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Harvard-MIT Math Tournament

Geometry Difficulty 5.0 AIME Prove it United States

Problem:

Start with an angle of $60^{\circ}$ and bisect it, then bisect the lower $30^{\circ}$ angle, then the upper $15^{\circ}$ angle, and so on, always alternating between the upper and lower of the previous two angles constructed. This process approaches a limiting line that divides the original $60^{\circ}$ angle into two angles. Find the measure (degrees) of the smaller angle.

Solution

Solution:

The fraction of the original angle is $\frac{1}{2} - \frac{1}{4} + \frac{1}{8} - \cdots$ . This is just a geometric series with first term $\frac{1}{2}$ and ratio $-\frac{1}{2}$ , so the sum is $\frac{1}{3}$ . Therefore the smaller angle is $20^{\circ}$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.