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Revision as of 02:10, 7 May 2020
- 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?
Solutions
Solution 1
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
Solution 2
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 3(When you don't have time)
Note that the altitude and side form a right triangle with
as the hypotenuse. We guess the altitude is then
as it may be a 45-45-90 triangle. Then we find the volume of the tetrahedron using (1/3)(base area)(altitude) = (1/3)(6)(12/5) =
and hence our answer is
-srisainandan6
See Also
Video Solution: https://www.youtube.com/watch?v=1J_P0tXszLQ
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 |
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