1995 AIME Problems/Problem 12
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
Solution
Solution 1 (trigonometry)
![[asy] import three; // calculate intersection of line and plane // p = point on line // d = direction of line // q = point in plane // n = normal to plane triple lineintersectplan(triple p, triple d, triple q, triple n) { return (p + dot(n,q - p)/dot(n,d)*d); } // projection of point A onto line BC triple projectionofpointontoline(triple A, triple B, triple C) { return lineintersectplan(B, B - C, A, B - C); } currentprojection=perspective(2,1,1); triple A, B, C, D, O, P; A = (sqrt(2 - sqrt(2)), sqrt(2 - sqrt(2)), 0); B = (-sqrt(2 - sqrt(2)), sqrt(2 - sqrt(2)), 0); C = (-sqrt(2 - sqrt(2)), -sqrt(2 - sqrt(2)), 0); D = (sqrt(2 - sqrt(2)), -sqrt(2 - sqrt(2)), 0); O = (0,0,sqrt(2*sqrt(2))); P = projectionofpointontoline(A,O,B); draw(D--A--B); draw(B--C--D,dashed); draw(A--O); draw(B--O); draw(C--O,dashed); draw(D--O); draw(A--P); draw(P--C,dashed); label("$A$", A, S); label("$B$", B, E); label("$C$", C, NW); label("$D$", D, W); label("$O$", O, N); dot("$P$", P, NE); [/asy]](http://latex.artofproblemsolving.com/3/9/8/39848850592bbca24d0dfb347065b97e2a7d50dc.png)
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
![\[8 - 4\sqrt {2} = 1 + 1 - 2\cos \theta \Longrightarrow \cos \theta = - 3 + 2\sqrt {2} = - 3 + \sqrt{8}.\]](http://latex.artofproblemsolving.com/0/1/b/01bbdfbd970c7c425dc80bcd12db6f34b0f64fae.png)
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
![\[\overrightarrow{OA}\cdot \overrightarrow{OB} = \parallel \overrightarrow{OA}\parallel \parallel \overrightarrow{OB}\parallel \cos 45^\circ.\]](http://latex.artofproblemsolving.com/1/5/d/15db9106c2b042c930c05370f965dabee7020550.png)
Since 
and 
this simplifies to
![\[z^{2}\sqrt {2} = 1 + z^{2}\implies z^{2} = 1 + \sqrt {2}.\]](http://latex.artofproblemsolving.com/7/a/f/7afe6d0412e092b1d3e6d1bf391817eee2512876.png)
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)
![\[\vec{u} = \overrightarrow{OA}\times \overrightarrow{OB} = \left| \begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & - z \\ 0 & 1 & - z \end{array}\right| =\, < z,z,1 > .\]](http://latex.artofproblemsolving.com/e/9/d/e9d2f5dcf8f658b9f4907dbf247f606072d8627e.png)
Similarly,
![\[\vec{v} = \overrightarrow{OB}\times \overrightarrow{OC} - \left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & - z \\ - 1 & 0 & - z \end{array}\right| =\, < - z,z,1 > .\]](http://latex.artofproblemsolving.com/6/d/4/6d425c167b9ceb268b8a611c473264ee4e51aae2.png)
Hence, taking the dot product of and yields
![\[\cos \theta = \frac{ \vec{u} \cdot \vec{v} }{ \parallel \vec{u} \parallel \parallel \vec{v} \parallel } = \frac{- z^{2} + z^{2} + 1}{(\sqrt {1 + 2z^{2}})^{2}} = \frac {1}{3 + 2\sqrt {2}} = 3 - 2\sqrt {2} = 3 - \sqrt {8}.\]](http://latex.artofproblemsolving.com/a/7/0/a70117d157744189dece6d79491c17a4b28ce679.png)
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 | ||
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