1999 AHSME Problems/Problem 29

Problem

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

$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }0.1 \qquad \mathrm{(C) \ }0.2 \qquad \mathrm{(D) \ }0.3 \qquad \mathrm{(E) \ }0.4$

Solution

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Let the radius of the large sphere be $R$ , and of the inner sphere $r$ . Label the vertices of the tetrahedron $ABCD$ , and let $O$ be the center. Then pyramid $[OABC] + [OABD] + [OACD] + [OBCD] = [ABCD]$ , where $[\ldots]$ denotes volume; thus $[OABC] = \frac{[ABCD]}{4}$ . Since $OABC$ and $ABCD$ are both pyramids that share a common face $ABC$ , the ratio of their volumes is the ratio of their altitudes to face $ABC$ , so $r = \frac {h_{ABCD}}4$ . However, $h_{ABCD} = r + R$ , so it follows that $r = \frac {R}{3}$ . Then the radius of an external sphere is $\frac{R-r}2 = \frac {R}{3} = r$ .

Since the five described spheres are non-intersecting, it follows that the ratio of the volumes of the spheres is $5 \cdot \left( \frac 13 \right)^2 = \frac{5}{27} \Longrightarrow \mathrm{(B)}$ .

1999 AHSME (Problems • Answer Key • Resources)
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1999 AHSME Problems/Problem 29

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