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. | ||
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| + | 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>. | ||
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| + | 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 14:14, 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 
:
![\[EG^2=(8\pm\sqrt{19})^2+3^2+6^2=64\pm16\sqrt{19}+19+9+36=128\pm16\sqrt{19}.\]](http://latex.artofproblemsolving.com/c/2/8/c2837512e57b1f04af3d61a3c99df136c26e83ac.png)
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 | ||
