Difference between revisions of "2001 AIME I Problems/Problem 12"
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== Solution == | == Solution == | ||
− | <center><asy> | + | <center><asy>import three; pointpen = black; pathpen = black+linewidth(0.7); currentprojection = perspective(5,-10,4); |
− | 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); | |
− | D( | + | draw(C--A--D--C--B--D--I--A--B--I--C); |
− | </asy></center> | + | label("$I$",I,S); |
+ | label("$A$",A,S); | ||
+ | label("$B$",B,E); | ||
+ | label("$C$",C,N); | ||
+ | label("$D$",D,W);</asy></center> | ||
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>. | 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>. |
Revision as of 10:42, 12 March 2014
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) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.