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Berkeley Math Circle

Geometry Difficulty 4.5 AIME Prove it United States

Problem:

Oscar draws a triangle $ABC$ on a sheet of paper. He finds that the side lengths of $ABC$ are all powers of $2$ (i.e. among $1, 2, 4, 8, \ldots$ ). Prove that Oscar's triangle is isosceles.

Solution

Solution:

Consider a longest side of the triangle, $2^{a}$ . We claim that another side must have this length too. Otherwise, suppose for contradiction they are $2^{b}$ and $2^{c}$ where $b, c < a$ . Then
$2^{b} + 2^{c} < 2^{a-1} + 2^{a-1} = 2^{a}$
which contradicts the triangle inequality.
Hence there must be a second side of length $2^{a}$ .

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