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1999 AHSME Problems/Problem 29

Revision as of 10:39, 3 October 2023 by Isabelchen (talk | contribs) (→‎Solution 2)
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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 1

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 $5 \cdot \left( \frac 13 \right)^3 = \frac{5}{27} \approx 0.2 \Longrightarrow \fbox{C}$.

Solution 2

Let the side length of the tetrahedron be $1$.

The altitude of the equilateral triangle base is $\frac{\sqrt{3}}{2}$. Thus, the distance from the center of the equilateral triangle to its vertex is $\frac{\sqrt{3}}{2} \cdot \frac 23 = \frac{\sqrt{3}}{3}$. Therefore, the altitude of the tetrahedron is $\sqrt{1^2-(\frac{\sqrt{3}}{3})^2} = \frac{\sqrt{6}}{3}$.

Let the radius of the large sphere be $R$. \[(\frac{\sqrt{3}}{3})^2 + (\frac{\sqrt{6}}{3} - R)^2 = R^2\] \[R = \frac{\sqrt{6}}{4}\]

The distance from the center of the tetrahedron to the center of one of the bases is $\frac{\sqrt{6}}{3} - R = \frac{\sqrt{6}}{3} - \frac{\sqrt{6}}{4} = \frac{\sqrt{6}}{12}$

The distance from the center of the tetrahedron to the center of one of the bases is also the same as the radius of the small sphere in the center of the tetrahedron.

The radius of the smaller spheres tangent to the large sphere and the tetrahedron is $\frac{\frac{\sqrt{6}}{4}-\frac{\sqrt{6}}{12}}{2} = \frac{\sqrt{6}}{12}$

Therefore, the probability that $P$ lies inside one of the five small spheres is $\frac{5 \cdot (\frac{\sqrt{6}}{12})^3 \cdot \frac43 \cdot \pi }{(\frac{\sqrt{6}}{4})^3 \cdot \frac43 \cdot \pi} = \frac{5}{27} \approx \boxed{\textbf{(C) } 0.2}$

~isabelchen

See also

1999 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 28 		Followed by
Problem 30
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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