Yet another way to reflect H to the circumcircle!

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Yet another way to reflect H to the circumcircle!

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Source: Baltic Way 2024, Problem 14

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#1 h Nov 16, 2024, 6:23 PM

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Let $ABC$ be an acute triangle with circumcircle $\omega$ . The altitudes $AD$ , $BE$ and $CF$ of the triangle $ABC$ intersect at point $H$ . A point $K$ is chosen on the line $EF$ such that $KH\parallel BC$ . Prove that the reflection of $H$ in $KD$ lies on $\omega$ .

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#2 h Nov 16, 2024, 6:57 PM • 1 Y

Y by Funcshun840

Solution

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#3 h Nov 17, 2024, 10:44 AM • 1 Y

Y by eddiekopp

Alternate super long solution:
Lemma: in triangle $ABC$ , let $E,F$ be the feet from $B,C$ . Let $A'$ be the reflection of $A$ across the perpendicular bisector of $BC$ . Let $Q$ be the $A$ queue point. Then, $QAEF\sim QA'CB$ .
Proof: Consider the spiral similarity that maps $(AEF)$ to $(ABC)$ . $F$ is sent to $B$ , $E$ to $C$ so we need that $A$ is mapped to $A'$ . as $HAA'=90^\circ$ , $AA'$ is tangent to $(AEF)$ , so by properties of spiral simiarities, the lemma is proven.

Now, let $H'$ be the intersection of the circle with radius $HK$ with center $K$ with $(ABC)$ . Let $A'$ be the point s.t. $AA'\parallel BC$ , with $A'\in (ABC)$ , and $Q$ be the $A$ -queue point.

Claim 1: $QK=KH$
proof. $KH$ is tangent to $(AEF)$ as $KH\perp AD$ .
Now, as $(Q,H;E,F)=-1$ and $K\in EF$ we have that $KQ$ has to be tangent as well, which implies the conclusion.

Now by claim $1$ , we have that $K$ is the center of $(QKH)'$ and $(QKH')$ is tangent to $AD$ .
Claim 2: $H,H',A'$ is collinear.
proof: By lemma, $\measuredangle QFA = \measuredangle QBA'$ , so $\measuredangle QHH' = \measuredangle QHA = \measuredangle QFA = \measuredangle QBA' =\measuredangle QH'A'$ as desired.

to finish, by claim 2 we have $\angle A'H'L = 90^\circ$ , where $L=(ABC)\cap AD$ , but by the reflection of orthocenter lemma, $D$ is the midpoint of $HL$ , so $DH'=DL=DH$ so as $KH'=KH$ , $KD$ is the perpendicular bisector of $HH'$ as desired.

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#4 h Nov 24, 2024, 7:26 PM

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