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Difference between revisions of "1980 AHSME Problems/Problem 26"

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== Solution ==
 
== Solution ==
  
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Let <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math> be the centers of the four spheres with radius <math>1</math>.  These points must form a regular tetrahedron of side length <math>2</math>.  Let <math>O</math> be the center of <math>\triangle ABC</math>. Then <math>OA = \frac{2\sqrt{3}}{3}</math>.  By the Pythagorean Theorem, the height <math>DO</math> of tetrahedron <math>ABCD</math> must be <math>\sqrt{2^2 - \left(\frac{2\sqrt{3}}{3}\right)^2} = \frac{2\sqrt{6}}{3}</math>.  Let <math>I</math> be the center of tetrahedron <math>ABCD</math>.  Then <math>IABC</math>, <math>IABD</math>, <math>IACD</math>, and <math>IBCD</math> are all congruent tetrahedra, each with <math>\frac{1}{4}</math> of the volume of <math>ABCD</math>.  Since tetrahedra <math>IABC</math> and <math>ABCD</math> share base <math>ABC</math>, the height <math>IO</math> of tetrahedron <math>IABC</math>, and therefore the inradius of tetrahedron <math>ABCD</math>, must be <math>\frac{1}{4} \cdot \frac{2\sqrt{3}}{3} = \frac{\sqrt{6}}{6}</math>.   
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Let <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math> be the centers of the four spheres of radius <math>1</math>.  These points must form a regular tetrahedron of side length <math>2</math>.  Let <math>O</math> be the center of <math>\triangle ABC</math>. Then <math>OA = \frac{2\sqrt{3}}{3}</math>.  By the Pythagorean Theorem, the height <math>DO</math> of tetrahedron <math>ABCD</math> must be <math>\sqrt{2^2 - \left(\frac{2\sqrt{3}}{3}\right)^2} = \frac{2\sqrt{6}}{3}</math>.  Let <math>I</math> be the center of tetrahedron <math>ABCD</math>.  Then <math>IABC</math>, <math>IABD</math>, <math>IACD</math>, and <math>IBCD</math> are all congruent tetrahedra, each with <math>\frac{1}{4}</math> of the volume of <math>ABCD</math>.  Since tetrahedra <math>IABC</math> and <math>ABCD</math> share base <math>ABC</math>, the height <math>IO</math> of tetrahedron <math>IABC</math>, and therefore the inradius of tetrahedron <math>ABCD</math>, must be <math>\frac{1}{4} \cdot \frac{2\sqrt{3}}{3} = \frac{\sqrt{6}}{6}</math>.   
  
 
Let <math>EFGH</math> be the tetrahedron circumscribed around the spheres, with <math>AE</math>, <math>BF</math>, <math>CG</math>, and <math>DH</math> intersecting at <math>I</math>.  Planes <math>ABC</math> and <math>EFG</math> must be parallel and a distance <math>1</math> from each other.  So the distance from <math>I</math> to plane <math>EFG</math>, and therefore the inradius of tetrahedron <math>EFGH</math>, must be <math>1 + \frac{\sqrt{6}}{6}</math>.  Therefore, the ratio of the inradius of tetrahedron <math>EFGH</math> to the inradius of tetrahedron <math>ABCD</math> is <math>\frac{1+\sqrt{6}/6}{\sqrt{6}/6} = 1 + \sqrt{6}</math>.  The ratio of the side length of tetrahedron <math>EFGH</math> to the side length of tetrahedron <math>ABCD</math> must also be <math>1+\sqrt{6}</math>, so <math>s = 2 (1+\sqrt{6}) = 2 + 2\sqrt{6}</math>. <math>\fbox{(E)}</math>
 
Let <math>EFGH</math> be the tetrahedron circumscribed around the spheres, with <math>AE</math>, <math>BF</math>, <math>CG</math>, and <math>DH</math> intersecting at <math>I</math>.  Planes <math>ABC</math> and <math>EFG</math> must be parallel and a distance <math>1</math> from each other.  So the distance from <math>I</math> to plane <math>EFG</math>, and therefore the inradius of tetrahedron <math>EFGH</math>, must be <math>1 + \frac{\sqrt{6}}{6}</math>.  Therefore, the ratio of the inradius of tetrahedron <math>EFGH</math> to the inradius of tetrahedron <math>ABCD</math> is <math>\frac{1+\sqrt{6}/6}{\sqrt{6}/6} = 1 + \sqrt{6}</math>.  The ratio of the side length of tetrahedron <math>EFGH</math> to the side length of tetrahedron <math>ABCD</math> must also be <math>1+\sqrt{6}</math>, so <math>s = 2 (1+\sqrt{6}) = 2 + 2\sqrt{6}</math>. <math>\fbox{(E)}</math>

Revision as of 05:03, 25 June 2025

Problem

Four balls of radius $1$ are mutually tangent, three resting on the floor and the fourth resting on the others. A tetrahedron, each of whose edges have length $s$, is circumscribed around the balls. Then $s$ equals

$\text{(A)} \ 4\sqrt 2 \qquad \text{(B)} \ 4\sqrt 3 \qquad \text{(C)} \ 2\sqrt 6 \qquad \text{(D)}\ 1+2\sqrt 6\qquad \text{(E)}\ 2+2\sqrt 6$

Solution

Let $A$, $B$, $C$, and $D$ be the centers of the four spheres of radius $1$. These points must form a regular tetrahedron of side length $2$. Let $O$ be the center of $\triangle ABC$. Then $OA = \frac{2\sqrt{3}}{3}$. By the Pythagorean Theorem, the height $DO$ of tetrahedron $ABCD$ must be $\sqrt{2^2 - \left(\frac{2\sqrt{3}}{3}\right)^2} = \frac{2\sqrt{6}}{3}$. Let $I$ be the center of tetrahedron $ABCD$. Then $IABC$, $IABD$, $IACD$, and $IBCD$ are all congruent tetrahedra, each with $\frac{1}{4}$ of the volume of $ABCD$. Since tetrahedra $IABC$ and $ABCD$ share base $ABC$, the height $IO$ of tetrahedron $IABC$, and therefore the inradius of tetrahedron $ABCD$, must be $\frac{1}{4} \cdot \frac{2\sqrt{3}}{3} = \frac{\sqrt{6}}{6}$.

Let $EFGH$ be the tetrahedron circumscribed around the spheres, with $AE$, $BF$, $CG$, and $DH$ intersecting at $I$. Planes $ABC$ and $EFG$ must be parallel and a distance $1$ from each other. So the distance from $I$ to plane $EFG$, and therefore the inradius of tetrahedron $EFGH$, must be $1 + \frac{\sqrt{6}}{6}$. Therefore, the ratio of the inradius of tetrahedron $EFGH$ to the inradius of tetrahedron $ABCD$ is $\frac{1+\sqrt{6}/6}{\sqrt{6}/6} = 1 + \sqrt{6}$. The ratio of the side length of tetrahedron $EFGH$ to the side length of tetrahedron $ABCD$ must also be $1+\sqrt{6}$, so $s = 2 (1+\sqrt{6}) = 2 + 2\sqrt{6}$. $\fbox{(E)}$

-j4andrews  

See also

1980 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 25 		Followed by
Problem 27
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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