# probability of rectangle lying completely inside circle

Let $$C$$ be a circle. Two points, $$P_1$$ and $$P_2$$ are randomly chosen, with $$P_1$$ on the circumference of $$C$$ and $$P_2$$ in the interior of $$C$$. What is the probability that the rectangle with diagonal $$P_1P_2$$ and sides parallel to the $$x$$-axis and $$y$$-axis lies entirely inside $$C$$?

The probability is apparently $$\frac{4}{\pi^2}.$$ WLOG, we can assume the circle has radius 1. One can pick a point $$P_1$$ on a circle and let $$O$$ be the origin. Then let $$\theta$$ be the angle formed by $$OP_1$$ and the diameter of $$C$$ parallel to the $$y$$-axis. So then a rectangle with both sides parallel to the $$x$$- and $$y$$-axes and diagonal $$P_1P_2$$ will lie entirely inside the circle iff $$P_2$$ is inside the rectangle inscribed by the circle with both sides parallel to the axes. This inscribed rectangle has area $$4\sin\theta \cos\theta.$$ Also, the angle $$\theta$$ varies from $$0$$ to $$\frac{\pi}2$$ by definition. So the required area is $$\frac{2}{\pi}\int_0^{\pi/2} \dfrac{4\sin\theta\cos\theta}{\pi}d\theta = \dfrac{4}{\pi^2}.$$

The question is, where does the $$\frac{2}{\pi}$$ come from in this answer? Clearly the $$\dfrac{4\sin\theta\cos\theta}{\pi}$$ comes from dividing the rectangle's area by the circle's area.

• The uniform distribution of $\theta$ comes with a factor of $2/\pi$. If the ratio of areas was always $1$ you’d want the probability to be $1$, not $\pi/2$. Sep 14, 2020 at 2:54
• @ErickWong thanks but how did you come up with that factor? Is it a ratio of average areas or something?
– user747916
Sep 14, 2020 at 3:15
• It is the reciprocal of the length of the interval $[0,\pi/2]$, for precisely the reason I gave at the end. Sep 14, 2020 at 3:18
• Is $Q$ the same point as $P_2$? Sep 14, 2020 at 4:28

To help you understand, you could alternatively allow the angle $$\theta$$ to range from $$0$$ to $$2\pi$$. Then the area of the inscribed rectangle is $$|4 \cos \theta \sin \theta|$$, since it must be positive (and in quadrants II and IV, exactly one of $$\cos \theta$$ and $$\sin \theta$$ is negative).
But if we recall that $$P_1$$ is uniformly chosen at random on the circumference of the circle, and $$\theta$$ is the angle that $$P_1 O$$ makes with the positive $$x$$-axis, then $$\theta$$ is uniformly distributed on $$[0, 2\pi)$$. That means the probability density of $$\theta$$ is given by $$f(\theta) = \frac{1}{2\pi}, \quad 0 \le \theta < 2\pi.$$
Then the desired integral is $$\frac{1}{2\pi} \int_{\theta = 0}^{2\pi} \frac{|4 \cos \theta \sin \theta|}{\pi} \, d\theta.$$ You can use a symmetry argument to complete the calculation.
So back to the original method of solution, all that is different is that by restricting our attention to quadrant I where both $$\sin$$ and $$\cos$$ are positive, and exploiting symmetry at an earlier stage, we can simplify the computation. But when we do this, $$\theta$$ is not uniformly distributed on $$[0, \pi/2)$$. So the probability density in this case would be $$f(\theta) = \frac{1}{\pi/2} = \frac{2}{\pi}, \quad 0 \le \theta < \pi/2.$$