Maths Olympiad Prep

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Geometry Difficulty 4.5 AIME Prove it Soviet Union

Problem:

Circles CC and C′C' intersect at OO and XX. A circle center OO meets CC at QQ and RR and meets C′C' at PP and SS. PRPR and QSQS meet at YY distinct from XX. Show that ∠YXO=90∘\angle YXO = 90^{\circ}.

Solution

Solution:

We show first that YRSXYRSX is cyclic.

Figure 1

It is sufficient to show that ∠RYS=∠RXS\angle RYS = \angle RXS. We have ∠RYS=∠PRQ−∠RQY=∠PSQ−∠RQY=∠OSQ−∠OSP−∠RQY=∠OQS−∠RQY−∠OSP=∠OQR−∠OSP\angle RYS = \angle PRQ - \angle RQY = \angle PSQ - \angle RQY = \angle OSQ - \angle OSP - \angle RQY = \angle OQS - \angle RQY - \angle OSP = \angle OQR - \angle OSP. But ∠RXS=∠RXO−∠OXS=∠OQR−∠OXS=∠OQR−∠OPS=∠OQR−∠OSP\angle RXS = \angle RXO - \angle OXS = \angle OQR - \angle OXS = \angle OQR - \angle OPS = \angle OQR - \angle OSP. So YRSXYRSX is cyclic.

Hence ∠YXO=∠YXS+∠SXO=∠PRS+∠SXO=∠PRS+2∠SPO=12∠POS+∠SPO=90∘\angle YXO = \angle YXS + \angle SXO = \angle PRS + \angle SXO = \angle PRS + 2\angle SPO = \frac{1}{2}\angle POS + \angle SPO = 90^{\circ}.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.