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Yet another way to reflect H to the circumcircle!
Tintarn   3
N Nov 24, 2024 by Alex-Five
Source: Baltic Way 2024, Problem 14
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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Tintarn
Nov 16, 2024
Alex-Five
Nov 24, 2024
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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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Tintarn
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Tintarn
#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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bin_sherlo
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bin_sherlo
#2 h  Nov 16, 2024, 6:57 PM • 1 Y
Y by Funcshun840
Solution
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Lil_flip38
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Lil_flip38
#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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Alex-Five
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Alex-Five
#4 h  Nov 24, 2024, 7:26 PM
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