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

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

@@ Line 5: / Line 5: @@
 <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==
+==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, 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
@@ Line 11: / Line 14: @@
 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>
-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>
+<math>= \boxed{\textbf{(C) } \dfrac{24}{5}}</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 ==
 {{AMC10 box|year=2015|ab=A|num-b=20|num-a=22}}
 {{AMC12 box|year=2015|ab=A|num-b=15|num-a=17}}
 {{MAA Notice}}
+[[Category: Introductory Geometry Problems]]

2015 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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All AMC 10 Problems and Solutions

2015 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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