This is my rewording of question B-4 of the 1985 Putnam competition
A point $P1$, is uniformly sampled from the unit disc, and another $P2$, from the unit circle that is the boundary of the disc. What is the probability that the rectangle $R(P1,P2)$ with sides parallel to the axes and with these two points as opposite vertices has points outside the disc.
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Fig 1. Diagram showing a pair of points and the
rectangle the problem refers to. |
Since we are not sitting in an exam room, nor am I going to look up the solution , and as usual when working on a problem all resources are deployed my first action when faced with a problem like this is to estimate the answer by simulation. That way I will have a check value when a more appropriate answer is found.
The first thing to observe is that if any points of the rectangle defined by $P1$ and $P2$ lie outside the unit disc then one of the other vertices of the rectangle must be outside the disc. This makes the simulation easier to implement as we need only check the other vertices.
The Gnu-Octave (Matlab clone-oid) simulation gives results:
Putnam1
number of replications = 100000
pp = 0.5971
se = 1.5510e-03
Where pp is the probability estimate, and se its standard error.
While we have a simulation available we might as well look at the scatter plot of points $P1$ that produce rectangles with points outside the disc...
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Fig 2. Scatter Plot of points P1 that result in a rectangle
with points outside the disc. The red marker indicates
position of P2 for this case. |
The scatter plot suggests, what if I had been in training for the competition should have been obvious, that given $P2$ the points $P1$ outside the inscribed rectangle with $P2$ as a vertex will produce a rectangle $R(P1,P2)$ with points outside the unit disc.
So if $P2=(x,y)$ the area of the region is $A(P2)=2x\times 2y$. Or if we write $P2=(\cos(\theta),\sin(\theta))$ we get $A(P2)=4\cos(\theta).\sin(\theta)$.
Hence the probability that R(P2) has points outside the unit disc is
$p(\theta)=(\pi-4 \cos(\theta) \sin(\theta))/ \pi$
Then the probability asked for in the question is:
$P=\frac{\int_0^{\pi/2} p(\theta) d\theta}{\pi/2} $
(the integral here only needs to extend over a quarter of the unit circle due to the symmetries of the geometry.
This integral is more or less text-book stuff, and can be done by using the double angle identity: $\sin(2\theta)=2 \sin(\theta).\cos(\theta)$ or observing that $\frac{d}{d\theta}\sin(\theta)=2 \sin(\theta).\cos(\theta)$.
Either way we find that:
$P= \left(\frac{\pi}{2}-\frac{2}{\pi}\right)/(\pi/2)\approx 0.5947$
The simulation estimate is less than 2 standard errors from this, so these are sufficiently in agreement for government purposes.
