Competition problem — Geometry — Maths Olympiad Prep

Library / /860 of 1394

Geometry Difficulty 5.4 AIME, harder Prove it United States

Problem:

Holden has a collection of polygons. He writes down a list containing the measure of each interior angle of each of his polygons. He writes down the list $30^{\circ}$ , $50^{\circ}$ , $60^{\circ}$ , $70^{\circ}$ , $90^{\circ}$ , $100^{\circ}$ , $120^{\circ}$ , $160^{\circ}$ , and $x^{\circ}$ , in some order. Compute $x$ .

Solution

Solution:

We work in degrees. The sum of all 9 angles is $680 + x$ . The sum of the angles in a polygon with $n$ sides is $180(n - 2) \equiv 180n$ mod $360$ . Since there are 9 angles, the polygons have a total of 9 sides, so the sum of the 9 angles must be $9 \cdot 180 \equiv 180$ mod $360$ . Thus $680 + x \equiv 180$ mod $360$ , so $x \equiv 220$ mod $360$ . Since $0 < x < 360$ , we know $x = \boxed{220}$ .

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