Difference between revisions of "2016 AIME I Problems/Problem 4"
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A right prism with height <math>h</math> has bases that are regular hexagons with sides of length <math>12</math>. A vertex <math>A</math> 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 <math>A</math> measures <math>60</math> degrees. Find <math>h^2</math>. | A right prism with height <math>h</math> has bases that are regular hexagons with sides of length <math>12</math>. A vertex <math>A</math> 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 <math>A</math> measures <math>60</math> degrees. Find <math>h^2</math>. | ||
− | == | + | ==Diagram== |
− | |||
− | (Solution by | + | <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> | ||
+ | |||
+ | ~gundraja | ||
+ | == Solution 1 == | ||
+ | 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>. | ||
+ | |||
+ | ~gundraja | ||
+ | |||
+ | == Solution 2 == | ||
+ | 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. Notice that we can already find some lengths. We have <math>AB=AC=12</math> (given) and <math>BC=BD=\sqrt{144+h^2}</math> by the Pythagorean Theorem. Let <math>M</math> be the midpoint of <math>BC</math>. Then, we have <math>AM=6</math> (<math>30-60-90</math>) triangles and <math>DM=\sqrt{36+h^2}</math> by the Pythagorean Theorem. Applying the Law of Cosines, since <math>\angle AMD=60^{\circ}</math>, we get <cmath>h^2=36+h^2+36-\frac12 \cdot 12 \sqrt{36+h^2} \implies h^2=\boxed{108},</cmath> as desired. | ||
+ | |||
+ | -A1001 | ||
== See also == | == See also == |
Latest revision as of 07:56, 28 November 2023
Contents
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
.
Diagram
~gundraja
Solution 1
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
.
~gundraja
Solution 2
Let and
be the vertices adjacent to
on the same base as
, and let
be the last vertex of the triangular pyramid. Notice that we can already find some lengths. We have
(given) and
by the Pythagorean Theorem. Let
be the midpoint of
. Then, we have
(
) triangles and
by the Pythagorean Theorem. Applying the Law of Cosines, since
, we get
as desired.
-A1001
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.