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

Revision as of 15:22, 29 May 2020

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

Solution 1

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 precisely $\frac{1}{3}$ of the sides of the net. 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 $a + b + c = 14\ \mathbf{(E)}$ . (Can somebody clarify this and provide a diagram?)

Alternate Solution

Note that an octahedron can be classified as an equilateral triangle antiprism. There are 6 triangle faces forming the middle of the antiprism that will be slices by this plane. Therefore, because the octahedron is regular, it will form a regular hexagon. It will be formed by 6 lines halfway along the height on each of the 6 triangles. Because these each lines are midlines, they are $\frac{1}{2}$ . So it is a regular hexagon with side length $\frac{1}{2}$ , so $6\frac{\sqrt{3}\left(\frac{1}{2}\right)^2}{4}=\frac{3\sqrt{3}}{8}, \:\:3+3+8=14 \:\mathbf{(E)}$

@@ Line 5: / Line 5: @@
 == Solution ==
+=== Solution 1 ===
 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 precisely <math>\frac{1}{3}</math> of the sides of the net. 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>.
 (Can somebody clarify this and provide a diagram?)

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