Problem 1. Points $D$ and $E$ are chosen on the sides $AB$ and $AC$, respectively, of a triangle $\triangle ABC$ such that $DE\parallel BC$. The circumcircle $k$ of triangle $\triangle ADE$ intersects the lines $BE$ and $CD$ at the points $M$ and $N$ (different from $E$ and $D$). The lines $AM$ and $AN$ intersect the side $BC$ at points $P$ and $Q$ such that $BC=2\cdot PQ$ and the point $P$ lies between $B$ and $Q$. Prove that the circle $k$ passes through the point of intersection of the side $BC$ and the angle bisector of $\angle BAC$. Nikolai Nikolov
Problem
Source: Bulgarian TST1/2006 Problem 1
Tags: geometry, circumcircle, parallelogram, symmetry, angle bisector, power of a point, geometry unsolved
Jumbler
22.09.2006 18:06
I got $QC = \sqrt{(QA)(QN)}$ and $PB = \sqrt{(PA)(PM)}$ What next?
jonfv
21.11.2006 01:28
Following the post of Jumbler: Let $F$ be the point on $BC$ between $Q$ and $P$ such that $BP=PF$ and $CQ=QF$. We have, $PF^{2}=BP^{2}=PM.PA$, $QF^{2}=CQ^{2}=QN.QA$. Then, $<PFM=<PAF$, $<QFN=<QAF$ and $<MAN+<MFN=180^{0}$. So, $F\in k$. Since $NADF$ is cyclic, then $<FAD=<FND=<QCN+<QFN=<CAQ+<FAQ=<CAF$. So $AF$ is the angle bisector of $<ABC$.
yetti
22.11.2006 00:57
Let A' be the midpoint of BC, so that AA' is the A-median. Since $DE \parallel BC,$ by Ceva's theorem, BE, CD meet at $X \in AA'.$ Let AA' meet the circumcircle (O) of the $\triangle ABC$ again at Y and let $X_{0}\in AA'$ (on the opposite side of BC than Y) be a point such that $A'X_{0}= A'Y.$ Let $CX_{0},\ BX_{0}$ meet AB, AC at $D_{0},\ E_{0}.$ Again, by Ceva's theorem, $D_{0}E_{0}\parallel BC.$
Even though not necessary to solve the problem, we can find the exact positions of $D_{0},\ E_{0}$ on AB, AC. Power of A' to (O) is $\frac{a^{2}}{4}= A'B \cdot A'C = A'Y \cdot A'A = A'X_{0}\cdot A'A = X_{0}A' \cdot m_{a},$ hence $X_{0}A' = \frac{a^{2}}{4m_{a}}$ and $\frac{AX_{0}}{X_{0}A'}= \frac{AA'-X_{0}A'}{X_{0}A'}= \frac{4m_{a}^{2}}{a^{2}}-1.$ By van Aubel's theorem,
$2\ \frac{AD_{0}}{D_{0}B}= \frac{AD_{0}}{D_{0}B}+\frac{AE_{0}}{E_{0}C}= \frac{AX_{0}}{X_{0}A'}= \frac{4m_{a}}{a^{2}}-1$
$\frac{AE_{0}}{AC}= \frac{AD_{0}}{AB}= \frac{AD_{0}}{AD_{0}+D_{0}B}= \frac{4m_{a}^{2}-a^{2}}{4m_{a}^{2}+a^{2}}= \frac{b^{2}+c^{2}-a^{2}}{b^{2}+c^{2}}= 1-\frac{a^{2}}{b^{2}+c^{2}}$
$BX_{0}CY$ is a parallelogram, its diagonals $BC,\ X_{0}Y$ cutting each other in half at A'. The angles $\angle D_{0}AX_{0}\equiv \angle BAY = \angle BCY = \angle CBE_{0}= \angle BE_{0}D_{0}\equiv \angle X_{0}E_{0}D_{0},$ are then equal and the quadrilateral $AD_{0}X_{0}E_{0}$ is cyclic. Consequently, $M_{0}\equiv X_{0}\equiv N_{0}$ are identical, $P_{0}\equiv A' \equiv Q_{0}$ are also identical, so that $P_{0}Q_{0}= 0.$ By the problem conditions (point P lies between B and Q), the intersection X of BE, CD lies on the segment $X_{0}A'.$ As the line $DE \parallel BC$ moves from $D_{0}E_{0}$ to BC, PQ monotonically increases from 0 to BC. Hence, a unique position of $X \in X_{0}A'$ exists, for which $PQ = \frac{BC}{2}.$ Assume now that the position of $X \in X_{0}A'$ is such that the circumcircle k of the triangle $\triangle ADE$ is tangent to BC at a point T. Then T is the midpoint of the arc DE of the circle k opposite to A, so that AT bisects the angle $\angle EAD \equiv \angle CAB \equiv \angle A.$ Let $AM \equiv AP$ meet the circumcircle of the triangle $\triangle ABT$ again at U. Then $\angle BTU = \angle BAU \equiv \angle DAM = \angle DEM \equiv \angle DEB = \angle CBE \equiv \angle TBM$ which implies $TU \parallel BM.$ Similarly, $\angle TBU = \angle TAU \equiv \angle TAM = \angle BMT$ which implies $BU \parallel MT.$ Thus BMTU is a parallelogram, its diagonals BT, MU cut each other in half at P and $PT = \frac{BT}{2}.$ In exactly the same way, we can find $QT = \frac{CT}{2}.$ Combining, $PQ = PT+QT = \frac{BT+CT}{2}= \frac{BC}{2}.$ But this is true for a unique position of $X \in X_{0}A',$ of the corresponding triangle $\triangle ADE,$ and of its circumcircle k.
Altheman
27.11.2006 09:00
Could anybody verify if my solution is correct? (sorry for no picture) Let $Y$, $Z$, be the midpoints of $AB$ and $AC$, respectively. $YZ=\frac{1}{2}*BC=PQ$, also $YZ//PQ$ by similarity, so $YZQP$ is a parallelogram, which gives $PY//QZ$. Let $l$ be a line such that $PY//l//QZ$ that passes through $A$, let $l\cap BC=X$, by similarity, we have that $BP=PX$ and $QC=QX$. Then $\angle BAP=\angle DEB=\angle MBP$ (by inscribed angles and parallel lines), so $\triangle PAB\sim\triangle PBM$, so $BP^{2}=PM\cdot PA=PX^{2}$, By the converse of power of a point, the circumcircle of $AMX$ is tangent to $BC$ at $X$, by symmetry, the circumcircle of $\triangle ANX$ is tangent to $BC$ at $X$, since the circle through a point, and tangent to a line at a point defines a unique circle, $AMXN$ is cyclic. Hence, the circumcircle of $AENXMD$ is tangent to $BC$ at $X$. Since $DE//BC$, $X$ is the midpoint of arc $DXE$, then $\angle DAX=\angle EAX$ by inscribed angles, Conclusion follows.