2009 AMC 12A Problems/Problem 22

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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,...

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 $a + b + c = 14\ \mathbf{(E)}$ .

2009 AMC 12A (Problems • 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

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2009 AMC 12A Problems/Problem 22

From AoPSWiki

Problem

Solution

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