Difference between revisions of "1995 AIME Problems/Problem 12"
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== Problem == | == Problem == | ||
− | Pyramid <math> | + | [[Pyramid]] <math>OABCD</math> has square base <math>ABCD,</math> congruent edges <math>\overline{OA}, \overline{OB}, \overline{OC},</math> and <math>\overline{OD},</math> and <math>\angle AOB=45^\circ.</math> Let <math>\theta</math> be the measure of the [[dihedral angle]] formed by faces <math>OAB</math> and <math>OBC.</math> Given that <math>\cos \theta=m+\sqrt{n},</math> where <math>m_{}</math> and <math>n_{}</math> are integers, find <math>m+n.</math> |
+ | __TOC__ | ||
== Solution == | == Solution == | ||
+ | === Solution 1 (trigonometry) === | ||
+ | <center><asy> | ||
+ | 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*(18^.5-2)/5; /* , P = foot(A, O, B) */ | ||
+ | draw(A--B--C--D--A--O--B--O--C--O--D); D(A--P--C); | ||
+ | </asy></center> | ||
+ | |||
+ | {{incomplete|Asymptote}} | ||
+ | |||
+ | The angle <math>\theta</math> is the angle formed by two [[perpendicular]]s drawn to <math>BO</math>, one on the plane determined by <math>OAB</math> and the other by <math>OBC</math>. Let the perpendiculars from <math>A</math> and <math>C</math> to <math>\overline{OB}</math> meet <math>\overline{OB}</math> at <math>P.</math> [[Without loss of generality]], let <math>AP = 1.</math> It follows that <math>OP = AP = 1,</math> <math>OB = OA = \sqrt {2},</math> and <math>AB = \sqrt {4 - 2\sqrt {2}}.</math> Therefore, <math>AC = \sqrt {8 - 4\sqrt {2}}.</math> | ||
+ | |||
+ | From the [[Law of Cosines]], <math>AC^{2} = AP^{2} + PC^{2} - 2(AP)(PC)\cos \theta,</math> so | ||
+ | |||
+ | <cmath> 8 - 4\sqrt {2} = 1 + 1 - 2\cos \theta \Longrightarrow \cos \theta = - 3 + 2\sqrt {2} = - 3 + \sqrt{8}.</cmath> | ||
+ | |||
+ | Thus <math>m + n = \boxed{005}</math>. | ||
+ | |||
+ | === Solution 2 (analytic/vectors) === | ||
+ | Without loss of generality, place the pyramid in a 3-dimensional coordinate system such that <math>A = (1,0,0),</math> <math>B = (0,1,0),</math> <math>C = ( - 1,0,0),</math> <math>D = (0, - 1,0),</math> and <math>O = (0,0,z),</math> where <math>z</math> is unknown. | ||
+ | |||
+ | We first find <math>z.</math> Note that | ||
+ | |||
+ | <cmath>\overrightarrow{OA}\cdot \overrightarrow{OB} = \parallel \overrightarrow{OA}\parallel \parallel \overrightarrow{OB}\parallel \cos 45^\circ.</cmath> | ||
+ | |||
+ | Since <math>\overrightarrow{OA} =\, <1,0, - z></math> and <math>\overrightarrow{OB} =\, <0,1, - z> ,</math> this simplifies to | ||
+ | |||
+ | <cmath>z^{2}\sqrt {2} = 1 + z^{2}\implies z^{2} = 1 + \sqrt {2}.</cmath> | ||
+ | |||
+ | Now let's find <math>\cos \theta.</math> Let <math>\vec{u}</math> and <math>\vec{v}</math> be normal vectors to the planes containing faces <math>OAB</math> and <math>OBC,</math> respectively. It follows that letting | ||
+ | |||
+ | <cmath>\vec{u}\cdot \vec{v} = \parallel \vec{u}\parallel \parallel \vec{v}\parallel \cos \theta</cmath> | ||
+ | |||
+ | will allow us to solve for <math>\cos \theta.</math> A cross product yields | ||
+ | |||
+ | <cmath>\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 > .</cmath> | ||
+ | |||
+ | Similarly, | ||
+ | |||
+ | <cmath>\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 > .</cmath> | ||
+ | |||
+ | Hence, taking the dot product of <math>\vec{u}</math> and <math>\vec{v}</math> yields | ||
+ | |||
+ | <cmath>- z^{2} + z^{2} + 1 = 1 = (\sqrt {1 + 2z^{2}})^{2}\cos \theta.</cmath> | ||
+ | |||
+ | Simplifying, | ||
+ | |||
+ | <cmath>\cos \theta = \frac {1}{3 + 2\sqrt {2}} = 3 - 2\sqrt {2} = 3 - \sqrt {8}.</cmath> | ||
+ | |||
+ | Flipping the signs (we found the cosine of the supplement angle) yields <math>\cos \theta = - 3 + \sqrt {8},</math> so the answer is <math>\boxed{005}</math>. | ||
== See also == | == See also == | ||
− | + | {{AIME box|year=1995|num-b=11|num-a=13|t=394527}} | |
− | + | ||
− | + | [[Category:Intermediate Geometry Problems]] |
Revision as of 20:35, 29 July 2008
Problem
Pyramid has square base congruent edges and and Let be the measure of the dihedral angle formed by faces and Given that where and are integers, find
Contents
[hide]Solution
Solution 1 (trigonometry)
The angle is the angle formed by two perpendiculars drawn to , one on the plane determined by and the other by . Let the perpendiculars from and to meet at Without loss of generality, let It follows that and Therefore,
From the Law of Cosines, so
Thus .
Solution 2 (analytic/vectors)
Without loss of generality, place the pyramid in a 3-dimensional coordinate system such that and where is unknown.
We first find Note that
Since and this simplifies to
Now let's find Let and be normal vectors to the planes containing faces and respectively. It follows that letting
will allow us to solve for A cross product yields
Similarly,
Hence, taking the dot product of and yields
Simplifying,
Flipping the signs (we found the cosine of the supplement angle) yields so the answer is .
See also
1995 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |