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> | ||
− | ==Solution== | + | ==Solution 1 (Three Right Triangles)== |
− | Drawing the tetrahedron out and testing side lengths, we realize that the | + | 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> |
− | + | ~Icewolf10 ~Bakedpotato66 ~MRENTHUSIASM | |
− | |||
− | == Video Solution (Using Pythagorean Theorem, 3D Geometry | + | ==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 \\ | ||
+ | &=\frac13\cdot\left(\frac12\cdot AC \cdot AD\right)\cdot |z| \\ | ||
+ | &=\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 AD\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 | + | ==Video Solution by TheBeautyofMath== |
− | https://youtu.be/ | + | https://youtu.be/t-EEP2V4nAE?t=813 |
− | + | ~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 19:12, 28 October 2022
Contents
Problem
What is the volume of tetrahedron with edge lengths
,
,
,
,
, and
?
Solution 1 (Three Right Triangles)
Drawing the tetrahedron out and testing side lengths, we realize that the and
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
as the base, then
must be the altitude. The volume of tetrahedron
is
~Icewolf10 ~Bakedpotato66 ~MRENTHUSIASM
Solution 2 (One Right Triangle)
We will place tetrahedron in the
-plane. By the Converse of the Pythagorean Theorem, we know that
is a right triangle. Without the loss of generality, let
and
We apply the Distance Formula to and
respectively:
Subtracting
from
gives
from which
Subtracting from
gives
from which
Substituting into
produces
or
Let the brackets denote areas. Finally, we find the volume of tetrahedron using
as the base:
~MRENTHUSIASM
Solution 3 (Trirectangular Tetrahedron)
https://mathworld.wolfram.com/TrirectangularTetrahedron.html
Given the observations from Solution 1, where and
are right triangles, the base is
We can apply the information about a trirectangular tetrahedron (all of the face angles are right angles), which states that the volume is
~AMC60 (Solution)
~MRENTHUSIASM (Revision)
Remark
Here is a similar problem from another AMC test: 2015 AMC 10A Problem 21.
Video Solution (Simple & Quick)
~ Education, the Study of Everything
Video Solution by Omega Learn (Using Pythagorean Theorem, 3D Geometry: Tetrahedron)
~ 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.