Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(fix)
(improved writeup)
Line 6: Line 6:
  
 
==Solution==
 
==Solution==
−
[[WLOG]], we let <math>AB</math> go between the <math>x</math> and <math>y</math> axes, <math>BC</math> between <math>y</math> and <math>z</math> axes, <math>CA</math> between <math>z</math> and <math>x</math> axes. Let <math>x,y,z</math> be <math>OA,OB,OC,</math> respectively. By the [[Pythagorean Theorem]], <math>x^2+y^2=25</math>, <math>y^2+z^2=36</math>, <math>z^2+x^2=49</math>. Thus, <math>x^2 = 30</math>, <math>y^2 = 19</math>, and <math>z^2 = 6</math>. Thus the volume of the tetraehdron is <math>\frac{\sqrt{30\cdot 19\cdot 6}}{6}=\sqrt{95}\Rightarrow \boxed{C}</math>.
+
 
 +
{{image}}
 +
 
 +
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]],
 +
<cmath> \begin{align*}
 +
a^2+b^2 &=5^2 , \
 +
b^2+c^2&=6^2, \
 +
c^2+a^2 &=7^2 ,
 +
\end{align*} </cmath>
 +
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 = \sqrt{a^2b^2c^2}/6 = \frac{\sqrt{19 \cdot 6 \cdot 30}}{6} = \sqrt{95}, </cmath>
 +
which is answer choice C.  <math>\blacksquare</math>
 +
 
 +
 
 +
{{alternate solutions}}
  
  

Revision as of 23:45, 18 February 2008

Problem

A triangle $\triangle ABC$ with sides $5$, $6$, $7$ is placed in the three-dimensional plane with 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 if $OABC$?

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

Solution

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

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 lenghts 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 \[abc/6 = \sqrt{a^2b^2c^2}/6 = \frac{\sqrt{19 \cdot 6 \cdot 30}}{6} = \sqrt{95},\] which is answer choice C. $\blacksquare$


Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.


2008 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12A_Problems/Problem_18&oldid=23460"