Let $A B C D$ be a square. We randomly select a point $P$ inside the square, uniformly and at random. What is the probability that $\angle B P C>90^{\circ}$ ?
Let $\omega$ be the semicircle whose diameter is $\overline{B C}$ and whose interior lies inside $A B C D$. Let $M$ be denote the center of $\omega$. We use the result that

- $\angle B P C90^{\circ}$ if $P$ is outside $\omega$.

In order to prove this, we first show the result when $P$ is on $\omega$ (this result is sometimes called Thales' Theorem.)

![](https://cdn.mathpix.com/cropped/2024_04_17_52952e9dd3498d52e48bg-1.jpg?height=244&width=415&top_left_y=729&top_left_x=844)

Compute

$$
\begin{aligned}
180^{\circ} & =360^{\circ}-(\angle P M B+\angle P M C) & \text { as } & \angle P M B+\angle P M C=180^{\circ} \\
& =\left(180^{\circ}-\angle P M B\right)+\left(180^{\circ}-\angle P M C\right) & & \\
& =2 \angle B P M+\left(180^{\circ}-\angle P M C\right) & & M C=M P \text { in } \triangle B M P \\
& =2 \angle B P M+2 \angle C P M & & \\
& =2 \angle B P C & &
\end{aligned}
$$

as claimed.

In the case that $P$ is outside $\omega$, we may intersect segment $M P$ with $\omega$ to obtain a point $K$. Then $90^{\circ}=\angle B K C>$ $\angle B P C$ (since, say $\angle B K M>\angle B P M$ and $\angle C K M>\angle C P M$ ).

![](https://cdn.mathpix.com/cropped/2024_04_17_52952e9dd3498d52e48bg-1.jpg?height=407&width=415&top_left_y=1599&top_left_x=844)

The case where $P$ is inside $\omega$ is dealt in the same way.

Thus, we see that we seek the probability that $P$ lies inside $\omega$. The area of $\omega$ inside $A B C D$ is simply $\frac{1}{2} \pi\left(\frac{1}{2}\right)^{2}=\frac{1}{8} \pi$ (choosing our units so that $A B=1$ ), and the area of $A B C D$ is 1 ; hence the answer is $\frac{1}{8} \pi$.