Given a fixed circle $(O)$ and two fixed points $B, C$ on that circle, let $A$ be a moving point on $(O)$ such that $\triangle ABC$ is acute and scalene. Let $I$ be the midpoint of $BC$ and let $AD, BE, CF$ be the three heights of $\triangle ABC$. In two rays $\overrightarrow{FA}, \overrightarrow{EA}$, we pick respectively $M,N$ such that $FM = CE, EN = BF$. Let $L$ be the intersection of $MN$ and $EF$, and let $G \neq L$ be the second intersection of $(LEN)$ and $(LFM)$. a) Show that the circle $(MNG)$ always goes through a fixed point. b) Let $AD$ intersects $(O)$ at $K \neq A$. In the tangent line through $D$ of $(DKI)$, we pick $P,Q$ such that $GP \parallel AB, GQ \parallel AC$. Let $T$ be the center of $(GPQ)$. Show that $GT$ always goes through a fixed point.
Problem
Source: Vietnam TST 2021 P5
Tags: geometry
02.04.2021 14:11
Here is my solution for this problem Solution a) Since $BM = BF + FM = EN + CE = CN$ and $G$ is Miquel point of completed quadrilateral $ANLF.EM;$ we have $(AMGN)$ passes through midpoint of arc $BAC$ of $(O)$ b) Let $H$ be orthocenter of $\triangle ABC;$ ray $IH$ intersects $(O)$ at $S$. Redefine $G$ be a point such as $\triangle ABC \cup S$ $\sim$ $\triangle AEF \cup G$. We have $\dfrac{GE}{GF} = \dfrac{HF}{HE} = \dfrac{BF}{CE} = \dfrac{EN}{FM}$. Then $\dfrac{GE}{EN} = \dfrac{GF}{FM}$. But $\angle{GEN} = \angle{GFM}$ then $\triangle GEN$ $\sim$ $\triangle GFM$ or $G$ is Miquel point of completed quadrilateral $ANLF.EM$. We also have $\angle{AGP} = 180^{\circ} - \angle{GAB} = 180^{\circ} - \angle{CAS} = \angle{GQP}$. Then $AG$ $\perp$ $GT$ or $H, T, G$ are collinear. Since $HG$ passes through midpoint of $EF,$ then $HG$ is $H$ - symmedian of $\triangle HBC$ or $\overline{H, T, G}$ passes through intersection of tangents at $B, C$ of $(HBC)$
02.01.2023 06:50
khanhnx wrote: Here is my solution for this problem Solution a) Since $BM = BF + FM = EN + CE = CN$ and $G$ is Miquel point of completed quadrilateral $ANLF.EM;$ we have $(AMGN)$ passes through midpoint of arc $BAC$ of $(O)$ b) Let $H$ be orthocenter of $\triangle ABC;$ ray $IH$ intersects $(O)$ at $S$. Redefine $G$ be a point such as $\triangle ABC \cup S$ $\sim$ $\triangle AEF \cup G$. We have $\dfrac{GE}{GF} = \dfrac{HF}{HE} = \dfrac{BF}{CE} = \dfrac{EN}{FM}$. Then $\dfrac{GE}{EN} = \dfrac{GF}{FM}$. But $\angle{GEN} = \angle{GFM}$ then $\triangle GEN$ $\sim$ $\triangle GFM$ or $G$ is Miquel point of completed quadrilateral $ANLF.EM$. We also have $\angle{AGP} = 180^{\circ} - \angle{GAB} = 180^{\circ} - \angle{CAS} = \angle{GQP}$. Then $AG$ $\perp$ $GT$ or $H, T, G$ are collinear. Since $HG$ passes through midpoint of $EF,$ then $HG$ is $H$ - symmedian of $\triangle HBC$ or $\overline{H, T, G}$ passes through intersection of tangents at $B, C$ of $(HBC)$ Hi, In part a, how did you conclude that $(AMGN)$ passes through the midpoint arc so quickly? And part b, how did you come up with the idea let $S$ be the intersection of $IH$ and $(O)$. Thank you.
06.01.2023 18:03
RopuToran wrote: khanhnx wrote: Here is my solution for this problem Solution a) Since $BM = BF + FM = EN + CE = CN$ and $G$ is Miquel point of completed quadrilateral $ANLF.EM;$ we have $(AMGN)$ passes through midpoint of arc $BAC$ of $(O)$ b) Let $H$ be orthocenter of $\triangle ABC;$ ray $IH$ intersects $(O)$ at $S$. Redefine $G$ be a point such as $\triangle ABC \cup S$ $\sim$ $\triangle AEF \cup G$. We have $\dfrac{GE}{GF} = \dfrac{HF}{HE} = \dfrac{BF}{CE} = \dfrac{EN}{FM}$. Then $\dfrac{GE}{EN} = \dfrac{GF}{FM}$. But $\angle{GEN} = \angle{GFM}$ then $\triangle GEN$ $\sim$ $\triangle GFM$ or $G$ is Miquel point of completed quadrilateral $ANLF.EM$. We also have $\angle{AGP} = 180^{\circ} - \angle{GAB} = 180^{\circ} - \angle{CAS} = \angle{GQP}$. Then $AG$ $\perp$ $GT$ or $H, T, G$ are collinear. Since $HG$ passes through midpoint of $EF,$ then $HG$ is $H$ - symmedian of $\triangle HBC$ or $\overline{H, T, G}$ passes through intersection of tangents at $B, C$ of $(HBC)$ Hi, In part a, how did you conclude that $(AMGN)$ passes through the midpoint arc so quickly? And part b, how did you come up with the idea let $S$ be the intersection of $IH$ and $(O)$. Thank you. In part a, I used the following lemma Lemma. Given $\triangle ABC$ and $M, N$ lie on ray $BA, CA$ such as $BM = CN$. Then $(AMN)$ passes through midpoint of arc $BAC$ of $(ABC)$ In part b, you can see that $HG$ passes through midpoint of $EF,$ so $HG$ must passes through orthocenter of $\triangle AEF$. But $\triangle AEF \sim \triangle ABC,$ so that we need to define the point $S$ as above
25.01.2024 17:14
Lol Vietnam TST geos are so long... (a) Notice by angle chase $G$ is miquel point of $ANEFLM,$ so by the following lemma this part is done, Lemma:In triangle $ABC$, if $D,E$ lie on $AB,AC$ s.t. $BD=CE$ then $(ADE)$ passes through the arch midpoint of larger arc $\widehat{BC}$ (b) Notice since $G$ is miquel point of $ANEFLM$, $AG \perp GT$, so $HG$ and the midpoint of $EF$ is collinear, then by the following we're done Lemma: If $M$ is midpoint of $EF$, then $HM$ is the symmedian in $\triangle HBC$. Hence the fixed point intersection of tangents to $(HBC)$ at $B$ and $C$.