Difference between revisions of "2017 AIME I Problems/Problem 4"
(→Solution) |
m (→Solution) |
||
Line 42: | Line 42: | ||
<math>V = (192)(25\sqrt{3}/2)/3</math>. This simplifies to <math>V = 800\sqrt {3}</math>, so <math>m+n = \boxed {803}</math>. | <math>V = (192)(25\sqrt{3}/2)/3</math>. This simplifies to <math>V = 800\sqrt {3}</math>, so <math>m+n = \boxed {803}</math>. | ||
− | |||
− |
Revision as of 17:05, 8 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,
(1)
Squaring both sides, we have
Substituting with equation (1):
.
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
.