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2008 AMC 10A Problems/Problem 21: Difference between revisions

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Asymptote by worthawholebean (presumably)
solution
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==Problem==
==Problem==
A cube with side length <math>1</math> is sliced by a plane that passes through two diagonally opposite vertices <math>A</math> and <math>C</math> and the midpoints <math>B</math> and <math>D</math> of two opposite edges not containing <math>A</math> or <math>C</math>, as shown. What is the area of quadrilateral <math>ABCD</math>?
A [[cube]] with side length <math>1</math> is sliced by a plane that passes through two diagonally opposite vertices <math>A</math> and <math>C</math> and the [[midpoint]]s <math>B</math> and <math>D</math> of two opposite edges not containing <math>A</math> or <math>C</math>, as shown. What is the area of [[quadrilateral]] <math>ABCD</math>?


<asy>
<asy>
Line 23: Line 23:


==Solution==
==Solution==
{{solution}}
<center><asy>
import three;
unitsize(3cm);
defaultpen(fontsize(8)+linewidth(0.7));
currentprojection=obliqueX;


pair A=(0.5,0,0),C=(0,1,1),D=(0.5,1,0.5),B=(0,0,0.5);
draw((0.5,0,0)--(0,0,0)--(0,0,1)--(0,0,0)--(0,1,0),linetype("4 4"));
draw((0.5,0,1)--(0,0,1)--(0,1,1)--(0.5,1,1)--(0.5,0,1)--(0.5,0,0)--(0.5,1,0)--(0.5,1,1));
draw((0.5,1,0)--(0,1,0)--(0,1,1));
dot((0.5,0,0));
label("$A$",A,WSW);
dot((0,1,1));
label("$C$",C,NE);
dot((0.5,1,0.5));
label("$D$",D,ESE);
dot((0,0,0.5));
label("$B$",B,NW);
draw(B--C--A--B--D,linetype("4 4"));
draw(A--D--C);
</asy></center>
Since <math>AB = AD = CB = CD = \sqrt{.5^2+1^2}</math>, it follows that <math>ABCD</math> is a [[rhombus]]. The area of the rhombus can be computed by the formula <math>A = \frac 12 d_1d_2</math>, where <math>d_1,\,d_2</math> are the diagonals of the rhombus (or of a [[kite]] in general). <math>BD</math> has the same length as a face diagonal, or <math>\sqrt{1^2 + 1^2} = \sqrt{2}</math>. <math>AC</math> is a space diagonal, with length <math>\sqrt{1^2+1^2+1^2} = \sqrt{3}</math>. Thus <math>A = \frac 12 \times \sqrt{2} \times \sqrt{3} = \frac{\sqrt{6}}{2}\ \mathrm{(A)}</math>.
==See also==
==See also==
{{AMC10 box|year=2008|ab=A|num-b=20|num-a=22}}
{{AMC10 box|year=2008|ab=A|num-b=20|num-a=22}}


[[Category:Introductory Geometry Problems]]
[[Category:Introductory Geometry Problems]]

Revision as of 10:36, 26 April 2008

Problem

A cube with side length $1$ is sliced by a plane that passes through two diagonally opposite vertices $A$ and $C$ and the midpoints $B$ and $D$ of two opposite edges not containing $A$ or $C$, as shown. What is the area of quadrilateral $ABCD$?

[asy] import three; unitsize(3cm); defaultpen(fontsize(8)+linewidth(0.7)); currentprojection=obliqueX; draw((0.5,0,0)--(0,0,0)--(0,0,1)--(0,0,0)--(0,1,0),linetype("4 4")); draw((0.5,0,1)--(0,0,1)--(0,1,1)--(0.5,1,1)--(0.5,0,1)--(0.5,0,0)--(0.5,1,0)--(0.5,1,1)); draw((0.5,1,0)--(0,1,0)--(0,1,1)); dot((0.5,0,0)); label("$A$",(0.5,0,0),WSW); dot((0,1,1)); label("$C$",(0,1,1),NE); dot((0.5,1,0.5)); label("$D$",(0.5,1,0.5),ESE); dot((0,0,0.5)); label("$B$",(0,0,0.5),NW);[/asy]

$\mathrm{(A)}\ \frac{\sqrt{6}}{2}\qquad\mathrm{(B)}\ \frac{5}{4}\qquad\mathrm{(C)}\ \sqrt{2}\qquad\mathrm{(D)}\ \frac{5}{8}\qquad\mathrm{(E)}\ \frac{3}{4}$

Solution

import three;
unitsize(3cm);
defaultpen(fontsize(8)+linewidth(0.7));
currentprojection=obliqueX;

pair A=(0.5,0,0),C=(0,1,1),D=(0.5,1,0.5),B=(0,0,0.5);
draw((0.5,0,0)--(0,0,0)--(0,0,1)--(0,0,0)--(0,1,0),linetype("4 4"));
draw((0.5,0,1)--(0,0,1)--(0,1,1)--(0.5,1,1)--(0.5,0,1)--(0.5,0,0)--(0.5,1,0)--(0.5,1,1));
draw((0.5,1,0)--(0,1,0)--(0,1,1));
dot((0.5,0,0));
label("$A$",A,WSW);
dot((0,1,1));
label("$C$",C,NE);
dot((0.5,1,0.5));
label("$D$",D,ESE);
dot((0,0,0.5));
label("$B$",B,NW);
draw(B--C--A--B--D,linetype("4 4"));
draw(A--D--C);
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Since $AB = AD = CB = CD = \sqrt{.5^2+1^2}$, it follows that $ABCD$ is a rhombus. The area of the rhombus can be computed by the formula $A = \frac 12 d_1d_2$, where $d_1,\,d_2$ are the diagonals of the rhombus (or of a kite in general). $BD$ has the same length as a face diagonal, or $\sqrt{1^2 + 1^2} = \sqrt{2}$. $AC$ is a space diagonal, with length $\sqrt{1^2+1^2+1^2} = \sqrt{3}$. Thus $A = \frac 12 \times \sqrt{2} \times \sqrt{3} = \frac{\sqrt{6}}{2}\ \mathrm{(A)}$.

See also

2008 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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
All AMC 10 Problems and Solutions
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