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
Problem
Source: Vietnam Mathematical Olympiad 2022 problem 3 day 1
Tags: vmo, geometry, circumcircle
799786
05.03.2022 03:34
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 $BCFE$
NTstrucker
08.03.2022 00:10
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 $PU=PV$.
trinhquockhanh
02.07.2023 15:49
khanhnx
12.07.2024 17:18
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.