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Question

In triangle ABC, AB = 3 and AC = 5. Let O be the circumcenter of triangle ABC. Find OA BC.

Accepted Answer

Let a = OA, b = OB, and c = OC. Then AB = OB - OA = b - a.Similarly, AC = c - a and BC = c - b. We then want to compute OA BC = a (c - b) = a c - a b.FIGURE Since AC = 5, AC^2 = 25. But align* AC^2 &= |c - a|^2 &= (c - a) (c - a) &= |c|^2 - 2 a c + |a|^2 &= 2R^2 - 2 a c, align*where R is the circumradius. Hence, a c = R^2 - AC^22.Similarly, we can prove that a b = R^2 - AB^22.Therefore, a c - a b = ( R^2 - AC^22 ) - ( R^2 - AB^22 ) = AB^2 - AC^22 = 3^2 - 5^22 = -8.