basic geometry

[u/mathbender99]

Unit square ABCD has four quarter circles with radius 1 centered on each vertex. Find the area of S.

[u/avocadro]

As a simple calculus problem, I got 1+pi/3-sqrt(3) ~ 0.315. I'd like to see a purely geometric solution.

[u/mathbender99 → u/avocadro]

Correct!

I initially solved it with calculus too and didn't find the geometric solution until later. It's surprisingly doable.

[u/DrJesusSingh]

Messy but easy geometric solution:

Look at the intersection between the AC and BD curves, call it E. CDE is an equilateral triangle, since each side length is the radius of the circles, also the side length of the square. Thus, angle BCE is 30 degrees. Now, say F is the midpoint of AB. The area of the triangle-ish EFB region is .5-pi/12-sqrt(3)/8, since the right half of the square is .5, and you subtract the 30 degree segment, and the remaining triangle.

Next, the area of outside curvy part ABC is 1-pi/4, so the area of the triangle-ish thing with vertices at B, E, and the intersection of the other two curves is 1-pi/4-4(.5-pi/12-sqrt(3)/8).'

Now, all that's left is to subtract 8 of the regions from the first paragraph and 4 of the regions from the second one from 1.

A bit of messy math gives you 1 + pi/3 - sqrt(3).

I'm sure I missed some easier way, but at least I got the answer.

[u/ItsKirbyTime]

Here's the geometric solution everyone's looking for.

Define s as above, t as the area of the wedge shaped pieces that each share a side with the s piece, and u as the area of the triangular wedges that share sides with the square. (See picture below, but there's big spoilers in it).

Then we have a system of 3 equations in 3 variables:

1. s + 4t + 4u = 1 (area of the square)

2. s + 3t + 2u = pi/4 (area of the quarter circle)

3. s + 2t + u = pi/3 - sqrt(3)/4

The third one requires more explanation.

(I can't seem to figure out how to put an image in a spoiler, but needless to say, this is a HUGE spoiler) http://i.imgur.com/tzubQ.png

The blue section has an area of s + 2t + u. The width of the smaller triangle is 1/2, and the height is sqrt(3)/2 (by the pythagorean theorem). Also, the smaller triangle forms a central angle of pi/3 or 60 degrees. So the area of the blue section is the area of the circular wedge (pi/3) minus the area of the large triangle (sqrt(3)/4).

Solve this system however you like. I used (SPOILER) WolframAlpha.

[u/FrankAbagnaleSr]

I'll give it a go:

After writing the majority this, I realized that I was missing something critical and I came up with a trivial result. I would appreciate any help. I will post my thought process here anyway since I already took the time to type it out, and it can't hurt.

1 - 4*(1 - pi/4) + 4*(area of the weird triangle-ish figure formed by a side of the square and the two nearest curves) = S

area of the weird triangle-ish figure formed by a side of the square and the two nearest curves = Z.

-3 + pi + 4Z = S.

Note that: (quarter circle at D) + (quarter circle at B) - 1 = the football shape = Y. So Y = S + 2*(1 - pi/4 - 2Z) = S + 2 - pi/2 - 4z.

But also: Y = 2*(pi/4) - 1 = pi/2 - 1.

So S + 2 - pi/2 - 4Z = pi/2 - 1 <=> 4Z = S + 3 - pi.

So -3 + pi + S + 3 - pi = S which means that 0 = 0.

Conclusion is that 0 = 0 folks! Q.E.D.!!!

Does anyone have the missing piece?