Problem 45: EGMO 2016, Day 1, Problem 2

by henderson, Aug 1, 2016, 7:41 PM

$\color{red}\bf{Problem \ 45}$ Let $ABCD$ be a cyclic quadrilateral, and let diagonals $AC$ and $BD$ intersect at $X.$ Let $C_1,D_1$ and $M$ be the midpoints of segments $CX,DX$ and $CD,$ respectively. Lines $AD_1$ and $BC_1$ intersect at $Y,$ and line $MY$ intersects diagonals $AC$ and $BD$ at different points $E$ and $F,$ respectively. Prove that line $XY$ is tangent to the circle through $E,F$ and $X.$ $($ My solution

$\color{red}\bf{My \ solution}$ $XY$ is tangent to the circumcircle of $\triangle EFX$ $\iff$ $\angle FEX=\angle FXY.$ So, I'll just prove that $\angle FEX=\angle FXY.$
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Since $ABCD$ is a cyclic quadrilateral $\Longrightarrow$ $AX\cdot CX=BX\cdot DX$ $\Longrightarrow$ $AX\cdot \frac{CX}{2}=BX\cdot \frac{DX}{2}$ $\Longrightarrow$ $AX\cdot C_1X=BX\cdot D_1X$ $\Longrightarrow$ $ABC_1D_1$ is a cyclic quadrilateral, too. $\Longrightarrow$ $\frac{AY}{BY}=\frac{YC_1}{YD_1}.$ $\quad \quad \color{red}{(1)}$

Let $\angle D_1YM=\alpha, \angle MYX=\theta$ and $\angle XYC_1=\beta.$ It's obvious that $D_1MC_1X$ is a parallelogram. Thus, $D_1M=XC_1$ and $D_1X=MC_1.$ By sines' law, we obtain
$\frac{D_1M}{\sin \alpha}=\frac{MY}{\sin(\angle YD_1M)} \quad(\triangle D_1YM)$ and
$\frac{C_1M}{\sin (\beta+\theta)}=\frac{MY}{\sin(\angle YC_1M)} \quad(\triangle C_1YM).$ Observe that $\angle YD_1M=\angle YC_1M:$

$\angle YD_1M=180^\circ-\angle AD_1X-\angle XD_1M$ and $\angle YC_1M=180^\circ-\angle BC_1X-\angle XC_1M.$ From the cyclic quadrilateral $ABC_1D_1$ we can obtain $\angle AD_1X=\angle BC_1X,$ and because $D_1MC_1X$ is a parallelogram, it's obvious that $\angle XD_1M=\angle XC_1M,$ so $\angle YD_1M=\angle YC_1M.$

Therefore, the above trigonometric proportions imply
$\frac{D_1M}{C_1M}=\frac{\sin \alpha}{\sin (\beta+\theta)}. \quad \quad \color{red}{(2)}$

On the other hand,
$\frac{D_1X}{\sin(\alpha+\theta)}=\frac{YX}{\sin(\angle XD_1Y)} \quad (\triangle D_1XY)$ and
$\frac{C_1X}{\sin \beta}=\frac{YX}{\sin(\angle XC_1Y)}\quad (\triangle C_1XY).$ Here notice that $\angle XD_1Y=\angle XC_1Y:$

Since $\angle XD_1Y=180^\circ-\angle AD_1X$ and $\angle XC_1Y=180^\circ-\angle BC_1X,$ and $\angle AD_1X=\angle BC_1X$ in the cyclic quadrilateral $ABC_1D_1,$ we are done.

So, $\frac{C_1X}{D_1X}=\frac{\sin \beta}{\sin(\alpha+\theta)}. \quad \quad \color{red}{(3)}$ We've mentioned before that $D_1M=XC_1$ and $D_1X=MC_1,$ since $D_1MC_1X$ is a parallelogram. Then, combining $\color{red}{(2)}$ and $\color{red}{(3)},$ we see that
$\frac{\sin \alpha}{\sin (\beta+\theta)}=\frac{\sin \beta}{\sin(\alpha+\theta)}. \quad \quad \color{red}{(4)}$ Let's use the sines' law again:
$\frac{AE}{\sin \alpha}=\frac{AY}{\sin(\angle AEY)} \quad (\triangle AEY) \quad \text{and} \quad \frac{C_1E}{\sin (\beta+\theta)}=\frac{C_1Y}{\sin(\angle C_1EY)} \quad (\triangle C_1EY)$ $\Longrightarrow$
$\frac{AE}{\sin \alpha} \cdot \frac{\sin (\beta+\theta)}{C_1E}=\frac{AY}{C_1Y}. \quad \text{(remember that} \ \angle AEY+\angle C_1EY=180^\circ ) \quad \quad \color{red}{(5)}$ Similarly, we can conclude that
$\frac{BX}{\sin \beta} \cdot \frac{\sin (\alpha+\theta)}{D_1X}=\frac{BY}{D_1Y}. \quad \quad \color{red}{(6)}$ Comparing $\color{red}{(5)}$ and $\color{red}{(6)},$ we get
$\frac{AE}{BX} \cdot \frac{\sin (\beta+\theta)}{\sin \alpha}\cdot \frac{\sin \beta}{\sin(\alpha+\theta)} \cdot \frac{D_1X}{C_1E}=\frac{AY}{BY}\cdot \frac{D_1Y}{C_1Y}. \quad \quad \color{red}{(7)}$ But, after using $\color{red}{(1)}$ and $\color{red}{(4)}$ ,
$\color{red}{(7)}$ will reduce to
$\frac{AE}{C_1E}=\frac{BX}{D_1X}.$ Now, consider the following process:
$\frac{AE}{C_1E}=\frac{BX}{D_1X}$ $\iff$
$\frac{AE+C_1E}{C_1E}=\frac{BX+D_1X}{D_1X}$ $\iff$
$\frac{AC_1}{C_1E}=\frac{BD_1}{D_1X}$ $\iff$
$\frac{AC_1}{BD_1}=\frac{C_1E}{D_1X}$ $\iff$
$\frac{C_1Y}{D_1Y}=\frac{C_1E}{D_1X} \quad \left(ABC_1D_1 \text{is cyclic} \Longrightarrow \triangle AC_1Y \sim \triangle BD_1Y \Longrightarrow \frac{AC_1}{BD_1}=\frac{C_1Y}{D_1Y}\right) .$ Note that, since $\angle AD_1B=\angle AC_1B,$ we have $\angle EC_1Y=\angle XD_1Y.$ Combining this fact with
$\frac{C_1Y}{D_1Y}=\frac{C_1E}{D_1X},$ we get that $\triangle EC_1Y \sim \triangle XD_1Y$ $\Longrightarrow$ $\angle C_1EY=\angle D_1XY$ $\Longrightarrow$ $\angle XEF=\angle FXY,$ and the solution is completed. $\square$
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