Let $ABC$ be an acute triangle with $AC > AB$ and circumcircle $\Gamma$. The tangent from $A$ to $\Gamma$ intersects $BC$ at $T$. Let $M$ be the midpoint of $BC$ and let $R$ be the reflection of $A$ in $B$. Let $S$ be a point so that $SABT$ is a parallelogram and finally let $P$ be a point on line $SB$ such that $MP$ is parallel to $AB$. Given that $P$ lies on $\Gamma$, prove that the circumcircle of $\triangle STR$ is tangent to line $AC$. [i]Proposed by Sam Bealing, United Kingdom[/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]

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.

It is known about the triangle $ABC$ that $3 BC = CA + AB$. Let the $A$-symmedian of triangle $ABC$ intersect the circumcircle of triangle $ABC$ at point $D$. Prove that $\frac{1}{BD}+ \frac{1}{CD}= \frac{6}{AD}$. (Ercole Suppa, Italy)

Circles $\omega_1$ and $\omega_2$ are externally tangent to each other. Circle $\Omega$ is placed such that $\omega_1$ is internally tangent to $\Omega$ at $X$ while $\omega_2$ is internally tangent to $\Omega$ at $Y$. Line $\ell$ is tangent to $\omega_1$ at $P$ and $\omega_2$ at $Q$ and furthermore intersects $\Omega$ at points $A$ and $B$ with $AP<AQ$. Suppose that $AP=2$, $PQ=4$, and $QB=3$. Compute the length of line segment $\overline{XY}$.

Let ABC with orthocenter $H$ and circumcenter $O$ be an acute scalene triangle satisfying $AB = AM$ where $M$ is the midpoint of $BC$. Suppose $Q$ and $K$ are points on $(ABC)$ distinct from A satisfying $\angle AQH = 90$ and $\angle BAK = \angle CAM$. Let $N$ be the midpoint of $AH$. • Let $I$ be the intersection of $B\text{-midline}$ and $A\text{-altitude}$ Prove that $IN = IO$. • Prove that there is point $P$ on the symmedian lying on circle with center $B$ and radius $BM$ such that $(APN)$ is tangent to $AB$. [i]Proposed by Krutarth Shah[/i]

Let triangle $ \vartriangle ABC$ have $AB = 17$, $BC = 14$, $CA = 12$. Let $M_A$, $M_B$, $M_C$ be midpoints of $\overline{BC}$, $\overline{AC}$, and $\overline{AB}$ respectively. Let the angle bisectors of $ A$, $ B$, and $C$ intersect $\overline{BC}$, $\overline{AC}$, and $\overline{AB}$ at $P$, $Q$, and $R$, respectively. Reflect $M_A$ about $\overline{AP}$, $M_B$ about $\overline{BQ}$, and $M_C$ about $\overline{CR}$ to obtain $M'_A$, $M'_B$, $M'_C$, respectively. The lines $AM'_A$, $BM'_B$, and $CM'_C$ will then intersect $\overline{BC}$, $\overline{AC}$, and $\overline{AB}$ at $D$, $E$, and $F$, respectively. Given that $\overline{AD}$, $\overline{BE}$, and $\overline{CF}$ concur at a point $K$ inside the triangle, in simplest form, the ratio $[KAB] : [KBC] : [KCA]$ can be written in the form $p : q : r$, where $p$, $q$ and $ r$ are relatively prime positive integers and $[XYZ]$ denotes the area of $\vartriangle XYZ$. Compute $p + q + r$.

Given a cyclic quadrilateral $ABCD$, let the diagonals $AC$ and $BD$ meet at $E$ and the lines $AD$ and $BC$ meet at $F$. The midpoints of $AB$ and $CD$ are $G$ and $H$, respectively. Show that $EF$ is tangent at $E$ to the circle through the points $E$, $G$ and $H$. [i]Proposed by David Monk, United Kingdom[/i]

Let $\triangle ABC$ be a triangle with $AB<AC$. Let the angle bisector of $\angle BAC$ meet $BC$ at $D$, and let $M$ be the midpoint of $\overline{BC}$. Let $P$ be the foot of the perpendicular from $B$ to $\overline{AD}$. Extend $\overline{BP}$ to meet $\overline{AM}$ at $Q$. Show that $\overline{DQ}$ is parallel to $\overline{AB}$.

Let $P$ and $Q$ be on segment $BC$ of an acute triangle $ABC$ such that $\angle PAB=\angle BCA$ and $\angle CAQ=\angle ABC$. Let $M$ and $N$ be the points on $AP$ and $AQ$, respectively, such that $P$ is the midpoint of $AM$ and $Q$ is the midpoint of $AN$. Prove that the intersection of $BM$ and $CN$ is on the circumference of triangle $ABC$. [i]Proposed by Giorgi Arabidze, Georgia.[/i]

Let $ABC$ be an acute triangle with $AB\ne AC$ and circumcircle $\omega$. The angle bisector of $BAC$ intersects $BC$ and $\omega$ at $D$ and $E$ respectively. Circle with diameter $DE$ intersects $\omega$ again at $F \ne E$. Point $P$ is on $AF$ such that $PB = PC$ and $X$ and $Y$ are feet of perpendiculars from $P$ to $AB$ and $AC$ respectively. Let $H$ and $H'$ be the orthocenters of $ABC$ and $AXY$ respectively. $AH$ meets $\omega$ again at $Q$ . If $AH'$ and $HH'$ intersect the circle with diameter $AH$ again at points $S$ and $T$, respectively, prove that the lines $AT , HS$ and $FQ$ are concurrent.

Let $w$-circumcircle of triangle $ABC$ with an obtuse angle $C$ and $C '$symmetric point of point $C$ with respect to $AB$. $M$ midpoint of $AB$. $C'M$ intersects $w$ at $N$ ($C '$ between $M$ and $N$). Let $BC'$ second crossing point $w$ in $F$, and $AC'$ again crosses the $w$ at point $E$. $K$-midpoint $EF$. Prove that the lines $AB, CN$ and$ KC'$are concurrent.

[color=blue][size=150]PERU TST IMO - 2006[/size] Saturday, may 20.[/color] [b]Question 04[/b] In an actue-angled triangle $ABC$ draws up: its circumcircle $w$ with center $O$, the circumcircle $w_1$ of the triangle $AOC$ and the diameter $OQ$ of $w_1$. The points are chosen $M$ and $N$ on the straight lines $AQ$ and $AC$, respectively, in such a way that the quadrilateral $AMBN$ is a parallelogram. Prove that the intersection point of the straight lines $MN$ and $BQ$ belongs the circumference $w_1.$ --- [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=88513]Spanish version[/url] $\text{\LaTeX}{}$ed by carlosbr

Let $ABC$ be a triangle with circumcircle $\omega$ and circumcenter $O.$ Suppose that $AB = 15$, $AC = 14$, and $P$ is a point in the interior of $\triangle ABC$ such that $AP = \frac{13}{2}$, $BP^2 = \frac{409}{4}$, and $P$ is closer to $\overline{AC}$ than to $\overline{AB}$. Let $E$, $F$ be the points where $\overline{BP}$, $\overline{CP}$ intersect $\omega$ again, and let $Q$ be the intersection of $\overline{EF}$ with the tangent to $\omega$ at $A.$ Given that $AQOP$ is cyclic and that $CP^2$ is expressible in the form $\frac{a}{b} - c \sqrt{d}$ for positive integers $a$, $b$, $c$, $d$ such that $\gcd(a, b) = 1$ and $d$ is not divisible by the square of any prime, compute $1000a+100b+10c+d.$ [i]Proposed by Edward Wan[/i]