vartriangle ABC$. Points $E'$ and $F'$ are the reflections of $E$ and $F$ over $AD$, respectively. The lines $BF'$ and $CE'$ intersect at $X$, while the lines $BE'$ and $CF'$ intersect at the point $Y$. Prove that if $H$ is the orthocenter of $\vartriangle ABC$, then the lines $AX, YH$, and $BC$ are concurrent.
Let $ABC$ be a non-equilateral triangle and let $M_a$, $M_b$, $M_c$ be the midpoints of the sides $BC$, $CA$, $AB$, respectively. Let $S$ be a point lying on the Euler line. Denote by $X$, $Y$, $Z$ the second intersections of $M_aS$, $M_bS$, $M_cS$ with the nine-point circle. Prove that $AX$, $BY$, $CZ$ are concurrent.
A circle $ \omega$ with center $ O$ is tangent to the rays of an angle $ BAC$ at $ B$ and $ C$. Point $ Q$ is taken inside the angle $ BAC$. Assume that point $ P$ on the segment $ AQ$ is such that $ AQ\perp OP$. The line $ OP$ intersects the circumcircles $ \omega_{1}$ and $ \omega_{2}$ of triangles $ BPQ$ and $ CPQ$ again at points $ M$ and $ N$. Prove that $ OM \equal{} ON$.
Consider the convex quadrilateral $ABCD$. The point $P$ is in the interior of $ABCD$. The following ratio equalities hold:
\[\angle PAD:\angle PBA:\angle DPA=1:2:3=\angle CBP:\angle BAP:\angle BPC\]
Prove that the following three lines meet in a point: the internal bisectors of angles $\angle ADP$ and $\angle PCB$ and the perpendicular bisector of segment $AB$.
[i]Proposed by Dominik Burek, Poland[/i]
Let $A_1C_2B_1A_2C_1B_2$ be an equilateral hexagon. Let $O_1$ and $H_1$ denote the circumcenter and orthocenter of $\triangle A_1B_1C_1$, and let $O_2$ and $H_2$ denote the circumcenter and orthocenter of $\triangle A_2B_2C_2$. Suppose that $O_1 \ne O_2$ and $H_1 \ne H_2$. Prove that the lines $O_1O_2$ and $H_1H_2$ are either parallel or coincide.
[i]Ankan Bhattacharya[/i]
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 $BDE$.
Let $\triangle ABC$ satisfy $AB = 17, AC = \frac{70}{3}$ and $BC = 19$. Let $I$ be the incenter of $\triangle ABC$ and $E$ be the excenter of $\triangle ABC$ opposite $A$. (Note: this means that the circle tangent to ray $AB$ beyond $B$, ray $AC$ beyond $C$, and side $BC$ is centered at $E$.) Suppose the circle with diameter $IE$ intersects $AB$ beyond $B$ at $D$. If $BD = \frac{a}{b}$ where $a, b$ are coprime positive integers, find $a + b$.
Let $ABC$ be a triangle in which $CA>BC>AB$. Let $H$ be its orthocentre and $O$ its circumcentre. Let $D$ and $E$ be respectively the midpoints of the arc $AB$ not containing $C$ and arc $AC$ not containing $B$. Let $D'$ and $E'$ be respectively the reflections of $D$ in $AB$ and $E$ in $AC$. Prove that $O, H, D', E'$ lie on a circle if and only if $A, D', E'$ are collinear.
Let $AX$ be a diameter of a circle $\Omega$ with radius $10$, and suppose that $C$ lies on $\Omega$ so that $AC = 16$. Let $D$ be the other point on $\Omega$ so $CX = CD$. From here, define $D'$ to be the reflection of $D$ across the midpoint of $AC$, and $X'$ to be the reflection of $X$ across the midpoint of $CD$. If the area of triangle $CD'X'$ can be written as $\frac{p}{q}$ , where $p, q$ are relatively prime, find $p + q$.
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 $\angle ACB = 90^{\circ}$.
Consider an isosceles triangle $ABC$ with $AB=AC$, and a circle $\omega$ which is tangent to the sides $AB$ and $AC$ of this triangle and intersects the side $BC$ at the points $K$ and $L$. The segment $AK$ intersects the circle $\omega$ at a point $M$ (apart from $K$). Let $P$ and $Q$ be the reflections of the point $K$ in the points $B$ and $C$, respectively. Show that the circumcircle of triangle $PMQ$ is tangent to the circle $\omega$.
The diagonals of a cyclic quadrilateral $ABCD$ intersect at $P$, and there exist a circle $\Gamma$ tangent to the extensions of $AB,BC,AD,DC$ at $X,Y,Z,T$ respectively. Circle $\Omega$ passes through points $A,B$, and is externally tangent to circle $\Gamma$ at $S$. Prove that $SP\perp ST$.
Let $P$ be a point inside a quadrilateral $ABCD$ such that $\angle APB+\angle CPD=180^{\circ}$. Points $P_{a}$, $P_{b}$, $P_{c},$ $P_{d}$ are isogonally conjugated to $P$ with respect to the triangles $BCD$, $CDA$, $DAB$, $ABC$ respectively. Prove that the diagonals of the quadrilaterals $ABCD$ and $P_{a}P_{b}P_{c}P_{d}$ concur.
Proposed by: G.Galyapin
The diagonals AC and BD of a cyclic ABCD intersect at E. Let O be circumcentre of ABCD. If midpoints of AB, CD, OE are collinear prove that AD=BC.
Bomb
[color=red][Moderator edit: The problem is wrong. See also http://www.mathlinks.ro/Forum/viewtopic.php?t=53090 .][/color]
Let $ABCD$ be a parallelogram such that $\angle BAD = 60^{\circ}.$ Let $K$ and $L$ be the midpoints of $BC$ and $CD,$ respectively. Assuming that $ABKL$ is a cyclic quadrilateral, find $\angle ABD.$
Let $ABC$ be an acute angled triangle with circumcircle $O$. Line $AO$ intersects the circumcircle of triangle $ABC$ again at point $D$. Let $P$ be a point on the side $BC$. Line passing through $P$ perpendicular to $AP$ intersects lines $DB$ and $DC$ at $E$ and $F$ respectively . Line passing through $D$ perpendicular to $BC$ intersects $EF$ at point $Q$. Prove that $EQ = FQ$ if and only if $BP = CP$.
Diagonals $AC,BD$ of cyclic quadrilateral $ABCD$ intersect at $P$.Point $Q$ is on$BC$ (between$B$ and $C$) such that $PQ \perp AC$.Prove that the line passes through the circumcenters of triangles $APD$ and $BQD$ is parallel to $AD$.(A.Kuznetsov)
Let $\triangle ABC$ be an acute-angled triangle with $AB \not= AC$. Let $H$ be the orthocenter of triangle $ABC$, and let $M$ be the midpoint of the side $BC$. Let $D$ be a point on the side $AB$ and $E$ a point on the side $AC$ such that $AE=AD$ and the points $D$, $H$, $E$ are on the same line. Prove that the line $HM$ is perpendicular to the common chord of the circumscribed circles of triangle $\triangle ABC$ and triangle $\triangle ADE$.
Let $G, H$ be the centroid and orthocentre of $\vartriangle ABC$ which has an obtuse angle at $\angle B$. Let $\omega$ be the circle with diameter $AG$. $\omega$ intersects $\odot(ABC)$ again at $L \ne A$. The tangent to $\omega$ at $L$ intersects $\odot(ABC)$ at $K \ne L$. Given that $AG = GH$, prove $\angle HKG = 90^o$
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[i]Sam Bealing, United Kingdom[/i]