Difference between revisions of "1995 AIME Problems/Problem 12"
(incomplete; both solutions written by 4everwise) |
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=== Solution 1 (trigonometry) === | === Solution 1 (trigonometry) === | ||
<center><asy> | <center><asy> | ||
+ | size(220); defaultpen(linewidth(0.7)); currentprojection = perspective(5,3,2); | ||
import three; | 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*( | + | 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); D(A--P--C); | + | 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)); | ||
</asy></center> | </asy></center> | ||
− | + | 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>\triangle OPA</math> is a <math>45-45-90</math> [[right triangle]], so <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> | |
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− | 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 | From the [[Law of Cosines]], <math>AC^{2} = AP^{2} + PC^{2} - 2(AP)(PC)\cos \theta,</math> so | ||
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Thus <math>m + n = \boxed{005}</math>. | Thus <math>m + n = \boxed{005}</math>. | ||
− | === Solution 2 ( | + | === Solution 2 (analytical/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. | 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. | ||
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<cmath>z^{2}\sqrt {2} = 1 + z^{2}\implies z^{2} = 1 + \sqrt {2}.</cmath> | <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. | + | 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. From the definition of the [[dot product]] as <math>\vec{u}\cdot \vec{v} = \parallel \vec{u}\parallel \parallel \vec{v}\parallel \cos \theta</math>, we will be able to solve for <math>\cos \theta.</math> A cross product yields (alternatively, it is simple to find the equation of the planes <math>OAB</math> and <math>OAC</math>, and then to find their normal vectors) |
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− | < | ||
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− | will | ||
<cmath>\vec{u} = \overrightarrow{OA}\times \overrightarrow{OB} = \left| \begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \ | <cmath>\vec{u} = \overrightarrow{OA}\times \overrightarrow{OB} = \left| \begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \ | ||
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Hence, taking the dot product of <math>\vec{u}</math> and <math>\vec{v}</math> yields | Hence, taking the dot product of <math>\vec{u}</math> and <math>\vec{v}</math> yields | ||
− | <cmath>- z^{2} + z^{2} + 1 | + | <cmath>\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}.</cmath> |
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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>. | 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>. |
Revision as of 13:20, 31 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)
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 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
is a
right triangle, so
and
Therefore,
From the Law of Cosines, so
Thus .
Solution 2 (analytical/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. From the definition of the dot product as
, we will be able to solve for
A cross product yields (alternatively, it is simple to find the equation of the planes
and
, and then to find their normal vectors)
Similarly,
Hence, taking the dot product of and
yields
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 |