4 Circles in space

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 $\sqrt{(u+v)^2 - \tfrac43u^2} \times \frac{u}{u+v} \; =\; \frac{70}{43}\sqrt{1826} \;=\; \boxed{69.5633}$

We notice that each of the four circles is on a face of a tetrahedron. Every circle is inscribed in the respective triangle and is internally tangent to the three sides of the triangle or the edges of the tetrahedron.

The base of the tetrahedron is an equilateral triangle $PQR$ inscribing a circle of radius $70$ and center $O$ (top view). $S$ is where the circle tangent to $PQ$ . Since $O$ is the centroid of $\triangle PQR$ , $\overline{RS} = 3 \overline{OS} = 210$ and the side length of $\triangle PQR$ , $\overline{PR} = \overline{RS} \csc 60^\circ = 140\sqrt 3$ .

The slanting faces of the tetrahedron are isosceles triangles congruent to $\triangle PQT$ with $PT=QT$ and $PQ=140\sqrt 3$ (normal to face view). $\triangle PQT$ has an incircle of radius $50$ and center $D$ , which is tangent to its sides at $A$ , $B$ , and $S$ . Let $PT=QT=a$ . Then the semi-perimeter of $\triangle PQT$ , $s= \frac {2a+140\sqrt 3}2 = a+70\sqrt 3$ , and area of $\triangle PQT$ is $s \times$ inradius or $[PQT]=50s$ and by Heron's formula , we have $[PQT] = \sqrt{s(s-a)^2(s-140\sqrt 3)}$ . Then we have,

$\begin{aligned} \sqrt{(a+70\sqrt 3)(70\sqrt 3)^2(a-70\sqrt3)} & = 50(a+70\sqrt 3) \\ 70\sqrt 3 \sqrt{(a+70\sqrt 3)(a-70\sqrt3)} & = 50(a+70\sqrt 3) \\ \sqrt{\frac {a-70\sqrt 3}{a+70\sqrt 3}} & = \frac 5{7\sqrt 3} \\ \frac {a-70\sqrt 3}{a+70\sqrt 3} & = \frac {25}{147} \\ (147-25)a & = (147+25)70\sqrt 3 \\ \implies a & = \frac {6020\sqrt 3}{61} \end{aligned}$

By Pythagorean theorem , $\overline{TS} = \sqrt{\overline{PT}^2-\overline{PS}^2} = \sqrt{\dfrac {6020^2\cdot 3}{61^2}-70^2\cdot 3} = \dfrac {7350}{61}$ . Since $\triangle PST$ and $\triangle ADT$ are similar, then $\dfrac {\overline{AT}}{\overline{AD}} = \dfrac {\overline{TS}}{\overline{PS}} \implies \overline{AT} = \dfrac {\overline{TS}}{\overline{PS}} \cdot \overline{AD} = \dfrac {7350\cdot 50}{61 \cdot 70\sqrt 3} = \dfrac {1750\sqrt 3}{61}$ .

Now the height of the tetrahedron $\overline{OT} = \sqrt{\overline{TS}^2-\overline{OS}^2} = \sqrt{\frac {7350^2}{61^2}-70^2} = \dfrac {140\sqrt{1826}}{61}$ . The height of points $A$ , $B$ , and $C$ above ground:

$\begin{aligned} h & = \frac {\overline{AP}}{\overline{PT}} \cdot \overline{OT} = \frac {\overline{PT}-\overline{AT}}{\overline{PT}} \cdot \overline{OT} =\left(1-\frac {\overline{AT}}{\overline{PT}}\right) \overline{OT} \\ & = \left(1-\frac {1750}{6020}\right) \dfrac {140\sqrt{1826}}{61} = \frac {70}{43}\sqrt{1826} \approx \boxed{69.56} \end{aligned}$