Question

A tetrahedron with four equilateral triangular faces has a sphere inscribed within it and a sphere circumscribed about it. For each of the four faces, there is a sphere tangent externally to the face at its center and to the circumscribed sphere. A point $P$ is selected at random inside the circumscribed sphere. The probability that $P$ lies inside one of the five small spheres is closest to

$0$

$0.1$

$0.2$

$0.3$

$0.4$

Answer: C

Concepts:3D geometry sphere geometric probability volume

Difficulty rating: 2380

Small Hint:

For a regular tetrahedron the circumradius is $3$ times the inradius

Big Hint:

Each of the five small spheres has the same volume as the inscribed sphere

Solution:

Let $O$ be the common center of the inscribed and circumscribed spheres. Splitting the tetrahedron into four congruent pieces from $O$ shows the circumradius is $3$ times the inradius, so the circumscribed sphere has $27$ times the inscribed sphere’s volume $V.$

If one of the four other small spheres has radius $s,$ its center is $r+s$ from $O$ and also $3r-s$ from $O,$ so $s=r.$ These five spheres have disjoint interiors (the central sphere is tangent to each of the other four), so their union has volume $5V.$ The probability is $\dfrac{5V}{27V} = \dfrac{5}{27} \approx 0.185,$ closest to $0.2.$

Thus, the correct answer is C.

1999 AMC 12 Problem 29

Answer: C

Difficulty rating: 2380

Small Hint:

Big Hint:

Solution:

Problem 29 in Other Years