circumcenter of XYM lies on BC, excircle and circumcircles related

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Source: Ukraine TST 2014 p4

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#1 h May 1, 2020, 1:28 PM

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The $A$ -excircle of the triangle $ABC$ touches the side $BC$ at point $K$ . The circumcircles of triangles $AKB$ and $AKC$ intersect for the second time with the bisector of angle $A$ at points $X$ and $Y$ respectively. Let $M$ be the midpoint of $BC$ . Prove that the circumcenter of triangle $XYM$ lies on $BC$ .

This post has been edited 1 time. Last edited by parmenides51, May 1, 2020, 1:28 PM

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#2 h May 1, 2020, 10:50 PM • 6 Y

Y by GeoMetrix, vuongnhattinpro, ayan_mathematics_king, Bumblebee60, Mango247, Mango247

Ukraine TST 2014 P4 wrote:

The $A$ -excircle of the triangle $ABC$ touches the side $BC$ at point $K$ . The circumcircles of triangles $AKB$ and $AKC$ intersect for the second time with the bisector of angle $A$ at points $X$ and $Y$ respectively. Let $M$ be the midpoint of $BC$ . Prove that the circumcenter of triangle $XYM$ lies on $BC$ .

Key Claim:- $M$ is the Incenter of $\Delta XKY$

Proof:- Notice that $\angle XKM=\angle CAB=\angle XAC=\angle YKM$ . Now $\frac{MD}{MK}=\frac{BM-BD}{CM-CK}=\frac{\left(\frac{a}{2}-\frac{ac}{b+c}\right)}{\left(\frac{b}{2}-\frac{c}{2}\right)}=\frac{a}{b+c}=\frac{1}{c}\cdot\left(\frac{ac}{b+c}\right)=\frac{DB}{AB}=\frac{DX}{XK}$ Hence, $M$ is the Incenter of $\Delta XKY$ . Hence Circumcenter of $\Delta XMY\in BC.$ $\blacksquare$

This post has been edited 9 times. Last edited by amar_04, May 1, 2020, 10:55 PM

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#3 h Oct 23, 2023, 3:20 AM

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Here’s a pure angle chasing solution. We’re going to prove that $M$ is the incenter of triangle $KXY$ .

Add incenter $I$ . Using the fact that $IM\parallel AK$ , we have
$\angle XIM=\angle XAK= \angle XBK =\angle XBM.$ Thus, $B,I,M,X$ are concyclic, implying that $\angle YXM=\angle IXM=\angle IBM=\frac{\angle ABC}{2}.$ However, $\angle YXK=\angle AXK=\angle ABK=\angle ABC$ . So, $XM$ bisects $\angle YXK$ . Similarly, $YM$ bisects $\angle XYK$ , as desired.

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