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Difference between revisions of "2005 AIME II Problems/Problem 10"

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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 and <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>.   
 
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 and <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 [[triangle median |medians]] 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>\left(\frac{2s^3\sqrt2}{27}\right)\big/\left(\frac{s^3\sqrt2}{3}\right) = \frac29</math> and so the answer is <math>\boxed{011}</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 [[triangle median |medians]] 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{\left(\frac{s^3\sqrt2}{3}\right)}{\left(\frac{2s^3\sqrt2}{27}\right)} = \frac92</math> and so the answer is <math>\boxed{011}</math>.
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edit: could someone put a,b,c,d,e,f by the respective vertices on the octohedron?
 
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thanks.
 
  
 
=== 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 lying in each octant is a [[tetrahedron]] with three edges mutually perpendicular and of length 3.  Thus the octahedron has volume <math>8 \cdot \left(\frac 16 \cdot3^3\right) = 36</math>, so the ratio is <math>\frac 8{36} = \frac 29</math> and so the answer is <math>\boxed{011}</math>.
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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 \left(\frac 16 \cdot3^3\right) = 36</math>, so the ratio is <math>\frac {36}{8} = \frac 92</math> and so the answer is <math>\boxed{011}</math>.
  
 
== See also ==
 
== See also ==
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[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 19:35, 13 August 2024

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

[asy] import three; currentprojection = perspective(4,-15,4); defaultpen(linewidth(0.7)); draw(box((-1,-1,-1),(1,1,1))); draw((-3,0,0)--(0,0,3)--(0,-3,0)--(-3,0,0)--(0,0,-3)--(0,-3,0)--(3,0,0)--(0,0,-3)--(0,3,0)--(0,0,3)--(3,0,0)--(0,3,0)--(-3,0,0)); [/asy]

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 and $AF = s\sqrt2$. 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 $\frac{\left(\frac{s^3\sqrt2}{3}\right)}{\left(\frac{2s^3\sqrt2}{27}\right)} = \frac92$ and so the answer is $\boxed{011}$.

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 $8 \cdot \left(\frac 16 \cdot3^3\right) = 36$, so the ratio is $\frac {36}{8} = \frac 92$ and so the answer is $\boxed{011}$.

See also

2005 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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