Probability that a Rectangle lies within a circle

[guccimane]

This is not really a homework problem, I'm just doing it as an exercise puzzle. I think I'm on the right track, but at this point I feel a little exhausted and would love a hint.

Homework Statement

Let C be a unit circle: x^2+y^2=1 . Let "p" be a point on the circumference and "q" be a point anywhere inside C. Finally, let the line p-q be the diagonal of a rectangle R, the sides of which are parallel to the x and y axes. What's the probability that ALL points within R lie within C?

Presumed steps: 1,2,3, ?, ... n) integrate some density function, get answer.
Given answer: 4/(pi^2)

The Attempt at a Solution

(t:="theta")

I'm pretty certain that 1) and 2) are right:

1) Find the Cartesian coordinate representation of the 4 vertices of R:

p= (cos(t), sin(t)); q= (a, b); p'= (cos(t), b); q'= (a, sin(t))2) We know that p and q are within the disk C, so we need the conditions under which p' and q' lie in C:

for p' : cos^2(t) + b^2 < 1 i.e |b|<|sin(t)|
for q' : a^2 + sin^2(t) < 1 i.e. |a|<|cos(t)|3) Next is my attempt to find the following probabilities:
Letting 0≤ t ≤ pi/2 -- trying to stay in the first quadrant;

P(|b|<|sin(t)|) = (∫[0..cos(t)] of √(1-x^2) dx) / (pi/4)
= (1/2*(cos(t)(sin(t)) + pi/2 - t)/ (pi/4)

P(|a|<|cos(t)|) = (∫[0..sin(t)] of √(1-y^2) dy) / (pi/4)
= (1/2*(cos(t)(sin(t)) + t) /(pi/4)4) One has to find the probability of the intersection of the two events in (3).

My initial though was to multiply the two probability function in (3) and then integrate from 0 to pi/2; Result: wrong answer.

In my second attempt, I tried first differentiating the two probability functions, then multiplying them together, then integrating from 0 to pi/2.
(Don't ask why - I guess I thought differentiating them would make them look like density functions ... or something). Result: Wrong answer.

==================================================================
My solution fell off the rails 4 - or perhaps at 3 in some ridiculously elementary way(s). This a somewhat basic problem that should conclude in the integration of a prob.-dens. function.

I'll be eternally grateful for any assistance on this!

Homework Statement

Homework Equations

The Attempt at a Solution

 

[Ray Vickson]

This is not really a homework problem, I'm just doing it as an exercise puzzle. I think I'm on the right track, but at this point I feel a little exhausted and would love a hint.

Homework Statement

Let C be a unit circle: x^2+y^2=1 . Let "p" be a point on the circumference and "q" be a point anywhere inside C. Finally, let the line p-q be the diagonal of a rectangle R, the sides of which are parallel to the x and y axes. What's the probability that ALL points within R lie within C?

Presumed steps: 1,2,3, ?, ... n) integrate some density function, get answer.
Given answer: 4/(pi^2)

The Attempt at a Solution

(t:="theta")

I'm pretty certain that 1) and 2) are right:

1) Find the Cartesian coordinate representation of the 4 vertices of R:

p= (cos(t), sin(t)); q= (a, b); p'= (cos(t), b); q'= (a, sin(t))


2) We know that p and q are within the disk C, so we need the conditions under which p' and q' lie in C:

for p' : cos^2(t) + b^2 < 1 i.e |b|<|sin(t)|
for q' : a^2 + sin^2(t) < 1 i.e. |a|<|cos(t)|


3) Next is my attempt to find the following probabilities:
Letting 0≤ t ≤ pi/2 -- trying to stay in the first quadrant;

P(|b|<|sin(t)|) = (∫[0..cos(t)] of √(1-x^2) dx) / (pi/4)
= (1/2*(cos(t)(sin(t)) + pi/2 - t)/ (pi/4)

P(|a|<|cos(t)|) = (∫[0..sin(t)] of √(1-y^2) dy) / (pi/4)
= (1/2*(cos(t)(sin(t)) + t) /(pi/4)


4) One has to find the probability of the intersection of the two events in (3).

My initial though was to multiply the two probability function in (3) and then integrate from 0 to pi/2; Result: wrong answer.

In my second attempt, I tried first differentiating the two probability functions, then multiplying them together, then integrating from 0 to pi/2.
(Don't ask why - I guess I thought differentiating them would make them look like density functions ... or something). Result: Wrong answer.

==================================================================
My solution fell off the rails 4 - or perhaps at 3 in some ridiculously elementary way(s). This a somewhat basic problem that should conclude in the integration of a prob.-dens. function.

I'll be eternally grateful for any assistance on this!

Assume the "outer" corner is at angle t, counterclockwise from the x-axis, and assume that t is uniformly distributed on (0,2π). Assume also that the point q is uniformly distributed within the circle. If F={rectangle fits in circle} we have
$$P\{F\} = \frac{1}{2 \pi} \int_{0}^{2 \pi} P\{ F|t\}\, dt.$$
It is clear that we can just integrate from 0 to π/2, then multiply by 4.

For 0 < t < π/2, how do we find the largest possible rectangle? Just draw a horizontal and vertical line through $p = (\cos(t),\sin(t))$ and see where these lines cut the circle again. That will give the largest rectangle R_t. If q lies inside R_t. the constructed rectangle with opposite corners p and q will lie completely inside the circle. The sides of R_t are of lengths 2*cos(t) and 2*sin(t), and so its area is easy to compute in terms of t. We have
$$P\{F|t\} = \frac{\text{Area of }R_t}{\text{Area of circle}}.$$

RGV