# A lawn, a flower, a pipe and the neighbours

You have a square lawn and a precious flower in the centre. You want to make sure you water the flower, and you don't particularly care how much of the lawn you water. To please your aleatory aesthetic sense, you uniformly randomly pick two points on the lawn as end points of an elevated irrigation pipe, which will rotate around its centre and irrigate the circle it covers. You keep picking points until the pipe waters the flower and none of the neighbours' plots.

What's the probability for one pick to yield admissible points? Suppose your costs are mainly determined by the material for the pipe; what is its expected length? Suppose your costs are mainly determined by the water used; what's the expected irrigated area? How do these compare to the corresponding values if you just picked two random points without worrying about the flower or the neighbours?

The background to this question is that I realized in thinking about this answer that a) calculations can actually be simplified by seemingly complicated conditions like the circle determined by the points being inside the square (the expected length of the segment is easier to determine with that condition than without), and b) some conditions, like the circle being inside the square, reduce the expected length and area, whereas others, like the centre being inside the circle, increase the expected length and area, and it would be interesting to see which of these effects is stronger. I'm hoping that perhaps it's different for the length and the area. Feel free to calculate higher moments.

I expect this to be a straightforward calculation with the approach I used in that answer; it's just that the integration limits are slightly complicated by the condition that the centre be in the circle. I'll try to perform the calculation and write an answer if noone else does, but I thought I'd share the problem with you first.

## 1 Answer

The integration limits do need some thought but the rest of the problem is reasonable.

Like the answers to the previous question, I will concentrate on cases where the midpoint $M$ of the two random points and coordinates $(x,y)$ is in the octant with $0 \le y \le x \le \frac12$.

The probability density of a midpoint $M$ at is proportional to the area of the blue rectangle in which the two random points must lie, i.e. proportional to $2x \times 2y$ in this octant. Since $8 \int_{x=0}^\frac12 \int_{y=0}^x 4xy \, dy \,dx = \frac14$, we know the constant of proportionality is $4$, which we need to use to multiply the area of the rectangle, and later to multiply the area we are interested in.

For the water not to hit the neighbours, we need the two random points to lie inside the outer purple circle. This has radius $y$ and so area $\pi y^2$. For the water to hit the square's centre $C$ at $\left( \frac12, \frac12 \right)$, we need the two random points to lie outside the inner green circle. This has radius $\sqrt{\left( \frac12 -x \right)^2 + \left( \frac12 -y \right)^2 }$ and so area $\pi \left(\left( \frac12 -x \right)^2 + \left( \frac12 -y \right)^2\right)$. We also require the radius of the inner circle to be no more than that of the outer circle which leads to $y \ge x^2-x+\frac12$ and $x \ge 1 - \frac{1}{\sqrt{2}}$ in this octant.

So the problem becomes one of integrating the area of the annulus. This gives $$8 \int_{x=1 - \frac{1}{\sqrt{2}}}^\frac12 \int_{y=x^2-x+\frac12}^x 4 \pi \left( y^2 -\left( \tfrac12 -x \right)^2 -\left( \tfrac12 -y \right)^2\right) \, dy \,dx$$ $$= \pi\frac{(\sqrt{2048}-43)}{30}$$ which is about $0.2361$.