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Source: Turkey TST 2025 P7

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#1 h Today at 6:56 AM

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Let $\omega$ be a circle on the plane. Let $\omega_1$ and $\omega_2$ be circles which are internally tangent to $\omega$ at points $A$ and $B$ respectively. Let the centers of $\omega_1$ and $\omega_2$ be $O_1$ and $O_2$ respectively and let the intersection points of $\omega_1$ and $\omega_2$ be $X$ and $Y$ . Assume that $X$ lies on the line $AB$ . Let the common external tangent of $\omega_1$ and $\omega_2$ that is closer to point $Y$ be tangent to the circles $\omega_1$ and $\omega_2$ at $K$ and $L$ respectively. Let the second intersection point of the line $AK$ and $\omega$ be $P$ and let the second intersection point of the circumcircle of $PKL$ and $\omega$ be $S$ . Let the circumcenter of $AKL$ be $Q$ and let the intersection points of $SQ$ and $O_1O_2$ be $R$ . Prove that
$\frac{\overline{O_1R}}{\overline{RO_2}}=\frac{\overline{AX}}{\overline{XB}}$

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#2 h an hour ago • 1 Y

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Let $AB$ and $KL$ meet at $H$ . By the Shooting Lemma, $AK$ and $BL$ intersect at the arc midpoint $P$ and furthermore $PK\cdot PA=PL\cdot PB$ so $ABKL$ is cyclic. By the Radical Axes Theorem applied to $(ABKL)$ , $w_1$ , and $w_2$ we get that $P$ also lies on $XY$ . Then $ABKL$ is a complete cyclic quadrilateral with Miquel Point $S$ . It is well known that $S$ lies on $PH$ , and that $Q$ , the circumcenter of $(ABKL)$ , lies on the perpendicular from $S$ to $PH$ .

Since $\angle AXK=180^{\circ}-\angle AKL=\angle XBL$ and similarly $\angle BXL=\angle XAK$ so triangles $\triangle AXK$ and $\triangle XBL$ are similar. Thus $H$ is the center of the homothety sending $w_1$ to $w_2$ . Thus $H$ , $O_1$ , and $O_2$ are collinear. This also means that $HX^2=HA\cdot HB=HP\cdot HS$ . Thus $\angle XSH=\angle PXH=\angle YXH=\angle XYH$ , so $XYSH$ is cyclic. Since $\angle RSH=90^{\circ}$ and $R$ lies on $O_1O_2$ it follows that $R$ is the antipode of $H$ on this circle. Then $\angle RXH=90^{\circ}$ . Now if we let $A'$ and $B'$ to be the midpoints of $XA$ and $XB$ then: $\frac{\overline{O_1R}}{\overline{RO_2}}=\frac{\overline{A'X}}{\overline{XB'}}=\frac{\overline{AX}}{\overline{XB}}$

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