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Difference between revisions of "2005 AIME II Problems/Problem 10"
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=== Solution 2 ===
 
=== Solution 2 ===
  
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Let the octahedron have vertices <math>(\pm 3, 0, 0), (0, \pm 3, 0), (0, 0, \pm 3)</math>.  Then the vertices of the cube lie at the centroids of the faces, which have coordinates <math>(\pm 1, \pm 1, \pm 1)</math>.  The cube has volume 8.  The region of the octahedron lieing in each octant is a [[tetrahedron]] with three edges mutually perpendicular and of length 3.  Thus the octahedron has volume <math>8 \cdot (\frac 16 \cdot3^3) = 36</math>, so the ratio is <math>\frac 8{36} = \frac 29</math> and so the answer is 011.
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Let the octahedron have vertices <math>(\pm 3, 0, 0), (0, \pm 3, 0), (0, 0, \pm 3)</math>.  Then the vertices of the cube lie at the centroids of the faces, which have coordinates <math>(\pm 1, \pm 1, \pm 1)</math>.  The cube has volume 8.  The region of the octahedron lying in each octant is a [[tetrahedron]] with three edges mutually perpendicular and of length 3.  Thus the octahedron has volume <math>8 \cdot (\frac 16 \cdot3^3) = 36</math>, so the ratio is <math>\frac 8{36} = \frac 29</math> and so the answer is 011.
  
 
== See Also ==
 
== See Also ==
  
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*[[2005 AIME II Problems/Problem 9| Previous problem]]
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*[[2005 AIME II Problems/Problem 11| Next problem]]
 
* [[2005 AIME II Problems]]
 
* [[2005 AIME II Problems]]
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]

Revision as of 20:05, 7 September 2006

Problem

Given that $O$ is a regular octahedron, that $C$ is the cube whose vertices are the centers of the faces of $O,$ and that the ratio of the volume of $O$ to that of $C$ is $\frac mn,$ where $m$ and $n$ are relatively prime integers, find $m+n.$

Solutions

Solution 1

Let the side of the octahedron be of length $s$. Let the vertices of the octahedron be $A, B, C, D, E, F$ so that $A$ and $F$ are opposite each other, sp $AF = s\sqrt2$. The height of the square pyramid $ABCDE$ is $\frac{AF}2 = \frac s{\sqrt2}$ and so it has volume $\frac 13 s^2 \cdot \frac s{\sqrt2} = \frac {s^3}{3\sqrt2}$ and the whole octahedron has volume $\frac {s^3\sqrt2}3$.

Let $M$ be the midpoint of $BC$, $N$ be the midpoint of $DE$, $G$ be the centroid of $\triangle ABC$ and $H$ be the centroid of $\triangle ADE$. Then $\triangle AMN \sim \triangle AGH$ and the symmetry ratio is $\frac 23$ (because the medians of a triangle are trisected by the centroid), so $GH = \frac{2}{3}MN = \frac{2s}3$. $GH$ is also a diagonal of the cube, so the cube has side-length $\frac{s\sqrt2}3$ and volume $\frac{2s^3\sqrt2}{27}$. The ratio of the volumes is then $(\frac{2s^3\sqrt2}{27})\big/(\frac{s^3\sqrt2}{3}) = \frac29$ and so the answer is 011.

Solution 2

Let the octahedron have vertices $(\pm 3, 0, 0), (0, \pm 3, 0), (0, 0, \pm 3)$. Then the vertices of the cube lie at the centroids of the faces, which have coordinates $(\pm 1, \pm 1, \pm 1)$. The cube has volume 8. The region of the octahedron lying in each octant is a tetrahedron with three edges mutually perpendicular and of length 3. Thus the octahedron has volume $8 \cdot (\frac 16 \cdot3^3) = 36$, so the ratio is $\frac 8{36} = \frac 29$ and so the answer is 011.

See Also

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