Difference between revisions of "2002 AIME I Problems/Problem 15"
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== Solution == | == Solution == | ||
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+ | Let's put the polyhedron onto a coordinate plane. For simplicity, let the origin be the center of the square: <math>A(-6,6,0)</math>, <math>B(-6,-6,0)</math>, <math>C(6,-6,0)</math> and <math>D(6,6,0)</math>. Since <math>ABFG</math> is an isosceles trapezoid and <math>CDE</math> is an isosceles triangle, we have symmetry about the <math>xz</math>-plane. | ||
+ | |||
+ | Therefore, the <math>y</math>-component of <math>E</math> is 0. We are given that the <math>z</math> component is 12, and it lies over the square, so we must have <math>E(2,0,12)</math> so <math>CE=\sqrt{4^2+6^2+12^2}=\sqrt{196}=14</math> (the other solution, <math>E(10,0,12)</math> does not lie over the square). Now let <math>F(a,-3,b)</math> and <math>G(a,3,b)</math>, so <math>FG=6</math> is parallel to <math>\overline{AB}</math>. We must have <math>BF=8</math>, so <math>(a+6)^2+b^2=8^2-3^2=55</math>. | ||
+ | |||
+ | The last piece of information we have is that <math>ADEG</math> (and its reflection, <math>BCEF</math>) are faces of the polyhedron, so they must all lie in the same plane. Since we have <math>A</math>, <math>D</math>, and <math>E</math>, we can derive this plane. First, <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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+ | Now that we have located <math>G</math>, we can calculate <math>EG^2</math>: | ||
+ | <cmath>EG^2=(8\pm\sqrt{19})^2+3^2+6^2=64\pm16\sqrt{19}+19+9+36=128\pm16\sqrt{19}.</cmath> Taking the negative root because the answer form asks for it, we get <math>128-16\sqrt{19}</math>, and <math>128+16+19=\fbox{163}</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}} |
Revision as of 15:14, 21 August 2008
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
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. First,
, so the plane is
. Since
lies on this plane, we must have
, so
. Therefore,
. So
.
Now that we have located , we can calculate
:
Taking the negative root because the answer form asks for it, we get
, 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 |