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

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<math>\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</math>
 
<math>\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</math>
  
==Solution 1==
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==Solutions==
 +
===Solution 1===
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Drop altitudes of triangle <math>ABC</math> and triangle <math>ABD</math> down from <math>C</math> and <math>D</math>, respectively. Both will hit the same point; let this point be <math>T</math>. Because both triangle <math>ABC</math> and triangle <math>ABD</math> are 3-4-5 triangles, <math>CT = DT = \dfrac{3\cdot4}{5} = \dfrac{12}{5}</math>. Because <math>CT^{2} + DT^{2} = 2\left(\frac{12}{5}\right)^{2} = \left(\frac{12}{5}\sqrt{2}\right)^{2} = CD^{2}</math>, it follows that the <math>CTD</math> is a right triangle, meaning that <math>\angle CTD = 90^\circ</math>, and it follows that planes <math>ABC</math> and <math>ABD</math> are perpendicular to each other. Now, we can treat <math>ABC</math> as the base of the tetrahedron and <math>TD</math> as the height. Thus, the desired volume is <cmath>V = \dfrac{1}{3} Bh = \dfrac{1}{3}\cdot[ABC]\cdot TD = \dfrac{1}{3} \cdot 6 \cdot \dfrac{12}{5} = \dfrac{24}{5}</cmath> which is answer <math>\boxed{\textbf{(C) } \dfrac{24}{5}}</math>
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===Solution 2===
 
Let the midpoint of <math>CD</math> be <math>E</math>. We have <math>CE = \dfrac{6}{5} \sqrt{2}</math>, and so by the Pythagorean Theorem <math>AE = \dfrac{\sqrt{153}}{5}</math> and <math>BE = \dfrac{\sqrt{328}}{5}</math>. Because the altitude from <math>A</math> of tetrahedron <math>ABCD</math> passes touches plane <math>BCD</math> on <math>BE</math>, it is also an altitude of triangle <math>ABE</math>. The area <math>A</math> of triangle <math>ABE</math> is, by Heron's Formula, given by
 
Let the midpoint of <math>CD</math> be <math>E</math>. We have <math>CE = \dfrac{6}{5} \sqrt{2}</math>, and so by the Pythagorean Theorem <math>AE = \dfrac{\sqrt{153}}{5}</math> and <math>BE = \dfrac{\sqrt{328}}{5}</math>. Because the altitude from <math>A</math> of tetrahedron <math>ABCD</math> passes touches plane <math>BCD</math> on <math>BE</math>, it is also an altitude of triangle <math>ABE</math>. The area <math>A</math> of triangle <math>ABE</math> is, by Heron's Formula, given by
  
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<cmath>V = \dfrac{1}{3} Bh = \dfrac{1}{3} h \cdot BE \cdot \dfrac{6\sqrt{2}}{5} = \dfrac{24}{5},</cmath>
 
<cmath>V = \dfrac{1}{3} Bh = \dfrac{1}{3} h \cdot BE \cdot \dfrac{6\sqrt{2}}{5} = \dfrac{24}{5},</cmath>
 
and so our answer is <math>\boxed{\textbf{(C) } \dfrac{24}{5}}</math>
 
and so our answer is <math>\boxed{\textbf{(C) } \dfrac{24}{5}}</math>
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===Solution 3 (When you don't have time)===
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Note that the altitude and side <math>CD</math> form a right triangle with <math>CD</math> as the hypotenuse. We guess the altitude is then <math>\dfrac{12}{5}</math> as it may be a 45-45-90 triangle. Then we find the volume of the tetrahedron using (1/3)(base area)(altitude) = (1/3)(6)(12/5) = <math>\dfrac{24}{5}</math> and hence our answer is <math>\boxed{\textbf{(C) } \dfrac{24}{5}}</math>  -srisainandan6
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=== Video Solution by Richard Rusczyk ===
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 +
https://www.youtube.com/watch?v=1J_P0tXszLQ
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 +
Similar to Solution 1. Using a similar idea, we find that the altitude of the tetrahedron to face <math>ABC</math> lies along ADB so the height of the tetrahedron is just the altitude to <math>AB</math> from <math>D</math>. And that's <math>\dfrac{12}{5}</math>.
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So the total area is <math>\dfrac{1}{3} \cdot \dfrac{(3)(4)}{2} \cdot \dfrac{12}{5}=\boxed{\dfrac{24}{5}}</math>
  
  
==Solution 2==
+
More info in the video and Solution 1
Drop altitudes of triangle <math>ABC</math> and triangle <math>ABD</math> down from <math>C</math> and <math>D</math>, respectively. Both will hit the same point; let this point be <math>T</math>. Because both triangle <math>ABC</math> and triangle <math>ABD</math> are 3-4-5 triangles, <math>CT = DT = \dfrac{3\cdot4}{5} = \dfrac{12}{5}</math>. Because <math>CT^{2} + DT^{2} = 2\left(\frac{12}{5}\right)^{2} = \left(\frac{12}{5}\sqrt{2}\right)^{2} = CD^{2}</math>, it follows that the <math>CTD</math> is a right triangle, which means <math>\angle CTD = \dfrac{\pi}{2}</math>, which means that planes <math>ABC</math> and <math>ABD</math> are perpendicular to each other. Now, we can treat <math>ABC</math> as the base of the tetrahedron and <math>TD</math> as the height. Thus, the desired volume is <cmath>V = \dfrac{1}{3} Bh = \dfrac{1}{3}\cdot[ABC]\cdot TD = \dfrac{1}{3} \cdot 6 \cdot \dfrac{12}{5} = \dfrac{24}{5}</cmath> which is answer <math>\boxed{\textbf{(C) } \dfrac{24}{5}}</math>
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~BakedPotato66
  
 
== See Also ==
 
== See Also ==
 +
 
{{AMC10 box|year=2015|ab=A|num-b=20|num-a=22}}
 
{{AMC10 box|year=2015|ab=A|num-b=20|num-a=22}}
 
{{AMC12 box|year=2015|ab=A|num-b=15|num-a=17}}
 
{{AMC12 box|year=2015|ab=A|num-b=15|num-a=17}}
  
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
 +
[[Category: Introductory Geometry Problems]]

Revision as of 16:05, 7 November 2020

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?

$\textbf{(A) }3\sqrt2\qquad\textbf{(B) }2\sqrt5\qquad\textbf{(C) }\dfrac{24}5\qquad\textbf{(D) }3\sqrt3\qquad\textbf{(E) }\dfrac{24}5\sqrt2$

Solutions

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 (When you don't have time)

Note that the altitude and side $CD$ form a right triangle with $CD$ as the hypotenuse. We guess the altitude is then $\dfrac{12}{5}$ as it may be a 45-45-90 triangle. Then we find the volume of the tetrahedron using (1/3)(base area)(altitude) = (1/3)(6)(12/5) = $\dfrac{24}{5}$ and hence our answer is $\boxed{\textbf{(C) } \dfrac{24}{5}}$ -srisainandan6

Video Solution by Richard Rusczyk

https://www.youtube.com/watch?v=1J_P0tXszLQ

Similar to Solution 1. Using a similar idea, we find that the altitude of the tetrahedron to face $ABC$ lies along ADB so the height of the tetrahedron is just the altitude to $AB$ from $D$. And that's $\dfrac{12}{5}$. So the total area is $\dfrac{1}{3} \cdot \dfrac{(3)(4)}{2} \cdot \dfrac{12}{5}=\boxed{\dfrac{24}{5}}$


More info in the video and Solution 1 ~BakedPotato66

See Also

2015 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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