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

m (→‎Used ABC instead of BCD)
(→‎Solution)
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== Solution ==
 
== Solution ==
−
<center><asy>import three; pointpen = black; pathpen = black+linewidth(0.7); currentprojection = perspective(-2,9,4);
+
$<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 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 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);
Line 14: Line 14:
 
label("$B$",B,S);
 
label("$B$",B,S);
 
label("$C$",C,W*-1);
 
label("$C$",C,W*-1);
−
label("$D$",D,W*-1);</asy></center>
+
label("$D$",D,W*-1);</asy></center><math>
  
−
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.
+
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>
+
</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.
+
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>
+
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>
+
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\\
 
<cmath>\begin{align*}\frac{(I-C) \cdot P}{|P|}&=-r\\
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−
Finally <math>2+3=\boxed{005}</math>
+
Finally </math>2+3=\boxed{005}$
  
 
==Solution 2==
 
==Solution 2==

Revision as of 17:47, 3 April 2016

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(-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); (Error making remote request. Unknown error_msg)

$The center$I$of the insphere must be located at$(r,r,r)$where$r$is the sphere's radius.$I$must also be a distance$r$from the plane$ABC$The signed distance between a plane and a point$I$can be calculated as$\frac{(I-G) \cdot P}{|P|}

$, where G is any point on the plane, and P is a vector perpendicular to ABC.

A vector$ (Error compiling LaTeX. Unknown error_msg)P$perpendicular to plane$ABC$can be found as$V=(A-C)\times(B-C)=\langle 8, 12, 24 \rangle$Thus$\frac{(I-C) \cdot P}{|P|}=-r$where the negative comes from the fact that we want$I$to be in the opposite direction of$P$<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$ (Error compiling LaTeX. Unknown error_msg)2+3=\boxed{005}$

Solution 2

Notice that we can split the tetahedron into $4$ smaller tetrahedrons such that the height of each tetrahedron is $r$ 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 $V$ and surface area be $F$, using the volume formula for each pyramid(base times height divided by 3) we have $\dfrac{rF}{3}=V$. The surface area of the pyramid is $\dfrac{6\cdot{4}+6\cdot{2}+4\cdot{2}}{2}+[ABC]=22+[ABC]$. We know triangle ABC's side lengths, $\sqrt{2^{2}+4^{2}}, \sqrt{2^{2}+6^{2}},$ and $\sqrt{4^{2}+6^{2}}$, so using the expanded form of heron's formula, $[ABC]=\sqrt{\dfrac{2(a^{2}b^{2}+b^{2}c^{2}+a^{2}c^{2})-a^{4}-b^{4}-c^{4}}{16}}=\sqrt{2(5\cdot{13}+10\cdot{5}+13\cdot{10})-5^{2}-10^{2}-13^{2}}=\sqrt{196}=14$. Therefore, the surface area is $14+22=36$, and the volume is $\dfrac{[BCD]\cdot{6}}{3}=\dfrac{4\cdot{2}\cdot{6}}{3\cdot{2}}=8$, and using the formula above that $\dfrac{rF}{3}=V$, we have $12r=8$ and thus $r=\dfrac{2}{3}$, so the desired answer is $2+3=\boxed{005}$.

(Solution by Shaddoll)

See also

  • <url>viewtopic.php?p=384205#384205 Discussion on AoPS</url>
2001 AIME I (Problems • Answer Key • Resources)
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