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: \[ \boxed{5} \]