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All-Soviet-Union Mathematical Olympiad

Geometry Difficulty 5.2 AIME, harder Prove it Soviet Union

Problem:

The sides and diagonals of $ABCD$ have rational lengths. The diagonals meet at $O$ . Prove that the length $AO$ is also rational.

Solution

Solution:

$AB = AO \cos OAB + BO \cos OBA$ . We can derive a rational expression for $\cos OAB$ using the cosine rule for triangle $ABC$ . Similarly for $\cos OBA$ using the cosine rule for triangle $DAB$ . So $OA = r_1 + r_2 OB$ , where $r_1$ denotes a rational number. Similarly, $OB = r_3 + r_4 OC$ , so $OA = r_5 + r_6 OC$ . But $OA + OC = AC = r_7$ . Hence $OA$ is rational.

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