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Mock AIME II 2012 Problems/Problem 13

Problem

Regular octahedron $ABCDEF$ (such that points $B$, $C$, $D$, and $E$ are coplanar and form the vertices of a square) is divided along plane $\mathcal{P}$, parallel to line $BC$, into two polyhedra of equal volume. The cosine of the acute angle plane $\mathcal{P}$ makes with plane $BCDE$ is $\frac{1}{3}$. Given that $AB=30$, find the area of the cross section made by plane $\mathcal{P}$ with octahedron $ABCDEF$.

Solution

Let $O$ be the center of the octahedron. The plane must pass through $O$ in order to bisect the area of the octahedron. We see that the cross-section will be a hexagon as it passes through six of the eight faces. By symmetry, the area of this hexagon is twice that of the trapezoid contained within square pyramid $ABCDE$.

We proceed by finding the height of the square pyramid, which is $AO$. The altitude from $A$ to $BC$ of $\Delta ABC$ is $15\sqrt{3}$. The distance from $O$ to the midpoint of $BC$ is half the side length of the square $BCDE$, so its $15$. The altitude of $\Delta ABC$, the segment joining $O$ and the midpoint of $BC$, and $AO$ make a right triangle. Then by the Pythagorean Theorem, $AO^2+15^2=(15\sqrt3)^2$ so $AO = 15\sqrt2$. Now consider $\angle A$ of this right triangle. We have $\sin A = \frac{15}{15\sqrt3} = \frac{1}{\sqrt3}$. This will be important later.

Back to the trapezoid. One of its bases is on $BCDE$ and the other base must be on $ABC$ or $ADE$ (it cannot be on the other two faces as it is parallel to $BC$). WLoG let the base be on $\Delta ABC$. Let $M$ be the midpoint of this base and consider $\Delta OAM$. We have that $\angle AOM$ is the complement of the angle between the plane $\mathcal{P}$ and the square $BCDE$. Then $\sin \angle AOM = \frac{1}{3}$. Now consider $\angle OAM$. This is the same angle as the $\angle A$ we had before. Then $\sin \angle OAM = \frac{1}{\sqrt 3}$.

Now by the Law of Sines $\frac{AM}{\sin \angle AOM} = \frac{OM}{\sin \angle OAM}$ so $AM = \frac{OM}{\sqrt3}$. By the Law of Cosines, $AM^2 = (15\sqrt2)^2+OM^2 -2(OM)(15\sqrt2)(\frac{1}{3})$. This solves to $OM = 15$ and $AM = 5\sqrt3$. We reject the solution $OM = 45$ as it is too large; $AO=15\sqrt2$ must be largest side in $\Delta AOM$ as $\angle M$ is the largest angle. We know this as the other angles have sines less that $\sin 45 = \frac{1}{\sqrt 2}$ and so have values less than $45$.

Now on $\Delta ABC$ we have that the second base is parallel to $BC$. Let $b$ be the length of the second base. Then by the two similar triangles, $\frac{b}{AM} = \frac{BC}{15\sqrt{3}}$ where we have compared the side length of the triangles to their heights. As $BC = 30$ is a given, this solves to $b = 10$.

Then the area of the trapezoid is $\frac{1}{2}(OM)(30+ b) = \frac{1}{2}(15)(30+10) = 300$. The area of the whole hexagon is twice this, so the final answer is $\boxed{600}$

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