The unchanging red ball

Geometry Level 4

Tetrahedron $ABCD$ has $AB = AC = 4$ , $AD = 8$ and the edges that meet at $A$ are perpendicular to each other $(\angle BAC = \angle CAD = \angle DAB = 90^\circ)$ .

The ${\color{#D61F06}{\text{red sphere}}}$ is inscribed in the tetrahedron.

Now, suppose we adjust the lengths of $AB$ , $AC$ and $AD$ in a way that does not affect the size of the inscribed sphere and manage to minimise the volume of the tetrahedron. If this minimal volume is ${V_{\min }} = m + n\sqrt p$ , where $m$ , $n$ , and $p$ are positive integers and $p$ is square free, submit $m + n + p$ .

The answer is 17.

1 solution

A Former Brilliant Member
Jun 8, 2020

The sides of the base of the tetrahedron are of lengths $4\sqrt 2,4\sqrt 5,4\sqrt 5$ , and the height is $\dfrac{8}{3}$ . The area of the faces are $24,16,16,8$ . Hence the radius of the inscribed sphere, which is the $z$ -coordinate of the incenter, is

$\dfrac{\frac{8}{3}\times 24}{24+8+16+16}=1$ .

The volume of the tetrahedron will be the minimum when it's lateral sides are equal in length. Let this length be $a$ . The the length of each side of the base is $a\sqrt 2$ . Since the inradius, which is the $z$ - coordinate of the incenter, is $1,a=3+\sqrt 3$ , and the volume of the tetrahedron is

$\dfrac{1}{3}\times \dfrac{\sqrt 3}{2}\times (3+\sqrt 3)^2\times \dfrac{1}{\sqrt 3}\times (3+\sqrt 3)=9+5\sqrt 3$ .

Hence, $m=9,n=5,p=3$ , and $m+n+p=\boxed {17}$ .

Why can I get the radius by the first formula(which is involving area of faces and height)?

wing yan yau - 1 year ago

This is a standard relation. Similar to the inradius of a triangle.

A Former Brilliant Member - 1 year ago

The unchanging red ball

The answer is 17.

1 solution

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