I've been trying to construct the following figure geometrically:
I've been tearing my hair out all afternoon. Because of the irrational radius lengths of the circles, this problem is (at least to me) incredibly difficult.
If I was given the triangle joining the centers of the congruent circles and asked to draw the circles and the large triangle, it would be easy. But given the large triangle how can I do this? Can I work it backwards somehow?
To clarify, the triangle is equilateral and the circles are tangent to each other, and to the triangle.
Please help!
Best Answer
For an equilateral triangle $ABC$, I believe the construction using the fewest elementary operations is:
Construct $CM$, the perpendicular bisector of $AB$, which will intersect $C$; $M$ is the midpoint of $AB$.
Construct the angle bisectors of angle $CAB$ and angle $CMA$; these meat at point $P$.
Drop a perpendicular from $P$ meeting $AB$ at $T$.
Construct $BN$, the perpendicular bisector of $AC$; $N$ is the midpoint of $AC$.
Construct $Q$ on $CM$ such that $CQ = AP$, and $R$ on $BN$ such that $BR = AP$.
The centers of the three circles will be at $P,Q, R$ and the radii will be the length of $PT$.