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Difference between revisions of "2001 AIME I Problems/Problem 12"

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== Problem ==
 
== Problem ==
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A sphere is inscribed in the tetrahedron whose vertices are <math>A = (6,0,0), B = (0,4,0), C = (0,0,2),</math> and <math>D = (0,0,0).</math>  The radius of the sphere is <math>m/n,</math> where <math>m</math> and <math>n</math> are relatively prime positive integers.  Find <math>m + n.</math>
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A [[sphere]] is inscribed in the [[tetrahedron]] whose vertices are <math>A = (6,0,0), B = (0,4,0), C = (0,0,2),</math> and <math>D = (0,0,0).</math>  The [[radius]] of the sphere is <math>m/n,</math> where <math>m</math> and <math>n</math> are relatively prime positive integers.  Find <math>m + n.</math>
  
 
== Solution ==
 
== Solution ==
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{{solution}}
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<center><asy>
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import three; pointpen = black; pathpen = black+linewidth(0.7); currentprojection = perspective(5,-10,4);
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pair A=(6,0,0), B=(0,4,0), C=(0,0,2), D=(0,0,0);
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D(MP("A",A)--MP("B",B)--MP("C",C,N)--A--MP("D",D)--B--D--C);
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</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 using 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>
  
 
== See also ==
 
== See also ==
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*<url>viewtopic.php?p=384205#384205 Discussion on AoPS</url>
 
{{AIME box|year=2001|n=I|num-b=11|num-a=13}}
 
{{AIME box|year=2001|n=I|num-b=11|num-a=13}}
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[[Category:Intermediate Geometry Problems]]

Revision as of 18:09, 12 June 2008

Problem

A sphere is inscribed in the tetrahedron whose vertices are $A = (6,0,0), B = (0,4,0), C = (0,0,2),$ and $D = (0,0,0).$ The radius of the sphere is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Solution

import three; pointpen = black; pathpen = black+linewidth(0.7); currentprojection = perspective(5,-10,4);
pair A=(6,0,0), B=(0,4,0), C=(0,0,2), D=(0,0,0);
D(MP("A",A)--MP("B",B)--MP("C",C,N)--A--MP("D",D)--B--D--C);
 (Error making remote request. Unknown error_msg)

Connect all four vertices of tetrahedron $ABCD$ to its incenter, $I$. This yields four tetrahedra $ABCI, ABDI, ACDI, BCDI$, all of which have height of $r$ (the radius of the inscribed sphere), and which together form $ABCD$. It follows that

$\begin{align*}V &= V_{ABCI}+V_{ABDI}+V_{ACDI}+V_{BCDI}\\ &= \frac 13 \cdot r \cdot \left([ABC] + [ABD] + [ACD] + [BCD]\right) = \frac{rS}{3}, \end{align*}$ (Error compiling LaTeX. Unknown error_msg)

where $S$ is the surface area of $ABCD$.

Since $\triangle ABD, ACD, BCD$ all lie on the planes containing the axes, their areas are straightforward to calculate; respectively $12,6,4$. To find $[ABC]$, we can using the 3-dimensional distance formula ($d=\sqrt{(\Delta x)^2+(\Delta y)^2+(\Delta z)^2}$) to find that $AB = \sqrt{52}, BC=\sqrt{20}, CA=\sqrt{40}$. From here, we can use the Law of Cosines and the sine area formula to compute $[ABC]$, or we can use a manipulated version of Heron's formula: $A = \frac{1}{4}\sqrt{4a^2b^2 - (a^2+b^2-c^2)^2} = 14$.[1]

Thus, $S = 14 + 12 + 6 + 4 = 36$. The volume of $ABCD$ we can compute by letting $AD$ to be the height to face $BCD$, so $V = \frac{1}{3} \cdot 6 \cdot \frac 12 \cdot 4 \cdot 2 = 8$. Therefore, $r = \frac{3V}{S} = \frac{24}{36} = \frac 23$, and $m+n = \boxed{005}$.


^ There are a couple of other ways to compute $[ABC]$, 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

  • <url>viewtopic.php?p=384205#384205 Discussion on AoPS</url>
2001 AIME I (Problems • Answer Key • Resources)
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Problem 11 		Followed by
Problem 13
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All AIME Problems and Solutions
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