Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.0 AIME, harder Find the answer

In triangle $ABC, \angle A=2 \angle C$ . Suppose that $AC=6, BC=8$ , and $AB=\sqrt{a}-b$ , where $a$ and $b$ are positive integers. Compute $100 a+b$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Let $x=AB$ , and $\angle C=\theta$ , then $\angle A=2 \theta$ and $\angle B=180-3 \theta$ . Extend ray $BA$ to $D$ so that $AD=AC$ . We know that $\angle CAD=180-2 \theta$ , and since $\triangle ADC$ is isosceles, it follows that $\angle ADC=\angle ACD=\theta$ , and so $\angle DCB=2 \theta=\angle BAC$ , meaning that $\triangle BAC \sim \triangle BCD$ . Therefore, we have $\frac{x+6}{8}=\frac{8}{x} \Longrightarrow x(x+6)=8^{2}$ Since $x>0$ , we have $x=-3+\sqrt{73}$ . So $100 a+b=7303$ .

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.