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High School Math Grades 9-12, Ages 13-18

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High School Math Grades 9-12, Ages 13-18

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#1 h May 18, 2025, 3:04 PM

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ChGiven an acute triangle ABC inscribed in circle $(O)$ The altitudes $BE, CF$ , intersect
each other at $H$ . The tangents at $B$ and $C$ of $(O)$ intersect at $S$ . Let $M$ be the midpoint of $BC$ . $EM$ intersects $SC$
at $I$ , $FM$ intersects $SB$ at $J.$
a) Prove that the points $I, S, M, J$ lie on the same circle.
b) The circle with diameter $AH$ intersects the circle $(O)$ at the second point $T.$ The line $AH$ intersects
$(O)$ at the second point $K$ . Prove that $S,K,T$ are collinear.

This post has been edited 2 times. Last edited by luckvoltia.112, May 18, 2025, 3:06 PM
Reason: latex error

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#2 h May 19, 2025, 11:37 AM • 1 Y

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My short solution

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#3 h May 21, 2025, 1:35 PM

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Thanks !

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#4 h Jun 1, 2025, 10:01 PM

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I skipped a), since it's an easy angle chase. Here's an alternate solution to b)

https://i.ibb.co/35nxv2mk/HM-lemma-zim.png

Claim 1 $T$ , $H$ and $M$ are collinear.

Notice that
$\angle BTH=\angle ATB-90^\circ=180^\circ-\angle ACB-90^\circ=\angle EBM$ and
$\angle ETH=\angle CAK=90^\circ-\angle ACB=\angle EBM$ Since $\angle BEC=90^\circ$ , one has $BM=CM=EM$ . Then
$\angle ETB=\angle ETH+\angle BTH=\angle EBM+\angle BEM=180^\circ-\angle BME\implies TBME - \text{cyclic}$ Then $\angle BTM=\angle BEM=\angle BTH$ , where then $T$ , $H$ and $M$ are collinear. $\square$

Claim 2 $\Delta TBS\sim\Delta THC$

Since $S$ is the intersection of tangents to $(ABC)$ , one has $BS=CS$ . Thus $MS$ is also an altitude and angle bisector. Then
$\angle CSM=90^\circ-\angle BCS=90^\circ-\angle BAC=\angle ECH\implies\Delta ECH\sim\Delta MSC$ Then, since $TBME$ - cyclic, one has $\Delta TBH\sim\Delta EMH$ . We need to prove
$\frac{BT}{BS}=\frac{TH}{CH}$ $\frac{CM}{EH}=\frac{EM}{EH}=\frac{BT}{TH}=\frac{BS}{CH}=\frac{CS}{CH}$ $\frac{CH}{EH}=\frac{CS}{CM}$ , which is true, since $\Delta ECH\sim\Delta MSC$ . Also notice, that
$\angle TBE=180^\circ-\angle TAC=\angle THE$ and
$\angle CBS=\angle BAC=\angle CHE$ Then
$\angle TBS=\angle TBC+\angle CBS=\angle THE+\angle CHE=\angle CTH$ Combining these results, we have $\Delta TBS\sim\Delta THC$ . $\square$

Claim 3 $T$ , $K$ and $S$ are collinear.

Since $\Delta TBS\sim\Delta THC$ , one has
$\angle BTS=\angle CTH=90^\circ-\angle ATC=90^\circ-\angle ABC=\angle BAK=\angle BTK$ Then $T$ , $K$ and $S$ are collinear. $\square$

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