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enter image description here$H$ is the orthocenter of acute triangle $A B C$. Let $\omega$ be the circumcircle of $B H C$ with center $O^{\prime}. \Omega$ is the nine-point circle of $A B C . X$ is an arbitrary point on arc $B H C$ of $\omega$ and $A X$ intersects $\Omega$ at $Y . P$ is a point on $\Omega$ such that $PX= PY$. Prove that $ \angle O^{\prime}PX=90°$.

I extended $XP$ to intersect circumcircle of $ \triangle BHC$ at $M$ and then tried proving $\angle XYM$ a right angle. Kept angle chasing but didn't get any satisfactory result.

user600016
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  • Would it be possible for you to create a diagram (e.g. jpeg file) of the problem, and embed it directly into your posting? For help managing an image, see the Images section, at the bottom of this article, and see also this article. – user2661923 May 19 '22 at 16:31

1 Answers1

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First, consider the homothety at $A$ with scale factor $2$. This maps three points on the nine-point circle (the midpoints of $AB$, $AC$, and $AH$) to three points on $(HBC)$, so it maps the nine-point circle to $(HBC)$. The image of $Y$ is thus a point on both line $AX$ and $(HBC)$; it isn't $X$ (based on the diagram; if it is $X$ the problem is in fact false). So, if $AX$ intersects $(HBC)$ again at $Z\neq X$, then $Y$ is the midpoint of $AZ$. Also, if $N$ is the center of the nine-point circle, then the homothety takes $N$ to $O'$, and so $N$ is the midpoint of $AO'$.

We can now "simplify" the problem as follows:

The points $X$ and $Z$ lie on a circle centered at $O'$, and a point $A$ lies in line $XZ$. $Y$ is the midpoint of $AZ$, and $N$ is the midpoint of $AO'$. Show that (1) the circle centered at $N$ passing through $Y$, (2) the circle with diameter $O'X$, and (3) the perpendicular bisector of $XY$ concur, as long as (1) and (3) intersect.

(We are given that (1) and (3) intersect since $P$ exists; this is the same as the original problem since $P$ is on the circle with diameter $O'X$ if and only if $\angle XPO'=90^\circ$.) To solve this restated problem, let $R$ be the midpoint of $O'X$, so that (1) and (2) are the circles centered at $N$ passing through $Y$ and centered at $R$ passing through $X$, respectively. Note that $$\measuredangle RXZ=\measuredangle O'XZ=\measuredangle XZO'=\measuredangle AZO'=\measuredangle AYN=\measuredangle XYN,$$ that $R$ and $N$ lie on the same side of line $XY$ as each other, and that $$XR=\frac{O'X}2=\frac{O'Z}2=NY,$$ so $XRNY$ is an isosceles trapezoid with $RN\|XY$. This means that $N$ and $R$ are reflections over (3), so circles (1) and (2) are as well; this means that any intersection of (1) and (3) also lies on (2), as desired.

Carl Schildkraut
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