Difference between revisions of "2008 AMC 12A Problems/Problem 18"
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==Problem== | ==Problem== | ||
+ | Triangle <math>ABC</math>, with sides of length <math>5</math>, <math>6</math>, and <math>7</math>, has one [[vertex]] on the positive <math>x</math>-axis, one on the positive <math>y</math>-axis, and one on the positive <math>z</math>-axis. Let <math>O</math> be the [[origin]]. What is the volume of [[tetrahedron]] <math>OABC</math>? | ||
− | + | <math>\mathrm{(A)}\ \sqrt{85}\qquad\mathrm{(B)}\ \sqrt{90}\qquad\mathrm{(C)}\ \sqrt{95}\qquad\mathrm{(D)}\ 10\qquad\mathrm{(E)}\ \sqrt{105}</math> | |
− | < | + | ==Solution== |
+ | <center><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></center> | ||
− | + | 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*} | |
+ | 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</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> | ||
+ | which is answer choice <math>\boxed{\text{C}}</math>. | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/MOcX5BFbcwU?t=1320 | ||
+ | ~ pi_is_3.14 | ||
− | {{AMC12 box|year=2008|num-b= | + | == See also == |
+ | {{AMC12 box|year=2008|num-b=17|num-a=19|ab=A}} | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 04:51, 16 January 2023
Problem
Triangle , with sides of length
,
, and
, has one vertex on the positive
-axis, one on the positive
-axis, and one on the positive
-axis. Let
be the origin. What is the volume of tetrahedron
?
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]](http://latex.artofproblemsolving.com/b/4/8/b48d9f2414f67d68d2e53886e0726e31626cd919.png)
Without loss of generality, let be on the
axis,
be on the
axis, and
be on the
axis, and let
have respective lengths of 5, 6, and 7. Let
denote the lengths of segments
respectively. Then by the Pythagorean Theorem,
so
; similarly,
and
. Since
,
, and
are mutually perpendicular, the tetrahedron's volume is
because we can consider the tetrahedron to be a right triangular pyramid.
which is answer choice
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.