2001 AIME I Problems/Problem 12
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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
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 ![\begin{align*}V &= V_{ABCI}+V_{ABDI}+V_{ACDI}+V_{BCDI}\\ &= \frac 13 \cdot r \cdot \left([ABC] + [ABD] + [ACD] + [BCD...](http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/6/3/d/63d3a1a208ec9e8a54f9c84077171dd7c611db6f.gif)
is the surface area of .
Since 
all lie on the planes containing the axes, their areas are straightforward to calculate; respectively 
. To find ![[ABC]](http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/8/b/7/8b78d271c3d197e48980793383e67d5fd4a27dfc.gif)
, we can using 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 
.
^ There are a couple of other ways to compute , including by vectors. In fact, it is known that in a trirectangular tetrahedron (one in which three edges are mutually perpendicular, as is the case here), the sum of the squares of the areas of the three smaller faces equals the square of the area of the larger face. See the thread below for details.
, including by vectors. In fact, it is known that in a trirectangular tetrahedron (one in which three edges are mutually perpendicular, as is the case here), the sum of the squares of the areas of the three smaller faces equals the square of the area of the larger face. See the thread below for details. See also
| 2001 AIME I (Problems • Resources) | ||
| Preceded by Problem 11 | Followed by Problem 13 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
