Difference between revisions of "2015 AMC 10A Problems/Problem 21"
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==Solution 2== | ==Solution 2== | ||
− | Drop altitudes of triangle <math>ABC</math> and triangle <math>ABD</math> down from <math>C</math> and <math>D</math>, respectively. Both will hit the same point; let this point be <math>T</math>. Because both triangle <math>ABC</math> and triangle <math>ABD</math> are 3-4-5 triangles, <math>CT = DT = \dfrac{3\cdot4}{5} = \dfrac{12}{5}</math>. Because <math>CT^{2} + DT^{2} = 2\left(\frac{12}{5}\right)^{2} = \left(\frac{12}{5}\sqrt{2}\right)^{2} = CD^{2}</math>, it follows that the <math>CTD</math> is a right triangle, | + | Drop altitudes of triangle <math>ABC</math> and triangle <math>ABD</math> down from <math>C</math> and <math>D</math>, respectively. Both will hit the same point; let this point be <math>T</math>. Because both triangle <math>ABC</math> and triangle <math>ABD</math> are 3-4-5 triangles, <math>CT = DT = \dfrac{3\cdot4}{5} = \dfrac{12}{5}</math>. Because <math>CT^{2} + DT^{2} = 2\left(\frac{12}{5}\right)^{2} = \left(\frac{12}{5}\sqrt{2}\right)^{2} = CD^{2}</math>, it follows that the <math>CTD</math> is a right triangle, meaning that <math>\angle CTD = 90^\circ</math>, and it follows that planes <math>ABC</math> and <math>ABD</math> are perpendicular to each other. Now, we can treat <math>ABC</math> as the base of the tetrahedron and <math>TD</math> as the height. Thus, the desired volume is <cmath>V = \dfrac{1}{3} Bh = \dfrac{1}{3}\cdot[ABC]\cdot TD = \dfrac{1}{3} \cdot 6 \cdot \dfrac{12}{5} = \dfrac{24}{5}</cmath> which is answer <math>\boxed{\textbf{(C) } \dfrac{24}{5}}</math> |
== See Also == | == See Also == |
Revision as of 17:17, 25 January 2016
- The following problem is from both the 2015 AMC 12A #16 and 2015 AMC 10A #21, so both problems redirect to this page.
Contents
[hide]Problem
Tetrahedron has
,
,
,
,
, and
. What is the volume of the tetrahedron?
Solution 1
Let the midpoint of be
. We have
, and so by the Pythagorean Theorem
and
. Because the altitude from
of tetrahedron
passes touches plane
on
, it is also an altitude of triangle
. The area
of triangle
is, by Heron's Formula, given by
Substituting
and performing huge (but manageable) computations yield
, so
. Thus, if
is the length of the altitude from
of the tetrahedron,
. Our answer is thus
and so our answer is
Solution 2
Drop altitudes of triangle and triangle
down from
and
, respectively. Both will hit the same point; let this point be
. Because both triangle
and triangle
are 3-4-5 triangles,
. Because
, it follows that the
is a right triangle, meaning that
, and it follows that planes
and
are perpendicular to each other. Now, we can treat
as the base of the tetrahedron and
as the height. Thus, the desired volume is
which is answer
See Also
2015 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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 10 Problems and Solutions |
2015 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
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.