An acute triangle $ABC$ is given and let $H$ be its orthocenter. Let $\omega$ be the circle through $B$, $C$ and $H$, and let $\Gamma$ be the circle with diameter $AH$. Let $X\neq H$ be the other intersection point of $\omega$ and $\Gamma$, and let $\gamma$ be the reflection of $\Gamma$ over $AX$. Suppose $\gamma$ and $\omega$ intersect again at $Y\neq X$, and line $AH$ and $\omega$ intersect again at $Z \neq H$. Show that the circle through $A,Y,Z$ passes through the midpoint of segment $BC$.
Problem
Source: Baltic Way 2020, Problem 14
Tags: geometry, geometry proposed
Kimchiks926
15.11.2020 00:01
Here is sketch of inversion solution, which I found during the contest. Let $\triangle DEF$ be orthic triangle of $\triangle ABC$ and $M$ is midpoint of $BC$. Step 1: $AX$ passes through point $M$ Consider inversion with centre $A$ with radius $\sqrt{AH \cdot AD}$. Under this inversion $\odot(BHXC)$ goes to the nine point circle of $\triangle ABC$ and $\odot(AH)$ goes to the line $BC$. Thus inverse of $X$ is intersection of nine point circle with line $BC$ different from $D$ which is point $M$. Thus $A, X,M$ are collinear as desired. Step 2: Find inverses of $Z$ and $Y$. Note that $AD = DZ$, thus inverse of $Z$ is point $Z'$, which is midpoint $AH$. Since $Y$ lies on $\odot(BHC)$ and $\gamma$, then inverse of $Y$ is intersection of nine point circle with line, which is obtained by reflecting line $BC$ over line $AM$. So now we define $Y'$ as intersection of nine - point circle and line $ZX$. If we show that $Y'M$ is reflection of $BC$ over $AM$ we are done. Note that this is equivalent to show that $\angle AMD = \angle AMY'$ Step 3: Angle chase Define point $U$ as intersection of nine point circle and line $AM$. Observe that $\angle AUZ'= \angle AXH =90^{\circ}$. Thus $ZU'$ is midline in $\triangle AXH$. This implies that $\triangle AZ'X$ is Isosceles. Now observe that: $$ \angle AMD = \angle AZ'U = \angle UZ'X = \angle AMY'$$as desired.
MarkBcc168
15.11.2020 11:24
Nice angle chasing exercise! We begin with the following two well-known observations: points $A,Z$ are symmetric w.r.t. $BC$, and $AX$ passes through the midpoint $M$ of $BC$. In addition, we let $P$ be the reflection of $H$ across $AX$, so that $P\in\gamma$. With these in mind, we are almost done; just angle chase to see that \begin{align*} \measuredangle YAM &= \measuredangle YAP + \measuredangle PAM \\ &= \measuredangle YXP + \measuredangle PAM \\ &= \measuredangle YXH + \measuredangle PAM \\ &= \measuredangle YZH + \measuredangle MAZ \\ &= \measuredangle YZH + \measuredangle HZM \\ &= \measuredangle YZM. \end{align*}
RodwayWorker
16.11.2020 18:32
Baltic Way 2020 Problem 14 wrote: An acute triangle $ABC$ is given and let $H$ be its orthocenter. Let $\omega$ be the circle through $B$, $C$ and $H$, and let $\Gamma$ be the circle with diameter $AH$. Let $X\neq H$ be the other intersection point of $\omega$ and $\Gamma$, and let $\gamma$ be the reflection of $\Gamma$ over $AX$. Suppose $\gamma$ and $\omega$ intersect again at $Y\neq X$, and line $AH$ and $\omega$ intersect again at $Z \neq H$. Show that the circle through $A,Y,Z$ passes through the midpoint of segment $BC$. My solution is similar to MarkBcc Solution but in more detail I suppose
We see that if $\Omega$ is the circumcircle of $\square BEFC$, then inverting $A$ about $\Omega$ will be a point $P$ on the $A$ - median of $\triangle ABC$ since $M$ the midpoint of segment $BC$ and the center of $\Omega$ and $A$ both lie on the $A$ - median, but moreover, we see that $\Gamma$ and $\Omega$ are orthogonal to each other by Three Tangents Lemma and this means that the inverted image of $A$ will be another point on $\Gamma$ distinct from $A$, and we know that the intersection of $A$ - median and $\Gamma$ distinct from $A$ is dearly called as $A$ - HM Point and so $P$ is the $A$ - HM Point. Now, we know that $P$ lies on $\Gamma$ but moreover, inverting points $B, C, A, Q$ about $\Omega$ where $Q$ is the point of intersection of line $MH$ and circumcircle of $\triangle ABC$, we see that the collinearity $A, Q, K$ where $K$ is the intersection of lines $EF, BC$ when inverted about $\Omega$ gives that $P, M, D, Q^\star$ are concyclic but since $Q$ lies on $\Gamma$ too, $Q^\star \in \Gamma$ and since $P$ is the $A$ - HM Point and $P \neq Q^\star$, we must have that $Q^\star = H$ and so $P$ lies on circumcircle of $\triangle BHC$ and so $P = X$ or $X$ is the $A$ - HM Point.
Now, we see that $\angle BAC + \angle BHC$ which is a well known fact and easily provable fact implies that the reflection of $A$ about line $BC$ lies on circumcircle of $\triangle BHC$ and hence it can be seen that $AD = DZ$ from this. Now, it is a known fact that $\angle AXH = 90^\circ$ which is again easy to see why and the reflection of $H$ about $AX$ that is let us say a point $T$ will lie on $\gamma$
Now, $$\measuredangle YAM = \measuredangle YAT + \measuredangle TAM$$$$= \measuredangle YXT + \measuredangle TAM$$$$= \measuredangle YZH + \measuredangle MAZ$$$$= \measuredangle YZM$$which basically proves our claim.
Mehrshad
22.05.2021 16:40
hello There is a LaTeX error in 2020 BALTIC WAY. Is it from this question? If it is please solve it so we'd be able to download the PDF.
Tintarn
20.07.2021 13:07
Mehrshad wrote: hello There is a LaTeX error in 2020 BALTIC WAY. Is it from this question? If it is please solve it so we'd be able to download the PDF. I'm not sure I understand. I don't see any LaTeX error in the problem statement. If you find one, you can tell me and I will try to fix it immediately.