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

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==Solution 1==
 
==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>
 +
 +
==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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Substituting <math>a = AE, b = BE, c = 5</math> and performing huge (but manageable) computations yield <math>A^2 = 18</math>, so <math>A = 3\sqrt{2}</math>. Thus, if <math>h</math> is the length of the altitude from <math>A</math> of the tetrahedron, <math>BE \cdot h = 2A = 6\sqrt{2}</math>. Our answer is thus
 
Substituting <math>a = AE, b = BE, c = 5</math> and performing huge (but manageable) computations yield <math>A^2 = 18</math>, so <math>A = 3\sqrt{2}</math>. Thus, if <math>h</math> is the length of the altitude from <math>A</math> of the tetrahedron, <math>BE \cdot h = 2A = 6\sqrt{2}</math>. Our answer is thus
 
<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>\textbf{(C)}</math>.
+
and so our answer is <math>\boxed{\textbf{(C) } \dfrac{24}{5}}</math>
 +
 
 +
==Solution 3==
 +
 
 +
Similar to solution 1, <math>\triangle CTD</math> is an isosceles right triangle. <math>AB</math> is perpendicular to the plane <math>CTD</math>. So, we can cut tetrahedron <math>ABCD</math> into 2 tetrahedrons <math>ACTD</math> and <math>BCTD</math> with <math>\triangle CTD</math> as their common base, <math>BT</math> and <math>AT</math> as their heights.
 +
 
 +
 
 +
<math>[CTD]=\frac{1}{2} \cdot CT \cdot DT=\frac{1}{2} \cdot \frac{12}{5} \cdot \frac{12}{5}= \frac{72}{25}</math>
  
 +
<math>V_{ABCD}=V_{ACDT}+V_{BCDT}=\frac{1}{3} \left([CTD] \cdot AT + [CTD] \cdot BT \right)=\frac{1}{3} \cdot [CTD] \cdot (AT+BT)=\frac{1}{3} \cdot [CTD] \cdot AB  =\frac{1}{3} \cdot \frac{72}{25} \cdot 5 </math>
  
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==Solution 2==
+
<math>= \boxed{\textbf{(C) } \dfrac{24}{5}}</math>
−
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, which we'll call <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>\angle CTD = \dfrac{\pi}{2}</math>, which means that planes <math>ABC</math> and <math>ABD</math> are perpendicular to each other. It follows that we can treat <math>ABC</math> as the base, and <math>CT</math> as the height. Thus the desired volume is <cmath>V = \dfrac{1}{3} Bh = \dfrac{1}{3}\cdot[ABC]\cdot CT = \dfrac{1}{3} \cdot 6 \cdot \dfrac{12}{5} = \dfrac{24}{5}</cmath> which is answer <math>\boxed{\textrm{C}}</math>.
+
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
 +
 
 +
 
 +
==Solution 4 (Francesca's Irregular Tetrahedron Formula)==
 +
Note: Please don't ever try doing this in an actual competition. It's fun to do, however.
 +
 
 +
Using Piero della Francesca's Theorem:
 +
<math>V = \frac{1}{12} \left( \sqrt{-a^2b^2c^2 - a^2d^2e^2 - b^2d^2f^2 - c^2e^2f^2 + a^2c^2d^2 + b^2c^2d^2 + a^2b^2e^2 + b^2c^2e^2 + b^2d^2e^2 + c^2d^2e^2 + a^2b^2f^2 + a^2c^2f^2 + a^2d^2f^2 + c^2d^2f^2 + a^2e^2f^2 + b^2e^2f^2 - c^4d^2 - c^2d^4 - b^4e^2 - b^2e^4 - a^4f^2 - a^2f^4} \right)</math>
 +
 
 +
Substituting this all in... You get
 +
<math>\boxed{\textbf{(C) } \dfrac{24}{5}}</math>
 +
 
 +
~Banspeedrun
 +
 
 +
== Video Solution by Richard Rusczyk ==
 +
 
 +
https://www.youtube.com/watch?v=1J_P0tXszLQ
 +
 
 +
==Video Solution by TheBeautyofMath==
 +
https://www.youtube.com/watch?v=ckRtrNuNgk4
 +
 
 +
~IceMatrix
  
 
== 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]]

Latest revision as of 19:16, 18 June 2024

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?

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

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

Similar to solution 1, $\triangle CTD$ is an isosceles right triangle. $AB$ is perpendicular to the plane $CTD$. So, we can cut tetrahedron $ABCD$ into 2 tetrahedrons $ACTD$ and $BCTD$ with $\triangle CTD$ as their common base, $BT$ and $AT$ as their heights.


$[CTD]=\frac{1}{2} \cdot CT \cdot DT=\frac{1}{2} \cdot \frac{12}{5} \cdot \frac{12}{5}= \frac{72}{25}$

$V_{ABCD}=V_{ACDT}+V_{BCDT}=\frac{1}{3} \left([CTD] \cdot AT + [CTD] \cdot BT \right)=\frac{1}{3} \cdot [CTD] \cdot (AT+BT)=\frac{1}{3} \cdot [CTD] \cdot AB =\frac{1}{3} \cdot \frac{72}{25} \cdot 5$

$= \boxed{\textbf{(C) } \dfrac{24}{5}}$

~isabelchen


Solution 4 (Francesca's Irregular Tetrahedron Formula)

Note: Please don't ever try doing this in an actual competition. It's fun to do, however.

Using Piero della Francesca's Theorem: $V = \frac{1}{12} \left( \sqrt{-a^2b^2c^2 - a^2d^2e^2 - b^2d^2f^2 - c^2e^2f^2 + a^2c^2d^2 + b^2c^2d^2 + a^2b^2e^2 + b^2c^2e^2 + b^2d^2e^2 + c^2d^2e^2 + a^2b^2f^2 + a^2c^2f^2 + a^2d^2f^2 + c^2d^2f^2 + a^2e^2f^2 + b^2e^2f^2 - c^4d^2 - c^2d^4 - b^4e^2 - b^2e^4 - a^4f^2 - a^2f^4} \right)$

Substituting this all in... You get $\boxed{\textbf{(C) } \dfrac{24}{5}}$

~Banspeedrun

Video Solution by Richard Rusczyk

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

Video Solution by TheBeautyofMath

https://www.youtube.com/watch?v=ckRtrNuNgk4

~IceMatrix

See Also

2015 AMC 10A (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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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