Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.7 AIME Prove it United States

Problem:

On the bases $AB$ and $CD$ of a trapezoid $ABCD$ draw two squares externally to $ABCD$ . Let $O$ be the intersection point of the diagonals $AC$ and $BD$ , and let $O_{1}$ and $O_{2}$ be the centers of the two squares. Prove that $O_{1}$ , $O$ and $O_{2}$ lie on a line (i.e. they are collinear; see Figure.)

Solution

Solution:

The idea is to show that $\triangle OCO_{2} \sim \triangle OAO_{1}$ . Indeed, first notice that $\triangle AO_{1}B \sim \triangle CO_{2}D$ — both are right isosceles triangles. Therefore, $AO_{1} : CO_{2} = AB : CD$ . But $\triangle AOB \sim \triangle COD$ ( $AB \parallel CD \Rightarrow$ all three angles are the same), so $AB : CD = AO : CO$ . This implies $AO_{1} : CO_{2} = AO : CO$ . Further, $\angle O_{1}AO = \angle O_{1}AB + \angle BAO = 45^{\circ} + \angle DCO = \angle OCO_{2}$ , and we finally conclude that $\triangle OCO_{2} \sim \triangle OAO_{1}$ .

Hence, $\angle AOO_{1} = \angle COO_{2}$ . Since $AOC$ is a line, then $O_{2}OO_{1}$ is also a line.

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