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Difference between revisions of "2015 AMC 10A Problems/Problem 21"

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==Solution 4 (Francesca's Irregular Tetrahedron Formula)==
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Note: This formula is very long and time consuming, but if you really wanted to do this, it's fun i guess
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Using Piero della Francesca's Theorem:
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<math>V = \frac{1}{12}(-\sqrt[2]{-a^2*b^2*c^2-a^2*d^2*e^2-b^2*d^2*f^2-c^2*e^2*f^2+a^2*c^2*d^2+b^2*c^2*d^2+a^2*b^2*e^2+b^2*c^2*e^2+b^2*d^2*e^2+c^2*d^2*e^2+a^2*b^2*f^2+a^2*c^2*f^2+a^2*d^2*f^2+c^2*d^2*f^2+a^2*e^2*f^2+b^2*e^2*f^2-c^4*d^2-c^2*d^4-b^4*e^2-b^2*e^4-a^4*f^2-a^2*f^4})
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Substituting this all in... You get </math>= \boxed{\textbf{(C) } \dfrac{24}{5}}$
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~Banspeedrun
  
 
== Video Solution by Richard Rusczyk ==
 
== Video Solution by Richard Rusczyk ==

Revision as of 16:12, 17 June 2024

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: This formula is very long and time consuming, but if you really wanted to do this, it's fun i guess

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

Substituting this all in... You get$ (Error compiling LaTeX. Unknown error_msg)= \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

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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