Difference between revisions of "1995 AIME Problems/Problem 12"
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== Problem == | == Problem == | ||
− | + | 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__ | __TOC__ | ||
+ | |||
== Solution == | == Solution == | ||
=== Solution 1 (trigonometry) === | === Solution 1 (trigonometry) === | ||
Line 89: | Line 90: | ||
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>. | ||
+ | |||
+ | === Solution 3 (bashy trig) === | ||
+ | |||
+ | <center><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></center> | ||
+ | |||
+ | Similar to Solution 1, <math>\angle APC</math> is the dihedral angle we want. WLOG, we will let <math>AB=1,</math> meaning <math>AC=\sqrt{2}</math>. | ||
+ | |||
+ | Because <math>\triangle OAB,\triangle OBC</math> are isosceles, <math>\angle ABP = 67.5^{\circ}</math> <math>PC=PA=\cos(\angle PAB)=\cos(22.5^{\circ})</math>. | ||
+ | |||
+ | Thus by the half-angle identity, | ||
+ | |||
+ | <cmath>PA=\cos\left(\frac{45}{2}\right) = \sqrt{\frac{1+\cos(45^{\circ})}{2}}</cmath> | ||
+ | <cmath>= \sqrt{\frac{2+\sqrt{2}}{4}}.</cmath> | ||
+ | |||
+ | Now looking at triangle <math>\triangle PAC,</math> we drop the perpendicular from <math>P</math> to <math>AC</math>, and call the foot <math>H</math>. Then <math>\angle CPH = \theta / 2.</math> By Pythagoreas, | ||
+ | <cmath>PH=\sqrt{\frac{2+\sqrt{2}}{4}-\frac{1}{2}}=\frac{\sqrt[4]{2}}{2}.</cmath> | ||
+ | |||
+ | <center><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></center> | ||
+ | |||
+ | We have that | ||
+ | <cmath>\cos\left(\frac{\theta}{2}\right)=\frac{\sqrt[4]{2}}{\sqrt{2+\sqrt{2}}},\text{ so}</cmath> | ||
+ | <cmath>\cos(\theta)=2\cos^{2}\left(\frac{\theta}{2}\right)-1</cmath> | ||
+ | <cmath>=2\left(\frac{\sqrt{2}}{2+\sqrt{2}}\right)-1</cmath> | ||
+ | <cmath>=2(\frac{2\sqrt{2}-2}{2})-1</cmath> | ||
+ | <cmath>=-3+\sqrt{8}.</cmath> | ||
+ | |||
+ | Because <math>m</math> and <math>n</math> can be negative integers, our answer is <math>(-3)+8=\boxed{005.}</math> | ||
+ | |||
+ | Notice that <math>-1\le \cos(\theta) \le 1</math> as well. | ||
+ | |||
+ | ~RubixMaster21 | ||
== See also == | == See also == |
Latest revision as of 20:57, 22 September 2022
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)
![[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
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
.
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]](http://latex.artofproblemsolving.com/3/9/8/39848850592bbca24d0dfb347065b97e2a7d50dc.png)
Similar to Solution 1, is the dihedral angle we want. WLOG, we will let
meaning
.
Because are isosceles,
.
Thus by the half-angle identity,
Now looking at triangle we drop the perpendicular from
to
, and call the foot
. Then
By Pythagoreas,
![[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]](http://latex.artofproblemsolving.com/7/a/2/7a2d91773d02d972fde87e3414ec5d098d3ab28b.png)
We have that
Because and
can be negative integers, our answer is
Notice that as well.
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.