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

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== Solution ==
 
== Solution ==
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<asy><center>
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Firstly, note that the intersection of the plane must be a hexagon. Consider the net of the octahedron. Notice that the hexagon becomes a line on the net. Also, notice that, given the parallel to the faces conditions, the line must be parallel to the sides of the net (precisely <math>\frac{1}{3}</math> of them). Now, notice that, through symmetry, 2 of the hexagon's vertexes lie on the midpoint of the side of the "square" in the octahedron. In the net, the condition gives you that one of the intersections of the line with the net have to be on the midpoint of the side. However, if one is on the midpoint, because of the parallel conditions, all of the vertices are on the midpoint of a side. Thus, we have a regular hexagon with a side length of the midline of an equilateral triangle with side length 1, which is <math>\frac{1}{2}</math>. Thus, the answer is<math> \frac {3\sqrt {3}}{8}</math>, and <math>a + b + c = 14\ \mathbf{(E)}</math>.
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import three; currentprojection = orthographic(0.5,-3,1.4); pen g = rgb(0.8,1,0.8);
 
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triple[ P = {(1,0,0),(0,1,0),(-1,0,0),(0,-1,0),(0,0,1),(0,0,-1)}];
 
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void drawFrontFace(int x, int y, int z){ draw(P[x] -- P[y] -- P[z] -- cycle, linewidth(0.7)); }  
 
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void drawBackFace(int x, int y, int z){ draw(P[x] -- P[y] -- P[z] -- cycle, linetype("2 6")); }  
 
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void fillFace(int x, int y, int z, pen c) {fill(P[x] -- P[y] -- P[z] -- cycle, c);}
 
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pair midpt(int x,int y){ return (P[x] + P[y])/2;}
 
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path planecut = midpt(1,0)--midpt(1,5)--midpt(2,5)--midpt(2,3)--midpt(4,3)--midpt(4,0)--cycle;
 
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fillFace(0,3,5,g);fillFace(1,2,4,g);fill(planecut,rgb(0.8,0.8,1));
 
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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);
 
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draw(planecut,linetype("4 4")+linewidth(0.7)); dot((0,0,0));
 
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</center></asy>
 
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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>\frac{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>.
 
  
 
== See also ==
 
== See also ==

Revision as of 13:37, 22 November 2015

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

Firstly, note that the intersection of the plane must be a hexagon. Consider the net of the octahedron. Notice that the hexagon becomes a line on the net. Also, notice that, given the parallel to the faces conditions, the line must be parallel to the sides of the net (precisely $\frac{1}{3}$ of them). Now, notice that, through symmetry, 2 of the hexagon's vertexes lie on the midpoint of the side of the "square" in the octahedron. In the net, the condition gives you that one of the intersections of the line with the net have to be on the midpoint of the side. However, if one is on the midpoint, because of the parallel conditions, all of the vertices are on the midpoint of a side. Thus, we have a regular hexagon with a side length of the midline of an equilateral triangle with side length 1, which is $\frac{1}{2}$. Thus, the answer is$\frac {3\sqrt {3}}{8}$, 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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