Hong Kong TST2 2016 P2

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133 posts

YanYau

#1 h Oct 25, 2015, 9:44 AM • 3 Y

Y by Davi-8191, Adventure10, Mango247

Let $\Gamma$ be a circle and $AB$ be a diameter. Let $l$ be a line outside the circle, and is perpendicular to $AB$ . Let $X$ , $Y$ be two points on $l$ . If $X'$ , $Y'$ are two points on $l$ such that $AX$ , $BX'$ intersect on $\Gamma$ and such that $AY$ , $BY'$ intersect on $\Gamma$ . Prove that the circumcircles of triangles $AXY$ and $AX'Y'$ intersect at a point on $\Gamma$ other than $A$ , or the three circles are tangent at $A$ .

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#2 h Oct 25, 2015, 5:25 PM • 5 Y

Y by YanYau, Stefan4024, mjuk, Adventure10, Mango247

Let $U \equiv AX \cap BX',$ $V \equiv AY \cap BY'$ and let $AX',AY'$ cut $\Gamma$ again at $N,M,$ respectively. Clearly, $A$ is center of inversion taking $\Gamma$ into $l$ and this carries $\odot(AXY)$ and $\odot(AX'Y')$ into $UV$ and $MN,$ respectively. Thus, it's enough to show that $l,MN,UV$ concur.

From the inversion with center $A,$ it follows that $M,N,X',Y'$ are concyclic and since $B$ is also center of inversion taking $\Gamma$ into $l,$ then $U,V,X',Y'$ are concyclic. Thus $UV,MN,l$ are pairwise radical axes of $\Gamma,$ $\odot(MNX'Y'),$ $\odot(UVX'Y')$ concurring at their radical center, as desired.

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YanYau

#3 h Oct 27, 2015, 8:19 AM • 2 Y

Y by Adventure10, Mango247

The solution I wrote during the test was along the lines of this, but in more detail:

Let $AX\cap BX'=X_0$ , $AY\cap BY'=Y_0$ , and $AB\cap l=I$

Notice that $YY_0BI$ and $XX_0BI$ are cyclic, then by PoP/Radical Lemma, $XX_0YY_0$ are cyclic.

Similarly, we can prove that $X'Y'X_0Y_0$ is cyclic.

Consider the radical center of $\Gamma$ , $XYX_0Y_0$ , and $AXY$ . The radical center will be the intersection of $XY$ and $X_0Y_0$ , Let this point be $C$ . Then the radical axis of $\Gamma$ and $AXY$ is $AC$ .

Similarly, consider the radical center of $\Gamma$ , $X'Y'X_0Y_0$ , and $AX'Y'$ , the radical center is the intersection of $X'Y'$ and $X_0Y_0$ , which is again $C$ . So the radical axis of $\Gamma$ and $AX'Y'$ is also $AC$

$\Gamma$ , $AXY$ and $AX'Y'$ all share the same radical axis. Therefore they must all intersect on another point on $\Gamma$ other than $A$ .

The 3 circles are tangent when $X_0Y_0$ and $l$ don't intersect. Which is achieved when $X,Y$ are reflections across $AB$ .

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