Pool Math

First, look here .

As shown in the figure above, we assume that, as the bar moves, it is always tangent to a given curve e inside of the original section - and this curve is not the curve described by point P. Let Q be the point at which the bar is tangent to e , A and B be the points at each end of the bar and r the distance from A to Q. We then use the fact that the area of a tangent sweep equals the area of a tangent cluster ( Mamikon's Theorem , probably not the most famous of theorems) to say that the area covered by a given tangent cluster is:

$\int_{0}^{2\pi}\int_{0}^{k(\theta)}RdRd\theta = \frac{1}{2}\int_{0}^{2\pi}k^{2}(\theta)d\theta$

Instead of solving for numbers, I'll suppose segment AP has length a and segment PB has segment b, for a total of a+b. We can then write three integrals:

1) The area covered by segment AQ

$I_{AQ}=\frac{1}{2}\int_{0}^{2\pi}r^{2}d\theta$

2) The area covered by segment BQ

$I_{BQ}=\frac{1}{2}\int_{0}^{2\pi}(a+b-r)^{2}d\theta$

3) The area covered by segment PQ

$I_{PQ}=\frac{1}{2}\int_{0}^{2\pi}(a-r)^{2}d\theta$

Notice that segments AQ and BQ sweep the same area, which means $I_{AQ}=I_{BQ}$ . Notice also that if we take the area swept by BQ minus the area swept by PQ, we get the area swept by PB - which is what we wanted in the first place. Therefore:

$A=I_{BQ}-I_{PQ}=\frac{1}{2}\int_{0}^{2\pi}(a+b-r)^{2}-(a-r)^{2}d\theta=\frac{1}{2}\int_{0}^{2\pi}(2ab+b^{2}-2br)d\theta$

However, r depends on \theta; we then use our previous relation

$I_{AQ}=I_{BQ}$

which yields

$\int_{0}^{2\pi}r d\theta = \pi(a+b)$ .

Finally, we substitute this last integral on A:

$A=\frac{1}{2}\int_{0}^{2\pi}(2ab+b^{2}-2br)d\theta=2\pi ab+\pi b^{2}-\pi ab-\pi b^{2}=\pi ab$

Therefore, $A=\pi ab$ . This is called Holditch's Theorem .

Choosing a=3 and b=7, we get $A=\pi\times 3\times 7=65.97344...\approx \boxed{66}$ .

Since the problem implies that the difference in area is constant no matter the shape of the pool, I envisioned a simplified pool to make the problem easier. It is very interesting that this is invariant, by the way.

I used a circular pool of diameter 10, so that the bar just fit across. Then the point P makes a circle of radius 2. The difference in area is just pi*(5^2 - 2^2). This is just a bit less than:

$\boxed{66}$