Balkan MO 2005

[Tahmid]

Let $ABC$ be an acute angled triangle whose inscribed circle touches $AB$ and $AC$ at $D$ and $E$ respectively . Let $X$ and $Y$ be the point of intersection of the bisectors of the angles $\angle ACB$ and $\angle ABC$ eith the line $DE$ and let $Z$ be the midpoint of $BC$. Prove that the triangle $XYZ$ is equilateral if and only if $\angle A=60$.

[Nirjhor]

All the angles are directed $\bmod ~\pi$ since there can be $2/3$ distinct configurations. $I$ is the incenter.

Since $AD=AE$ we have $\measuredangle DEA = \measuredangle ADE$. Then notice that \[\begin{eqnarray} \measuredangle DXI &=&\measuredangle DXC=\measuredangle EXC=\measuredangle ECX+\measuredangle XEC=\dfrac 1 2 \measuredangle ACB+\measuredangle DEA \\ &=&\dfrac 1 2 \measuredangle ACB+\dfrac 1 2 \left(\measuredangle DEA+\measuredangle ADE\right)=\dfrac 1 2 \measuredangle ACB+\dfrac 1 2 \measuredangle DAE \\ &=&\dfrac 1 2\left(\measuredangle ACB+\measuredangle BAC\right)=\dfrac 1 2 \measuredangle ABC=\measuredangle DBI \end{eqnarray}\] hence points $B, D, X, I$ are concyclic, so are $C, E, Y, I$. Then $\measuredangle BXI=\measuredangle BDI=\pi/2=\measuredangle IYC$. It follows that the points $B,X,Y,C$ lie on the circle with center $Z$ and radius $ZB=ZX=ZY=ZC$. Hence $\measuredangle ZXC=\measuredangle XCZ=\measuredangle ACX$ implying $CX\parallel AC$ and similarly $CY\parallel AB$. It quickly follows that $\triangle ADE\sim \triangle XYZ$. Being isosceles triangle, $\triangle ADE$ is equilateral if and only if $\angle A=\pi/3$ and by similarity, so is $\triangle XYZ$. $\square$

[tanmoy]

All the angles are directed $\bmod ~\pi$ since there can be $2/3$ distinct configurations. $I$ is the incenter.

Since $AD=AE$ we have $\measuredangle DEA = \measuredangle ADE$. Then notice that \[\begin{eqnarray} \measuredangle DXI &=&\measuredangle DXC=\measuredangle EXC=\measuredangle ECX+\measuredangle XEC=\dfrac 1 2 \measuredangle ACB+\measuredangle DEA \\ &=&\dfrac 1 2 \measuredangle ACB+\dfrac 1 2 \left(\measuredangle DEA+\measuredangle ADE\right)=\dfrac 1 2 \measuredangle ACB+\dfrac 1 2 \measuredangle DAE \\ &=&\dfrac 1 2\left(\measuredangle ACB+\measuredangle BAC\right)=\dfrac 1 2 \measuredangle ABC=\measuredangle DBI \end{eqnarray}\] hence points $B, D, X, I$ are concyclic, so are $C, E, Y, I$. Then $\measuredangle BXI=\measuredangle BDI=\pi/2=\measuredangle IYC$. It follows that the points $B,X,Y,C$ lie on the circle with center $Z$ and radius $ZB=ZX=ZY=ZC$. Hence $\measuredangle ZXC=\measuredangle XCZ=\measuredangle ACX$ implying $CX\parallel AC$ and similarly $CY\parallel AB$. It quickly follows that $\triangle ADE\sim \triangle XYZ$. Being isosceles triangle, $\triangle ADE$ is equilateral if and only if $\angle A=\pi/3$ and by similarity, so is $\triangle XYZ$. $\square$

I have got the same solution.

[Tahmid]

well, i have proved the first part in differrent way . here it is ,

let $BX\cap AC={X}'$ and the incircle touches $BC$ at $F$
now , apply menelus's
$ \frac{AE}{E{X}'}\frac{{X}'X}{XB}\frac{BD}{DA}=1 \Leftrightarrow \frac{{X}'X}{XB}=\frac{E{X}'}{BD}=\frac{E{X}'}{BF}$
as , $\frac{{X}'X}{XB}=\frac{CX}{CB}$ [angle bisector theorem]
$\therefore \frac{CX}{CB}=\frac{E{X}'}{BF} \leftrightarrow B{X}'\parallel EF $
$\therefore CX\perp BX$
samely we can show that $BY\perp CY$

so, $B,C,Y,X$ lie on a circle with diameter $BC$

and the rest part is as same as nirjhor.