### Circular Window

**Click
here for a problem description**

There are two aspects to this problem:

- Construct it using only a straight edge and compass, and
- Determine the area of the interior space as a function of the area
of the whole circle.

The construction relies on first "seeing" the underlying problem
and then on translating that to a construction that has already been dealt
with.

The underlying issue in this problem is to construct three circles interior
to the large circle each of them tangent to each other and tangent to the
large circle.

Consider only one of these circle, it must be tangent to two radii of
the circle (see solid lines) and must meet the large circle where the angle
bisector (see dashed lines) of these radii meets the large circle. In our
thinking we can really reduce this to constructing a circle through the
point where the angle bisector meets the large circle and tangent to ONE
of the radii. This final analysis reduces the problem to the construction
of a circle tangent to a line and passing throug two (equal) points - a
construction dealt with in the **Apollonius
problem.**

Following the lead from the circle through two points and tangent to
a line construction:

Construct a line perpendicular to OA and let it meet the line of interest
at B. On OB, mark C so that AB = CB, the center of one of our circles is
then found by determining the intersection of the perpendicular to OB through
C and the radius OA. The other circles follow in a similar manner.

To determine the area of the interior region it may be helpful to think
of that in terms of the triangle O'O''O'''.

In particular, if we can determine the radius of the interior circles
then we will know the sides of the triangle and the area ratio should follow.

To begin, note that triangle OAB is similar to triangle OCO', from which
it follows that OA / OC = AB / CO' (ie R / OC = AB / r),

and that:

The unshaded interior interior of the triangle is derived as follows:

And finally the ratio of the interior space to the whole circle is given
by:

**Return**