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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 this pyramid must be <math>\sqrt{2^2 - \left(\frac{2\sqrt{3}}{3}\right)^2} = \frac{2\sqrt{6}}{3}</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 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 this pyramid 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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-j4andrews
 
 
 
 
 
<math>\fbox{E}</math>
 
<math>\fbox{E}</math>

Revision as of 04:47, 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 with 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 this pyramid 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 $\frac{1}{4} \cdot \frac{2\sqrt{3}}{3} = \frac{\sqrt{6}}/{6}$.

-j4andrews   $\fbox{E}$

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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