The non-isosceles triangle $ABC$ is inscribed in the circle $\omega$. The tangent line to this circle at the point $C$ intersects the line $AB$ at the point $D$. Let the bisector of the angle $CDB$ intersect the segments $AC$ and $BC$ at the points $K$ and $L$, respectively. The point $M$ is on the side $AB$ such that $\frac{AK}{BL} = \frac{AM}{BM}$. Let the perpendiculars from the point $M$ to the straight lines $KL$ and $DC$ intersect the lines $AC$ and $DC$ at the points $P$ and $Q$ respectively. Prove that $2\angle CQP=\angle ACB$
Problem
Source: Kazakhstan National Olympiad 2017, Final Round, 11-Grade, P1
Tags: geometry
Anar24
28.05.2018 15:51
This problem was proposed by Kungozhin.
ThE-dArK-lOrD
08.06.2018 18:50
Either I miss some nice synthetic way or this definitely shouldn't be P1. WLOG that $AC>BC$, i.e. $D,B,A$ collinear in this order. Let $A,B,C$ denote $\angle{BAC},\angle{CAB},\angle{ACB}$, respectively. Angle chasing gives us $\angle{BDC}=\frac{B-A}{2}$ and so $LK\perp \ell$ where $\ell$ is the bisector of $\angle{ACB}$. This means $CL=CK$. Menelaus's gives us $1=\frac{AD}{DB}\times \frac{BL}{KA}=\frac{AD\times BM}{DB\times MA}$. Hence, $(D,M;B,A)=-1$. Let $CM$ intersects $\omega$ again at $E$. We get that $DE$ tangent to $\omega$ at $E$. Other consequence of $LK\perp \ell$ is that $\ell \parallel MP$, and so $\angle{MPC}=\frac{C}{2}$.
Let $\angle{CQP}=x$ and $\angle{ECB}=\angle{BED}=t$. We want to show that $x=\frac{C}{2}$.
Using Ceva's with cevians $PM,QM,CM$ of $\triangle{PCQ}$, we get$$\frac{\sin(B+x+\frac{C}{2})}{\sin (\frac{C}{2})} \times \frac{\sin (C-t)}{\sin (A+t)} \times \frac{1}{\cos (x)}=1.$$As usual of this kind of problem, we'll prove first that $\frac{\sin(A+t)}{\sin (C-t)} =\frac{2\sin(A)}{\sin(C)}$.
We've
\begin{align*}
\frac{\sin(A+t)}{\sin (C-t)} =\frac{2\sin(A)}{\sin(C)} &\iff 2\sin (A+t)\sin(C) =2\times 2\sin(A)\sin (C-t)\\
& \iff \cos (A)\sin (t)\sin (C)=\sin (C)\cos (t)\sin (A)-2\cos (C)\sin (t)\sin (A).
\end{align*}But from $\frac{DE}{DB}=\frac{DC}{DB}$, using law of sines, we get
$$\frac{\sin (C-t)}{\sin (t)} =\frac{\sin (A+C)}{\sin (A)}\implies \cos (A)\sin (t)\sin (C)=\sin (C)\cos (t)\sin (A)-2\cos (C)\sin (t)\sin (A).$$Hence, $\frac{\sin(A+t)}{\sin (C-t)} =\frac{2\sin(A)}{\sin(C)}$ is true.
Now, we've $\frac{\sin(B+x+\frac{C}{2})}{\cos (x)}=\frac{2\sin (A)}{\sin (C)}\times \sin (\frac{C}{2})=\frac{\sin (B+C)}{\cos (\frac{C}{2})}$.
So, $$\sin (B+x+C)+\sin (B+x)=\sin (B+x+C)+\sin (B+C-x)\implies 2\cos (B+\frac{C}{2})\sin (\frac{C}{2}-x)=0.$$Note that $\pi >B+\frac{C}{2}>\frac{A+B+C}{2}=\frac{\pi}{2}$ and $\frac{C}{2},x\in (0,\frac{\pi}{2})\implies \frac{C}{2}-x\in (-\frac{\pi}{2},\frac{\pi}{2})$.
This means the only possible solution is $\frac{C}{2}-x=0\implies x=\frac{C}{2}$, done.
sbealing
08.06.2018 20:45
WLOG $AC>AB$. Let $R=BC \cap MP$. Let $F$ be the second intersection of $\odot MQD$ and $\odot ABC$. As above we see $(D,M;B,A)=-1$ and $MP$ is parallel to the $C$ angle bisector. Due to harmonic bundles and right angles $\angle AFM=\angle MFB=\frac{\angle C}{2}$ and from this $AMFP$ and $RFBM$ are cyclic. Now angle chasing gives: $$\measuredangle QFR= \measuredangle QFM+\measuredangle MFR=\measuredangle CDB+\measuredangle DBC=\measuredangle QCR$$So $CRFQ$ is cyclic $$\measuredangle CPF=\measuredangle APF=\measuredangle DMF=\measuredangle CQF$$So $CPQF$ is cyclic. Combining these two results gives $CPQFR$ is cyclic and now using $MP$ parallel to the $C$ angle bisector we get: $$\frac{\angle C}{2}=\measuredangle CPR=\measuredangle CQR$$As desired.