Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.9 AIME Find the answer

3. Given that $AB$ is a chord of the circle $\odot O$ with radius 1, and the length of $AB$ is the positive root of the equation $x^{2}+x-1=0$ . Then the degree of $\angle AOB$ is $\qquad$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

3. $36^{\circ}$ .

From $x^{2}+x-1=0$ , we know $x=1-x^{2}<0$ , so $AB < OB$ . Therefore, on $OB$ , we take $OC = AB = x$ . Also, from $x^{2}+x-1=0$ , we can get $\frac{x}{1-x}=\frac{1}{x}$ . As shown in the figure, $\frac{AB}{BC} = \frac{OA}{AB}$ , thus $\triangle OAB \sim \triangle ABC$ . Since $\triangle OAB$ is an isosceles triangle, we have $AC = AB = OC$ . Given $\angle 1 = \angle 2 = \angle 3$ , we have $5 \angle AOB = 180^{\circ}$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.