Difference between revisions of "2009 AMC 12A Problems/Problem 22"
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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 | + | 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 09:34, 9 February 2015
Problem
A regular octahedron has side length 
. 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 
, where 
, 
, and 
are positive integers, and are relatively prime, and is not divisible by the square of any prime. What is 
?
Solution
<center>
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));
</center> (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 
(by midline and so forth). Thus, the intersection of the plane and the octahedron is a regular hexagon, and the answer is 
, and 
.
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
