AIME 1995 · 第 12 题

Solution

Solution 1 (trigonometry)

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 $\boxed{005}$.

Solution 3 (bashy trig)

Similar to Solution 1, $\angle APC$ is the dihedral angle we want. WLOG, we will let $AB=1,$ meaning $AC=\sqrt{2}$.

Because $\triangle OAB,\triangle OBC$ are isosceles, $\angle ABP = 67.5^{\circ}$ $PC=PA=\cos(\angle PAB)=\cos(22.5^{\circ})$.

Thus by the half-angle identity,

$$PA=\cos\left(\frac{45}{2}\right) = \sqrt{\frac{1+\cos(45^{\circ})}{2}}$$ $$= \sqrt{\frac{2+\sqrt{2}}{4}}.$$ Now looking at triangle $\triangle PAC,$ we drop the perpendicular from $P$ to $AC$, and call the foot $H$. Then $\angle CPH = \theta / 2.$ By Pythagoreas,

$$PH=\sqrt{\frac{2+\sqrt{2}}{4}-\frac{1}{2}}=\frac{\sqrt[4]{2}}{2}.$$

We have that

$$\cos\left(\frac{\theta}{2}\right)=\frac{\sqrt[4]{2}}{\sqrt{2+\sqrt{2}}},\text{ so}$$ $$\cos(\theta)=2\cos^{2}\left(\frac{\theta}{2}\right)-1$$ $$=2\left(\frac{\sqrt{2}}{2+\sqrt{2}}\right)-1$$ $$=2(\frac{2\sqrt{2}-2}{2})-1$$ $$=-3+\sqrt{8}.$$ Because $m$ and $n$ can be negative integers, our answer is $(-3)+8=\boxed{005.}$

Notice that $-1\le \cos(\theta) \le 1$ as well.

~RubixMaster21