A diameter $AK$ is drawn for the circumscribed circle $\omega$ of an acute-angled triangle $ABC$, an arbitrary point $M$ is chosen on the segment $BC$, the straight line $AM$ intersects $\omega$ at point $Q$. The foot of the perpendicular drawn from $M$ on $AK$ is $D$, the tangent drawn to the circle $\omega$ through the point $Q$, intersects the straight line $MD$ at $P$. A point $L$ (different from $Q$) is chosen on $\omega$ such that $PL$ is tangent to $\omega$. Prove that points $L$, $M$ and $K$ lie on the same line.

Let $\gamma$ be the circumscribed circle about a right-angled triangle $ABC$ with right angle $C$. Let $\delta$ be the circle tangent to the sides $AC$ and $BC$ and tangent to the circle $\gamma$ internally. (a) Find the radius $i$ of $\delta$ in terms of $a$ when $AC$ and $BC$ both have length $a$. (b) Show that the radius $i$ is twice the radius of the inscribed circle of $ABC$.

Let $ABCD$ be a quadrilateral inscribed in a circle $\omega$. The lines $AB$ and $CD$ meet at $P$, the lines $AD$ and $BC$ meet at $Q$, and the diagonals $AC$ and $BD$ meet at $R$. Let $M$ be the midpoint of the segment $PQ$, and let $K$ be the common point of the segment $MR$ and the circle $\omega$. Prove that the circumcircle of the triangle $KPQ$ and $\omega$ are tangent to one another.

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

Prove this proposition: Center the sphere circumscribed around a tetrahedron which coincides with the center of a sphere inscribed in that tetrahedron if and only if the skew edges of the tetrahedron are equal.

Let $AX$, $AY$ be tangents to circle $\omega$ from point $A$. Le $B$, $C$ be points inside $AX$ and $AY$ respectively, such that perimeter of $\triangle ABC$ is equal to length of $AX$. $D$ is reflection of $A$ over $BC$. Prove that circumcircle $\triangle BDC$ and $\omega$ are tangent to each other.

Let $ABC$ an acute triangle and $\Gamma$ its circumcircle. The bisector of $BAC$ intersects $\Gamma$ at $M\neq A$. A line $r$ parallel to $BC$ intersects $AC$ at $X$ and $AB$ at $Y$. Also, $MX$ and $MY$ intersect $\Gamma$ again at $S$ and $T$, respectively. If $XY$ and $ST$ intersect at $P$, prove that $PA$ is tangent to $\Gamma$.

Lines $a_1, b_1, c_1$ pass through the vertices $A, B, C$ respectively of a triange $ABC$; $a_2, b_2, c_2$ are the reflections of $a_1, b_1, c_1$ about the corresponding bisectors of $ABC$; $A_1 = b_1 \cap c_1, B_1 = a_1 \cap c_1, C_1 = a_1 \cap b_1$, and $A_2, B_2, C_2$ are defined similarly. Prove that the triangles $A_1B_1C_1$ and $A_2B_2C_2$ have the same ratios of the area and circumradius (i.e. $\frac{S_1}{R_1} = \frac{S_2}{R_2}$, where $S_i = S(\triangle A_iB_iC_i)$, $R_i = R(\triangle A_iB_iC_i)$)

Let $N$ be a point on the longest side $AC$ of a triangle $ABC$. The perpendicular bisectors of $AN$ and $NC$ intersect $AB$ and $BC$ respectively in $K$ and $M$. Prove that the circumcenter $O$ of $\triangle ABC$ lies on the circumcircle of triangle $KBM$.

Let $ABC$ be a triangle, let $H$ be the orthocentre and $L,M,N$ the midpoints of the sides $AB, BC, CA$ respectively. Prove that \[HL^{2} + HM^{2} + HN^{2} < AL^{2} + BM^{2} + CN^{2}\] if and only if $ABC$ is acute-angled.

