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

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Problem # 9 + 10 :D.
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==Problem 21==
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Tetrahedron <math>ABCD</math> has <math>AB=5</math>, <math>AC=3</math>, <math>BC=4</math>, <math>BD=4</math>, <math>AD=3</math>, and <math>CD=\tfrac{12}5\sqrt2</math>.  What is the volume of the tetrahedron?
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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>
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==Solution==
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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 of triangle <math>ABE</math> is, by Heron's Formula, given by
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<cmath>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.</cmath>
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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
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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>
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and so our answer is <math>\textbf{(C)}</math>.

Revision as of 20:49, 4 February 2015

Problem 21

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

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 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 $\textbf{(C)}$.

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