equal angles wanted, circle of diameter AB and touchpoints with incircle

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equal angles wanted, circle of diameter AB and touchpoints with incircle

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Source: 2013 Swiss IMO TST p11

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#1 h May 21, 2020, 8:00 AM

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Let $ABC$ be a triangle with $\angle ACB \ge 90^o$ and $k$ the circle with diameter $AB$ and center $O$ . The incircle of $ABC$ touches $AC ,BC$ at $M,N$ respectively. $MN$ intersects $k$ at points $X$ and $Y$ . Show that $\angle XOY = \angle ACB$ .

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#2 h May 21, 2020, 10:12 AM • 2 Y

Y by Mango247, Mango247

By Iran's lemma, we know that $X= k\cap CI, Y=k\cap BI$ . Angle chasing gives $\angle XOA=\angle B\implies XO\| CB$ and similarly, $YO\| CA$ . Thus, $\angle XOY=\angle ACB$ .

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#3 h May 21, 2020, 10:51 AM • 2 Y

Y by Mango247, Mango247

Let $AI$ and $BI$ meet $(k)$ at $X'$ and $Y'$ . Since $\angle AY'B=\angle IY'B=\angle INB=\angle ILB=90^{\circ}$ it follows that $LINY'B$ is cyclic. Similarly $LIMX'A$ is cyclic.

It is easy to see that $OY'\parallel AC$ and $OX'\parallel BC$ , which implies $\angle X'OY'=\angle ACB$

Now, since $\triangle CMN$ is isosceles, $\angle CMN=\angle CNM=\frac{\angle A+\angle B}{2}=\angle Y'IB=\angle Y'NB$

$\therefore Y'$ lies on $MN$ and similarly, $X'$ lies on $MN$ , which means $X'=X$ and $Y'=Y$ and hence $\angle XOY=\angle ACB$ . $\blacksquare$

edit:
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This post has been edited 5 times. Last edited by Greenleaf5002, May 21, 2020, 11:46 AM
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