## Saturday, May 21, 2011

### n circles with radius r ...

... are inscribed in a circle with radius 1

Determine the radius r and the area A of the shaded region between the circles for n = 4 and n = 3

Hint: In an equilateral triangle the medians divide each other in the ratio of 2:1

1. for n=3 the radius of the circles are 2 Sqrt[3]-3 and area of shaded section would be
21 sqrt[3] - 36 - 21/2 Pi + 6 sqrt[3] Pi which is approx 0.0347327.

For n=4 the radius would be sqrt[2] - 1 and the shaded secion works out at 12 - 8 sqrt[2] + 2 sqrt[3]* Pi - 3 * Pi which is approx 0.147279

Paul

2. Nice Paul. Now, let's contemplate with n > 4

3. There are some quite long expressions when n=5. We use sin 36 degs as Sqrt[10 - 2 Sqrt[5]]/4 to find r which equates to, -5 + 2 Sqrt[5] + Sqrt[50 - 22 Sqrt[5]] or 0.370192.

We then have a Pentagon whos side length is 2r and area of, 5 (30 - 14 Sqrt[5] + Sqrt[1885 - 842 Sqrt[5]]) or 0.943111.

We have a combines circle area to remove of 3/2 * Pi r^2. (3/2 because the area of 1 circle section is (108/360) and we have 5 of them). This area is quite nasty giving
(285 \[Pi])/2 - 63 Sqrt[5] \[Pi] - 15 Sqrt[50 - 22 Sqrt[5]] \[Pi] +
6 Sqrt[5 (50 - 22 Sqrt[5])] \[Pi]. or 0.64579

and finaly the remaining area is

0.297316

The expresion is
5 (30 - 14 Sqrt[5] + Sqrt[1885 - 842 Sqrt[5]]) -
3/2 (95 - 42 Sqrt[5] + 4 Sqrt[250 - 110 Sqrt[5]] -
10 Sqrt[50 - 22 Sqrt[5]]) \[Pi].

Working on n=6 to follow.

Paul

4. n=6 is an easy one. The radius of small circle is 1/3. the area of the small hex formed is 2/Srqt[3]. The section of the circles removed is 2/9 * Pi leaving the shaded area as
0.4565688

Paul.