Given: A square and a circle tangent to two adjacent sides of the square Construct: Another circle tangent to the two remaining sides of the square that is also tangent to the given circle
In making this construction, I observed that the green circle disappears if its radius becomes larger than half the distance of the side of the square. Why did this happen? If we look at the range in which arbitrary point H can travel we see that it can travel all the way down to the endpoint of the square. When this happens, the red circle becomes undefined which forces the green circle go beyond its points of tangency with the square. Click here to review this construction.
My next task was to try and refine this construction
so that the green circle does not disappear from the square at
any given time. In other words, I needed to adjust the range (i.e.
make it smaller) in which H can travel. How did I figure this
out? Well, I know that the red circle is always fixed within the
circle. What is the size of the green circle when the red circle
is at its maximum size? For starters the green circle exists,
which means that point H cannont travel all the way to the vertex
of the square (verify by the picture below). The second boundary
must fall somewhere between the midpoint of the side of the square
and the vertex. So, how are we going to locate this fixed point?
When the red circle is at its maximum size,
point H is coincident with the midpoint O of side AC. Another
way of thinking about this is that the radius of the maximum circle
has the same distance as the distance from the vertex to midpoint
O. At the same time, the green circle is going to be at its smallest
and still have all three three given points of tangency on the
red circle hold. Therefore, if we were to reverse the order (i.e.
green circle at its maximum and red circle at its minimum), then
H will be coincident with a point that is the same distance away
from the vertex as the radius of the minimum circle (see picture
below). Therefore, if we can construct a fixed point on that side
of the square that is equal to the distance of the radius of the
minimum red (or green) circle, then we have our two boundary points.
So, how do we create this fixed point?
One observation made was that the points of tangency, namely E on side CD and S on side AB, on opposite sides of the square and the point of tangency, J, between the two circles are collinear. Since they are (although not proven), if we can recreate these three collinear points on the other pair of opposite sides, AC and BD, we can construct a fixed point which will serve as our other boundary point (or end point) for which H can travel. In other words, the two boundary points will serve as the segment that H can travel to keep not only the red circle in tact but the green one as well.
With any fixed point, this point must be totally independent from any of the current constructions needed to make our figure. So, in essence, I need to start over, by constructing an inscribed (blue) circle in a square. Please link on here for the rest of the discussion on this construction.
I can construct the point of tangency of the two circles easily because I already know that it lies on the diagonal which the circle I've just created intersects. By definition of inscribe circles, I already have the point of tangency to the square I need to create third point of tangency on the opposite side of the square. This third point is the fixed boundary point I'm looking for. Once the segment that H can travel on is constructed, I can quickly go back and reconstruct both the red and green circles again.