2001 AIME I Problems/Problem 12
Contents
[hide]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
First let us find the equation of the plane passing through . The "point-slope form" is
Plugging in
gives
Plugging in
gives
We can then use Cramer's rule/cross multiplication to get
Solve for A, B, C to get
respectively. We can then get
Cancel out k on both sides. Next, let us substitute
. We can then get
as the equation of the plane. We can divide the equation by its magnitude to get the normal form of the plane. We get
to be the normal form. Note that the point is going to be at
We find the distance from
to the plane as
, which is
. We take the negative value of this because if we plug in
to the equation of the plane we get a negative value. We equate that value to r and we get the equation
to solve
, so the answer is
.
Solution 5 (Vectors)
Note that the center is obviously at some point since it must be equidistant from the x-y, x-z, and y-z planes. Now note that the point-plane distance formula is
(the direction vector of the normal to the plane dotted with the distance between a point on the plane and the point we're interested in). By inspection we know plane
's cartesian form is
thus its normal vector is
Let the "arbitrary point on the plane be
Then
which implies
so
and
giving us
as the answer.
See also
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 | ||
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