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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==
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==Solution 1 (Three Right Triangles)==
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Drawing the tetrahedron out and testing side lengths, we realize that the <math>\triangle ABD, \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>AB</math> must be the height. <math>\dfrac{1}{3} \cdot \dfrac{3 \cdot 4}{2} \cdot 2</math>, so we have an answer of <math>\boxed{\textbf{(C) } 4}</math>.
+
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==
+
~Icewolf10 ~Bakedpotato66 ~MRENTHUSIASM
−
https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_21
+
 
 +
==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) \\
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(x-3)^2+y^2+z^2&=\sqrt{13}^2, &(2) \\
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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>
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~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 \\
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&=\frac16\cdot2\cdot4\cdot3 \\
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&=\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)==
 
==Video Solution (Simple & Quick)==
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~ Education, the Study of Everything
 
~ Education, the Study of Everything
  
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== Video Solution (Using Pythagorean Theorem, 3D Geometry - Tetrahedron) ==
+
== Video Solution by Omega Learn (Using Pythagorean Theorem, 3D Geometry: Tetrahedron) ==
 
https://youtu.be/i4yUaXVUWKE
 
https://youtu.be/i4yUaXVUWKE
  

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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