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Difference between revisions of "2008 AMC 12A Problems/Problem 18"

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==Solution==
 
==Solution==
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{{image}}
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<center><asy>
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defaultpen(fontsize(8));
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draw((0,10)--(0,0)--(8,0));draw((-3,-4)--(0,0));draw((0,10)--(-3,-4)--(8,0)--cycle);
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label("A",(8,0),(1,0));label("B",(0,10),(0,1));label("C",(-3,-4),(-1,-1));label("O",(0,0),(1,1));
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label("$a$",(4,0),(0,1));label("$b$",(0,5),(1,0));label("$c$",(-1.5,-2),(1,0));
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label("$5$",(4,5),(1,1));label("$6$",(-1.5,3),(-1,0));label("$7$",(2.5,-2),(1,-1));
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</asy></center>
  
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Without loss of generality, let <math>A</math> be on the <math>x</math> axis, <math>B</math> be on the <math>y</math> axis, and <math>C</math> be on the <math>z</math> axis, and let <math>AB, BC, CA</math> have respective lenghts of 5, 6, and 7.  Let <math>a,b,c</math> denote the lengths of segments <math>OA,OB,OC,</math> respectively.  Then by the [[Pythagorean Theorem]],
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Without loss of generality, let <math>A</math> be on the <math>x</math> axis, <math>B</math> be on the <math>y</math> axis, and <math>C</math> be on the <math>z</math> axis, and let <math>AB, BC, CA</math> have respective lengths of 5, 6, and 7.  Let <math>a,b,c</math> denote the lengths of segments <math>OA,OB,OC,</math> respectively.  Then by the [[Pythagorean Theorem]],
 
<cmath> \begin{align*}
 
<cmath> \begin{align*}
 
a^2+b^2 &=5^2 , \\  
 
a^2+b^2 &=5^2 , \\  
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c^2+a^2 &=7^2 ,
 
c^2+a^2 &=7^2 ,
 
\end{align*} </cmath>
 
\end{align*} </cmath>
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so <math>a^2 = (5^2+7^2-6^2)/2 = 19</math>; similarly, <math>b^2 = 6</math> and <math>c^2 = 30</math>. Since <math>OA</math>, <math>OB</math>, and <math>OC</math> are mutually perpendicular, the tetrahedron's volume is
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so <math>a^2 = (5^2+7^2-6^2)/2 = 19</math>; similarly, <math>b^2 = 6</math> and <math>c^2 = 30</math>. Since <math>OA</math>, <math>OB</math>, and <math>OC</math> are mutually perpendicular, the tetrahedron's volume is <cmath> abc/6</cmath> because we can consider the tetrahedron to be a right triangular pyramid.
 
<cmath> abc/6 = \sqrt{a^2b^2c^2}/6 = \frac{\sqrt{19 \cdot 6 \cdot 30}}{6} = \sqrt{95}, </cmath>
 
<cmath> abc/6 = \sqrt{a^2b^2c^2}/6 = \frac{\sqrt{19 \cdot 6 \cdot 30}}{6} = \sqrt{95}, </cmath>
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which is answer choice C.  <math>\blacksquare</math>
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which is answer choice <math>\boxed{\text{C}}</math>.
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== Video Solution by OmegaLearn ==
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https://youtu.be/MOcX5BFbcwU?t=1320
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~ pi_is_3.14
  
 
== See also ==
 
== See also ==
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[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 03:51, 16 January 2023

Problem

Triangle $ABC$, with sides of length $5$, $6$, and $7$, has one vertex on the positive $x$-axis, one on the positive $y$-axis, and one on the positive $z$-axis. Let $O$ be the origin. What is the volume of tetrahedron $OABC$?

$\mathrm{(A)}\ \sqrt{85}\qquad\mathrm{(B)}\ \sqrt{90}\qquad\mathrm{(C)}\ \sqrt{95}\qquad\mathrm{(D)}\ 10\qquad\mathrm{(E)}\ \sqrt{105}$

Solution

[asy] defaultpen(fontsize(8)); draw((0,10)--(0,0)--(8,0));draw((-3,-4)--(0,0));draw((0,10)--(-3,-4)--(8,0)--cycle); label("A",(8,0),(1,0));label("B",(0,10),(0,1));label("C",(-3,-4),(-1,-1));label("O",(0,0),(1,1)); label("$a$",(4,0),(0,1));label("$b$",(0,5),(1,0));label("$c$",(-1.5,-2),(1,0)); label("$5$",(4,5),(1,1));label("$6$",(-1.5,3),(-1,0));label("$7$",(2.5,-2),(1,-1)); [/asy]

Without loss of generality, let $A$ be on the $x$ axis, $B$ be on the $y$ axis, and $C$ be on the $z$ axis, and let $AB, BC, CA$ have respective lengths of 5, 6, and 7. Let $a,b,c$ denote the lengths of segments $OA,OB,OC,$ respectively. Then by the Pythagorean Theorem, \begin{align*} a^2+b^2 &=5^2 , \\ b^2+c^2&=6^2, \\ c^2+a^2 &=7^2 , \end{align*} so $a^2 = (5^2+7^2-6^2)/2 = 19$; similarly, $b^2 = 6$ and $c^2 = 30$. Since $OA$, $OB$, and $OC$ are mutually perpendicular, the tetrahedron's volume is \[abc/6\] because we can consider the tetrahedron to be a right triangular pyramid. \[abc/6 = \sqrt{a^2b^2c^2}/6 = \frac{\sqrt{19 \cdot 6 \cdot 30}}{6} = \sqrt{95},\] which is answer choice $\boxed{\text{C}}$.

Video Solution by OmegaLearn

https://youtu.be/MOcX5BFbcwU?t=1320

~ pi_is_3.14

See also

2008 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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