Difference between revisions of "1995 AIME Problems/Problem 12"
Made in 2016 (talk | contribs) (Fixed the Asymptote error by copying the diagram from the AIME A class.) |
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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>. | ||
+ | |||
+ | === 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> are integers, our answer is <math>(-3)+8=\boxed{005.}</math> | ||
+ | |||
+ | ~RubixMaster21 | ||
== See also == | == See also == |
Revision as of 19:48, 26 October 2021
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)
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)
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,
We have that
Because and are integers, our answer is
~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.