Difference between revisions of "2015 AMC 10A Problems/Problem 21"
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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 | + | 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 |
<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> | <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> |
Revision as of 20:49, 4 February 2015
Problem 21
Tetrahedron has
,
,
,
,
, and
. What is the volume of the tetrahedron?
Solution
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
.