2026-04-11

Problem — 2026-04-11

Medium

Mediumprobability

Three points are chosen uniformly at random on a circle. What is the probability that the resulting triangle contains the center of the circle?

Express as a fraction

Show solution

Fix one point

A

. The triangle contains the center iff the other two points

B

and

C

lie on opposite sides of the diameter through

A

. Equivalently, the arc from

B

C

(not containing

A

) must be more than a semicircle — but we need all three arcs less than a semicircle. By symmetry, for each of the three arcs, there is a

\frac{1}{2}

chance it exceeds a semicircle (and exactly one arc can). So

P(\text{contains center}) = 1 - 3 \cdot \frac{1}{4} = \frac{1}{4}