Kazakhstan National Olympiad 2017, Final Round, 11-Grade, P1

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Kazakhstan National Olympiad 2017, Final Round, 11-Grade, P1

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Source: Kazakhstan National Olympiad 2017, Final Round, 11-Grade, P1

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#1 h Mar 18, 2017, 9:31 PM • 1 Y

Y by Adventure10

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 $2\angle CQP=\angle ACB$

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#3 h May 28, 2018, 12:51 PM • 2 Y

Y by Adventure10, Mango247

This problem was proposed by Kungozhin.

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#4 h Jun 8, 2018, 3:50 PM • 3 Y

Y by ThE-dArK-IOrD, Adventure10, Mango247

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 $\angle{MPC}=\frac{C}{2}$ .
Then, have fun bashing.

This post has been edited 3 times. Last edited by ThE-dArK-lOrD, Jun 9, 2018, 1:47 AM

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#5 h Jun 8, 2018, 5:45 PM • 3 Y

Y by alisherktlktl, Adventure10, Mango247

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.

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