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== | + | ==Solution 1== |
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>\textbf{(C)}</math>. | and so our answer is <math>\textbf{(C)}</math>. | ||
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+ | ==Solution 2== | ||
+ | Drop altitudes down from <math>C</math> and <math>D</math>, which will both hit a point 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} = 12/5</math>. Because <math>CT^{2} + DT^{2} = (\frac{12}{5})^{2}\cdot2 = (\frac{12}{5}\cdot\sqrt{2})^{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>. | ||
== See Also == | == See Also == |
Revision as of 00:18, 5 February 2015
- The following problem is from both the 2015 AMC 12A #16 and 2015 AMC 10A #21, so both problems redirect to this page.
Contents
Problem
Tetrahedron has
,
,
,
,
, and
. What is the volume of the tetrahedron?
Solution 1
Let the midpoint of be
. We have
, and so by the Pythagorean Theorem
and
. Because the altitude from
of tetrahedron
passes touches plane
on
, it is also an altitude of triangle
. The area
of triangle
is, by Heron's Formula, given by
Substituting
and performing huge (but manageable) computations yield
, so
. Thus, if
is the length of the altitude from
of the tetrahedron,
. Our answer is thus
and so our answer is
.
Solution 2
Drop altitudes down from and
, which will both hit a point we'll call
. Because both triangle
and triangle
are 3-4-5 triangles,
. Because
, it follows that the
, which means that planes
and
are perpendicular to each other. It follows that we can treat
as the base, and
as the height. Thus the desired volume is
which is answer
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.