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.

Problem

Tetrahedron $ABCD$ has $AB=5$ , $AC=3$ , $BC=4$ , $BD=4$ , $AD=3$ , and $CD=\tfrac{12}5\sqrt2$ . What is the volume of the tetrahedron?

$\textbf{(A) }3\sqrt2\qquad\textbf{(B) }2\sqrt5\qquad\textbf{(C) }\dfrac{24}5\qquad\textbf{(D) }3\sqrt3\qquad\textbf{(E) }\dfrac{24}5\sqrt2$

Solution

Let the midpoint of $CD$ be $E$ . We have $CE = \dfrac{6}{5} \sqrt{2}$ , and so by the Pythagorean Theorem $AE = \dfrac{\sqrt{153}}{5}$ and $BE = \dfrac{\sqrt{328}}{5}$ . Because the altitude from $A$ of tetrahedron $ABCD$ passes touches plane $BCD$ on $BE$ , it is also an altitude of triangle $ABE$ . The area $A$ of triangle $ABE$ is, by Heron's Formula, given by

$16A^2 = 2a^2 b^2 + 2b^2 c^2 + 2c^2 a^2 - a^4 - b^4 - c^4 = -(a^2 + b^2 - c^2)^2 + 4a^2 b^2.$ Substituting $a = AE, b = BE, c = 5$ and performing huge (but manageable) computations yield $A^2 = 18$ , so $A = 3\sqrt{2}$ . Thus, if $h$ is the length of the altitude from $A$ of the tetrahedron, $BE \cdot h = 2A = 6\sqrt{2}$ . Our answer is thus $V = \dfrac{1}{3} Bh = \dfrac{1}{3} h \cdot BE \cdot \dfrac{6\sqrt{2}}{5} = \dfrac{24}{5},$ and so our answer is $\textbf{(C)}$ .

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

Page

Toolbox

Search

2015 AMC 10A Problems/Problem 21

Problem

Solution

See Also