Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.4 AIME Find the answer Slovenia

The lengths $a$ , $b$ and $c$ of the sides of the triangle $ABC$ satisfy $c^2 = 2ab$ and $a^2 + c^2 = 3b^2$ . The inner angles of the triangle $ABC$ measure

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Solution

From $c^2 = 2ab$ and $a^2 + c^2 = 3b^2$ we get $a^2 + 2ab = 3b^2$ or $(a + b)^2 = 4b^2$ . It follows that $(a + b - 2b)(a + b + 2b) = 0$ . Since $a$ and $b$ are positive, the only possibility is that $a = b$ . Then $c^2 = 2a^2 = a^2 + b^2$ . Hence, we have a right isosceles triangle and the inner angles measure $45^\circ$ , $45^\circ$ and $90^\circ$ .

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