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Difference between revisions of "2009 AMC 12A Problems/Problem 22"

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(→‎Solution)
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<center><asy>
 
<center><asy>
 
import three; currentprojection = orthographic(0.5,-3,1.4); pen g = rgb(0.8,1,0.8);
 
import three; currentprojection = orthographic(0.5,-3,1.4); pen g = rgb(0.8,1,0.8);
−
triple[] P = {(1,0,0),(0,1,0),(-1,0,0),(0,-1,0),(0,0,1),(0,0,-1)};
+
triple[ P = {(1,0,0),(0,1,0),(-1,0,0),(0,-1,0),(0,0,1),(0,0,-1)]};
  
 
void drawFrontFace(int x, int y, int z){ draw(P[x] -- P[y] -- P[z] -- cycle, linewidth(0.7)); }  
 
void drawFrontFace(int x, int y, int z){ draw(P[x] -- P[y] -- P[z] -- cycle, linewidth(0.7)); }  
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If the plane divides the octahedron into two congruent solids, it goes through the center of the octahedron. As it is parallel to two opposite faces (colored above in green), it passes through the midpoints of the edges connecting the corresponding vertices of the faces. The distance between the center and any of the midpoints, as well as the distance between any consecutive midpoints, is found to be <math>1/2</math> (by midline and so forth). Thus, the intersection of the plane and the octahedron is a regular hexagon, and the answer is <math>6 \times \left(\frac {\left(\frac {1}{2}\right)^2 \sqrt {3}}{4}\right) = \frac {3\sqrt {3}}{8}</math>, and <math>a + b + c = 14\ \mathbf{(E)}</math>.
 
If the plane divides the octahedron into two congruent solids, it goes through the center of the octahedron. As it is parallel to two opposite faces (colored above in green), it passes through the midpoints of the edges connecting the corresponding vertices of the faces. The distance between the center and any of the midpoints, as well as the distance between any consecutive midpoints, is found to be <math>1/2</math> (by midline and so forth). Thus, the intersection of the plane and the octahedron is a regular hexagon, and the answer is <math>6 \times \left(\frac {\left(\frac {1}{2}\right)^2 \sqrt {3}}{4}\right) = \frac {3\sqrt {3}}{8}</math>, and <math>a + b + c = 14\ \mathbf{(E)}</math>.
−
 
−
Since the code above was unparseable, this diagram should be more useful: [http://a.images.memegenerator.net/instances/500x/14091624.jpg]
 
  
 
== See also ==
 
== See also ==

Revision as of 01:34, 17 February 2012

Problem

A regular octahedron has side length $1$. A plane parallel to two of its opposite faces cuts the octahedron into the two congruent solids. The polygon formed by the intersection of the plane and the octahedron has area $\frac {a\sqrt {b}}{c}$, where $a$, $b$, and $c$ are positive integers, $a$ and $c$ are relatively prime, and $b$ is not divisible by the square of any prime. What is $a + b + c$?

$\textbf{(A)}\ 10\qquad \textbf{(B)}\ 11\qquad \textbf{(C)}\ 12\qquad \textbf{(D)}\ 13\qquad \textbf{(E)}\ 14$

Solution

import three; currentprojection = orthographic(0.5,-3,1.4); pen g = rgb(0.8,1,0.8);
triple[ P = {(1,0,0),(0,1,0),(-1,0,0),(0,-1,0),(0,0,1),(0,0,-1)]};

void drawFrontFace(int x, int y, int z){ draw(P[x] -- P[y] -- P[z] -- cycle, linewidth(0.7)); } 
void drawBackFace(int x, int y, int z){ draw(P[x] -- P[y] -- P[z] -- cycle, linetype("2 6")); } 
void fillFace(int x, int y, int z, pen c) {fill(P[x] -- P[y] -- P[z] -- cycle, c);}
pair midpt(int x,int y){ return (P[x] + P[y])/2;}

path planecut = midpt(1,0)--midpt(1,5)--midpt(2,5)--midpt(2,3)--midpt(4,3)--midpt(4,0)--cycle; 
fillFace(0,3,5,g);fillFace(1,2,4,g);fill(planecut,rgb(0.8,0.8,1));
drawFrontFace(0,1,4);drawFrontFace(1,2,4);drawFrontFace(0,1,5);drawFrontFace(1,2,5);drawBackFace(2,3,4);drawBackFace(3,0,4);drawBackFace(2,3,5);drawBackFace(3,0,5);
draw(planecut,linetype("4 4")+linewidth(0.7)); dot((0,0,0));
 (Error making remote request. Unknown error_msg)

If the plane divides the octahedron into two congruent solids, it goes through the center of the octahedron. As it is parallel to two opposite faces (colored above in green), it passes through the midpoints of the edges connecting the corresponding vertices of the faces. The distance between the center and any of the midpoints, as well as the distance between any consecutive midpoints, is found to be $1/2$ (by midline and so forth). Thus, the intersection of the plane and the octahedron is a regular hexagon, and the answer is $6 \times \left(\frac {\left(\frac {1}{2}\right)^2 \sqrt {3}}{4}\right) = \frac {3\sqrt {3}}{8}$, and $a + b + c = 14\ \mathbf{(E)}$.

See also

2009 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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
All AMC 12 Problems and Solutions
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