Difference between revisions of "2006 AMC 10A Problems/Problem 24"

Revision as of 14:03, 23 January 2012

Problem

Centers of adjacent faces of a unit cube are joined to form a regular octahedron. What is the volume of this octahedron?

$\mathrm{(A) \ } \frac{1}{8}\qquad\mathrm{(B) \ } \frac{1}{6}\qquad\mathrm{(C) \ } \frac{1}{4}\qquad\mathrm{(D) \ } \frac{1}{3}\qquad\mathrm{(E) \ } \frac{1}{2}\qquad$

Solution

We can break the octahedron into two square pyramids by cutting it along a plane perpendicular to one of its internal diagonals. $[asy] import three; real r = 1/2; triple A = (-0.5,1.5,0); size(400); currentprojection=orthographic(1,1/4,1/2); draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--(0,0,0)^^(0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--(0,0,1)^^(0,0,0)--(0,0,1)^^(1,0,0)--(1,0,1)^^(0,1,0)--(0,1,1)^^(1,1,0)--(1,1,1),gray(0.8)); draw((0,r,r)--(r,1,r)--(1,r,r)--(r,0,r)--cycle^^(r,r,0)--(0,r,r)--(r,r,1)--(r,1,r)--(r,r,0)--(1,r,r)--(r,r,1)--(r,0,r)--(r,r,0)); draw((0,r,r)+A--(r,1,r)+A--(1,r,r)+A--(r,0,r)+A--cycle^^(0,r,r)+A--(r,r,1)+A--(1,r,r)+A^^(r,1,r)+A--(r,r,1)+A--(r,0,r)+A); [/asy]$ The cube has edges of length 1 so all edges of the regular octahedron have length $\frac{\sqrt{2}}{2}$ . Then the square base of the pyramid has area $\left(\frac{1}{2}\sqrt{2}\right)^2 = \frac{1}{2}$ . We also know that the height of the pyramid is half the height of the cube, so it is $\frac{1}{2}$ . The volume of a pyramid with base area $B$ and height $h$ is $A=\frac{1}{3}Bh$ so each of the pyramids has volume $\frac{1}{3}\left(\frac{1}{2}\right)\left(\frac{1}{2}\right) = \frac{1}{12}$ . The whole octahedron is twice this volume, so $\frac{1}{12} \cdot 2 = \frac{1}{6} \Longrightarrow \mathrm{(B)}$ .

2006 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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
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Difference between revisions of "2006 AMC 10A Problems/Problem 24"

Revision as of 14:03, 23 January 2012

Problem

Solution

See also

@@ Line 4: / Line 4: @@
 <math>\mathrm{(A) \ } \frac{1}{8}\qquad\mathrm{(B) \ } \frac{1}{6}\qquad\mathrm{(C) \ } \frac{1}{4}\qquad\mathrm{(D) \ } \frac{1}{3}\qquad\mathrm{(E) \ } \frac{1}{2}\qquad</math>
 == Solution ==
-We can break the octahedron into two [[square pyramid]]s by cutting it along a [[plane]] [[perpendicular]] to one of its internal [[diagonal]]s.  The cube has [[edge]]s of [[length]] one so the edge of the octahedron is of length <math>\frac{\sqrt{2}}{2}</math>.  Then the [[square (geometry) | square]] base of the [[pyramid]] has [[area]] <math>\left(\frac{1}{2}\sqrt{2}\right)^2 = \frac{1}{2}</math>.
+We can break the octahedron into two [[square pyramid]]s by cutting it along a [[plane]] [[perpendicular]] to one of its internal [[diagonal]]s.
+<asy>
+import three;
+real r = 1/2;
+triple A = (-0.5,1.5,0);
+size(400);
+currentprojection=orthographic(1,1/4,1/2);
+draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--(0,0,0)^^(0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--(0,0,1)^^(0,0,0)--(0,0,1)^^(1,0,0)--(1,0,1)^^(0,1,0)--(0,1,1)^^(1,1,0)--(1,1,1),gray(0.8));
+draw((0,r,r)--(r,1,r)--(1,r,r)--(r,0,r)--cycle^^(r,r,0)--(0,r,r)--(r,r,1)--(r,1,r)--(r,r,0)--(1,r,r)--(r,r,1)--(r,0,r)--(r,r,0));
+draw((0,r,r)+A--(r,1,r)+A--(1,r,r)+A--(r,0,r)+A--cycle^^(0,r,r)+A--(r,r,1)+A--(1,r,r)+A^^(r,1,r)+A--(r,r,1)+A--(r,0,r)+A);
+</asy>
+The cube has [[edge]]s of [[length]] 1 so all edges of the regular octahedron have length <math>\frac{\sqrt{2}}{2}</math>.  Then the [[square (geometry) | square]] base of the [[pyramid]] has [[area]] <math>\left(\frac{1}{2}\sqrt{2}\right)^2 = \frac{1}{2}</math>.
 We also know that the height of the pyramid is half the height of the cube, so it is <math>\frac{1}{2}</math>.  The volume of a pyramid with base area <math>B</math> and height <math>h</math> is <math>A=\frac{1}{3}Bh</math> so each of the pyramids has volume <math>\frac{1}{3}\left(\frac{1}{2}\right)\left(\frac{1}{2}\right) = \frac{1}{12}</math>. The whole octahedron is twice this volume, so <math>\frac{1}{12} \cdot 2 = \frac{1}{6} \Longrightarrow \mathrm{(B)}</math>.