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1995 AIME Problems/Problem 12

Revision as of 19:48, 26 October 2021 by Rubixmaster21 (talk | contribs) (Added Solution 3 (bashy trig))

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 $m+n.$

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]

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)

[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]

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

[asy] // if you see this // hello // gap for label on P--H: https://tex.stackexchange.com/questions/475945/asymptote-how-do-i-make-a-gap-in-a-segment-to-include-a-label pair P,C,A,H; H = (0, 0); C = (-0.71, 0); A = (0.71, 0); P = (0,0.59); draw(P--C--A--cycle); draw(P--H); label("$A$", A, SE); label("$C$", C, SW); label("$P$", P, N); label("$H$", H, S); label("$\sqrt{\frac{2+\sqrt{2}}{4}}$",align=NE,point(P--A,0.5)); label("$\sqrt{\frac{2+\sqrt{2}}{4}}$",align=NW,point(P--C,0.5)); label("$\frac{\sqrt{2}}{2}$",align=S,point(C--H,0.5)); label("$\frac{\sqrt{2}}{2}$",align=S,point(A--H,0.5)); pen fillpen = white; Label mylabel = Label("$\frac{\sqrt[4]{2}}{2}$", align=(0,0), position=MidPoint, filltype=Fill(fillpen)); draw(P--H, L=mylabel); [/asy]

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 $n$ are integers, our answer is $(-3)+8=\boxed{005.}$

~RubixMaster21

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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