Tangent circles and a fixed point

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Tangent circles and a fixed point

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Source: Bulgaria Winter Mathematical Competition 2025 12.2

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#1 h Jan 27, 2025, 10:23 AM

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In the plane are fixed two internally tangent circles $\omega$ and $\Omega$ , so that $\omega$ is inside $\Omega$ . Denote their common point by $T$ . The point $A \neq T$ moves on $\Omega$ and point $B$ on $\Omega$ is such that $AB$ is tangent to $\omega$ . The line through $B$ , perpendicular to $AB$ , meets the external angle bisector of $\angle ATB$ at $P$ . Prove that, as $A$ varies on $\Omega$ , the line $AP$ passes through a fixed point.

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#2 h Jan 27, 2025, 5:58 PM • 1 Y

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Let $E$ be the tangency point of $AB$ with $\omega$ . It is well-known that $TE$ bisects the angle $\angle{ATB}$ and since $TP$ is the external angle bisector of $\angle{ATB}$ , we have $\angle{ETP} = 90^{\circ}$ . Let $T \neq F = TP \cap \omega$ and $O$ be the center of $\omega$ . Then, clearly $O$ is the midpoint of $EF$ . Let $X = TP \cap AB$ . Since $\angle{XTE} = 90^{\circ}$ and $\angle{BTE} = \angle{ETA}$ , $-1 = (E, X, B, A) = (E, F, P_{\infty}, PA \cap EF)$ by projecting from $P$ onto the line $EF$ . Thus, $PA \cap EF$ is the midpoint of $EF$ which is $O$ and since it is fixed, we are done.

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#3 h Jan 28, 2025, 1:41 PM

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We claim that $AP$ passes through the center $O$ of $\omega$ . Suppose that $AB$ touches $\omega$ at $D$ . It is well known (as the Shooting lemma) that $TD$ is the angle bisector of $\angle ATB$ (to prove this, either use homothety or angle chase as follows: if $TX$ is the common tangent so that $B$ and $X$ are on different sides of $AT$ , then $\angle ADT = \angle XTD = \angle ATD + \angle ATX = \angle ATD + \angle ABT$ , so $\angle ATD = \angle BTD$ ) and hence $\angle DTP = 90^{\circ}$ .

Since $OD \perp AB$ by the tangency and $BT \perp AB$ by the problem condition, by Thales' theorem it suffices to prove
$\frac{AD}{AB} = \frac{OD}{PB}.$ Denote $\angle BAT = x$ and $\angle ATD = \angle BTD = \varphi$ . Then $\angle ADT = 180^{\circ}- x - \varphi$ , $\angle ODT = \varphi + x - 90^{\circ}$ , so from the isosceles triangle $OTD$ we get $\displaystyle OD = \frac{TD}{2\sin(\varphi + x)}$ . Since the quadrilateral $BPTD$ is cyclic (with diameter $PD$ ), we obtain $\angle DPT = \angle ABT = 180^{\circ} - 2\varphi - x$ , hence $DT = DP\sin(2\varphi + x)$ , as well as $\angle BPD = \angle BTD = \varphi$ with $BP = PD\cos\varphi$ . Thus
$\frac{OD}{PB} = \frac{\sin(2\varphi+x)}{2\cos\varphi\sin(\varphi+x)}.$ On the other hand, the Sine Law in triangles $ATB$ and $ATD$ with common side $AT$ yields
$\frac{AD}{AB} = \frac{AT\sin \varphi}{\sin (\varphi + x)} \cdot \frac{\sin(2\varphi + x)}{AT\sin 2\varphi} = \frac{\sin(2\varphi+x)}{2\cos\varphi\sin(\varphi+x)}$ as $\sin 2\varphi = 2\sin\varphi \cos\varphi$ . This completes the proof.

This post has been edited 1 time. Last edited by Assassino9931, Jan 28, 2025, 1:42 PM

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