JBMO Shortlist 2021 G3

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Source: JBMO Shortlist 2021

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267 posts

#1 h Jul 2, 2022, 9:09 PM • 1 Y

Y by Kingsbane2139

Let $ABC$ be an acute triangle with circumcircle $\omega$ and circumcenter $O$ . The perpendicular from $A$ to $BC$ intersects $BC$ and $\omega$ at $D$ and $E$ , respectively. Let $F$ be a point on the segment $AE$ , such that $2 \cdot FD = AE$ . Let $l$ be the perpendicular to $OF$ through $F$ . Prove that $l$ , the tangent to $\omega$ at $E$ , and the line $BC$ are concurrent.

Proposed by Stefan Lozanovski, Macedonia

This post has been edited 1 time. Last edited by Lukaluce, Jul 2, 2022, 10:07 PM

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#2 h Jul 3, 2022, 6:58 AM

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Nice problem! Let $T\in BC$ be such that $ET$ is tangent to $\omega$ - we want $T \in \ell$ . The crucialty is to see that $F$ is the midpoint of $CH$ , where $H$ is the orthocenter of $ABC$ , then use $CH = 2OM$ , where $M$ is the midpoint of $AB$ and angle chase.

This post has been edited 1 time. Last edited by Assassino9931, Jul 3, 2022, 6:59 AM

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#3 h Mar 20, 2023, 4:31 PM • 1 Y

Y by rafigamath

<Proof>

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Nartku

#4 h Oct 4, 2024, 2:45 PM

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Trigon bash

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OronSH

#6 h Oct 4, 2024, 3:18 PM

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Define orthocenter $H$ and queue point $Q$ then $F$ is the midpoint of $AH$ and $OF\perp AQ$ . Now $l$ is the $H$ midline of $\triangle AHQ$ so $l\cap BC=Y$ is equidistant to $Q,H$ and thus $E$ . Now $(B,C;Q,E)=-1$ so since the perpendicular bisector of $QE$ meets $BC$ at $Y$ we have $YE,YQ$ tangent to $\omega$ as desired.

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#7 h Mar 13, 2025, 10:17 AM

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Beautiful problem.

Let $H$ be the orthocenter of $\triangle ABC$ .Easy to see that $F$ is the midpoint $AH$ and it is well-known $H$ and $E$ are reflections of each other respect to $BC$ i.e. $D$ is midpoint of $HE$ .Let $K$ is the intersection of $BC$ and the tangent to $\omega$ at $E$ and we show that $K$ lies on $l$ i.e. $\angle OFK = 90$ .Let $M$ is midpoint of $BC$ so we will show that $K,F,O,M,E$ are cyclic because $OM \perp BC$ .

And we know that firstly $AH \parallel OM$ , $AH = 2HF = 2OM$ i.e. $HF = OM$ and this means $HFOM$ is parallelogram.

$\textbf{Claim-1:}$ $KOME$ is cyclic quadrilateral.

$\textbf{Proof:}$ : We know that $OE \perp KE$ and $OM \perp BC$ and these implies $KOME$ is cyclic. $\square$ .

$\textbf{Claim-2:}$ $FOME$ is cyclic quadrilateral.

$\textbf{Proof:}$ : Let $\angle HFO = a$ . We know that $\angle HFO = \angle OMH = a$ . And $\angle HMD = \angle DME = 90 - a$ . $\angle HFO + \angle EMO = \angle HFO + (\angle OMH + \angle HMD + \angle DME) = a + a + (90 - a) + (90 - a) = 180$ . $\square$ .

With $\textbf{Claim-1}$ and $\textbf{Claim-2}$ $K,F,O,M,E$ are cyclic.

Soo we are done. $\blacksquare$ .

This post has been edited 9 times. Last edited by Frd_19_Hsnzde, Mar 15, 2025, 11:26 PM

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