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Source: 2020 Serbian MO, Problem 3

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#1 h Sep 26, 2020, 9:18 PM

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We are given a triangle $ABC$ . Points $D$ and $E$ on the line $AB$ are such that $AD=AC$ and $BE=BC$ , with the arrangment of points $D - A - B - E$ . The circumscribed circles of the triangles $DBC$ and $EAC$ meet again at the point $X\neq C$ , and the circumscribed circles of the triangles $DEC$ and $ABC$ meet again at the point $Y\neq C$ . Find the measure of $\angle ACB$ given the condition $DY+EY=2XY$ .

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#2 h Nov 16, 2020, 2:25 PM

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WLOG $CA\ge CB$ .

$\textbf{Claim 1:}$ $X$ is the $C$ -excenter of $\triangle ABC$ .
$\textit{Proof:}$ Define $X'$ as the $C$ -excenter. From easy angle chasing, we have $\angle DCX'+\angle CX'A=90^{\circ}$ which implies $X'A\perp CD$ . We also have $CA=AD$ , hence $X'A$ is the perpendicular bisector of $CD$ . Therefore,
$\angle X'DB=\angle X'DA=\angle X'CA=\angle X'CB$ So, $DCBX'$ is cyclic. Similarly, we can prove that $ECAX'$ is cyclic. Therefore, $X=X'$ .

$\textbf{Claim 2:}$ $X$ is the circumcenter of $DCE$
$\textit{Proof:}$ Just note that as $X$ is the $C$ -excenter, we have $XA$ is perpendicular bisector of $CD$ and $XB$ is perpendicular bisector of $CE$ .

Now, define $O$ as the circumcenter of $ABC$ , $T$ as the midpoint of arc $\widehat{AB}$ not containing $C$ and $Z$ as the reflection of $X$ wrt line $AB$ . By claim 2, we have $XY=XC$ , hence $XO$ is the perpendicular bisector of $CY$ .

$\textbf{Claim 3:}$ $T,Z,Y$ are collinear.
$\textit{Proof:}$ Define $F=ZT\cap AB$ and $G$ such that $GT$ is tangent to $\odot (ABC)$ and $A,G$ lies on the same side of $CX$ . To prove the claim, we only need to prove that $\angle ZTG=\angle YTG$ . Note that
$\angle ZTG=\angle ZFA=90^{\circ}-\angle XZT$ and we also have
$\angle YTG=\angle YCT=\angle YCX=90^{\circ}-\angle OXC$ So, we just need to prove that $\angle XZT=\angle OXC$ . To do this, we will use complex number. As usual, we will use lowescript of a letter as it's complex coordinate. WLOG $\odot (ABC)$ is a unit circle. Assign $a=p^2, b=q^2,$ and $c=r^2$ . Hence, we will have $|p|=|q|=|r|=1$ . It's easy to check that $t=-pq$ and $x=qr+pr-pq$ .
$Z$ is the reflection of $X$ wrt to line $AB$ , so
$z=a+b-ab\overline{x}=p^2+q^2-p^2q^2(\frac{1}{qr}+\frac{1}{pr}-\frac{1}{pq})=\frac{p^2r+q^2r-p^2q-pq^2+pqr}{r}$ It suffices to prove that $\measuredangle CXO=\measuredangle XZT$ which equivalent to $\frac{o-x}{c-x}: \frac{t-z}{x-z}\in \mathbb R$ . But we have
$\frac{o-x}{c-x}:\frac{t-z}{x-z}=\frac{pq-pr-qr}{r^2-pr-qr+pq}\times \frac{qr+pr-pq-\frac{p^2r+q^2r-p^2q-pq^2+pqr}{r}}{-pq-\frac{p^2r+q^2r-p^2q-pq^2+pqr}{r}}=\frac{pq-pr-qr}{(r-p)(r-q)}\times \frac{\frac{(p+q)(r-p)(r-q)}{r}}{\frac{(p+q)(pq-pr-qr)}{r}}=1$ Hence, the claim is true.

Define $F=\overline{TZY}\cap AB$ . It's easy to see that we must have $Y,Z$ lies on the same side of $AB$ . Moreover, because $T,Y\in \odot (ABC)$ and they are on different side of $AB$ , we get that $F$ must lie on segment $AB$ .

$\textbf{Claim 4:}$ The condition $DY+EY=2XY$ implies $Y=Z$
$\textit{Proof:}$ Notice that $XY=XE=XD=ZE=ZD$ . So, $DY+EY=2XY\implies DY+EY=DZ+EZ$ .
Assume the contrary that $Y\neq Z$ . Then from $F,Z,Y$ collinear, $F$ lies on segment $AB$ , and $Y,Z$ lies on the same side of $AB$ , we must have either $Z$ lies strictly inside $\triangle DYE$ or $Y$ lies strictly inside $\triangle DZE$ .
If $Z$ lies strictly inside $\triangle DYE$ , it's well known that $DY+EY>DZ+EZ$ . A contradiction
Similarly, if $Y$ lies strictly inside $\triangle DZE$ , we have $DZ+EZ>DY+EY$ . Again another contradiction
Therefore, we must have $Y=Z$ . Claim 4 is now proven

Now notice that
$\angle DYE=\angle DCE=\angle DCA+\angle ACB+\angle BCE=\frac{\angle CAB}{2}+\angle ACB+\frac{\angle CBA}{2}=90^{\circ}+\frac{\angle ACB}{2}$ and we have
$\angle DZE=\angle DXE=2(180^{\circ}-\angle DCE)=180^{\circ}-\angle ACB$ Because $Y=Z$ , then $90^{\circ}+\frac{\angle ACB}{2}=180^{\circ}-\angle ACB\implies \angle ACB=\boxed{60^{\circ}}$ .

This post has been edited 8 times. Last edited by Sugiyem, Nov 24, 2020, 3:35 PM
Reason: typo mulu tai

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#3 h Nov 17, 2020, 3:39 AM

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Sugiyem wrote:

$\textbf{Claim 4:}$ The condition $DY+EY=2XY$ implies $Y=Z$

Here's a much fancier proof: Assume that $Y \not= Z$ . We know that we have $DY + EY = DZ + EZ$ . Notice that $F,Z,Y$ is collinear, where $F$ lies on segment $AB$ ; and $Z,Y$ lies on the same side of $DE \equiv AB$ . However, consider the ellipse with foci $D$ and $E$ . Notice that it is well known that $Y,Z$ lies on this ellipse as well. However, as $Y,Z$ lies on a line as well: they must be the two points of intersection of the ellipse and the line; which must lie on different side of the ellipse as the line intersects segment $AB$ , which forces a contradiction.

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