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 1

Drop altitudes of triangle $ABC$ and triangle $ABD$ down from $C$ and $D$ , respectively. Both will hit the same point; let this point be $T$ . Because both triangle $ABC$ and triangle $ABD$ are 3-4-5 triangles, $CT = DT = \dfrac{3\cdot4}{5} = \dfrac{12}{5}$ . Because $CT^{2} + DT^{2} = 2\left(\frac{12}{5}\right)^{2} = \left(\frac{12}{5}\sqrt{2}\right)^{2} = CD^{2}$ , it follows that the $CTD$ is a right triangle, meaning that $\angle CTD = 90^\circ$ , and it follows that planes $ABC$ and $ABD$ are perpendicular to each other. Now, we can treat $ABC$ as the base of the tetrahedron and $TD$ as the height. Thus, the desired volume is $V = \dfrac{1}{3} bh = \dfrac{1}{3}\cdot[ABC]\cdot TD = \dfrac{1}{3} \cdot 6 \cdot \dfrac{12}{5} = \dfrac{24}{5}$ which is answer $\boxed{\textbf{(C) } \dfrac{24}{5}}$

Solution 2

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 $\boxed{\textbf{(C) } \dfrac{24}{5}}$

Solution 3

Similar to solution 1, $\triangle CTD$ is an isosceles right triangle. $AB$ is perpendicular to the plane $CTD$ . So, we can cut tetrahedron $ABCD$ into 2 tetrahedrons $ACTD$ and $BCTD$ with $\triangle CTD$ as their common base, $BT$ and $AT$ as their heights.

$[CTD]=\frac{1}{2} \cdot CT \cdot DT=\frac{1}{2} \cdot \frac{12}{5} \cdot \frac{12}{5}= \frac{72}{25}$

$V_{ABCD}=V_{ACDT}+V_{BCDT}=\frac{1}{3} \left([CTD] \cdot AT + [CTD] \cdot BT \right)=\frac{1}{3} \cdot [CTD] \cdot (AT+BT)=\frac{1}{3} \cdot [CTD] \cdot AB =\frac{1}{3} \cdot \frac{72}{25} \cdot 5$

$= \boxed{\textbf{(C) } \dfrac{24}{5}}$

~isabelchen

Solution 4 (Francesca's Irregular Tetrahedron Formula)

Note: Please don't ever try doing this in an actual competition. It's fun to do, however.

Using Piero della Francesca's Theorem: $V = \frac{1}{12} \left( \sqrt{-a^2b^2c^2 - a^2d^2e^2 - b^2d^2f^2 - c^2e^2f^2 + a^2c^2d^2 + b^2c^2d^2 + a^2b^2e^2 + b^2c^2e^2 + b^2d^2e^2 + c^2d^2e^2 + a^2b^2f^2 + a^2c^2f^2 + a^2d^2f^2 + c^2d^2f^2 + a^2e^2f^2 + b^2e^2f^2 - c^4d^2 - c^2d^4 - b^4e^2 - b^2e^4 - a^4f^2 - a^2f^4} \right)$

Substituting this all in... You get $\boxed{\textbf{(C) } \dfrac{24}{5}}$

~Banspeedrun

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

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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