2015 AMC 10A Problems/Problem 21
- The following problem is from both the 2015 AMC 12A #16 and 2015 AMC 10A #21, so both problems redirect to this page.
Contents
Problem
Tetrahedron has , , , , , and . What is the volume of the tetrahedron?
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
Similar to solution 1, is an isosceles right triangle. is perpendicular to the plane . So, we can cut tetrahedron into 2 tetrahedrons and with as their common base, and as their heights.
Video Solution by Richard Rusczyk
https://www.youtube.com/watch?v=1J_P0tXszLQ
Video Solution by TheBeautyofMath
https://www.youtube.com/watch?v=ckRtrNuNgk4
~IceMatrix
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 |
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