Ukrainian NMO 2016, problem 11.7

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Ukrainian NMO 2016, problem 11.7

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Source: Ukrainian NMO 2016

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#1 h Apr 11, 2016, 9:49 PM • 3 Y

Y by doxuanlong15052000, Adventure10, Rounak_iitr

Problem 11.7. Triangle $ABC$ is given. Circle $\omega$ with center $Q$ is tangent to side $BC$ and touches the circumcircle of triangle $ABC$ internally at point $A.$ Let $M$ be the midpoint of $BC$ and $N$ be the midpoint of arc $BAC$ of the circumcircle of triangle $ABC.$ Point $S$ is chosen on the segment $BC$ such that $\angle BAM=\angle SAC$ . Prove that points $N,$ $Q$ and $S$ are collinear. (M. Plotnikov)

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#2 h Apr 11, 2016, 10:56 PM • 2 Y

Y by Adventure10, Mango247

My solution:
Let $X$ be a point such that $BX$ and $CX$ are tangent to $\odot (ABC)$ and $Y$ be a point in $BC$ such that $AY$ is tangent to $\odot (ABC)$ $\Longrightarrow$ $S=AX\cap BC$ . Let $P=MN\cap \odot (ABC)$ and $R=YP\cap \odot (ABC)$ and $S'=RN\cap BC$ $\Longrightarrow$ $\angle PRN=90^{\circ}$ and $\angle BRS'=\angle CRS'$ $\Longrightarrow$ $RS$ and $PR$ are the internal and external bisector of $\angle BRC$ $\Longrightarrow$ $(Y,S',B,C)=-1$ $\Longrightarrow$ the polar of $Y$ respect to $\odot (ABC)$ through $S'$ , but the polar of $Y$ respect to $\odot (ABC)$ is $AX$ $\Longrightarrow$ $A,X,S'$ are collinear hence $S'=S$ $\Longrightarrow$ $NS\perp YP$ in $R$ . Let $\Omega$ the circumfence of center $N$ and radius $NB$ and let $D=\omega\cap BC$ $\Longrightarrow$ $P,D,A$ are collinear, since $\angle BAP=\angle PBC$ $\Longrightarrow$ $PB^2=PD.PA$ since $PB$ and $PC$ are tangents to $\Omega$ $\Longrightarrow$ the power of point of $P$ respec to $\Omega$ is $PB^2$ and the power of point of $P$ respect to $\omega$ is $PD.PA$ , but $PB^2=PD.PA$ hence $P$ it belongs in the radical axis of $\omega$ and $\Omega$ . Similarly the power of point of $Y$ respect to $\omega$ is $PA ^2$ and the power of point of $Y$ respect to $\Omega$ is $YB.YC$ , but $YA^2=YB.YC$ hence $Y$ it belongs in the radical axis of $\omega$ and $\Omega$ $\Longrightarrow$ $PY$ is the radical axis of $\omega$ and $\Omega$ $\Longrightarrow$ then the line passing through the centers of $\Omega$ and $\omega$ is perpendicular to radical axis $\Longrightarrow$ $NQ\perp YP$ , but also that $NS\perp YP$ hence $N,S,Q$ are collinear.

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#3 h Apr 12, 2016, 8:54 AM • 4 Y

Y by mjuk, enhanced, Adventure10, Mango247

Let $T$ $\in$ $AS$ be the intersection of the tangents of the circumcircle $\odot (O)$ of $\triangle ABC$ passing through $B,$ $C,$ respectively. Since $O$ is the midpoint of arc $BC$ in $\odot (BTC),$ so $N$ is the T-excenter of $\triangle BTC,$ hence by Monge & D'Alembert theorem ( $\odot (N),$ $\odot (O),$ $\odot (Q)$ ) we conclude that $S$ is the exsimilicenter of $\odot (N)$ $\sim$ $\odot (Q)$ $\Longrightarrow$ $N,$ $Q,$ $S$ are collinear.

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#4 h Apr 12, 2016, 3:42 PM • 3 Y

Y by Adventure10, Mango247, soryn

Let k be middle point of arc $BC$ which $A$ is not on it. Let $T$ be the intersection of $AS$ and $MK$ .circle $Q$ touches the segment $BC$ at $D$ . for triangle $ATK$ and line $MDS$ . We use Menelaus's theorem And we can prove $\frac{TN}{NO}=\frac{TM}{MK}$ . (Very easy!) And consider triangle $AOT$ and points $N$ . $Q$ . $S$ . Using Menelaus's theorem ,then we are done

This post has been edited 1 time. Last edited by Stranger8, Sep 26, 2016, 1:20 AM

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#5 h May 14, 2016, 12:02 PM • 2 Y

Y by Adventure10, Mango247

It is well known that $\omega$ touches $BC$ at a point $K$ where $AK$ bisects $\angle BAC$ (trivial to prove by homothety argument). Denote $O$ the circumcenter of $\triangle ABC$ , of course $A, Q, O$ colinear. Let $KN\cap AO=P$ , then note $\frac{QP}{PO}=\frac{QK}{NO}=\frac{QA}{AO} \Rightarrow (NA, NP;NQ, NO)=-1$ .

But trivially $NA$ is external angle bisector of $\angle BAC$ , and thus is external angle bisector of $\angle SAM$ , so $(NA, NK; NS, NM)=-1$ . Since $N, P, K$ are colinear by definition, and of course $N, M, O$ colinear, then lines $NQ, NS$ must coincide, so $N, Q, S$ colinear.

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#6 h Oct 19, 2024, 2:59 AM

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Force overlay invert at $A$ and its a China geo https://artofproblemsolving.com/community/c6h3310538p30571594

This post has been edited 1 time. Last edited by TestX01, Oct 19, 2024, 2:59 AM

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