Difference between revisions of "1995 AIME Problems/Problem 12"

Revision as of 19:30, 4 July 2013

Problem

Pyramid $OABCD$ has square base $ABCD,$ congruent edges $\overline{OA}, \overline{OB}, \overline{OC},$ and $\overline{OD},$ and $\angle AOB=45^\circ.$ Let $\theta$ be the measure of the dihedral angle formed by faces $OAB$ and $OBC.$ Given that $\cos \theta=m+\sqrt{n},$ where $m_{}$ and $n_{}$ are integers, find $m+n.$

Solution

Solution 1 (trigonometry)

size(220); defaultpen(linewidth(0.7)); currentprojection = perspective(5,3,2);
import three;
triple A = (1,0,0), B=(0,0,0), C=(0,1,0), D=(1,1,0), O=(1,1,(1+2^.5)^.5)/2^.5, P=O*(1-.5^.5); /* , P = foot(A, O, B) */
draw(A--B--C--D--A--O--B--O--C--O--D); draw(A--C,linewidth(0.6)); D(A--P--C); MP("A",A);MP("B",B);MP("C",C);MP("D",D);MP("P",P,NW);MP("\theta",P,(0.5,-4));MP("45^{\circ}",O,(2,-6)); MP("O",O,N); 
draw(rightanglemark(C,P,O,2)); draw(anglemark(A,P,C,3)); draw(anglemark(D,O,C,3));
 (Error making remote request. Unknown error_msg)

The angle $\theta$ is the angle formed by two perpendiculars drawn to $BO$ , one on the plane determined by $OAB$ and the other by $OBC$ . Let the perpendiculars from $A$ and $C$ to $\overline{OB}$ meet $\overline{OB}$ at $P.$ Without loss of generality, let $AP = 1.$ It follows that $\triangle OPA$ is a $45-45-90$ right triangle, so $OP = AP = 1,$ $OB = OA = \sqrt {2},$ and $AB = \sqrt {4 - 2\sqrt {2}}.$ Therefore, $AC = \sqrt {8 - 4\sqrt {2}}.$

From the Law of Cosines, $AC^{2} = AP^{2} + PC^{2} - 2(AP)(PC)\cos \theta,$ so

$8 - 4\sqrt {2} = 1 + 1 - 2\cos \theta \Longrightarrow \cos \theta = - 3 + 2\sqrt {2} = - 3 + \sqrt{8}.$

Thus $m + n = \boxed{005}$ .

Solution 2 (analytical/vectors)

Without loss of generality, place the pyramid in a 3-dimensional coordinate system such that $A = (1,0,0),$ $B = (0,1,0),$ $C = ( - 1,0,0),$ $D = (0, - 1,0),$ and $O = (0,0,z),$ where $z$ is unknown.

We first find $z.$ Note that

$\overrightarrow{OA}\cdot \overrightarrow{OB} = \parallel \overrightarrow{OA}\parallel \parallel \overrightarrow{OB}\parallel \cos 45^\circ.$

Since $\overrightarrow{OA} =\, <1,0, - z>$ and $\overrightarrow{OB} =\, <0,1, - z> ,$ this simplifies to

$z^{2}\sqrt {2} = 1 + z^{2}\implies z^{2} = 1 + \sqrt {2}.$

Now let's find $\cos \theta.$ Let $\vec{u}$ and $\vec{v}$ be normal vectors to the planes containing faces $OAB$ and $OBC,$ respectively. From the definition of the dot product as $\vec{u}\cdot \vec{v} = \parallel \vec{u}\parallel \parallel \vec{v}\parallel \cos \theta$ , we will be able to solve for $\cos \theta.$ A cross product yields (alternatively, it is simple to find the equation of the planes $OAB$ and $OAC$ , and then to find their normal vectors)

$\vec{u} = \overrightarrow{OA}\times \overrightarrow{OB} = \left| \begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & - z \\ 0 & 1 & - z \end{array}\right| =\, < z,z,1 > .$

Similarly,

$\vec{v} = \overrightarrow{OB}\times \overrightarrow{OC} - \left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & - z \\ - 1 & 0 & - z \end{array}\right| =\, < - z,z,1 > .$

Hence, taking the dot product of $\vec{u}$ and $\vec{v}$ yields

$\cos \theta = \frac{ \vec{u} \cdot \vec{v} }{ \parallel \vec{u} \parallel \parallel \vec{v} \parallel } = \frac{- z^{2} + z^{2} + 1}{(\sqrt {1 + 2z^{2}})^{2}} = \frac {1}{3 + 2\sqrt {2}} = 3 - 2\sqrt {2} = 3 - \sqrt {8}.$

Flipping the signs (we found the cosine of the supplement angle) yields $\cos \theta = - 3 + \sqrt {8},$ so the answer is $\boxed{005}$ .

Revision as of 14:20, 31 July 2008 (view source) Azjps (talk \| contribs) (finishing) ← Older edit		Revision as of 19:30, 4 July 2013 (view source) Nathan wailes (talk \| contribs) Newer edit →
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	[[Category:Intermediate Geometry Problems]]		[[Category:Intermediate Geometry Problems]]
		+	{{MAA Notice}}

1995 AIME (Problems • Answer Key • Resources)
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