Difference between revisions of "2005 AIME II Problems/Problem 10"
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== Solution == | == Solution == | ||
| + | Without coordinate geometry: | ||
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| + | Let the side of the [[octahedron]] be of length <math>s</math>. Let the [[vertex |vertices]] of the octahedron be <math>A, B, C, D, E, F</math> so that <math>A</math> and <math>F</math> are opposite each other, sp <math>AF = s\sqrt2</math>. The height of the square pyramid <math>ABCDE</math> is <math>\frac{AF}2 = \frac s{\sqrt2}</math> and so it has volume <math>\frac 13 s^2 \cdot \frac s{\sqrt2} = \frac {s^3}{3\sqrt2}</math> and the whole octahedron has volume <math>\frac {s^3\sqrt2}3</math>. | ||
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| + | Let <math>M</math> be the midpoint of <math>BC</math>, <math>N</math> be the midpoint of <math>DE</math>, <math>G</math> be the [[centroid]] of <math>\triangle ABC</math> and <math>H</math> be the centroid of <math>\triangle ADE</math>. Then <math>\triangle AMN \sim \triangle AGH</math> and the symmetry ratio is <math>\frac 23</math> (because the [[median]]s of a triangle are trisected by the centroid), so <math>GH = \frac{2}{3}MN = \frac{2s}3</math>. <math>GH</math> is also a diagonal of the cube, so the cube has side-length <math>\frac{s\sqrt2}3</math> and volume <math>\frac{2s^3\sqrt2}{27}</math>. The ratio of the volumes is then <math>(\frac{2s^3\sqrt2}{27})\big/(\frac{s^3\sqrt2}{3}) = \frac29</math> and so the answer is 011. | ||
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| + | With coordinate geometry: | ||
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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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== See Also == | == See Also == | ||
*[[2005 AIME II Problems]] | *[[2005 AIME II Problems]] | ||
Revision as of 10:01, 21 July 2006
Problem
Given that 
is a regular octahedron, that 
is the cube whose vertices are the centers of the faces of 
and that the ratio of the volume of to that of is 
where 
and 
are relatively prime integers, find 
Solution
Without coordinate geometry:
Let the side of the octahedron be of length 
. Let the vertices of the octahedron be 
so that 
and 
are opposite each other, sp 
. The height of the square pyramid 
is 
and so it has volume 
and the whole octahedron has volume 
.
Let 
be the midpoint of 
, 
be the midpoint of 
, 
be the centroid of 
and 
be the centroid of 
. Then 
and the symmetry ratio is 
(because the medians of a triangle are trisected by the centroid), so 
. 
is also a diagonal of the cube, so the cube has side-length 
and volume 
. The ratio of the volumes is then 
and so the answer is 011.
With coordinate geometry:
Let the octahedron have vertices 
. Then the vertices of the cube lie at the centroids of the faces, which have coordinates 
. 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 
, so the ratio is 
and so the answer is 011.
