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TwoEllipses

Problem: Two congruent ellipses $\Gamma_1$ and $\Gamma_2$ have centers $X$ and $Y$, respectively. $\Gamma_1$ and $\Gamma_2$ don't meet each other, and the major axis of $\Gamma_1$ and the minor axis of $\Gamma_2$ lie on the same line. Denote two common internal tangents of $(\Gamma_1, \Gamma_2)$ as $l_1$ and $l_2$, and one common external tangents of $(\Gamma_1, \Gamma_2)$ as $l_3$. Let $l_1 \cap l_2=P$, $l_1 \cap l_3=Q$, $l_2 \cap l_3=R$ $(XQ<XR)$. If $l_1 \perp l_2$, show that $\triangle PXQ \sim \triangle PRY$.

I found this property using Geogebra, and I have a proof using coordinates. I was wondering if there exists a synthetic proof.

  • It is well-known that if $l_1 \perp l_2$, then $XP = YP = \sqrt{a^2 + b^2}$ where $a$ and $b$ are the major radius and the minor radius, respectively.

  • Since $\angle XPQ = \angle RPY = \pi/4$, it is sufficient to show that the angle between $XQ$ and $YR$ is $\pi/4$.

  • I've tried to use the center of the rotation, but I couldn't get any conclusion with this approach.

YNK
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anonymous
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  • Isn't any solution of this problem incomplete unless it first shows that the point $P$ lies on the line $XY$? – YNK Aug 25 '22 at 16:47
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    I think we don't need additional proof for that since $XY$ is the line of symmetry for both ellipses. – anonymous Aug 25 '22 at 18:59

1 Answers1

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Let $Z$ be the reflection of $X$ in $PQ$. Easy to see that $Z$ is symmetric to $Y$ wrt $PR$. We have to prove that $\angle PXQ+\angle RYP =135^\circ$. This is equivalent to $\angle QZR =135^\circ$.

Let $K, L, M, N$ be the points on segments $PQ, PR, QR, QR$, respectively, such that $MK \perp QR$, $LN\perp QR$, $MK$ is tangent to $\Gamma_1$, and $LN$ is tangent to $\Gamma_2$.

Note that the clockwise rotation about $90^\circ$ around $Z$ maps $Y$ to $X$, $\Gamma_2$ to $\Gamma_1$, $PR$ to $PQ$, $QR$ to $MK$ and $NL$ to $QR$. It follows it maps $L$ to $Q$, $R$ to $K$, and $N$ to $M$. In particular $ZYX$, $ZLQ$, $ZRK$, and $ZNM$ are 90-45-45 triangles.

Since the tangents from $M$ to $\Gamma_1$ are perpendicular, we have $XM=XP$. Similarly $YP=YN$. Since the rotation we considered maps $N$ to $M$ and $Y$ to $X$, it follows that $NY\perp XM$. Hence $\angle NYP +\angle PXM=90^\circ$. Since triangles $PXM$ and $NYP$ are isosceles, it follows easily that $\angle NPM=45^\circ$.

Note that $Z,M,P$ lie on the circle with diameter $RK$. Hence $\angle ZRM=\angle ZPM$. Similarly, $Z,N,P$ lie on the circle with diameter $QL$, hence $\angle NPZ=\angle NQZ$. It follows that $$\angle QZR=180^\circ-\angle ZRQ-\angle RQZ=180^\circ-\angle ZPM-\angle NPZ=180^\circ-\angle NPM=135^\circ.$$

timon92
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    What a wonderful solution! I defined the center of rotation $Z$ and the intersection point $N$ and $L$ from the rotated tangent as you did, but never thought about defining $M$ and $K$. I've learned a lot from your answer. – anonymous Aug 25 '22 at 18:45
  • @anonymous Thanks. Really nice problem. Had fun solving it. – timon92 Aug 25 '22 at 19:07