Let $ AB$ be the diameter of semicircle $O$ , $C, D $ be points on the arc $AB$, $P, Q$ be respectively the circumcenter of $\triangle OAC $ and $\triangle OBD $ . Prove that:$CP\cdot CQ=DP \cdot DQ$.[asy] import cse5; import olympiad; unitsize(3.5cm); dotfactor=4; pathpen=black; real h=sqrt(55/64); pair A=(-1,0), O=origin, B=(1,0),C=shift(-3/8,h)*O,D=shift(4/5,3/5)*O,P=circumcenter(O,A,C), Q=circumcenter(O,D,B); D(arc(O,1,0,180),darkgreen); D(MP("A",A,W)--MP("C",C,N)--MP("P",P,SE)--MP("D",D,E)--MP("Q",Q,E)--C--MP("O",O,S)--D--MP("B",B,E)--cycle,deepblue); D(O); [/asy]

Let $O$ be the center of the circle $\omega$ circumscribed around the acute-angled triangle $\vartriangle ABC$, and $W$ be the midpoint of the arc $BC$ of the circle $\omega$, which does not contain the point $A$, and $H$ be the point of intersection of the heights of the triangle $\vartriangle ABC$. Find the angle $\angle BAC$, if $WO = WH$. (O. Clurman)

If the circumradius of any three consecutive vertices of a convex polygon is at most $ 1, $ show that the discs of radius $ 1 $ centered at each vertex cover the polygon and its interior.

Let $ ABCDE$ be a convex pentagon such that \[ \angle BAC \equal{} \angle CAD \equal{} \angle DAE\qquad \text{and}\qquad \angle ABC \equal{} \angle ACD \equal{} \angle ADE. \]The diagonals $BD$ and $CE$ meet at $P$. Prove that the line $AP$ bisects the side $CD$. [i]Proposed by Zuming Feng, USA[/i]

In a triangle with sides $a,b,c,t_a,t_b,t_c$ are the corresponding medians and $D$ the diameter of the circumcircle. Prove that \[\frac{a^2+b^2}{t_c}+\frac{b^2+c^2}{t_a}+\frac{c^2+a^2}{t_b}\le 6D\]

Let $ABC$ be a triangle such that $|AC|=1$ and $|AB|=\sqrt 2$. Let $M$ be a point such that $|MA|=|AB|$, $m(\widehat{MAB}) = 90^\circ$, and $C$ and $M$ are on the opposite sides of $AB$. Let $N$ be a point such that $|NA|=|AX|$, $m(\widehat{NAC}) = 90^\circ$, and $B$ and $N$ are on the opposite sides of $AC$. If the line passing throung $A$ and the circumcenter of triangle $MAN$ meets $[BC]$ at $F$, what is $\dfrac {|BF|}{|FC|}$? $ \textbf{(A)}\ 2\sqrt 2 \qquad\textbf{(B)}\ 2\sqrt 3 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 3\sqrt 2 $

Let $ABC$ be an obtuse triangle with $\angle A > \angle B > \angle C$, and let $M$ be a midpoint of the side $BC$. Let $D$ be a point on the arc $AB$ of the circumcircle of triangle $ABC$ not containing $C$. Suppose that the circle tangent to $BD$ at $D$ and passing through $A$ meets the circumcircle of triangle $ABM$ again at $E$ and $\overline{BD}=\overline{BE}$. $\omega$, the circumcircle of triangle $ADE$, meets $EM$ again at $F$. Prove that lines $BD$ and $AE$ meet on the line tangent to $\omega$ at $F$.

Let $ABC$ an acute triangle and $H$ its orthocenter. Let $E$ and $F$ be the intersection of lines $BH$ and $CH$ with $AC$ and $AB$ respectively, and let $D$ be the intersection of lines $EF$ and $BC$. Let $\Gamma_1$ be the circumcircle of $AEF$, and $\Gamma_2$ the circumcircle of $BHC$. The line $AD$ intersects $\Gamma_1$ at point $I \neq A$. Let $J$ be the feet of the internal bisector of $\angle{BHC}$ and $M$ the midpoint of the arc $\stackrel{\frown}{BC}$ from $\Gamma_2$ that contains the point $H$. The line $MJ$ intersects $\Gamma_2$ at point $N \neq M$. Show that the triangles $EIF$ and $CNB$ are similar.

