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1995 AIME Problems/Problem 12

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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.

Contents

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=(...

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 ...

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 &a...

Similarly,

\vec{v} = \overrightarrow{OB}\times \overrightarrow{OC} - \left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\0 &am...

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^{...

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

See also

1995 AIME (ProblemsResources)
Preceded by
Problem 11
Followed by
Problem 13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Want to learn how to tackle those tough AMC/AIME/Olympiad algebra problems? Check out Art of Problem Solving's Intermediate Algebra by Richard Rusczyk and Mathew Crawford. Over 1600 problems!
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