Circles $O_1, O_2$ intersects at $A, B$. The circumcircle of $O_1BO_2$ intersects $O_1, O_2$ and line $AB$ at $R, S, T$ respectively. Prove that $TR = TS$
Problem
Source: 2017 Thailand October Camp 1.3
Tags: geometry, circumcircle
EmilXM
16.10.2020 01:57
Denote $\angle TSR = \alpha , \angle TRS = \beta , \angle O_1BR = \phi , \angle O_2BS = \theta, K = O_1 \cap TR, M = O_2\cap TS$
Claim: $\alpha+\theta +\phi = 90^o$ Proof$TR \cdot TK = TA\cdot TB = TS\cdot TM \Longrightarrow RSMK$ is cyclic$\Longrightarrow \angle RBA = \angle RKA = \angle RST = \alpha$. Similarly $\angle ABS = \beta$,
$O_1RSB$ is cyclic $\Longrightarrow RSO_1 = RBO_1 = \phi$.
$O_2B = O_2A \Longrightarrow \angle O_2BA = O_2AB = \beta + \theta \Longrightarrow \angle AO_2B = 180^o-2(\beta+\theta)$.
$\angle ASB = \frac{\angle AO_2B}{2} = \frac{180^o-2(\beta + \theta)}{2} = 90^o-\beta-\theta$.
$O_2S = O_2B \Longrightarrow \angle O_2SB = \angle O_2BS = \theta$.
$O_2STB$ is cyclic $\Longrightarrow O_2SM = O_2BT = \beta + \theta$.
$180^o = \angle TSM = \angle TSR + \angle RSB + \angle BSO_2 + O_2SM = \alpha + \phi + 90^o-\beta-\theta + \theta + \beta + \theta \Longrightarrow \alpha+\theta +\phi = 90^o$.
Analogously $\beta + \theta + \phi = 90^o$. So $\alpha = \beta \Longrightarrow \boxed{TR = TS}$
dgreenb801
16.10.2020 03:22
Suppose $T$ lies outside both circles. Since $O_1R=O_1B$, we have $\angle RTO_1 =\angle BTO_1 =\angle ATO_1$, since they intercept equal arcs in the circle $(O_1RTSO_2B)$. By symmetry, $\triangle TRO_1 \cong \triangle TAO_1$, so $TR=TA$. Similarly, $TS=TA$, so $TR=TS$. If $T$ lies inside both circles, the argument is similar.