http://www.mathisfunforum.com/viewtopic.php?id=19346

But I daresay when you give a better answer then you did in the best quote thread you will know.

]]>Using phrontister's coloring and layering techniques, I could make this drawing.

]]>Post 1209

This one gave me great joy to solve because although there are several people here who can and did solve it, no one could where it was posed?! (Update: Someone else has submitted a solution there too. Wunderbar!) A triumph for experimental math!

1) Let us start with an accurate drawing, so we use geogebra and create two points (0,0) and (10,0). They will be marked as A and B.

2) Use the regular polygon tool and select A and then B and enter 4 as the number of vertices and a square with sides 10 is created.

3) Use the midpoint tool and click on all 4 sides of the square and points E,F,G and H are created.

4) Use the circle with 3 points tool and click on E,F and G and the inscribed circle is created.

5) Use the circle with radius tool and click on B and enter 10 and the arc is created.

6) We can see the two red regions are same ( should be easy enough to prove. What we need is the intersection of the big circle and the little one marked on the drawing. Of course geogebra can get it easily but we will use M for the job.

7) Enter

Solve[{(x - 5)^2 + (y - 5)^2 == 25, (x - 10)^2 + y^2 == 100}, {x, y}]//First

You will get:

that is the intersection of the big and the little one.

8) Now we just need to get that area on the left by integration.

Integrate[5 - Sqrt[10 x - x^2], {x, 0, 5/4 (3 - Sqrt[7])}] -

Integrate[Sqrt[20 x - x^2], {x, 0, 5/4 (3 - Sqrt[7])}]

yields

We are done!

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