Difference between revisions of "2021 AMC 10A Problems/Problem 13"

Revision as of 18:45, 28 August 2021

Problem

What is the volume of tetrahedron $ABCD$ with edge lengths $AB = 2$ , $AC = 3$ , $AD = 4$ , $BC = \sqrt{13}$ , $BD = 2\sqrt{5}$ , and $CD = 5$ ?

$\textbf{(A)} ~3 \qquad\textbf{(B)} ~2\sqrt{3} \qquad\textbf{(C)} ~4\qquad\textbf{(D)} ~3\sqrt{3}\qquad\textbf{(E)} ~6$

Solution 1 (Three Right Triangles)

Drawing the tetrahedron out and testing side lengths, we realize that the $\triangle ACD, \triangle ABC,$ and $\triangle ABD$ are right triangles by the Converse of the Pythagorean Theorem. It is now easy to calculate the volume of the tetrahedron using the formula for the volume of a pyramid. If we take $\triangle ADC$ as the base, then $\overline{AB}$ must be the altitude. The volume of tetrahedron $ABCD$ is $\dfrac{1}{3} \cdot \dfrac{3 \cdot 4}{2} \cdot 2=\boxed{\textbf{(C)} ~4}.$

~Icewolf10 ~Bakedpotato66 ~MRENTHUSIASM

Solution 2 (Bash: One Right Triangle)

We will place tetrahedron $ABCD$ in the $xyz$ -plane. By the Converse of the Pythagorean Theorem, we know that $\triangle ACD$ is a right triangle. Without the loss of generality, let $A=(0,0,0), C=(3,0,0), D=(0,4,0),$ and $B=(x,y,z).$

We apply the Distance Formula to $\overline{BA},\overline{BC},$ and $\overline{BD},$ respectively: $\begin{align*} x^2+y^2+z^2&=2^2, &(1) \\ (x-3)^2+y^2+z^2&=\sqrt{13}^2, &(2) \\ x^2+(y-4)^2+z^2&=\left(2\sqrt5\right)^2. &\hspace{1mm} (3) \end{align*}$ Subtracting $(1)$ from $(2)$ gives $-6x+9=9,$ from which $x=0.$

Subtracting $(1)$ from $(3)$ gives $-8y+16=16,$ from which $y=0.$

Substituting $(x,y)=(0,0)$ into $(1)$ produces $z^2=4,$ or $|z|=2.$

Let the brackets denote areas. Finally, we find the volume of tetrahedron $ABCD$ using $\triangle ACD$ as the base: $\begin{align*} V_{ABCD}&=\frac13\cdot[ACD]\cdot h_B \\ &=\frac13\cdot\left(\frac12\cdot AC \cdot AD\right)\cdot |z| \\ &=\boxed{\textbf{(C)} ~4}. \end{align*}$ ~MRENTHUSIASM

Solution 3 (Trirectangular Tetrahedron)

https://mathworld.wolfram.com/TrirectangularTetrahedron.html

Given the observations from Solution 1, where $\triangle ACD, \triangle ABC,$ and $\triangle ABD$ are right triangles, the base is $\triangle ABD.$ We can apply the information about a trirectangular tetrahedron (all of the face angles are right angles), which states that the volume is $\begin{align*} V&=\frac16\cdot AB\cdot AD\cdot BD \\ &=\frac16\cdot2\cdot4\cdot3 \\ &=\boxed{\textbf{(C)} ~4}. \end{align*}$ ~AMC60 (Solution)

~MRENTHUSIASM (Revision)

Solution 4

Notice that $\angle ADC,\angle DAB,$ and $\angle BAC$ are right angles. Let the base of the tetrahedron be $\triangle DAC$ , so the height is $AB$ . $A_{\triangle CAD}=\frac{4 \cdot 3}{2}=6$ Therefore, $\frac{1}{3} \cdot 6 \cdot 2=\boxed{4}$

~ kante314

Remark

Here is a similar problem from another AMC test: 2015 AMC 10A Problem 21.

Video Solution (Simple & Quick)

https://youtu.be/bRrchiDCrKE

~ Education, the Study of Everything

Video Solution (Using Pythagorean Theorem, 3D Geometry: Tetrahedron)

https://youtu.be/i4yUaXVUWKE

~ pi_is_3.14

Video Solution by TheBeautyofMath

https://youtu.be/t-EEP2V4nAE?t=813

~IceMatrix

@@ Line 44: / Line 44: @@
 ~MRENTHUSIASM (Revision)
+==Solution 4==
+Notice that <math>\angle ADC,\angle DAB,</math> and <math>\angle BAC</math> are right angles. Let the base of the tetrahedron be <math>\triangle DAC</math>, so the height is <math>AB</math>. <cmath>A_{\triangle CAD}=\frac{4 \cdot 3}{2}=6</cmath>Therefore, <cmath>\frac{1}{3} \cdot 6 \cdot 2=\boxed{4}</cmath>
+~ kante314
 ==Remark==

2021 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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