In a right triangle rectangle $ABC$ such that $AB = AC$, $M$ is the midpoint of $BC$. Let $P$ be a point on the perpendicular bisector of $AC$, lying in the semi-plane determined by $BC$ that does not contain $A$. Lines $CP$ and $AM$ intersect at $Q$. Calculate the angles that form the lines $AP$ and $BQ$.

A fixed point $A$ inside a circle is given. Consider all chords $XY$ of the circle such that $\angle XAY$ is a right angle, and for all such chords construct the point $M$ symmetric to $A$ with respect to $XY$ . Find the locus of points $M$.

$\triangle{ABC}$ is isosceles with $AB = AC >BC$. Let $D$ be a point in its interior such that $DA = DB+DC$. Suppose that the perpendicular bisector of $AB$ meets the external angle bisector of $\angle{ADB}$ at $P$, and let $Q$ be the intersection of the perpendicular bisector of $AC$ and the external angle bisector of $\angle{ADC}$. Prove that $B,C,P,Q$ are concyclic.

The vertices of a square sheet of paper are $ A$, $ B$, $ C$, $ D$. The sheet is folded in a way that the point $ D$ is mapped to the point $ D'$ on the side $ BC$. Let $ A'$ be the image of $ A$ after the folding, and let $ E$ be the intersection point of $ AB$ and $ A'D'$. Let $ r$ be the inradius of the triangle $ EBD'$. Prove that $ r\equal{}A'E$.

Let $ABC$ be a triangle such that $AB<AC$. The perpendicular bisector of the side $BC$ meets the side $AC$ at the point $D$, and the (interior) bisectrix of the angle $ADB$ meets the circumcircle $ABC$ at the point $E$. Prove that the (interior) bisectrix of the angle $AEB$ and the line through the incentres of the triangles $ADE$ and $BDE$ are perpendicular.

A convex quadrilateral $ABCD$ has perpendicular diagonals. The perpendicular bisectors of the sides $AB$ and $CD$ meet at a unique point $P$ inside $ABCD$. Prove that the quadrilateral $ABCD$ is cyclic if and only if triangles $ABP$ and $CDP$ have equal areas.

In a triangle $ABC$, let $M$ be the midpoint of $BC$ and $I$ the incenter of $ABC$. If $IM$ = $IA$, find the least possible measure of $\angle{AIM}$.

Let $AD,BF,CE$ be altitudes of triangle $ABC$.$Q$ is a point on $EF$ such that $QF=DE$ and $F$ is between $E,Q$.$P$ is a point on $EF$ such that $EP=DF$ and $E$ is between $P,F$.Perpendicular bisector of $DQ$ intersect with $AB$ at $X$ and perpendicular bisector of $DP$ intersect with $AC$ at $Y$.Prove that midpoint of $BC$ lies on $XY$.

In the right triangle $ABC,$ $CF$ is the altitude based on the hypotenuse $AB.$ The circle centered at $B$ and passing through $F$ and the circle with centre $A$ and the same radius intersect at a point of $CB.$ Determine the ratio $FB : BC.$

Let $AA',BB',CC'$ be three diameters of the circumcircle of an acute triangle $ABC$. Let $P$ be an arbitrary point in the interior of $\triangle ABC$, and let $D,E,F$ be the orthogonal projection of $P$ on $BC,CA,AB$, respectively. Let $X$ be the point such that $D$ is the midpoint of $A'X$, let $Y$ be the point such that $E$ is the midpoint of $B'Y$, and similarly let $Z$ be the point such that $F$ is the midpoint of $C'Z$. Prove that triangle $XYZ$ is similar to triangle $ABC$.

