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Difference between revisions of "1999 AHSME Problems/Problem 29"

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Label the vertices of the tetrahedron <math>ABCD</math>, and let <math>O</math> be the center. Then pyramid <math>[OABC] + [OABD] + [OACD] + [OBCD] = [ABCD]</math>, where <math>[\ldots]</math> denotes volume; thus <math>[OABC] = \frac{[ABCD]}{4}</math>. Since <math>OABC</math> and <math>ABCD</math> are both pyramids that share a common face <math>ABC</math>, the ratio of their volumes is the ratio of their altitudes to face <math>ABC</math>, so <math>r = \frac {h_{ABCD}}4</math>. However, <math>h_{ABCD} = r + R</math>, so it follows that <math>r = \frac {R}{3}</math>. Then the radius of an external sphere is <math>\frac{R-r}2 = \frac {R}{3} = r</math>.
 
Label the vertices of the tetrahedron <math>ABCD</math>, and let <math>O</math> be the center. Then pyramid <math>[OABC] + [OABD] + [OACD] + [OBCD] = [ABCD]</math>, where <math>[\ldots]</math> denotes volume; thus <math>[OABC] = \frac{[ABCD]}{4}</math>. Since <math>OABC</math> and <math>ABCD</math> are both pyramids that share a common face <math>ABC</math>, the ratio of their volumes is the ratio of their altitudes to face <math>ABC</math>, so <math>r = \frac {h_{ABCD}}4</math>. However, <math>h_{ABCD} = r + R</math>, so it follows that <math>r = \frac {R}{3}</math>. Then the radius of an external sphere is <math>\frac{R-r}2 = \frac {R}{3} = r</math>.
  
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Since the five described spheres are non-intersecting, it follows that the ratio of the volumes of the spheres is <math>5 \cdot \left( \frac 13 \right)^3 = \frac{5}{27} \Longrightarrow \mathrm{(C)}</math>.
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Since the five described spheres are non-intersecting, it follows that the ratio of the volumes of the spheres is <math>5 \cdot \left( \frac 13 \right)^3 = \frac{5}{27} \Longrightarrow \fbox{C}</math>.
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<math>\fbox{C}</math>
 
  
 
== See also ==
 
== See also ==

Revision as of 00:40, 27 January 2019

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

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

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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