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Problem 977

AMC 12 late, AIME early

Geometry Difficulty 4.4 Prove it All-Soviet-Union Mathematical Olympiad · Soviet Union

A circle center $O$ is inscribed in $ABCD$ (touching every side). Prove that $\angle AOB + \angle COD$ equals $180$ degrees.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Solution:
Let $AB$ touch the circle at $W$ , $BC$ at $X$ , $CD$ at $Y$ , and $DA$ at $Z$ . Then $AO$ bisects angle $ZOW$ and $BO$ bisects angle $XOW$ . So $\angle AOB$ is half angle $ZOX$ . Similarly $\angle COD$ is half angle $XOZ$ and hence $\angle AOB + \angle COD$ equals $180$ .

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