Problem 18: APMO 2016, Problem 3

by henderson, Jun 6, 2016, 2:37 PM

$\bf\color{red}{Problem \ 18 }$ Let $AB$ and $AC$ be two distinct rays not lying on the same line, and let $\omega$ be a circle with center $O$ that is tangent to ray $AC$ at $E$ and ray $AB$ at $F$ . Let $R$ be a point on segment $EF$ . The line through $O$ parallel to $EF$ intersects line $AB$ at $P$ . Let $N$ be the intersection of lines $PR$ and $AC$ , and let $M$ be the intersection of line $AB$ and the line through $R$ parallel to $AC$ . Prove that line $MN$ is tangent to $\omega$ .
$\bf\color{red}Solution \ 1 \$ Let $X$ be the second tangency point from $N$ to $\omega$ and $Y$ be the reflection of $F$ over $OP.$ Clearly $Y$ lies on $\omega$ and $EY$ is the diameter of $\omega,$ moreover $Y$ is the second tangency point form $P$ to $\omega.$

Let $R^*$ be the intersection of $\overline{XY}$ and $\overline{EF}.$ It's easy to see that $N,P,R^*$ lies on the polar of $EX\cap YF,$ hence we conclude that $R^*\equiv R.$

Therefore $\angle RXF=\tfrac{1}{2}\angle A=\tfrac{1}{2}\angle RMF$ (here $\angle RMF=\angle A$ follows by parallel condition), which amounts to $M$ is the center of $\odot(XRF)\implies MX=MF\implies MX$ is tangent to $\omega.$ The statement follows. $\square$
$[asy] size(7cm); pointpen=black; pathpen=black; pointfontpen=fontsize(9pt); void b(){ pair O=D("O",origin,S); pair E=D("E",dir(150),dir(150)); pair F=D("F",dir(30),dir(-100)*1.5); pair A=D("A",IP(L(-E*dir(90)+E,E,5,5),L(-F*dir(-90)+F,F,5,5)),N); pair P=D("P",IP(L(F-E,O,5,5),L(A,F,5,5)),dir(45)); pair R=D("R",6*F/11+5*E/11,dir(-130)); pair N=D("N",extension(P,R,A,E),dir(135)); pair X=D("X",IP(CP(N,E),unitcircle),dir(60)); pair M=D("M",extension(N,X,A,F),dir(45)); pair Y=D("Y",extension(X,R,E,O),dir(-60)); D(unitcircle); D(A--E--F--cycle); D(N--M,blue+dashed); D(X--Y--E,red); D(O--P--F); D(M--R); D(CP(M,R),dotted); D(X--O--F); D(Y--P,dashed); } b(); pathflag=false; b(); [/asy]$

$\bf\color{red}Solution \ 2 \$ Let $T_{\infty}$ be the infinity point on $AE.$ From $OP$ $\parallel$ $EF$ we get that the second tangent from $P$ to $\odot (O)$ is parallel to $AE,$ so from the converse of Brianchon's theorem for the degenerate hexagon $MFPT_{\infty}EN$ we conclude that $MN$ is tangent to $\omega.$

This post has been edited 6 times. Last edited by henderson, Sep 11, 2016, 7:52 PM

Math Path

Problem 18: APMO 2016, Problem 3

by henderson, Jun 6, 2016, 2:37 PM

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