Difference between revisions of "2002 AIME I Problems/Problem 15"
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Polyhedron <math>ABCDEFG</math> has six faces. Face <math>ABCD</math> is a square with <math>AB = 12;</math> face <math>ABFG</math> is a trapezoid with <math>\overline{AB}</math> parallel to <math>\overline{GF},</math> <math>BF = AG = 8,</math> and <math>GF = 6;</math> and face <math>CDE</math> has <math>CE = DE = 14.</math> The other three faces are <math>ADEG, BCEF,</math> and <math>EFG.</math> The distance from <math>E</math> to face <math>ABCD</math> is 12. Given that <math>EG^2 = p - q\sqrt {r},</math> where <math>p, q,</math> and <math>r</math> are positive integers and <math>r</math> is not divisible by the square of any prime, find <math>p + q + r.</math> | Polyhedron <math>ABCDEFG</math> has six faces. Face <math>ABCD</math> is a square with <math>AB = 12;</math> face <math>ABFG</math> is a trapezoid with <math>\overline{AB}</math> parallel to <math>\overline{GF},</math> <math>BF = AG = 8,</math> and <math>GF = 6;</math> and face <math>CDE</math> has <math>CE = DE = 14.</math> The other three faces are <math>ADEG, BCEF,</math> and <math>EFG.</math> The distance from <math>E</math> to face <math>ABCD</math> is 12. Given that <math>EG^2 = p - q\sqrt {r},</math> where <math>p, q,</math> and <math>r</math> are positive integers and <math>r</math> is not divisible by the square of any prime, find <math>p + q + r.</math> | ||
− | == Solution == | + | == Solution 1 == |
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*One may also do this by vectors; <math>\vec{AD}\times\vec{DE}=(12,0,0)\times(-4,-6,12)=(0,-144,-72)=-72(0,2,1)</math>, so the plane is <math>2y+z=2\cdot6=12</math>. Since <math>G</math> lies on this plane, we must have <math>2\cdot3+b=12</math>, so <math>b=6</math>. Therefore, <math>a=-6\pm\sqrt{55-6^2}=-6\pm\sqrt{19}</math>. So <math>G(-6\pm\sqrt{19},-3,6)</math>. | *One may also do this by vectors; <math>\vec{AD}\times\vec{DE}=(12,0,0)\times(-4,-6,12)=(0,-144,-72)=-72(0,2,1)</math>, so the plane is <math>2y+z=2\cdot6=12</math>. Since <math>G</math> lies on this plane, we must have <math>2\cdot3+b=12</math>, so <math>b=6</math>. Therefore, <math>a=-6\pm\sqrt{55-6^2}=-6\pm\sqrt{19}</math>. So <math>G(-6\pm\sqrt{19},-3,6)</math>. | ||
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+ | ==Solution 2== | ||
+ | |||
+ | We let <math>A</math> be the origin, or <math>(0,0,0)</math> and <math>B = (0,0,12)</math> | ||
== See also == | == See also == | ||
{{AIME box|year=2002|n=I|num-b=14|after=Last Question}} | {{AIME box|year=2002|n=I|num-b=14|after=Last Question}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 10:06, 15 February 2017
Contents
Problem
Polyhedron has six faces. Face
is a square with
face
is a trapezoid with
parallel to
and
and face
has
The other three faces are
and
The distance from
to face
is 12. Given that
where
and
are positive integers and
is not divisible by the square of any prime, find
Solution 1
![[asy] size(200); import three; import graph; defaultpen(linewidth(0.7)+fontsize(8)); currentprojection=orthographic(-30,50,40); triple A=(-6,-6,0), B = (-6,6,0), C = (6,6,0), D = (6,-6,0), E = (2,0,12), H=(-6+2*sqrt(19),0,12), H1=(-6-2*sqrt(19),0,12), F, G, E1 = (6,0,12); F = 1/2*H+1/2*B; G = 1/2*H+1/2*A; draw((A--B--C--D--A)^^(D--E--C)^^(A--G--F--B)^^(G--E--F));draw((G--H--F)^^(H--E1),gray(0.6)); dot(H1^^H,linewidth(2)); label("$A$",A,( 0,-1, 0)); label("$B$",B,( 0, 1, 0)); label("$C$",C,( 0, 1, 0)); label("$D$",D,( 0,-1, 0)); label("$E$",E,(-1,-1, 1)); label("$F$",F,( 0, 1, 0)); label("$G$",G,(-1,-1, 1)); label("$H$",H,( 1,-1, 1)); label("$H'$",H1,(-1,-1, 1)); [/asy]](http://latex.artofproblemsolving.com/1/6/7/167e69f38f11e4b219976293223c73b2bdec1e72.png)
Let's put the polyhedron onto a coordinate plane. For simplicity, let the origin be the center of the square: ,
,
and
. Since
is an isosceles trapezoid and
is an isosceles triangle, we have symmetry about the
-plane.
Therefore, the -component of
is 0. We are given that the
component is 12, and it lies over the square, so we must have
so
(the other solution,
does not lie over the square). Now let
and
, so
is parallel to
. We must have
, so
.
The last piece of information we have is that (and its reflection,
) are faces of the polyhedron, so they must all lie in the same plane. Since we have
,
, and
, we can derive this plane.* Let
be the extension of the intersection of the lines containing
. It follows that the projection of
onto the plane
must coincide with the
, where
is the projection of
onto the plane
.
by a ratio of
, so the distance from
to the plane
is
and by the similarity, the distance from
to the plane
is
. The altitude from
to
has height
. By similarity, the x-coordinate of
is
. Then
.
Now that we have located , we can calculate
:
Taking the negative root because the answer form asks for it, we get
, and
.
- One may also do this by vectors;
, so the plane is
. Since
lies on this plane, we must have
, so
. Therefore,
. So
.
Solution 2
We let be the origin, or
and
See also
2002 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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