A Triangle. A Circle

By extension of Thales Theorem , if any angle in the triangle is obtuse, the triangle will not contain the center. Therefore, we are looking for the percentage of possible triangles that are not obtuse triangles.

Let $A$ , $B$ , and $C$ be the points on the circle and let $O$ be the center of the circle. Then without loss of generality, the points can be rotated so that $A$ is on the positive $x$ -axis. Then let $x$ be the central angle between $A$ and $B$ and $y$ be the central angle between $A$ and $C$ . Then both $x$ and $y$ are such that $0 \leq x < 360°$ and $0 \leq y < 360°$ .

$\triangle AOB$ , $\triangle BOC$ , and $\triangle AOC$ are isosceles triangles because each have two sides that are radii of the circle.

If $y > x$ , the base angles of these triangles are $90° - \frac{x}{2}$ , $90° - \frac{y}{2} + \frac{x}{2}$ , and $90° - \frac{y}{2}$ , which makes $\angle BAC = \frac{y - x}{2}$ , $\angle ABC = 180° - \frac{y}{2}$ , and $\angle ACB = \frac{x}{2}$ . For any one of these angles to be obtuse, either $\frac{y - x}{2} \geq 90°$ or $y \geq x + 180°$ , $180° - \frac{y}{2} \geq 90°$ or $y \leq 180°$ , or $\frac{x}{2} \geq 90°$ or $x \geq 180°$ , which is shaded in red in the graph below:

By symmetry, if $x > y$ , for any one of these angles to be obtuse, either $y \leq x - 180°$ , $x \leq 180°$ , or $y \geq 180°$ , which is shaded in red in the graph below:

Combining these graphs gives a visual representation of all the possible triangles that can be made by three points on the circumference of a circle, with the red regions showing the triangles that are obtuse, and the green regions showing the triangles that are not obtuse:

The percentage of the green regions represents the probability of a triangle covering the center of the circle, which is $\frac{2}{8} = \frac{1}{4} = \boxed{25\%}$ .

This 3Blue1Brown video beautifully explains the answer along with another cool fact

This problem has been posted before. Here's the solution I submitted last time.

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Given two points $A, B$ on a circle, the third point $C$ must lie in the red region below if $\triangle ABC$ is to contain the circle's center:

Now, we find the probability of this happening. Fix $A$ and $B$ on the circle, and let $\theta$ be the angle between them ( $\angle AOB$ ). Then $0<\theta\leq \pi$ . Note that the angle intercepting the red region in the diagram is precisely $\theta$ . Thus the probability of a random point $C$ being placed in the red area is $\frac{\theta}{2\pi}$ . We integrate to compute the average probability over all possible $\theta$ : $\frac{1}{\pi-0}\int_0^\pi \frac{\theta}{2\pi}\,\mathrm{d}\theta = \frac{1}{\pi}\left[\frac{\theta^2}{4\pi}\right]_{\theta=0}^\pi = \frac{\pi^2}{4\pi^2} = \frac{1}{4}=\boxed{25\%}$