Kazakhstan National Olympiad 2014 P2 D2 10 grade

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Kazakhstan National Olympiad 2014 P2 D2 10 grade

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Source: Kazakhstan National Olympiad 2014 P2 D2 10 grade

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ts0_9

#1 h Mar 27, 2014, 4:53 PM • 1 Y

Y by Adventure10

About not isosceles triangle $ABC$ the circle $w$ , a point $M$ is described —middle $AC$ . The tangent to $w$ in a point $B$ crosses a straight line $AC$ in a point $N$ , and a straight line $BM$ repeatedly crosses $w$ in a point $L$ . Let the point $P$ be symmetric to a point $L$ relatively $M$ . The circle, circumscribed about a triangle $BPN$ , repeatedly crosses line $AN$ in a point $Q$ . Prove that $\angle ABP = \angle QBC.$

This post has been edited 1 time. Last edited by ts0_9, Mar 27, 2014, 5:44 PM

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#2 h Mar 27, 2014, 5:35 PM • 2 Y

Y by Adventure10, Mango247

Let $Q'$ be the foot of the symmedian issued from $B$ in $\triangle ABC$ . Then, its well known that $(NAQ'C)=-1$ , so $MA^2=MQ'.MN=MA.MC=ML.MB=MP.MB$ , so $Q'$ is the second intersection of $\odot(BPN)$ with $AN$ and so $Q'=Q$ , and so $\angle ABP=\angle QBC$ as the symmedian and median are isogonal lines.

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#3 h Mar 27, 2014, 7:29 PM • 1 Y

Y by Adventure10

Let $BQ_1$ be symmedian and $BC>AB$ then $\frac{AQ_1}{Q_1C}=(\frac{AB}{BC})^2=(\frac{LC}{LA})^2=(\frac{AP}{PC})^2$ hence $PQ_1$ is symmedian of $\triangle{APC}$ then $\angle{BNA}=\angle{BAC}-\angle{BCA}=\angle{MPA}-\angle{Q_1PA}=\angle{MPQ_1}$ so we are done!

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