VMO 2022 problem 3 day 1

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Source: Vietnam Mathematical Olympiad 2022 problem 3 day 1

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799786

#1 h Mar 4, 2022, 10:11 AM • 1 Y

Y by loveGeometrytothemoon

Let $ABC$ be a triangle. Point $E,F$ moves on the opposite ray of $BA,CA$ such that $BF=CE$ . Let $M,N$ be the midpoint of $BE,CF$ . $BF$ cuts $CE$ at $D$
a) Suppost that $I$ is the circumcenter of $(DBE)$ and $J$ is the circumcenter of $(DCF)$ , Prove that $MN \parallel IJ$
b) Let $K$ be the midpoint of $MN$ and $H$ be the orthocenter of triangle $AEF$ . Prove that when $E$ varies on the opposite ray of $BA$ , $HK$ go through a fixed point

This post has been edited 1 time. Last edited by 799786, Mar 4, 2022, 10:31 AM
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#2 h Mar 5, 2022, 12:34 AM

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The idea of this problem is complete quadrilateral
a) Use similar triangle and a bit of angle chasing
b) note that $HK$ is the Steiner line of quadrilateral $BCFE$

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#3 h Mar 7, 2022, 9:10 PM

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a) Let $MI$ intersects $NJ$ at $P$ . Note that $MI$ and $NJ$ are perpendicular bisectors of $BE$ and $CF$ , we have $PB=PE$ and $PC=PF$ . Combining with $BF=CE$ we get $PEC \cong PBF$ . Hence, $PEB \sim PCF$ and $(I,J),(M,N)$ are corresponding points. So $\tfrac{PI}{PM}=\tfrac{PJ}{PN}$ and therefore $MN$ is parallel to $IJ$ .

b) We claim that $HK$ passes through $T$ , the orthocenter of $ABC$ .
Let $U,V$ be the midpoints of $BF$ and $CE$ . Observe that $K=\tfrac{M+N}{2}=\tfrac{1}{2} \left( \tfrac{B+E}{2} +\tfrac{C+F}{2} \right)=\tfrac{1}{2} \left( \tfrac{B+F}{2} +\tfrac{C+E}{2} \right)=\tfrac{U+V}{2}$ , so $K$ is the midpoint of $UV$ . Also note that $(CE)$ and $(BF)$ have the same radius, so $K$ lies on the radical axis of $(CE)$ and $(BF)$ .

As $H,T$ are orthocenter of $AEF$ and $ABC$ , it's easy to prove that $H,T$ lie on the radical axis of $(CE)$ and $(BF)$ (by introducing the feet of altitudes). So $H,T,K$ are collinear.

Remark. $P$ also lies on the radical axis, because $PEC \cong PBF$ implies $PU=PV$ .

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#4 h Jul 2, 2023, 12:49 PM

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#5 h Jul 12, 2024, 2:18 PM

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a) Let $S$ be second intersection of $(DBE)$ and $(DCF)$ . We have $\angle{SFD} = \angle{SCD}$ and $\angle{SED} = \angle{SBD}$ . Combine with $BF = CE,$ we have $\triangle SBF = \triangle SEC$ . Hence $SB = SE$ and $SC = SF$ . From this, we have $S, I, M$ are collinear and $S, J, N$ are collinear. We also have $\angle{SBE} = \angle{SDE} = \angle{SFC} = \angle{SCF} = \angle{SDF} = \angle{SEB}$ . Then $\triangle SBE \cup I \cup M \sim \triangle SFC \cup J \cup N$ . So $\dfrac{SI}{SM} = \dfrac{SJ}{SN}$ or $IJ \parallel MN$ .
b) Suppose that $P, Q$ are midpoints of $BF, CE$ . Then it's easy to see that $MPNQ$ is parallelogram. So $K$ is midpoint of $EF$ . But $BF = CE$ then $K$ lies on radical axis of $\bigodot(BF)$ and $\bigodot(CE)$ . Hence $HK$ is Steiner line of completed quadrilateral formed by $BC, CE, EF, FB$ . This means $HK$ passes through orthocenter of $\triangle ABC,$ which is a fixed point.

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