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'm doing this backwards, ie I'm starting with the circles rather than the triangle.

I am considering my circles to be 4 coins of the same denominator.

I arrange the 4 coins so that A', B' and C' touch D', ie, are tangent to D'. None of the 'planetary' coins touch each other.  Drawing the triangle so that each side is tangent to two coins fulfills the appearance of the intended outcome but does NOT fulfill the construction requirement.

I note that if centre D' is on the line joining centres A' and B', and the line joining centres C' and D' is perpendicular to A'B', then the triangle is a right angled isosceles. 
If the line A'B' is not perpendicular we have a scalene triangle.

That this arrangement should produce right angled triangles relates to the fact that A' an B' are points on the diameter of a circle of which D' is the centre.  C' is a point on the circumference of that circle.

Posted by brianjn on 2006-06-14 19:58:40