A well-known geo configuration revisited

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A well-known geo configuration revisited

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Source: Dutch TST 2024, 2.1

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#1 h Jun 28, 2024, 3:35 PM • 1 Y

Y by ehuseyinyigit

Let $ABC$ be a triangle with orthocenter $H$ and circumcircle $\Gamma$ . Let $D$ be the reflection of $A$ in $B$ and let $E$ the reflection of $A$ in $C$ . Let $M$ be the midpoint of segment $DE$ . Show that the tangent to $\Gamma$ in $A$ is perpendicular to $HM$ .

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#2 h Jun 28, 2024, 4:20 PM

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Let $N$ be the midpoint of $BC$ . Then reflecting about $N$ sends $M$ to $A$ and $H$ to the $A$ antipode, proving the claim.

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#3 h Jun 29, 2024, 5:25 PM • 2 Y

Y by sami1618, ehuseyinyigit

$a, b, c$ on unit circle, then $h = a + b + c, d = 2 \cdot b - a, e = 2 \cdot c - a, m = \frac{d+e}{2} = b + c - a, h - m = 2a$ .

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lian_the_noob12

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lian_the_noob12

#4 h Aug 1, 2024, 8:38 PM

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Solution by Sir Jawad(TanR314) IMO gold orz orz orz
$I$ be the midpoint of $BC.$ Reflect $H$ over $I$ to $H'$ and Reflecting $A$ over $I$ sends to $M$ . Hence $AHMH'$ is a parallelogram.
$\angle XAH+\angle XHA=\angle XAH +\angle HAH'=90° \blacksquare$
Another solution is by noting $BHCM$ is the nine-point circle of $\triangle ADC$ then showing $\angle XHA=|\angle D -\angle E|$

This post has been edited 1 time. Last edited by lian_the_noob12, Aug 2, 2024, 4:21 PM

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#5 h Sep 7, 2024, 11:36 AM

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idk why i couldn't see the synthetic solution....

Computations left to reader.
Let $N$ be midpoint of $BC$ . Take $N$ as the origin. $N=(0,0), B=(-1,0), C=(1,0), A=(a,b)$ . From $NA=NM$ $\implies$ $M=(-a,-b)$ . From $AO=BO$ $\implies$ $O=(0,\frac{a^2+b^2-1}{2b})$ . Take altitudes from $B$ and $C$ to $AC$ and $AB$ respectively and get equations of lines and their intersection which is $H$ . So, $H=(a,\frac{a^2-1}{-b})$ . We need to prove $AO$ is parallel to $HM$ . Which means their slopes should be equal.
Which is true since, $\frac{b-\frac{a^2+b^2-1}{2b}}{a}=\frac{\frac{a^2-1}{-b}+b}{2a}$ .

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#6 h Mar 26, 2025, 1:04 PM

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Let $O$ and $G$ be the circumcentre and centroid of $\triangle ABC$ respectively.
Note that the homothety with centre G and ratio $-\dfrac{1}{2}$ maps $H \mapsto O, M \mapsto A$ so $HM \parallel AO$ . $\blacksquare$

This post has been edited 1 time. Last edited by iced_tea, Mar 26, 2025, 1:08 PM
Reason: typo

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