CEMC Cayley, 2014

We call Clarise’s spot $C$ and Abe’s spot $A$.

Consider a circle centred at $C$ with radius 10 m. Since $A$ is 10 m from $C$, then $A$ is on this circle.

Bob starts at $C$ and picks a direction to walk, with every direction being equally likely to be chosen. We model this by having Bob choose an angle $\theta$ between $0^\circ$ and $360^\circ$ and walk 10 m along a segment that makes this angle when measured counterclockwise from $CA$.

Bob ends at point $B$, which is also on the circle.

We need to determine the probability that $\text{AB lt;AC}$.

Since the circle is symmetric above and below the diameter implied by $CA$, we can assume that $\theta$ is between $0^\circ$ and $180^\circ$ as the probability will be the same below the diameter.

Consider $\triangle CAB$ and note that $CA=CB=10$ m.

It will be true that $\text{AB lt;AC}$ whenever $AB$ is the shortest side of $\triangle ABC$.

$AB$ will be the shortest side of $\triangle ABC$ whenever it is opposite the smallest angle of $\triangle ABC$. (In any triangle, the shortest side is opposite the smallest angle and the longest side is opposite the largest angle.)

Since $\triangle ABC$ is isosceles with $CA=CB$, then $\angle CAB = \angle CBA$.

We know that $\theta=\angle ACB$ is opposite $AB$ and $\angle ACB + \angle CAB + \angle CBA = 180^\circ$.

Since $\angle CAB = \angle CBA$, then $\angle ACB + 2\angle CAB = 180^\circ$ or $\angle CAB = 90^\circ - \frac{1}{2}\angle ACB$.

If $\theta = \angle ACB$ is smaller than $60^\circ$, then $\angle CAB = 90^\circ - \frac{1}{2}\theta$ will be greater than $60^\circ$.

Similarly, if $\angle ACB$ is greater than $60^\circ$, then $\angle CAB = 90^\circ - \frac{1}{2}\theta$ will be smaller than $60^\circ$.

Therefore, $AB$ is the shortest side of $\triangle ABC$ whenever $\theta$ is between $0^\circ$ and $60^\circ$.

Since $\theta$ is uniformly chosen in the range $0^\circ$ to $180^\circ$ and $60^\circ = \frac{1}{3}(180^\circ)$, then the probability that $\theta$ is in the desired range is $\frac{1}{3}$.

Therefore, the probability that Bob is closer to Abe than Clarise is to Abe is $\frac{1}{3}$.

(Note that we can ignore the cases $\theta = 0^\circ$, $\theta = 60^\circ$ and $\theta = 180^\circ$ because these are only three specific cases out of an infinite number of possible values for $\theta$.)