# 1985 Putnam B4 Integration in Rectangular Coordinates

Let C be the circle radius 1, center the origin. A point P is chosen at random on the circumference of C, and another point Q is chosen at random in the interior of C. What is the probability that the rectangle with diagonal PQ, and sides parallel to the x-axis and y-axis, lies entirely inside (or on) C?

The answer is $4/\pi^2$ and is gotten by integrating with polar coordinates like so:

Given P, let R be the rectangle inscribed in C with sides parallel to the axes. Then the rectangle with diagonal PQ lies entirely inside C iff Q lies inside R.

Let O be the center of C. Let the diameter of C parallel to the y-axis make an angle θ with OP. Then the area of R is 4 sin θ cos θ. So the required probability is $2/\pi\int_0^{\pi/2} (4\sin\theta\cos\theta)/\pi\,d\theta = 4/\pi^2\int_0^{\pi/2}\sin 2\theta\,d\theta = 4/\pi^2$.

When I was trying to solve this I got the first part but tried to integrate in rectangular coordinates: Let $P=(x,y)$ then the area of the rectangle is $4x\sqrt{1-x^2}$ and integrating $4\int_0^14x\sqrt{1-x^2}/\pi$ doesn't give $4/\pi^2$.

I then have 3 questions:

1) Why does the integral in the solution have a $2/\pi$ coefficient?

2) Why does the integral in the solution go only to $\pi/2$, I thought its limits would be $2\pi$ or $4$ times the given boudaries.

3) How would you properly do the integral in rectangular coordinates?

• In your integral in Cartesian coordinates you are implicitly assuming that $x$ is uniform in $[0,1]$. An $x$ uniform on $[0,1]$ doesn't correspond to a uniform point on the circle. – Bettybel Jul 6 '17 at 20:48
• Could I do a double integral on x then y? I don't know how to set it up. – mtheorylord Jul 6 '17 at 20:54
• You can cheat. Change variable in the integral in polar coordinates to Cartesian. The Jacobian in the change of variable is the pull back of the uniform probability on the circle to the interval [0,1] of the x coordinate. – Bettybel Jul 6 '17 at 21:08

## 1 Answer

For 1 and 2:

You could integrate from 0 to 2PI. In this case, to take the average, we would need to multiply by 1/(2PI). However, by symmetry, you can also integrate from 0 to PI/2. In this case, to get expected value, you have to multiply the integral by 2/PI. This is what they are doing in the problem. It also works better because doing it this way, one does not have to deal with negative values cancelling things out.

Not so sure about #3