Difference between revisions of "2001 AIME I Problems/Problem 12"
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== Solution == | == Solution == | ||
| − | <center><asy>import three; pointpen = black; pathpen = black+linewidth(0.7); currentprojection = perspective( | + | <center><asy>import three; pointpen = black; pathpen = black+linewidth(0.7); currentprojection = perspective(-2,9,4); |
triple A = (6,0,0), B = (0,4,0), C = (0,0,2), D = (0,0,0); | triple A = (6,0,0), B = (0,4,0), C = (0,0,2), D = (0,0,0); | ||
| − | triple I = (3/2, | + | 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); |
| − | draw(C--A--D--C--B--D--I-- | + | triple I = (2/3,2/3,2/3); |
| − | label("$I$",I, | + | 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("$A$",A,S); | ||
| − | label("$B$",B, | + | label("$B$",B,S); |
| − | label("$C$",C, | + | label("$C$",C,W*-1); |
| − | label("$D$",D,W);</asy></center> | + | label("$D$",D,W*-1);</asy></center> |
| + | |||
| + | The center <math>I</math> of the insphere must be located at <math>(r,r,r)</math> where <math>r</math> is the sphere's radius. | ||
| + | <math>I</math> must also be a distance <math>r</math> from the plane <math>ABC</math> | ||
| + | |||
| + | The signed distance between a plane and a point <math>I</math> can be calculated as <math>frac{(I-G) \cdot P}{|P|}</math>, where G is any point on the plane, and P is a vector perpendicular to ABC. | ||
| + | |||
| + | A vector <math>P</math> perpendicular to plane <math>ABC</math> can be found as <math>V=(A-C)\times(B-C)=\langle 8, 12, 24 \rangle</math> | ||
| + | |||
| + | Thus <math>\frac{(I-C) \cdot P}{|P|}=-r</math> where the negative comes from the fact that we want <math>I</math> to be in the opposite direction of <math>P</math> | ||
| + | |||
| + | <cmath>\begin{align*}\frac{(I-C) \cdot P}{|P|}&=-r\\ | ||
| + | \frac{(\langle r, r, r \rangle-\langle 0, 0, 2 \rangle) \cdot P}{|P|}&=-r\\ | ||
| + | \frac{\langle r, r, r-2 \rangle \cdot \langle 8, 12, 24 \rangle}{\langle 8, 12, 24 \rangle}&=-r\\ | ||
| + | \frac{44r -48}{28}&=-r\\ | ||
| + | 44r-48&=-28r\\ | ||
| + | 72r&=48\\ | ||
| + | r&=\frac{2}{3} | ||
| + | \end{align*}</cmath> | ||
| + | |||
| + | |||
| + | Finally <math>2+3=\boxed{005}</math> | ||
== See also == | == See also == | ||
Revision as of 15:11, 8 May 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(-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/b/2/2/b22f739c87d34ee814fc3d229fdade0498040c90.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 
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 | |
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