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Difference between revisions of "1990 AHSME Problems/Problem 25"

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Let <math>r</math> be the radius, let <math>C</math> be the center of the cube, and let <math>P</math> be the center of one of the eight outer spheres, noting that <math>PC=2r</math>.
 
Let <math>r</math> be the radius, let <math>C</math> be the center of the cube, and let <math>P</math> be the center of one of the eight outer spheres, noting that <math>PC=2r</math>.
  
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Pack, in the corner of the unit cube, a smaller cube whose inner corner coincides with <math>P</math>. The cube is of dimensions <math>1\times 1\times 1</math> and its space diagonal is of length <math>\sqrt3</math>. Thus the space diagonal of the unit cube is <math>2(r\sqrt3+2r)=\sqrt3</math>. Solving for <math>r</math>, we get <math>\fbox{B}</math>
+
Back, in the corner of the unit cube, a smaller cube whose inner corner coincides with <math>P</math>. The cube is of dimensions <math>1\times 1\times 1</math> and its space diagonal is of length <math>\sqrt3</math>. Thus the space diagonal of the unit cube is <math>2(r\sqrt3+2r)=\sqrt3</math>. Solving for <math>r</math>, we get <math>\fbox{B}</math>
 
<asy>
 
<asy>
 
draw((0,0)--(0,1)--(1,1)--(1,0)--cycle);
 
draw((0,0)--(0,1)--(1,1)--(1,0)--cycle);

Latest revision as of 18:36, 6 August 2019

Problem

Nine congruent spheres are packed inside a unit cube in such a way that one of them has its center at the center of the cube and each of the others is tangent to the center sphere and to three faces of the cube. What is the radius of each sphere?

$\text{(A) } 1-\frac{\sqrt{3}}{2}\quad \text{(B) } \frac{2\sqrt{3}-3}{2}\quad \text{(C) } \frac{\sqrt{2}}{6}\quad \text{(D) } \frac{1}{4}\quad \text{(E) } \frac{\sqrt{3}(2-\sqrt2)}{4}$

Solution

Let $r$ be the radius, let $C$ be the center of the cube, and let $P$ be the center of one of the eight outer spheres, noting that $PC=2r$.

Back, in the corner of the unit cube, a smaller cube whose inner corner coincides with $P$. The cube is of dimensions $1\times 1\times 1$ and its space diagonal is of length $\sqrt3$. Thus the space diagonal of the unit cube is $2(r\sqrt3+2r)=\sqrt3$. Solving for $r$, we get $\fbox{B}$ [asy] draw((0,0)--(0,1)--(1,1)--(1,0)--cycle); for(int i=0;i<4;++i)draw(rotate(i*90,(.5,.5))*circle((0.25,0.25),0.225)); //draw(circle((0.75,0.25),0.225)); draw(circle((0.5,0.5),0.19),dashed); draw((0,0)--(.5,.5),dotted); dot((.25,.25));dot((.5,.5)); label("P",(.25,.25),S);label("C",(.5,.5),N); [/asy]

See also

1990 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Problem 26
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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