Not so classic orthocenter problem

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Not so classic orthocenter problem

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#1 h Mar 28, 2025, 4:59 PM • 1 Y

Y by ehuseyinyigit

Let $O$ be circumcenter of a non-isosceles triangle $ABC$ and $H$ be a point in the interior of $\triangle ABC$ . Let $E,F$ be foots of perpendicular lines from $H$ to $AC,AB$ . Suppose that $BCEF$ is cyclic and $M$ is the circumcenter of $BCEF$ , $HM\cap AB=K,AO\cap BE=T$ . Prove that $KT$ bisects $EF$

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#2 h Mar 29, 2025, 5:59 AM • 1 Y

Y by Funcshun840

You can actually MMP this twice and since its late I'll only give a sketch.
Let $A'$ antipode of $A$ on $\triangle ABC$ then $H,M,A'$ can be easly MMP'd as $H,M$ boh have degree one by simply checking where your cases are $F=B, E=C, E=F=A$ and $H$ orthocenter. Now from this notice $K$ also has degree one snd so does $T$ now which means that $K,N,T$ colinear is degree 4 and the cases are $E=A, E=C, F=B$ and $H$ orthocenter of $\triangle ABC$ , where on this case its just trivial by Pascal on $(AH)$ so you are done.

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#3 h Mar 29, 2025, 7:57 AM

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Let $Q = (AEF) \cap (ABC)$ which is the Miquel point of $BFEC$ . Note that $\angle MQA = 90$ so $MHQ$ collinear and passes through the $A$ antipode $Z$ . From Menelaus theorem it suffices to show $\frac{BT}{TE} = \frac{BK}{KF}$ Note that $\frac{BT}{TE} = \frac{BA}{AE} \cdot \frac{BZ}{CZ} = \frac{BA}{AE} \cdot \frac{EH}{HF}$ and
$\frac{BK}{KF} = \frac{BH}{HF} \cdot \frac{\sin \angle BHQ}{\sin \angle QHF}$ .
Hence we need $\frac{BA}{AE} \cdot EH = \frac{\sin \angle BHQ}{\sin \angle QHF} \cdot BH \iff BA \cdot \sin \angle BAQ \cdot EH = AE \cdot BH \cdot \sin \angle BHQ$ Since $\frac{BH \sin \angle BHQ}{BA \sin \angle BAQ} = \frac{BH \sin \angle BHQ}{BQ \sin \angle BQA} = \frac{\sin \angle BQH}{\sin \angle BQA} = \frac{BZ}{BA} = \frac{EH}{AE}$ we are done.

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#4 h Mar 29, 2025, 10:19 AM

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A solution without any antipode or MMP (because I haven't knew those :blush:

:blush:

)
Sketch
Let $AR$ be diameter of $(ABC)$ , $N$ be midpoint of $EF$ , $I$ be foot of perpendicular line from $M$ to $BF$ . Thus, it remains to prove that $K,N,T$ is collinear, but it is easily proven that $K,M,R$ is collinear and $MN\parallel AO$ (Because $MN\perp EF,EF\perp AO$ ). Thus, it suffices to prove that $\frac{MN}{TR}=\frac{KM}{KR}$ . But notice that $\frac{KM}{KR}=\frac{KI}{KB}=\frac{NI}{BT}$ and $\triangle NIM \sim \triangle TBR$ . Thus, $\frac{KM}{KR}=\frac{KI}{KB}=\frac{NI}{BT}=\frac{MN}{TR}$ and we're done

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#5 h Mar 29, 2025, 4:24 PM

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MathLuis wrote:

You can actually MMP this twice and since its late I'll only give a sketch.
Let $A'$ antipode of $A$ on $\triangle ABC$ then $H,M,A'$ can be easly MMP'd as $H,M$ boh have degree one by simply checking where your cases are $F=B, E=C, E=F=A$ and $H$ orthocenter. Now from this notice $K$ also has degree one snd so does $T$ now which means that $K,N,T$ colinear is degree 4 and the cases are $E=A, E=C, F=B$ and $H$ orthocenter of $\triangle ABC$ , where on this case its just trivial by Pascal on $(AH)$ so you are done.

sorry my english is not good, so what do antipode and MMP mean?

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#6 h Mar 29, 2025, 5:15 PM

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Its a projective solution... Unless you are good in transformations I think you arent familiar with the terms even in your native language.

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#7 h Mar 29, 2025, 5:17 PM • 1 Y

Y by hanzo.ei

hanzo.ei wrote:

sorry my english is not good, so what do antipode and MMP mean?

Antipode is diametrically opposite point and MMP is method of moving points.

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