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 11:01, 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.