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Source: Vietnam TST 2021 P5

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#1 h Apr 2, 2021, 8:20 AM • 5 Y

Y by Maths_geometryTPC2023, Brian_Xu, Mango247, Mango247, Mango247

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.

This post has been edited 2 times. Last edited by CheshireOrb, Apr 2, 2021, 8:32 AM

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#2 h Apr 2, 2021, 11:11 AM • 1 Y

Y by Brian_Xu

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 $(HBC)$

This post has been edited 1 time. Last edited by khanhnx, Apr 2, 2021, 11:11 AM

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#3 h Jan 2, 2023, 3:50 AM

Y by

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.

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#4 h Jan 6, 2023, 3:03 PM • 2 Y

Y by GeoKing, Mango247

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

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#5 h Jan 25, 2024, 2:14 PM

Y by

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 $C$ .

This post has been edited 2 times. Last edited by math_comb01, Jan 25, 2024, 2:17 PM

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