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All-Soviet-Union Mathematical Olympiad

Geometry Difficulty 4.2 AIME Prove it Soviet Union

Problem:
Given the lengths $AB$ and $BC$ and the fact that the medians to those two sides are perpendicular, construct the triangle $ABC$ .

Solution

Solution:
Let $M$ be the midpoint of $AB$ and $X$ the midpoint of $MB$ . Construct the circle center $B$ , radius $BC/2$ and the circle diameter $AX$ . If they do not intersect (so $BC < AB/2$ or $BC > AB$ ) then the construction is not possible. If they intersect at $N$ , then take $C$ so that $N$ is the midpoint of $BC$ . Let $CM$ meet $AN$ at $O$ . Then $AO/AN = AM/AX = 2/3$ , so the triangles $AOM$ and $ANX$ are similar. Hence $\angle AOM = \angle ANX = 90^{\circ}$ .

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