Three circles of radius $50$ and one big circle of radius $70$ are to be arranged such that the big circle lies in the horizontal $xy$ plane (the ground) and each of the three smaller circles, are tangent to the big circle and tangent to the other two small circles.

How high above the ground is the tangency point between two small circles? (these are points A, B, and C in the attached figure)

The answer is 69.56.

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The four circles will be the incircles of a triangular pyramid with equilateral triangle base, and with three congruent isosceles triangles as the other three sides. The incircle of the base will have radius $70$ , while the inradius of the isosceles triangles will be $50$ . The incircles will touch each other at their points of tangency with the edges of the tetrahedron. Thus the equilateral triangle base will have sides $2u$ , and the isosceles triangles will have sides of length $2u,u+v,u+v$ , where $u,v$ are the distances from the vertices of the triangular faces to the points of tangency with the incircles, as shown below.

Since the equilateral triangle with side length $2u$ has inradius $70$ , we deduce that $u = 70\sqrt{3}$ . Using Heron's formula, the isosceles triangle has area $\sqrt{u^2v(2u+v)}$ . and hence has inradius $\sqrt{\frac{u^2v}{2u+v}}$ Since this equals $50$ , we deduce that $v = \tfrac{1750}{61}\sqrt{3}$ . The height of $A,B,C$ above the base is then equal to $\tfrac{u}{u+v}$ times the height of the pyramid, and so is equal to $\sqrt{(u+v)^2 - \tfrac43u^2} \times \frac{u}{u+v} \; =\; \frac{70}{43}\sqrt{1826} \;=\; \boxed{69.5633}$