# Squares in a quadrant: How big is the pool?

The local council is building a new pool complex. A one hectare ($$100\times100$$m) block of land has ground suitable for pool-digging, but a local ring road runs through it, leaving only an exact quarter-circle available.

They want a children's wade pool and a deeper adults' pool, both square. They want to utilize all the suitable land they can, so the adults’ pool must reach the suitable land's edge at three of its corners, where the mandatory three lifeguards will be stationed (dots). Half the children's pool's perimeter must touch the edge, where a stand (L-shape) will be built for parents and lifeguards. The pools must meet at their corners so people can go between them easily.

After seeing the architect’s design (below), the grinchy councillors think the children’s pool is too big. "Exactly how many hectares is this pool?" they ask. The architect says their specifications left him no choice.

How many hectares does the children’s pool occupy?

This is a purely geometrical puzzle of my own construction. Do not assume the image is to scale. The setup is flavourtext and non-essential. The puzzle boils down to: Two squares are inscribed in a unit quadrant as shown in this simplified diagram: what is the blue square’s area? While it is simple and straightforward, and has several possible methods of solution (any of which are welcome), and the answer is not unexpected, there is a simple elegant way to solve it, which is (in my opinion) nice enough to qualify it for this site, rather than being just a maths problem.

• Two adults' lifeguards have to stand on the ring road? That kinda defeats the purpose of, y'know, being there to save lives. Oct 12, 2021 at 14:39
• The councillors are only concerned with maximizing the size of their pools given the suitable land and meeting the 3-lifeguard requirement of their local legal system. They don't care how they go 'bout meeting those requirements...
– Anon
Oct 12, 2021 at 15:39

## 1 Answer

Let O be the centre of the circle, A,B ccw the two points where the adults' pool touches the circle's perimeter and X the point where it touches the straight enclosure. Let Y be the point where the two pools touch and r be the side length of the children's pool.

Consider the perpendicular bisector M of A and B. M is an axis of symmetry for both the adult's pool and the (full) circle. Taking the mirror image wrt M therefore leaves O where it is and swaps X and Y. It follows that the distance OX is $$\sqrt 2 r$$.

Using some of the eight congruent right triangles indicated in orange we find: The coordinates of A relative to O are $$x,y = r \times (\sqrt 2 + 1), r \times (\sqrt 2 - 1)$$. Applying Pythagoras and taking advantage of the mixed terms in the squares cancelling we get $$(100 \mathrm m)^2 = r^2 \times 2 \times (2+1)$$. Alternatively, and even simpler, we can obtain the same identity from the coordinates of B which are $$x,y = r\times 2,r \times \sqrt 2$$.

The children's pool's area is exactly

1/6 ha.

• +1 A nice simple argument. I'll wait a little before accepting just in case an even nicer derivation is posted.
– Anon
Oct 12, 2021 at 8:30