Reflecting a circle, you find another one

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Tintarn

9037 posts

Tintarn

#1 h Nov 14, 2020, 3:12 PM • 3 Y

Y by DapperPeppermint, centslordm, Mango247

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 $BC$ .

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Kimchiks926

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Kimchiks926

#3 h Nov 14, 2020, 9:01 PM • 2 Y

Y by mkomisarova, centslordm

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.

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MarkBcc168

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MarkBcc168

#4 h Nov 15, 2020, 8:24 AM • 1 Y

Y by centslordm

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
$\begin{align*} \measuredangle YAM &= \measuredangle YAP + \measuredangle PAM \\ &= \measuredangle YXP + \measuredangle PAM \\ &= \measuredangle YXH + \measuredangle PAM \\ &= \measuredangle YZH + \measuredangle MAZ \\ &= \measuredangle YZH + \measuredangle HZM \\ &= \measuredangle YZM. \end{align*}$

This post has been edited 1 time. Last edited by MarkBcc168, Nov 15, 2020, 8:24 AM

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RodwayWorker

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RodwayWorker

#5 h Nov 16, 2020, 3:32 PM • 1 Y

Y by centslordm

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

Baltic Way 2020 Problem 14 Solution

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Mehrshad

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Mehrshad

#6 h May 22, 2021, 1:40 PM • 1 Y

Y by centslordm

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.

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Tintarn

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Tintarn

#7 h Jul 20, 2021, 10:07 AM

Y by

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.

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Nari_Tom

106 posts

Nari_Tom

#8 h Mar 29, 2025, 8:55 AM

Y by

Lemma-1: Let $ABC$ be a triangle and $H$ be it's orthocenter. $M$ be the midpoint of $BC$ . Let $X$ be the projection of $H$ to $AM$ . Then lemma states that $BHCX$ is cyclic.

Lemma-2: Let $\omega$ be a circle and $B,C$ be the two fixed points on it. When point $A$ is moving on $\omega$ then length of $AH$ is constant, were $H$ denotes the orthocenter.

Let's solve the main problem. Let $M$ be the midpoint of $BC$ , then by Lemma-1 we have that $A-X-M$ are collinear. Let $E$ be the point that $AHXE$ is parallelogram. It's easy to prove that $(AHX), (AEX)$ are symmetric about $AX$ . Then by Lemma-2 we have $EABC$ , $BCXY$ and $EAYX$ is cyclic. Now some angle chase gives $AYMZ$ is cyclic.

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Primeniyazidayi

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Primeniyazidayi

#9 h Mar 29, 2025, 12:16 PM

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MarkBcc168 wrote:

AX passes through the midpoint M of BC.

I didn't understand how we get this.Can you explain it,please?

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