In an acute-angled $ABC$, $\angle A>60^{\circ}$, $H$ is its orthocenter. $M,N$ are two points on $AB,AC$ respectively, such that $\angle HMB=\angle HNC=60^{\circ}$. Let $O$ be the circumcenter of triangle $HMN$. $D$ is a point on the same side with $A$ of $BC$ such that $\triangle DBC$ is an equilateral triangle. Prove that $H,O,D$ are collinear.

Triangle $ABC$ is given with its centroid $G$ and cicumcentre $O$ is such that $GO$ is perpendicular to $AG$. Let $A'$ be the second intersection of $AG$ with circumcircle of triangle $ABC$. Let $D$ be the intersection of lines $CA'$ and $AB$ and $E$ the intersection of lines $BA'$ and $AC$. Prove that the circumcentre of triangle $ADE$ is on the circumcircle of triangle $ABC$.

$ P$ is the intersection point of diagonals of cyclic $ ABCD$. The circumcenters of $ \triangle APB$ and $ \triangle CPD$ lie on circumcircle of $ ABCD$. If $ AC \plus{} BD \equal{} 18$, then area of $ ABCD$ is ? $\textbf{(A)}\ 36 \qquad\textbf{(B)}\ \frac {81}{2} \qquad\textbf{(C)}\ \frac {36\sqrt 3}{2} \qquad\textbf{(D)}\ \frac {81\sqrt 3}{4} \qquad\textbf{(E)}\ \text{None}$

A convex quadrilateral $ABCD$ is circumscribed about circle $\omega$. A tangent to $\omega$ parallel to $AC$ intersects $BD$ at a point $P$ outside of $\omega$. The second tangent from $P$ to $\omega$ touches $\omega$ at a point $T$. Prove that $\omega$ and circumcircle of $ATC$ are tangent. [i]Proposed by Nikolai Beluhov, Bulgaria[/i]

Let $ABC$ be a triangle and $L$, $M$, $N$ be the midpoints of $BC$, $CA$ and $AB$, respectively. The tangent to the circumcircle of $ABC$ at $A$ intersects $LM$ and $LN$ at $P$ and $Q$, respectively. Show that $CP$ is parallel to $BQ$.

Let $ABC$ be a triangle with incenter $I$. Let $U$, $V$ and $W$ be the intersections of the angle bisectors of angles $A$, $B$, and $C$ with the incircle, so that $V$ lies between $B$ and $I$, and similarly with $U$ and $W$. Let $X$, $Y$, and $Z$ be the points of tangency of the incircle of triangle $ABC$ with $BC$, $AC$, and $AB$, respectively. Let triangle $UVW$ be the [i]David Yang triangle[/i] of $ABC$ and let $XYZ$ be the [i]Scott Wu triangle[/i] of $ABC$. Prove that the David Yang and Scott Wu triangles of a triangle are congruent if and only if $ABC$ is equilateral. [i]Proposed by Owen Goff[/i]

Given is an acute triangle \( \triangle ABC \) with \( AB < AC \). Let \( M \) be the midpoint of side \( BC \), and let \( X \) and \( Y \) be points on segments \( BM \) and \( CM \), respectively, such that \( BX = CY \). Let \( \omega_1 \) be the circumcircle of \( \triangle ABX \), and \( \omega_2 \) the circumcircle of \( \triangle ACY \). The common tangent \( t \) to \( \omega_1 \) and \( \omega_2 \), which lies closer to point \( A \), touches \( \omega_1 \) and \( \omega_2 \) at points \( P \) and \( Q \), respectively. Let the line \( MP \) intersect \( \omega_1 \) again at \( U \), and the line \( MQ \) intersect \( \omega_2 \) again at \( V \). Prove that the circumcircle of triangle \( \triangle MUV \) is tangent to both \( \omega_1 \) and \( \omega_2 \).