Two circles $\Gamma_1$ and $\Gamma_2$are given with centres $O_1$ and $O_2$ and common exterior tangents $\ell_1$ and $\ell_2$. The line $\ell_1$ intersects $\Gamma_1$ in $A$ and $\Gamma_2$ in $B$. Let $X$ be a point on segment $O_1O_2$, not lying on $\Gamma_1$ or $\Gamma_2$. The segment $AX$ intersects $\Gamma_1$ in $Y \ne A$ and the segment $BX$ intersects $\Gamma_2$ in $Z \ne B$. Prove that the line through $Y$ tangent to $\Gamma_1$ and the line through $Z$ tangent to $\Gamma_2$ intersect each other on $\ell_2$.
Problem
Source: 2022 Dutch IMO TST 1.2
Tags: geometry, concurrency, common tangents
CristianTurcan
30.03.2023 00:40
Let $M$ be the intersection of the line tangent to $\Gamma_1$ through $Y$ and $AB$. Define $\ell_1\cap\ell_2=K$. Let $\ell_2\cap\Gamma_1=D$, $\ell_2\cap\Gamma_2=C$ and $MY\cap\ell_2=T$. We will prove now that $TZ$ is tangent to $\Gamma_2$: Let $\angle AXK=\alpha$, $\angle BXO_{2}=\beta$ and $\angle BKC=2\gamma$. $$\angle KAY=\angle KAX=180^{\circ} - \gamma - \alpha$$$$\angle MAY=180^{\circ} -\angle KAY=\gamma +\alpha$$$$\angle AMY=180^{\circ}-2\gamma -2\alpha$$$$\angle DAY=\angle KAY -\angle KAD=180^{\circ} -\gamma-\alpha-90^{\circ} + \gamma=90^{\circ} -\alpha$$$DT$ is tangent to $\Gamma_1$, therefore: $$\angle YDT=\angle DAY=90^{\circ} -\alpha$$$$\angle YTD=180^{\circ}-2\angle YDT=180^{\circ}-180^{\circ}+2\alpha=2\alpha$$$KX$ is median and altitude from $X$ in $\triangle AXD$, therefore $\triangle AXD$ is isoscel and $KX$ is the bisector of $\angle AXD$, so: $$\angle AXD= 2\angle AXK=2\alpha=\angle YTD$$$\angle YTD=\angle YXD$ so $YXTD$ is cyclic, therefore: $$\angle DXT=\angle DYT=\angle DAY=\angle ADX$$$\angle ADX=\angle DXT$ so $BC\|AD\|XT$. $$XT\|BC\implies \angle TXB=180^{\circ} -\angle XBC$$$$KC\perp\Gamma_2\implies\angle ZCT=\angle XBC\implies \angle TXZ+\angle ZCT=180^{\circ}\implies TXZC - cylcic$$$\triangle XBC$ is isoscel so $\angle BXC= 2\beta$ and $\angle XBC =90^{\circ} - \beta$. $$\angle TZC=\angle TXC = \angle TXZ - \angle ZXC= 180^{\circ} -90^{\circ} + \beta -2\beta=90^{\circ} -\beta=\angle ZBC$$$\angle TZC= \angle ZBC$ therefore $TZ$ is tangent to $\Gamma_2$.
Ferum_2710
22.04.2023 10:23
Let $C$ be the reflection of. So, $$\angle PYX = 180^{\circ} - \angle AYP = 180^{\circ} - \angle PAY = 180^{\circ} - \angle PCX$$. Which means $CPYX$ is cyclic. Let $S$ be the point of intersection of tangent to circle $\Gamma_1$ at $Y$ with the line $\ell_2$. As, $$\angle PYS = \angle PCS = 90^{\circ} $$, then $CSYP$ is cyclic. Then, pentagon $CSXYP$ is cyclic. $$\angle SYP = 90^{\circ} = \angle SXP$$. Which means $SX$ is perpendicular to $PQ$. The same way we can call the point of intersection of tangent to circle $\Gamma_2$ at $Z$ with the line $\ell_2$ $D$. We will get that line $DX$ is perpendicular to $PQ$. From here we get that $S = D$.