It is given a convex quadrilateral $ ABCD$ in which $ \angle B\plus{}\angle C < 180^0$. Lines $ AB$ and $ CD$ intersect in point E. Prove that $ CD*CE\equal{}AC^2\plus{}AB*AE \leftrightarrow \angle B\equal{} \angle D$

Triangle ABC has incircle w centered as S that touches the sides BC,CA and AB at P,Q and R respectively. AB isn't equal AC, the lines QR and BC intersects at point M, the circle that passes through points B and C touches the circle w at point N, circumcircle of triangle MNP intersects with line AP at L (L isn't equal to P). Then prove that S,L and M lie on the same line

Triangle $GRT$ has $GR=5,$ $RT=12,$ and $GT=13.$ The perpendicular bisector of $GT$ intersects the extension of $GR$ at $O.$ Find $TO.$

Let $H$ be the orthocenter of a triangle $ABC$, and let $D$ be the midpoint of the segment $AH$. The altitude $BH$ of triangle $ABC$ intersects the perpendicular to the line $AB$ through the point $A$ at the point $M$. The altitude $CH$ of triangle $ABC$ intersects the perpendicular to the line $CA$ through the point $A$ at the point $N$. The perpendicular bisector of the segment $AB$ intersects the perpendicular to the line $BC$ through the point $B$ at the point $U$. The perpendicular bisector of the segment $CA$ intersects the perpendicular to the line $BC$ through the point $C$ at the point $V$. Finally, let $E$ be the midpoint of the side $BC$ of triangle $ABC$. Prove that the points $D$, $M$, $N$, $U$, $V$ all lie on one and the same perpendicular to the line $AE$. [i]Extensions.[/i] In other words, we have to show that the points $M$, $N$, $U$, $V$ lie on the perpendicular to the line $AE$ through the point $D$. Additionally, one can find two more points on this perpendicular: [b](a)[/b] The nine-point circle of triangle $ABC$ is known to pass through the midpoint $E$ of its side $BC$. Let $D^{\prime}$ be the point where this nine-point circle intersects the line $AE$ apart from $E$. Then, the point $D^{\prime}$ lies on the perpendicular to the line $AE$ through the point $D$. [b](b)[/b] Let the tangent to the circumcircle of triangle $ABC$ at the point $A$ intersect the line $BC$ at a point $X$. Then, the point $X$ lies on the perpendicular to the line $AE$ through the point $D$. [i]Comment.[/i] The actual problem was created by Victor Thébault around 1950 (cf. Hyacinthos messages #1102 and #1551). The extension [b](a)[/b] initially was a (pretty trivial) lemma in Thébault's solution of the problem. Extension [b](b)[/b] is rather new; in the form "prove that $X\in UV$", it was [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=3659]proposed by Valentin Vornicu for the Balkan MO 2003[/url], however it circulated in the Hyacinthos newsgroup before (Hyacinthos messages #7240 and #7242), where different solutions of the problem were discussed as well. Hereby, "Hyacinthos" always refers to the triangle geometry newsgroup "Hyacinthos", which can be found at http://groups.yahoo.com/group/Hyacinthos . I proposed the problem for the QEDMO math fight wishing to draw some attention to it. It has a rather short and elementary solution, by the way (without using radical axes or inversion like the standard solutions). Darij

Aaron takes a square sheet of paper, with one corner labeled $A$. Point $P$ is chosen at random inside of the square and Aaron folds the paper so that points $A$ and $P$ coincide. He cuts the sheet along the crease and discards the piece containing $A$. Let $p$ be the probability that the remaining piece is a pentagon. Find the integer nearest to $100p$. [i]Proposed by Aaron Lin[/i]

Rhombus $ABCD$ has side length $2$ and $\angle B = 120 ^\circ$. Region $R$ consists of all points inside the rhombus that are closer to vertex $B$ than any of the other three vertices. What is the area of $R$? $ \textbf{(A)}\ \frac{\sqrt{3}}{3} \qquad \textbf{(B)}\ \frac{\sqrt{3}}{2} \qquad \textbf{(C)}\ \frac{2\sqrt{3}}{3} \qquad \textbf{(D)}\ 1+\frac{\sqrt{3}}{3} \qquad \textbf{(E)}\ 2 $

The vertices $ A$ and $ B$ of an equilateral triangle $ ABC$ lie on a circle $k$ of radius $1$, and the vertex $ C$ is in the interior of the circle $ k$. A point $ D$, different from $ B$, lies on $ k$ so that $ AD\equal{}AB$. The line $ DC$ intersects $ k$ for the second time at point $ E$. Find the length of the line segment $ CE$.

