Weekly Challenge 2.1: A problem motivated by B4 Putnam 1985

Putnam original problem statement : 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 with sides parallel to the x-axis and y-axis, lies entirely inside (or on) C ?

This problem is straightforward, in fact it has very elegant geometric proof

Let's draw a rectangle , in fact a square inscribed the circle from random point P , then the probability of all point in rectangle inside the circle is the probability the Q inscribed the square which is 4/π². QED

That’s is satisfying isn’t it. Now, let’s move on to the real weekly challenges which is the problem I have made.

Weekly challenge 2.1

Using the idea of geometric and probability, I created this problem

Problem statement

Given a C area bounded by the equation x² + y² ≤ 1

1. Take a two random point P,Q in this area
2. From P and Q , draw the perpendicular line to PQ, at P,Q.
3. This two line will intersect the circle at X,Y and M,N
4. The effective area A = |PQ|*min(|XY|,|MN|)

Calculating the value of E(A).

Note : This is equivalent to calculate the rectangle with the length |PQ| and the first side when extending to touch the circumference first.

Take your time and try to solve this problem, hope you have fun next part is the solution

Solution

At first glance, this looks like a straightforward geometric probability problem. But a direct approach quickly becomes messy.

Setting up

Using the same idea of the Putnam problem where we need to generalise and try to view the problem from different angle , we can let

θ : is the direction of the line PQ

r: distance from origin to PQ

u1,u2 : position of P, Q on the line

Now using wedge product we can obtain

Proving this we need to reparameterize P,Q

Calculating with applying the wedge anticommutativity

Formula for effective Area

Key insight

Now comes the key geometric observation.

For fixed u1​,u2​, the line containing PQ can slide up and down (changing r) while keeping both points inside the disk.

At this image PQ can be slide on the XY, MN to any new P’Q’ while preserving effective area

Hence we have

Or more explicitly the with this problem

This cancels the square root in h, turning the integrand into a polynomial.

Hence we plug in the expectated value formula and obtain

Symetry arguement

Notice that if we swap P -> Q, Q -> P, the effective area being preserved

Moreover , if we map u1->-u1 and u2-> -u2 the integration over (-1,0 ) equal the integration over (0,1)

Hence setting h² = 4(1-u1)² with |u1| > |u2|, u1>0

Master equation

We obtained this master equation

Integrating …

Hence the expected effective area is 32/9pi which means the effective area is covering ~36.0% of the unit circle !

If you want to submit any problem to weekly challenge , please email me at hokageoftheleaf205@gmail.com . Thank you very much for reading until here , have a good day !