Show how to construct four congruent circles, inside an acute triangle ABC, with centers A', B', C', and D' such that

1. circle with center A' is tangent to sides AB and AC,
2. circle with center B' is tangent to sides BC and BA,
3. circle with center C' is tangent to sides CA and CB, and
4. circle with center D' is externally tangent to the other three circles.

I begin with any given acute triangle ABC, construct a bigger triangle A"B"C", then use similar relationships to find the smaller triangle A'B'C'.

Note that when finished, triangle A'B'C' is similar to ABC by coresponding sides being parallel.

I will begin with ABC as if it is the inner triangle to some larger A"B"C".

Construct perpendicular bisectors of AB, BC, and AC. These intersect at a point X (the circumcenter of ABC) which is equidistant from A, B, and C. Let AX=BX=CX=2R (R is the radius of my big circles). Bisect any one of these segments to get the length of R and construct three circles centered at A, B, and C with radius R. Now construct external tangents to the circles parallel to AB, BC, and AC. These will intersect at points A",B", and C"; the verticies of my larger similar triangle.

Now I can use properties of similarity to construct A'B'C' as it is to ABC as ABC is to A"B"C". For example A,A', and A" are collinear and all lie on the angle bisector of A. Also the angle AA"B is similar to A'AB'. Constructing these similar angles will triangulate the positions of A',B', and C'. D' will simply be the new X' (or circumcenter) of triangle A'B'C'.

Once we have A',B',C', and D', the construction of the four congruent circles is trivial as the radius r is simply half the distance from D' to A',B', or C'.

Edited on June 14, 2006, 11:48 pm

Edited on June 15, 2006, 12:19 am

Posted by Eric on 2006-06-14 23:43:16