Trapezium inscribed in a circle

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Trapezium inscribed in a circle

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Source: Thirty Third Irish Mathematical Olympiad 2020 P9/10

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#1 h Jul 20, 2020, 5:44 PM

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A trapezium $A B C D,$ in which $A B$ is parallel to $D C,$ is inscribed in a circle of radius $R$ and centre $O .$ The non-parallel sides $D A$ and $C B$ are extended to meet at $P$ while diagonals $A C$ and $B D$ intersect at $E .$ Prove that $|O E| \cdot|O P|=R^{2}$ .

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#2 h Jul 20, 2020, 5:46 PM

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Solution

https://i.imgur.com/1TlrGRN.png

We wish to prove $|OE| \cdot |OP| = R^2$ . Let $R^2 = |OD| \cdot |OD|$ , so we want to show that
$\begin{align*} |OE| \cdot |OP| &= |OD| \cdot |OD| \\ \iff \frac{|OE|}{|OD|} &= \frac{|OD|}{|OP|}. \end{align*}$ This would follow immediately if $\triangle OED \sim \triangle ODP$ .

First we show $O$ , $E$ and $P$ are collinear.
We have $ABCD$ is an isosceles trapezium, as it's cyclic (well known) then $E, O$ and $P$ lie on the same line by symmetry.

Then we have the following. Let $PO$ intersect the circle at $D'$ . Then
$\angle EDO = \angle BDD' = \angle BAD.$ As $DD'$ is a diameter (passes through $O$ ), we have $\angle D'AD = D'AP = 90^\circ$ by Thales theorem.

Then as $AB \perp OP$ , let $X$ be the intersection of $PO$ and $AB$ and $Y$ be the intersection of $PO$ and $AD'$ . We have $\triangle PAY \sim \triangle AXY$ as shared $\angle AXY$ and right angle, thus $\angle YAX = \angle APX = \angle ODE$ .

Therefore $\triangle OED \sim \triangle ODP$ due to $\angle OPD = \angle EDO$ and shared angle $\angle EDO$ , which implies the result.

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#3 h Jul 20, 2020, 6:17 PM • 1 Y

Y by circlethm

Let $AB\cap CD=P_{\infty}$ be the point of infinity along $AB$ . It's well known that the inverse of $E$ with respect to $(ABCD)$ is the foot of the perpendicular from $E$ to $PP_{\infty}$ . By symmetry, $EP\perp AB$ , implying $EP\perp PP_{\infty}$ meaning $P$ must be the inverse of $E$ . This immediately gives the conclusion.

This post has been edited 1 time. Last edited by franchester, Jul 20, 2020, 6:17 PM

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zuss77

#4 h Jul 20, 2020, 6:23 PM

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$\angle AOC=2\angle B=\pi-\angle APC$ , so $P\in (AOC)$ . The same $P\in (BOD)$
With inversion about $(O)$ we have $AC\cap BD \mapsto (AOC)\cap (BOD)$ , i.e. $E\mapsto P$ .

This post has been edited 2 times. Last edited by zuss77, Jul 20, 2020, 8:49 PM

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#5 h Jul 20, 2020, 6:25 PM • 1 Y

Y by Mango247

zuss77 wrote:

$O$ is intersection of perp bisector of $AD$ and bisector of $\angle APC$ , so $P\in (AOC)$ . The same $P\in (BOD)$
With inversion about $(O)$ we have $AC\cap BD \mapsto (AOC)\cap (BOD)$ , i.e. $E\mapsto P$ .

I expect that the $OE \cdot OP = R^2$ gives away that there's a clean inversion soln.... If only I knew how to do that

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#6 h Jan 21, 2021, 4:34 PM

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Notice that the circumcircles of triangles $\triangle PAB$ and $\triangle PCD$ are tangent at $P$ since $\overline{AB} \parallel \overline{CD}$ , thus $P$ is the Miquel point of $ABCD$ , so it's well-known that if $E=\overline{AC} \cap \overline{BD}$ , then $OE\cdot OP=R^2$ . $\square$

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FadhbSolutionneur42

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FadhbSolutionneur42

#7 h May 3, 2021, 1:37 PM

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What does OE•OP = R^2 mean in this question?

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