Solution to Problem #8

James D. Beltz of St. Charles MO, Robin Stokes of the University of New England (Australia), and Raymond Roan of Los Gatos CA solved the first part. Ross Millikan of San Mateo CA solved both parts and extended the result to the four-dimensional case [see below]. Here is his solution to the first two parts followed by Raymond Roan's elementary solution to the first part:

We will solve the first part for a general radius of the circle, as this
will be helpful for the following part.  If we put the origin at the
lower left corner of the square, we can write an integral which
represents the upper right quarter of the intersection region.  Then
multiplying by 4 gives the total intersection area. The quarter
intersection area is the integral of sqrt(r^2-x^2)-1/2 from 1/2 to
sqrt(r^2-1/4).  This is 

1/4(1-sqrt(1-4r^2)-2r^2arctan(1/sqrt(4r^2-1))+2r^2arctan(sqrt(4r^2-1)))

For r=1, this gives (pi+3-3sqrt(3))/12, or about .078767

The total intersection area is four times this or about .315147

For the three dimensional volume, if we put the origin at the bottom of
the cube, at any z we will have an area of intersection as calculated
above with r=sqrt(1-z^2).  So the total volume is twice the integral of
the above expression from 1/2 to sqrt(2)/2, which is about .0152055.

[The exact value is: Sqrt[2] + 9*ArcTan[Sqrt[2]/5] - 1 -11*Pi/12]

He continues:

As a follow-up question, what happens in four dimensions?

Click on the button below for the answer.

Here is Raymond Roan's solution to the first part:



*	Notice that the region in question is a square with "bulging" sides.
My approach is to find the area of the square and the area of the "bulges".
*	The side of the square is a chord of a circle of radius one. The
"bulge" is the region between the chord and the circle. Or, viewed another
way, the bulge is the region left over after removing a triangle from a
sector of the circle.
*	The next key is to find the central angle for the sector
corresponding to the chord and "bulge". By using equilateral triangles
(formed from two radial lines of adjacent circles and the edge of the square
connecting their two centers), one can show that the sector angle is 30
degrees. The rest is mechanical.
*	The area of the sector is pi/12.
*	The area of the triangle is 1/4 (drop a foot from one of the
vertices on the circle to the side opposite to get an altitude of 1/2).
*	Therefore, the area of one bulge is [pi/12 - 1/4] and the area of
four bulges is [pi/3 - 1]
*	The length of the chord is sqrt(2 - sqrt(3)) (Pythagorean Theorem).
*	Therefore, the area of the square is [2 - sqrt(3)].
*	Lastly, the area of the region in question is [1 + pi/3 - sqrt(3)].

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