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| | label("$C$",C,N); | | label("$C$",C,N); |
| | label("$D$",D,W);</asy></center> | | label("$D$",D,W);</asy></center> |
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| − | Connect all four vertices of tetrahedron <math>ABCD</math> to its [[incenter]], <math>I</math>. This yields four tetrahedra <math>ABCI, ABDI, ACDI, BCDI</math>, all of which have height of <math>r</math> (the radius of the inscribed sphere), and which together form <math>ABCD</math>. It follows that <center><math>\begin{align*}V &= V_{ABCI}+V_{ABDI}+V_{ACDI}+V_{BCDI}\\ &= \frac 13 \cdot r \cdot \left([ABC] + [ABD] + [ACD] + [BCD]\right) = \frac{rS}{3}, \end{align*}</math></center> where <math>S</math> is the [[surface area]] of <math>ABCD</math>.
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| − | Since <math>\triangle ABD, ACD, BCD</math> all lie on the planes containing the axes, their areas are straightforward to calculate; respectively <math>12,6,4</math>. To find <math>[ABC]</math>, we can use the 3-dimensional [[distance formula]] (<math>d=\sqrt{(\Delta x)^2+(\Delta y)^2+(\Delta z)^2}</math>) to find that <math>AB = \sqrt{52}, BC=\sqrt{20}, CA=\sqrt{40}</math>. From here, we can use the [[Law of Cosines]] and the sine area formula to compute <math>[ABC]</math>, or we can use a manipulated version of [[Heron's formula]]: <math>A = \frac{1}{4}\sqrt{4a^2b^2 - (a^2+b^2-c^2)^2} = 14</math>.{{ref|1}}
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| − | Thus, <math>S = 14 + 12 + 6 + 4 = 36</math>. The volume of <math>ABCD</math> we can compute by letting <math>AD</math> to be the height to face <math>BCD</math>, so <math>V = \frac{1}{3} \cdot 6 \cdot \frac 12 \cdot 4 \cdot 2 = 8</math>. Therefore, <math>r = \frac{3V}{S} = \frac{24}{36} = \frac 23</math>, and <math>m+n = \boxed{005}</math>.
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| − | <div style="font-size:80%">{{note|1}} There are a couple of other ways to compute <math>[ABC]</math>, including by [[vector]]s. 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. </div>
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| | == See also == | | == See also == |
and 
The radius of the sphere is 
where 
and 
are relatively prime positive integers. Find 
![[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)
