Triangle and Three Circles Puzzle

Consider a circle O with a diameter AB, shown here in green. Draw a second circle (red) with diameter AC, such that C is on AB.  Draw an isosceles triangle with base CB and third vertex D on circle O. Draw a third circle (X), tangent to the first three figures. Prove that the line from C to the center of circle X is perpendicular to AB.

Solution:

I put the puzzle on a coordinate grid. I let the red circle have a radius of r, and the large green circle have a radius of R.

I put the center of the red circle (radius r) at (-r,0), so the origin is on the red circle.

Then the large green circle (radius R) is centered at (R-2r,0)

The apex of triangle is (R-r, sqrt(R²-r²))

Circle "X" is internally tangent to the green circle and externally tangent to the red circle. Letting x be the radius of circle x, we see the distance from the center of X to the center of the green circle is R-x. The distance from the center of X to the center of the red circle is r+x. So the sum of the distances from X to the centers of the other two circles is R-x+r+x, or simply R+r. The locus of points the sum of whose distances is R+r to the centers of the red and green circles is an ellipse with the centers of these circles as foci.

The center of the ellipse is (0,(R-3r)/2) which is locus of centers of circles tangent to r and R. The foci of the ellipse are centers of the two circles. The distance of an ellipse from foci equals sum of radii of circles.

Now, consider point "Y", the intersection of the ellipse and the y-axis, obtained from the equation of the ellipse by setting x to zero, is
(0,sqrt(2R(R-r)) 2r/(R+r)). The distance of point "Y" from the red circle is the distance between their centers minus the radius of the red circle,

sqrt(2R(R-r) (2r/(R+r))² + r²)-r
= r(3R-r)/(R+r)-r
= 2r(R-r)/(R+r)

and, of course, this is also the distance from "Y" to the green circle. Now, if this is also the distance from "Y" to the side of the isosceles triangle, then "Y" is the center of circle "X", putting it squarely on the y axis.

Consider the triangle whose vertices are the origin, "Y", and the apex of the triangle. Considering the y-axis the base, and the x-coordinate of the triangle's apex as the altitude, we see the area of this triangle is

sqrt(2R(R-r)) r(R-r)/(R+r).

Now, tilting your head a different way, considering the side of the triangle as the base, we find the perpendicular distance of point "Y" from this base by dividing twice the area by the length of the base.

Thus the distance from "Y" to the side of the triangle is

sqrt(2R(R-r)) 2r(R-r)/(R+r)/sqrt((R-r)²+R²-r²),
which simplifies to 2r(R-r)/(R+r)

So we see that "Y" is equidistant from the red circle, the green circle, and the side of the triangle. Thus it is the center of circle "X", and it is on the y axis, which completes the proof.

But is there a simpler, perhaps more geometric, solution?

Internet references

Ken Duisenberg's Problem Of The Week, February 14, 2002

Cut the knot presents an 1803 Sangaku problem: Circles and Regular Triangle

Jean Constant −> Wasan Geometry −> Sangaku problems  

Related pages in this website

Puzzles

Summary of Geometrical Theorems

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