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

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<math>\textbf{(A)} ~3 \qquad\textbf{(B)} ~2\sqrt{3} \qquad\textbf{(C)} ~4\qquad\textbf{(D)} ~3\sqrt{3}\qquad\textbf{(E)} ~6</math>
 
<math>\textbf{(A)} ~3 \qquad\textbf{(B)} ~2\sqrt{3} \qquad\textbf{(C)} ~4\qquad\textbf{(D)} ~3\sqrt{3}\qquad\textbf{(E)} ~6</math>
  
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==Solution==
+
==Solution 1 (Three Right Triangles)==
−
Drawing the tetrahedron out and testing side lengths, we realize that the triangles ABD and ABC are right triangles. It is now easy to calculate the volume of the tetrahedron using the formula for the volume of a pyramid: <math>\frac{3\cdot4\cdot2}{3\cdot2}=4</math>, so we have an answer of <math>\boxed{C}</math>. ~IceWolf10
+
Drawing the tetrahedron out and testing side lengths, we realize that the <math>\triangle ACD, \triangle ABC,</math> and <math>\triangle ABD</math> 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 <math>\triangle ADC</math> as the base, then <math>\overline{AB}</math> must be the altitude. The volume of tetrahedron <math>ABCD</math> is <math>\dfrac{1}{3} \cdot \dfrac{3 \cdot 4}{2} \cdot 2=\boxed{\textbf{(C)} ~4}.</math>
  
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==Similar Problem==
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~Icewolf10 ~Bakedpotato66 ~MRENTHUSIASM
−
https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_21
 
  
−
== Video Solution (Using Pythagorean Theorem, 3D Geometry - Tetrahedron) ==
+
==Solution 2 (One Right Triangle)==
 +
We will place tetrahedron <math>ABCD</math> in the <math>xyz</math>-plane. By the Converse of the Pythagorean Theorem, we know that <math>\triangle ACD</math> is a right triangle. Without the loss of generality, let <math>A=(0,0,0), C=(3,0,0), D=(0,4,0),</math> and <math>B=(x,y,z).</math>
 +
 
 +
We apply the Distance Formula to <math>\overline{BA},\overline{BC},</math> and <math>\overline{BD},</math> respectively:
 +
<cmath>\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*}</cmath>
 +
Subtracting <math>(1)</math> from <math>(2)</math> gives <math>-6x+9=9,</math> from which <math>x=0.</math>
 +
 
 +
Subtracting <math>(1)</math> from <math>(3)</math> gives <math>-8y+16=16,</math> from which <math>y=0.</math>
 +
 
 +
Substituting <math>(x,y)=(0,0)</math> into <math>(1)</math> produces <math>z^2=4,</math> or <math>|z|=2.</math>
 +
 
 +
Let the brackets denote areas. Finally, we find the volume of tetrahedron <math>ABCD</math> using <math>\triangle ACD</math> as the base:
 +
<cmath>\begin{align*}
 +
V_{ABCD}&=\frac13\cdot[ACD]\cdot h_B \\
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&=\frac13\cdot\left(\frac12\cdot AC \cdot AD\right)\cdot |z| \\
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&=\boxed{\textbf{(C)} ~4}.
 +
\end{align*}</cmath>
 +
~MRENTHUSIASM
 +
 
 +
==Solution 3 (Trirectangular Tetrahedron)==
 +
https://mathworld.wolfram.com/TrirectangularTetrahedron.html
 +
 
 +
Given the observations from Solution 1, where <math>\triangle ACD, \triangle ABC,</math> and <math>\triangle ABD</math> are right triangles, the base is <math>\triangle ABD.</math> We can apply the information about a trirectangular tetrahedron (all of the face angles are right angles), which states that the volume is
 +
<cmath>\begin{align*}
 +
V&=\frac16\cdot AB\cdot AC\cdot BD \\
 +
&=\frac16\cdot2\cdot4\cdot3 \\
 +
&=\boxed{\textbf{(C)} ~4}.
 +
\end{align*}</cmath>
 +
~AMC60 (Solution)
 +
 
 +
~MRENTHUSIASM (Revision)
 +
 
 +
==Remark==
 +
Here is a similar problem from another AMC test: [[2015_AMC_10A_Problems/Problem_21|2015 AMC 10A Problem 21]].
 +
 
 +
==Video Solution (Simple & Quick)==
 +
https://youtu.be/bRrchiDCrKE
 +
 
 +
~ Education, the Study of Everything
 +
 
 +
== Video Solution by Omega Learn (Using Pythagorean Theorem, 3D Geometry: Tetrahedron) ==
 
https://youtu.be/i4yUaXVUWKE
 
https://youtu.be/i4yUaXVUWKE
  
 
~ pi_is_3.14
 
~ pi_is_3.14
  
−
==Video Solution (Simple and Quick)==
+
==Video Solution by TheBeautyofMath==
−
https://youtu.be/bRrchiDCrKE
+
https://youtu.be/t-EEP2V4nAE?t=813
  
−
~ Education, the Study of Everything
+
~IceMatrix
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2021|ab=A|num-b=12|num-a=14}}
 
{{AMC10 box|year=2021|ab=A|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 19:22, 10 November 2024

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 (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 AC\cdot BD \\ &=\frac16\cdot2\cdot4\cdot3 \\ &=\boxed{\textbf{(C)} ~4}. \end{align*} ~AMC60 (Solution)

~MRENTHUSIASM (Revision)

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 by Omega Learn (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

See also

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

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