2001 AIME I Problems/Problem 12
Contents
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; currentprojection = perspective(-2,9,4); triple A = (6,0,0), B = (0,4,0), C = (0,0,2), D = (0,0,0); triple E = (2/3,0,0), F = (0,2/3,0), G = (0,0,2/3), L = (0,2/3,2/3), M = (2/3,0,2/3), N = (2/3,2/3,0); triple I = (2/3,2/3,2/3); triple J = (6/7,20/21,26/21); draw(C--A--D--C--B--D--B--A--C); draw(L--F--N--E--M--G--L--I--M--I--N--I--J); label("$I$",I,W); label("$A$",A,S); label("$B$",B,S); label("$C$",C,W*-1); label("$D$",D,W*-1); [/asy]](http://latex.artofproblemsolving.com/a/d/5/ad592fa5d796f97d2c229a092105e33dc3dfc1ed.png)
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 ![$\dfrac{6\cdot{4}+6\cdot{2}+4\cdot{2}}{2}+[ABC]=22+[ABC]$](http://latex.artofproblemsolving.com/b/a/0/ba0528715eb4dac76614710e6a3b9bb5a8ccc64a.png)
. We know triangle ABC's side lengths, 
and 
, so using the expanded form of heron's formula, ![$[ABC]=\sqrt{\dfrac{2(a^{2}b^{2}+b^{2}c^{2}+a^{2}c^{2})-a^{4}-b^{4}-c^{4}}{16}}=\sqrt{2(5\cdot{13}+10\cdot{5}+13\cdot{10})-5^{2}-10^{2}-13^{2}}=\sqrt{196}=14$](http://latex.artofproblemsolving.com/1/8/a/18a9e2184b9e80f2cf58dbefedf6be443e76017b.png)
. Therefore, the surface area is 
, and the volume is ![$\dfrac{[BCD]\cdot{6}}{3}=\dfrac{4\cdot{2}\cdot{6}}{3\cdot{2}}=8$](http://latex.artofproblemsolving.com/3/8/b/38ba0074d10f2be358b0ddc91d100bce3cb8e212.png)
, and using the formula above that , we have 
and thus 
, so the desired answer is .
(Solution by Shaddoll)
See also
- <url>viewtopic.php?p=384205#384205 Discussion on AoPS</url>
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