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
![[asy]import three; pointpen = black; pathpen = black+linewidth(0.7); currentprojection = perspective(5,-10,4); triple A = (6,0,0), B = (0,4,0), C = (0,0,2), D = (0,0,0); triple I = (3/2,1,1/2); draw(C--A--D--C--B--D--I--A--B--I--C); label("$I$",I,S); label("$A$",A,S); label("$B$",B,E); label("$C$",C,N); label("$D$",D,W);[/asy]](http://latex.artofproblemsolving.com/d/7/c/d7cfeed8b9d8a63a7d63621747ffb9d2cb2ee98a.png)
Connect all four vertices of tetrahedron to its incenter,
. This yields four tetrahedra
, all of which have height of
(the radius of the inscribed sphere), and which together form
. It follows that
where is the surface area of
.
Since all lie on the planes containing the axes, their areas are straightforward to calculate; respectively
. To find
, we can use the 3-dimensional distance formula (
) to find that
. From here, we can use the Law of Cosines and the sine area formula to compute
, or we can use a manipulated version of Heron's formula:
.[1]
Thus, . The volume of
we can compute by letting
to be the height to face
, so
. Therefore,
, and
.
![$[ABC]$](http://latex.artofproblemsolving.com/d/3/3/d33cc80fa8f093e155c5be46d2e5d9da3d7e1ef5.png)
See also
- <url>viewtopic.php?p=384205#384205 Discussion on AoPS</url>
2001 AIME I (Problems • Answer Key • Resources) | ||
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