Prove four points are concyclic in an ellipse

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Prove four points are concyclic in an ellipse

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Source: China second round 2006 p1

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#1 h Mar 10, 2012, 6:00 AM • 2 Y

Y by Adventure10, Mango247

An ellipse with foci $B_0,B_1$ intersects $AB_i$ at $C_i$ $(i=0,1)$ . Let $P_0$ be a point on ray $AB_0$ . $Q_0$ is a point on ray $C_1B_0$ such that $B_0P_0=B_0Q_0$ ; $P_1$ is on ray $B_1A$ such that $C_1Q_0=C_1P_1$ ; $Q_1$ is on ray $B_1C_0$ such that $B_1P_1=B_1Q_1$ ; $P_2$ is on ray $AB_0$ such that $C_0Q_1=C_0Q_2$ . Prove that $P_0=P_2$ and that the four points $P_0,Q_0,Q_1,P_1$ are concyclic.

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#2 h May 23, 2016, 4:53 AM • 2 Y

Y by Adventure10, Mango247

Suppose that $C_0B_1$ and $C_1B_O$ meets at C and Consider the incircle of $AC_1CC_0$ now the rest part is obvious

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#3 h Aug 17, 2020, 9:32 PM • 1 Y

Y by Mango247

The statement seems to be inaccurate, because even substituting $P_2=Q_2$ the first part doesn't hold:

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#4 h Apr 7, 2022, 1:18 PM

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We have $C_0P_2=C_0Q_1=B_1Q_1-B_1C_0=B_1P_1-B_1C_0=(B_1C_1+C_1P_1)-B_1C_0=B_1C_1+C_1Q_0-B_1C_0=B_1C_1+(C_1B_0+B_0Q_0)-B_1C_0.$ Since $C_0,C_1$ lie on the ellipse whose focuses are $B_1,B_0$ , we can know that $B_1C_1+C_1B_0=B_1C_0+C_0B_0$ .
So $C_0P_2=B_1C_0+C_0B_0+B_0Q_0-B_1C_0=C_0B_0+B_0Q_0=C_0P_0$ .
Hence $P_0=P_2$ since $P_0,P_2$ both lie on the extension of $AB_0$ .

We will show that $\angle P_1Q_1P_0=\angle P_1Q_0P_0$ .
In fact, $\angle P_1Q_1P_0=\angle P_1Q_1B_1+\angle P_0Q_1B_1=\dfrac{\pi -\angle P_1B_1Q_1}{2}+\dfrac{\pi-\angle Q_1C_0P_1}{2}=\dfrac{\pi +\angle B_1AC_0}{2}$ .
Similarly, $\angle P_1Q_0P_0=\dfrac{\pi +\angle C_1AB_0}{2}$ .
So $\angle P_1Q_1P_0=\angle P_1Q_0P_0$ , which indicates that $P_0,Q_0,Q_1,P_1$ are concyclic.

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#5 h Apr 7, 2022, 3:29 PM

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1. $|B_0Q_1|=|B_0C_0|+|C_0B_1|-|C_1B_1|-|C_1P_1|=|B_0C_0|-|C_0Q_0|=|B_0P_0|\implies P_0=P_2.$
2. Let $C=B_0C_1\cap C_0B_1.$ $|B_0C_0|+|B_1C_0|=|B_0C_1|+|B_1C_1|\implies AC_0CC_1$ is circumscribed. Therefore perpendicular bisectors of $P_0Q_0,Q_0P_1,P_1Q_1,Q_1P_0$ concur at incenter of this quadrilateral, we are done.

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