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Difference between revisions of "Mock AIME II 2012 Problems/Problem 13"

(Created page with "==Problem== Regular octahedron <math>ABCDEF</math> (such that points <math>B</math>, <math>C</math>, <math>D</math>, and <math>E</math> are coplanar and form the vertices of a s...")
 
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==Solution==
 
==Solution==
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There is currently no solution to this problem. If you discover one, please update this to the solution. You can check your answers with the answer key provided on the Mock AIME II 2012 Page.
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Let <math>O</math> be the center of the octahedron. The plane must pass through <math>O</math> 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 <math>ABCDE</math>.
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We proceed by finding the height of the square pyramid, which is <math>AO</math>. The altitude from <math>A</math> to <math>BC</math> of <math>\Delta ABC</math> is <math>15\sqrt{3}</math>. The distance from <math>O</math> to the midpoint of <math>BC</math> is half the side length of the square <math>BCDE</math>, so its <math>15</math>. The altitude of <math>\Delta ABC</math>, the segment joining <math>O</math> and the midpoint of <math>BC</math>, and <math>AO</math> make a right triangle. Then by the Pythagorean Theorem, <math>AO^2+15^2=(15\sqrt3)^2</math> so <math>AO = 15\sqrt2</math>. Now consider <math>\angle A</math> of this right triangle. We have <math>\sin A = \frac{15}{15\sqrt3} = \frac{1}{\sqrt3}</math>. This will be important later.
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Back to the trapezoid. One of its bases is on <math>BCDE</math> and the other base must be on <math>ABC</math> or <math>ADE</math> (it cannot be on the other two faces as it is parallel to <math>BC</math>). WLoG let the base be on <math>\Delta ABC</math>. Let <math>M</math> be the midpoint of this base and consider <math>\Delta OAM</math>. We have that <math>\angle AOM</math> is the complement of the angle between the plane <math> \mathcal{P} </math> and the square <math>BCDE</math>. Then <math>\sin \angle AOM = \frac{1}{3}</math>. Now consider <math>\angle OAM</math>. This is the same angle as the <math>\angle A</math> we had before. Then <math>\sin \angle OAM = \frac{1}{\sqrt 3}</math>.
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Now by the Law of Sines <math>\frac{AM}{\sin \angle AOM} = \frac{OM}{\sin \angle OAM}</math> so <math>AM = \frac{OM}{\sqrt3}</math>. By the Law of Cosines, <math>AM^2 = (15\sqrt2)^2+OM^2 -2(OM)(15\sqrt2)(\frac{1}{3})</math>. This solves to <math>OM = 15</math> and <math>AM = 5\sqrt3</math>. We reject the solution <math>OM = 45</math> as it is too large; <math>AO=15\sqrt2</math> must be largest side in <math>\Delta AOM</math> as <math>\angle M</math> is the largest angle. We know this as the other angles have sines less that <math>\sin 45 = \frac{1}{\sqrt 2}</math> and so have values less than <math>45</math>.
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Now on <math>\Delta ABC</math> we have that the second base is parallel to <math>BC</math>. Let <math>b</math> be the length of the second base. Then by the two similar triangles, <math>\frac{b}{AM} = \frac{BC}{15\sqrt{3}}</math> where we have compared the side length of the triangles to their heights. As <math>BC = 30</math> is a given, this solves to <math>b = 10</math>.
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Then the area of the trapezoid is <math>\frac{1}{2}(OM)(30+ b) = \frac{1}{2}(15)(30+10) = 300</math>. The area of the whole hexagon is twice this, so the final answer is <math>\boxed{600}</math>

Latest revision as of 18:32, 27 December 2012

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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