2002 AIME I Problems/Problem 15
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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.* 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 
.
See also
| 2002 AIME I (Problems • Resources) | ||
| Preceded by Problem 14 | Followed by Last Question | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
