Woodworking

Insphere is the largest sphere that can fit inside a given tetrahedron. First we are going to calculate radius of the insphere of the original tetrahedron. The formula is given as

$r=\frac{3V}{SA}$

where $SA$ is a Surface Area and $V=\frac{1}{3!}\sqrt{\frac{1}{2^3} det {a_{ij}^2, 1 \choose 1, 0}}$

$a_{ij}$ denotes an edge connecting vertex $A_{i}$ and $A_{j}$ and

$det {a_{ij}^2, 1 \choose 1, 0} = det \begin{pmatrix} 0 & a_{1,2}^2 & a_{1,3}^2 & a_{1,4}^2 & 1 \\ a_{1,2}^2 & 0 & a_{2,3}^2 & a_{2,4}^2 & 1 \\ a_{1,3}^2 & a_{2,3}^2 & 0 & a_{3,4}^2 & 1 \\ a_{1,4}^2 & a_{2,4}^2 & a_{3,4}^2 & 0 & 1 \\ 1 & 1 & 1 & 1 & 0 \end{pmatrix}$

Our determinant:

$det \begin{pmatrix} 0 & 8^2 & 8^2 & 8^2 & 1 \\ 8^2 & 0 & 5^2 & 6^2 & 1 \\ 8^2 & 5^2 & 0 & 7^2 & 1 \\ 8^2 & 6^2 & 7^2 & 0 & 1 \\ 1 & 1 & 1 & 1 & 0 \end{pmatrix} = 354168$

which gives volume of $\sqrt{4919}/2$ .

$SA \approx 81.12200460734$

$r \approx 1.296853732595$

After removing $0.5$ cm off each face the insphere's radius will be:

$r_2 = r - 0.5 \approx 0.796853732595$ .

The answer is $\boxed{7968}$ .

I kept thinking there was some subtle trickery involved in shaving off that 0.5 cm from the faces.