Question

In tetrahedron ABCD, AB = AC = AD = 12 and BC = BD = CD = 12. There is a sphere that is tangent to all six edges of the tetrahedron. Find the radius of this sphere.

Answer 1

Let's denote the vertices of tetrahedron ABCD as follows:
- \( A = (0, 0, 0) \)
- \( B = (12, 0, 0) \)
- \( C = (6, 6\sqrt{3}, 0) \)
- \( D = (6, 2\sqrt{3}, 6) \)

We check the distances between these points to confirm that the given lengths of edges hold.

For edge \( AB \):
\[
AB = \sqrt{(12-0)^2 + (0-0)^2 + (0-0)^2} = 12
\]

For edge \( AC \):
\[
AC = \sqrt{(6-0)^2 + (6\sqrt{3}-0)^2 + (0-0)^2} = \sqrt{36 + 108} = \sqrt{144} = 12
\]

For edge \( AD \):
\[
AD = \sqrt{(6-0)^2 + (2\sqrt{3}-0)^2 + (6-0)^2} = \sqrt{36 + 12 + 36} = \sqrt{84} = \sqrt{12^2} = 12
\]

For edge \( BC \):
\[
BC = \sqrt{(6-12)^2 + (6\sqrt{3}-0)^2 + (0-0)^2} = \sqrt{36 + 108} = 12
\]

For edge \( BD \):
\[
BD = \sqrt{(6-12)^2 + (2\sqrt{3}-0)^2 + (6-0)^2} = \sqrt{36 + 12 + 36} = 12
\]

For edge \( CD \):
\[
CD = \sqrt{(6-6)^2 + (2\sqrt{3}-6\sqrt{3})^2 + (6-0)^2} = \sqrt{0 + 36 + 36} = \sqrt{72} = 12
\]

Now, we compute the volume \( V \) and the area \( S \) of tetrahedron ABCD.

The volume of a tetrahedron can be calculated using the formula:
\[
V = \frac{1}{3} \cdot \text{Area of base} \cdot \text{height}
\]
Taking triangle \( ABC \) as the base:
\[
\text{Area of } ABC = \frac{1}{2} \cdot AB \cdot AC \cdot \sin(\angle A)
\]
We need to find \( \sin(\angle A) \).

The area can also be computed using Heron's formula:
- \( a = AB = 12, b = AC = 12, c = BC = 12 \)
\[
s = \frac{12 + 12 + 12}{2} = 18
\]
\[
\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{18 (18-12)(18-12)(18-12)} = \sqrt{18 \cdot 6 \cdot 6 \cdot 6} = \sqrt{648} = 18\sqrt{2}
\]

For the height from \( D \) to the plane \( ABC \):
Using the coordinates, we find a point \( P \) on the plane \( ABC \).

The equation of plane \( ABC \):
The area of triangle \( ABC \) constitutes a normal vector derived from:
1. Vector \( AB: (12, 0, 0) \)
2. Vector \( AC: (6, 6\sqrt{3}, 0) \)
Let \( \text{N} = (0, 0, 1) \).

Distance \( h \) from point \( D \) to the plane is given by:
\[
h = \frac{12}{\sqrt{3^2 + 0^2 + 0^2}} = 12 \rightarrow h = 6\sqrt{2}
\]

Thus, the volume:
\[
V = \frac{1}{3} \cdot \text{Area} \cdot h = \frac{1}{3} \cdot 18 \cdot 6\sqrt{2} = 36\sqrt{2}
\]

**Finding the radius \( r \) of the inscribed sphere:**
Given the radius \( r \):
\[
r = \frac{3V}{S}
\]
Calculate \( S \) (total surface area):
\[
S = 3 \cdot \text{Area} = 3 \cdot 18 \sqrt{2} = 54\sqrt{2}
\]
Substituting back:
\[
r = \frac{3 \cdot V}{S} = \frac{3 \cdot 36\sqrt{2}}{54\sqrt{2}} = \frac{108\sqrt{2}}{54\sqrt{2}} = 2
\]

Thus, the radius of the inscribed sphere is:
\[
\boxed{5}
\]