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 $\omega$.
Problem
Source: Baltic Way 2024, Problem 14
Tags: geometry, circumcircle, geometry proposed, geometric transformation, reflection
bin_sherlo
16.11.2024 21:57
If $P$ is the altitude from $H$ to $KD$, then $KP.KD=KH^2=KF.KE$ hence $P$ lies on the nine point circle. Applying homothety centered at $H$ with radius $\frac{1}{2}$ finishes.$\blacksquare$
Lil_flip38
17.11.2024 13:44
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.
Alex-Five
24.11.2024 22:26
For storage
First, notice that it suffices to show that the foot from $H$ to $DK$ lies on the nine point circle of $\bigtriangleup ABC$ due to a homothety of factor 2 in $H$.
We now prove that this is the case. Let $G$ be the foot from $H$ to $DK$.
Since $\angle AHK = 90^\circ $ we find $HK$ is tangent to $(AFHE)$ (remember that $AH$ must be the diameter of this circle), we get $KE \cdot KF =KH^2$.
Next, we show that $HK$ is tangent to $(HDG)$. Indeed, as $HD$ is the diameter (since $\angle HGD =90^\circ $) and $\angle KHD =90^\circ $ we now get $KH^2 =KG\cdot KD$ and $G$ lies on the nine point circle.