A circle with center $O$ and a point $A$ in this circle are given. Let $P_{B}$ is the intersection point of $[AB]$ and the internal bisector of $\angle AOB$ where $B$ is a point on the circle such that $B$ doesn't lie on the line $OA$, Find the locus of $P_{B}$ as $B$ varies.

It is given a convex quadrilateral $ ABCD$ in which $ \angle B\plus{}\angle C < 180^0$. Lines $ AB$ and $ CD$ intersect in point E. Prove that $ CD*CE\equal{}AC^2\plus{}AB*AE \leftrightarrow \angle B\equal{} \angle D$

Let $ABC$ be triangle with circumcircle $(O)$. Let $AO$ cut $(O)$ again at $A'$. Perpendicular bisector of $OA'$ cut $BC$ at $P_A$. $P_B,P_C$ define similarly. Prove that : I) Point $P_A,P_B,P_C$ are collinear. II ) Prove that the distance of $O$ from this line is equal to $\frac {R}{2}$ where $R$ is the radius of the circumcircle.

Let $AL$ and $BK$ be angle bisectors in the non-isosceles triangle $ABC$ ($L$ lies on the side $BC$, $K$ lies on the side $AC$). The perpendicular bisector of $BK$ intersects the line $AL$ at point $M$. Point $N$ lies on the line $BK$ such that $LN$ is parallel to $MK$. Prove that $LN = NA$.

$ABC$ is an acute triangle. (angle $C$ is bigger than angle $B$) Let $O$ be a center of the circle which passes $B$ and tangents to $AC$ at $C$. $O$ meets the segment $AB$ at $D$. $CO$ meets the circle $(O)$ again at $P$, a line, which passes $P$ and parallel to $AO$, meets $AC$ at $E$, and $EB$ meets the circle $(O)$ again at $L$. A perpendicular bisector of $BD$ meets $AC$ at $F$ and $LF$ meets $CD$ at $K$. Prove that two lines $EK$ and $CL$ are parallel.

Rhombus $ABCD$ has side length $2$ and $\angle B = 120 ^\circ$. Region $R$ consists of all points inside the rhombus that are closer to vertex $B$ than any of the other three vertices. What is the area of $R$? $ \textbf{(A)}\ \frac{\sqrt{3}}{3} \qquad \textbf{(B)}\ \frac{\sqrt{3}}{2} \qquad \textbf{(C)}\ \frac{2\sqrt{3}}{3} \qquad \textbf{(D)}\ 1+\frac{\sqrt{3}}{3} \qquad \textbf{(E)}\ 2 $

Let $\alpha, \beta$ and $\gamma$ denote the angles of the triangle $ABC$. The perpendicular bisector of $AB$ intersects $BC$ at the point $X$, the perpendicular bisector of $AC$ intersects it at $Y$. Prove that $\tan(\beta) \cdot \tan(\gamma) = 3$ implies $BC= XY$ (or in other words: Prove that a sufficient condition for $BC = XY$ is $\tan(\beta) \cdot \tan(\gamma) = 3$). Show that this condition is not necessary, and give a necessary and sufficient condition for $BC = XY$.

Given triangle $ABC$, let $D$, $E$, $F$ be the midpoints of $BC$, $AC$, $AB$ respectively and let $G$ be the centroid of the triangle. For each value of $\angle BAC$, how many non-similar triangles are there in which $AEGF$ is a cyclic quadrilateral?