there is circle passing S_1,S_2,T_1,T_2 tangent to circumcircle iff <ACB = 90^o

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there is circle passing S_1,S_2,T_1,T_2 tangent to circumcircle iff <ACB = 90^o

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Source: 2020 Thailand TST 4.1

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#1 h Oct 14, 2020, 6:41 PM

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Let $ABC$ be a triangle with circumcircle $\Gamma$ . Let $\omega_0$ be a circle tangent to chord $AB$ and arc $ACB$ . For each $i = 1, 2$ , let $\omega_i$ be a circle tangent to $AB$ at $T_i$ , to $\omega_0$ at $S_i$ , and to arc $ACB$ . Suppose $\omega_1 \ne \omega_2$ . Prove that there is a circle passing through $S_1, S_2, T_1$ , and $T_2$ , and tangent to $\Gamma$ if and only if $\angle ACB = 90^o$ .
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#2 h Oct 19, 2020, 9:39 AM • 3 Y

Y by Mango247, Mango247, Mango247

$\textbf{Lemma:}$ In $\triangle ABC$ , $P$ is an interior point and $D=AP\cap BC$ , $E=BP\cap AC$ and $F=CP\cap AB$ . Let $Q\in AC$ such that $PQ\parallel BC$ , $X=EF\cap AP$ and $Y=CX\cap BE$ . Points $F,Y,Q$ are collinear if and only if $D$ is the midpoint of $BC$ .
$\textbf{Proof:}$ Just simple application of Ceva and Menelaus theorems.

$\textbf{Notation:}$
Perpendicular bisector of $AB$ meet $AB$ at $M$ and $(ABC)$ at $D,E$ with $E$ in the same side of $\omega$ wrt $AB$ .
$\omega$ meet $AB, (ABC)$ at $T, R$ respectively and let $W$ be the center of $\omega$ .
$F=ET\cap (ABC)$ , $K=ER\cap AB$ and the tangent at $(ABC)$ at $R$ meet $AB$ at $L$ .
$S$ is the antipode of $T$ in $\omega$ .
$P=S_1S_2\cap RD$ .
$Q=RF\cap AB$

Back to the problem, I will skip many details.

$\textbf{Solution:}$
1- Inversion at $D$ with radius $DA$ shows that $D,T,R$ are collinear and also $S_1, S_2$ remains fixed so they belongs to inversion circle.

2- Facts:
* $E,S,R$ are collinear.
* $D,F,K$ are collinear.
* $L$ is the midpoint of $KT$ and $LF$ is tangent to $(ABC)$ .

3- Inversion at $L$ with radius $LR$ interchanges $\omega_1$ and $\omega_2$ , then $L,S_1,S_2$ are collinear and $T_1S_1S_2T_2$ is cyclic.

4- With the previous inversion $(FS_1S_2)$ is tangent to $(ABC)$ , let this circle meet $AB$ at $S_1^*,S_2^*$ with $S_1^*$ closer to $B$ . Now $RP$ is the polar of $L$ wrt $\omega$ then $FP$ is the polar of $L$ wrt $(FS_1S_2)$ then by Brocard $S_1^*S_1$ , $S_2^*S_2$ and $FP$ concur.

5- $E,W,P,Q$ are collinear. From $E$ , $(R,P,T,D)$ are harmonic and $TS\parallel DE$ then midpoint of $TS$ is on $EP$ . In $KRTF$ , $EP$ is the polar of $D$ then by Brocard $Q$ is on $EP$ .

6- $S_1,T_1$ and $S_2,T_2$ meet at $S$ .

with the previous and using the lemma we have $(S_1^*,S_2^*) = (S_1, S_2)$ iff $S\in FP$ iff $O$ is the midpoint of $DE$ iff $AB$ is diameter of $(ABC)$ .

$\textbf{Remark:}$ Point $C$ is a red herring, everything can be deduce just from chord $AB$ .

This post has been edited 3 times. Last edited by hectorraul, Oct 19, 2020, 9:45 AM

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#3 h Oct 19, 2020, 10:33 AM

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Point $C$ is irrelevent so we ignore it. We invert the entire problem around point $S$ , the tangency point between $\omega_0$ and $\Gamma$ , dropping $^*,'$ , etc. We get the following new problem.

Inverted Problem wrote:

Let $\omega_0$ be a circle passing through $S,A,B$ . Let $M$ be the midpoint of minor arc $\widehat{AB}$ . Let $\ell$ denote the line tangent to $\omega_0$ at $M$ . For $i=1,2$ , let $\omega_i$ be the circle tangent to $\omega_0$ at $T_i$ , $\ell$ at $S_i$ , and $AB$ . If $\omega_1$ and $\omega_2$ are distinct, prove that there exists a circle through $S_1,S_2,T_1,T_2$ and tangent to $\overline{AB}$ iff $\angle ASB = 90^{\circ}$ .

The advantage of this inversion is the obvious symmetry across the perpendicular bisector of $AB$ ; in particular, it already implies that $S_1,S_2,T_1,T_2$ are concyclic, and moreover, $\odot(S_1S_2T_1T_2)$ , if tangent to $\Gamma$ , must be tangent at the midpoint $K$ of $AB$ .

In addition, note by homothety that the lines $S_1T_1$ and $S_2T_2$ meet at the antipode $X$ of $M$ .

Now, we set $P=XS_1\cap AB$ , and let $Q$ be the tangency point of $\omega_1$ and $AB$ . We finish the problem by noting that
$\begin{align*} K\in\odot(S_1T_1S_2T_2) &\iff PT_1\cdot PS_1 = PK^2 \\ &\iff PQ = PK \\ &\iff P\text{ is midpoint of }\overline{S_1X}\\ &\iff \text{radius}(\omega_1) = \frac{\text{radius}(\omega_0)}{2} \\ &\iff K\text{ is the center of }\omega_0 \\ &\iff \angle ASB = 90^{\circ} \end{align*}$

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