Difference between revisions of "2017 AIME I Problems/Problem 4"
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<cmath>V = (192)(25\sqrt{3}/2)/3</cmath> | <cmath>V = (192)(25\sqrt{3}/2)/3</cmath> | ||
This simplifies to <math>V = 800\sqrt {3}</math>, so <math>m+n = \boxed {803}</math>. | This simplifies to <math>V = 800\sqrt {3}</math>, so <math>m+n = \boxed {803}</math>. | ||
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Solution by Zeroman | Solution by Zeroman | ||
Revision as of 16:48, 9 March 2017
Problem 4
A pyramid has a triangular base with side lengths ,
, and
. The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length
. The volume of the pyramid is
, where
and
are positive integers, and
is not divisible by the square of any prime. Find
.
Solution
Let the triangular base be , with
. Using Simplified Heron's formula for the area of an isosceles triangle gives
.
Let the fourth vertex of the tetrahedron be , and let the midpoint of
be
. Since
is equidistant from
,
, and
, the line through
perpendicular to the plane of
will pass through the circumcenter of
, which we will call
. Note that
is equidistant from each of
,
, and
. We find that
. Then,
Let .
Equation
:
Squaring both sides, we have
Substituting with equation :
We now find that .
Let the distance . Using the Pythagorean Theorem on triangle
,
, or
(all three are congruent by SSS):
Finally, by the formula for volume of a pyramid,
This simplifies to
, so
.
Solution by Zeroman
See Also
2017 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.