circumcircle of KHM, tangent to altitude, orthocenter and midpoints related

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circumcircle of KHM, tangent to altitude, orthocenter and midpoints related

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Source: Russian Olympiad 2017 HSO 11.6 High Standards Olympiad - Высшая проба

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#1 h Jun 8, 2020, 7:18 AM

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The heights $AA_1, BB_1, CC_1$ of the acute-angled triangle $ABC$ intersect at point $H$ . Let $M$ be the midpoint of $BC, K$ the midpoint of $B_1C_1$ . Prove that the circle passing through $K, H$ and $M$ is tangent to $AA_1$ .

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This post has been edited 2 times. Last edited by parmenides51, Jun 8, 2020, 8:50 PM

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#2 h Jun 8, 2020, 8:51 AM • 1 Y

Y by Mango247

15 minute bash
Here's a nice solution:
Note that the problem is equivalent to this (taking the $AB1C1$ as the starting triangle):
Given triangle $ABC$ and circumcenter $O$ and A-Antipode $A'$ . Let $S$ be the intersection of the tangets at $B$ and $C$ and $M$ be the midpoint of $BC$ . Note that $OM=R \cdot cos\alpha$ and $OS=\frac{R}{cos\alpha}$ thus $OA'^2=R^2=OS*OM$ so basically we could pick any point not just A-Antipode.

This post has been edited 2 times. Last edited by Steve12345, Jun 8, 2020, 9:04 AM

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#3 h Jun 8, 2020, 7:23 PM • 3 Y

Y by Mango247, Mango247, Mango247

First let's prove something, let $N$ be the midpoint of $AH$ , then $M, K, N$ are collinear, indeed let $K'=MN\cap B_1C_1$ and by since $M, N$ are the centers of $(BCB_1C_1), (AB_1HC_1)$ resp. by radical axis we have that $MN$ is the perpendicular bisector of $B_1C_1$ and $K\equiv K'$ now let $D=AM\cap B_1C_1$ and $E=AK\cap BC$ , notice that $AM, AK$ are isogonal about $\angle BAC$ , then $\angle ADK=\angle DAB_1+\angle DB_1A=\angle EAB+\angle EBA=\angle KEM$ and $KDME$ is cyclic and $\angle DEM=90^{\circ}$ . With this we see that $\angle A_1AM=\angle EDM=\angle EKM =\angle NKA$ and $\angle NAK=\angle KED=\angle KMD$ then $NA^2=NK\cdot NM=NH^2$ and we are done.

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#4 h Jun 8, 2020, 8:48 PM

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a figure by zuss77 from here

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jayme

#5 h Jun 24, 2020, 8:49 AM

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Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/Orthique%20encyclopedie%206.pdf p. 38...

Sincerely
Jean-Louis

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#6 h Jun 24, 2020, 9:33 AM • 1 Y

Y by amar_04

15 seconds solution

Let $N$ be the midpoint of $AH$ , i.e. the center of $\omega = \odot(AB_1C_1H)$ . By a well-known fact, $MB_1$ and $MC_1$ are tangent to $\omega$ . Thus $\overline{NMK}$ is the perpendicular bisector of $B_1C_1$ and $NK\cdot KM = NB_1^2 = NH^2$ , done.

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khina

#7 h Jun 25, 2020, 5:18 PM • 4 Y

Y by kiyoras_2001, Mango247, Mango247, Mango247

wait isn't this just a specific version of isl 2009 g4?

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