Difference between revisions of "2016 AIME I Problems/Problem 4"
Made in 2016 (talk | contribs) m (Fixed a grammar mistake.) |
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== Solution == | == Solution == | ||
Let <math>B</math> and <math>C</math> be the vertices adjacent to <math>A</math> on the same base as <math>A</math>, and let <math>D</math> be the last vertex of the triangular pyramid. Then <math>\angle CAB = 120^\circ</math>. Let <math>X</math> be the foot of the altitude from <math>A</math> to <math>\overline{BC}</math>. Then since <math>\triangle ABX</math> is a <math>30-60-90</math> triangle, <math>AX = 6</math>. Since the dihedral angle between <math>\triangle ABC</math> and <math>\triangle BCD</math> is <math>60^\circ</math>, <math>\triangle AXD</math> is a <math>30-60-90</math> triangle and <math>AD = 6\sqrt{3} = h</math>. Thus <math>h^2 = \boxed{108}</math>. | Let <math>B</math> and <math>C</math> be the vertices adjacent to <math>A</math> on the same base as <math>A</math>, and let <math>D</math> be the last vertex of the triangular pyramid. Then <math>\angle CAB = 120^\circ</math>. Let <math>X</math> be the foot of the altitude from <math>A</math> to <math>\overline{BC}</math>. Then since <math>\triangle ABX</math> is a <math>30-60-90</math> triangle, <math>AX = 6</math>. Since the dihedral angle between <math>\triangle ABC</math> and <math>\triangle BCD</math> is <math>60^\circ</math>, <math>\triangle AXD</math> is a <math>30-60-90</math> triangle and <math>AD = 6\sqrt{3} = h</math>. Thus <math>h^2 = \boxed{108}</math>. | ||
+ | |||
+ | <asy> | ||
+ | import bsp; | ||
+ | |||
+ | typedef path3[] shape; | ||
+ | |||
+ | shape operator *(transform3 T, shape p){ | ||
+ | shape os; | ||
+ | for(path3 g:p) os.push(T*g); | ||
+ | return os; | ||
+ | } | ||
+ | |||
+ | |||
+ | path3 path(triple[] T){ | ||
+ | path3 P; | ||
+ | for(triple i:T) P=P--i; | ||
+ | return P; | ||
+ | } | ||
+ | |||
+ | void addshapes(face[] F, shape[] shp, pen drawpen=currentpen, pen fillpen=white) | ||
+ | { | ||
+ | for(int i=0; i < shp.length; ++i) | ||
+ | for(int j=0; j < shp[i].length; ++j) { | ||
+ | path3 g=shp[i][j]; | ||
+ | picture pic=F.push(g); | ||
+ | if(fillpen != nullpen) filldraw(pic,project(g),fillpen, drawpen); | ||
+ | else draw(pic,project(g),drawpen); | ||
+ | // filldraw(pic,g,currentlight.intensity(F[F.length-1].point)*fillpen, drawpen); | ||
+ | } | ||
+ | } | ||
+ | |||
+ | shape cylinder(real R=1, real H=1, int n=6){ | ||
+ | shape Cyl; | ||
+ | triple[] CTop; | ||
+ | triple[] CBot; | ||
+ | for(int i=0; i <= n; ++i) | ||
+ | CBot.push((R*cos(2pi*i/n), R*sin(2pi*i/n),0)); | ||
+ | CTop = CBot+(0,0,H); | ||
+ | for(int i=0; i < n; ++i) | ||
+ | Cyl.push(CBot[i]--CBot[i+1]--CTop[i+1]--CTop[i]--cycle); | ||
+ | |||
+ | path3 P=path(CBot)--cycle; | ||
+ | Cyl.push(P); | ||
+ | Cyl.push(shift(H*Z)*P); | ||
+ | |||
+ | return Cyl; | ||
+ | } | ||
+ | |||
+ | |||
+ | size(10cm,0); | ||
+ | |||
+ | currentprojection=orthographic(1,1,1); | ||
+ | |||
+ | shape cyl1 = cylinder(R=1, H=2); | ||
+ | |||
+ | shape[] group={cyl1}; | ||
+ | |||
+ | face[] hidden, visible; | ||
+ | addshapes(hidden, group, drawpen=linewidth(bp)); | ||
+ | addshapes(visible, group, drawpen=dotted, fillpen=nullpen); | ||
+ | add(hidden); | ||
+ | add(visible); | ||
+ | |||
+ | //shipout(format="pdf"); | ||
+ | </asy> | ||
(Solution by gundraja) | (Solution by gundraja) |
Revision as of 14:28, 18 February 2019
Problem
A right prism with height has bases that are regular hexagons with sides of length . A vertex of the prism and its three adjacent vertices are the vertices of a triangular pyramid. The dihedral angle (the angle between the two planes) formed by the face of the pyramid that lies in a base of the prism and the face of the pyramid that does not contain measures degrees. Find .
Solution
Let and be the vertices adjacent to on the same base as , and let be the last vertex of the triangular pyramid. Then . Let be the foot of the altitude from to . Then since is a triangle, . Since the dihedral angle between and is , is a triangle and . Thus .
(Solution by gundraja)
See also
2016 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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.