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Harvard-MIT Mathematics Tournament

Geometry Difficulty 5.5 AIME, harder Prove it United States

Problem:

At time $0$ , an ant is at $(1,0)$ and a spider is at $(-1,0)$ . The ant starts walking counterclockwise along the unit circle, and the spider starts creeping to the right along the $x$ -axis. It so happens that the ant's horizontal speed is always half the spider's. What will the shortest distance ever between the ant and the spider be?

Solution

Solution:

Picture an instant in time where the ant and spider have $x$ -coordinates $a$ and $s$ , respectively. If $1 \leq s \leq 3$ , then $a \leq 0$ , and the distance between the bugs is at least $1$ . If $s > 3$ , then, needless to say, the distance between the bugs is at least $2$ . If $-1 \leq s \leq 1$ , then $s = 1 - 2a$ , and the distance between the bugs is
$\sqrt{(a - (1 - 2a))^2 + (1 - a^2)} = \sqrt{8a^2 - 6a + 2} = \sqrt{\frac{(8a - 3)^2 + 7}{8}}$
which attains the minimum value of $\sqrt{7/8}$ when $a = 3/8$ .

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