2001 AIME I Problems/Problem 12
Problem
A sphere is inscribed in the tetrahedron whose vertices are and
The radius of the sphere is
where
and
are relatively prime positive integers. Find
Solution
The center of the insphere must be located at
where
is the sphere's radius.
must also be a distance
from the plane
The signed distance between a plane and a point can be calculated as
, where G is any point on the plane, and P is a vector perpendicular to ABC.
A vector perpendicular to plane
can be found as
Thus where the negative comes from the fact that we want
to be in the opposite direction of
Finally
Solution 2
Notice that we can split the tetrahedron into smaller tetrahedrons such that the height of each tetrahedron is
and the base of each tetrahedron is one of the faces of the original tetrahedron. This is because the bases of the spheres are tangent to the sphere, so the line from the center to the foot of the perpendicular to the bases hits the tangency points. Letting volume be
and surface area be
, using the volume formula for each pyramid(base times height divided by 3) we have
. The surface area of the pyramid is
. We know triangle ABC's side lengths,
and
, so using the expanded form of heron's formula,
. Therefore, the surface area is
, and the volume is
, and using the formula above that
, we have
and thus
, so the desired answer is
.
(Solution by Shaddoll)
Solution 3
The intercept form equation of the plane is
Its normal form is
(square sum of the coefficients equals 1). The distance from
to the plane is
. Since
and
are on the same side of plane, the value in the absolute value sign is negative (same as the one by plugging in
). Therefore we have
So
which solves the problem.
Additionally, if is on the other side of
, we have
, which yields
corresponding an "ex-sphere" that is tangent to face
as well as the extensions of the other 3 faces.
-JZ
Solution 4
Notice that because three faces of the tetrahedron are the ,
, and
planes we know the location of the center:
.
Now we can calculate the plane of the last face, plane . We know that the general formula for a plane face is
so we can plug in the three points to find it.
Plugging in , we have that
. Similarly for
and
, we can obtain
and that
. It may seem that there are 4 variables and only three equations but that is because we can scale each variable up by any value and it would still be the same plane (it's equal to 0).
So, notice that and so
. We can let C be the least common multiple of 2 and 3, 6 to make things nicer. This gives
.
All that is left is to find the distance from to this plane. Using the point to plane formula (almost the same as point to line) which is
where
is the point. Our point is
so plugging in it is easy to calculate
as the distance. This is also equal to r so setting them equal,
. We have two solutions,
and
but three makes the sphere outside the tetrahedron, which means it cannot be tangent to the other three sides. Therefore the answer is
as the answer. ~Leonard_my_dude~
See also
- <url>viewtopic.php?p=384205#384205 Discussion on AoPS</url>
2001 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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.