Difference between revisions of "2017 AIME I Problems/Problem 4"
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A pyramid has a triangular base with side lengths <math>20</math>, <math>20</math>, and <math>24</math>. The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length <math>25</math>. The volume of the pyramid is <math>m\sqrt{n}</math>, where <math>m</math> and <math>n</math> are positive integers, and <math>n</math> is not divisible by the square of any prime. Find <math>m+n</math>. | A pyramid has a triangular base with side lengths <math>20</math>, <math>20</math>, and <math>24</math>. The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length <math>25</math>. The volume of the pyramid is <math>m\sqrt{n}</math>, where <math>m</math> and <math>n</math> are positive integers, and <math>n</math> is not divisible by the square of any prime. Find <math>m+n</math>. | ||
==Solution== | ==Solution== | ||
− | Let the triangular base be <math>\triangle ABC</math>. Using Simplified Heron's formula for the area of an isosceles triangle gives <math>12\sqrt{32(8)}=192</math>. | + | Let the triangular base be <math>\triangle ABC</math>, with <math>\overline {AB} = 24</math>. Using Simplified Heron's formula for the area of an isosceles triangle gives <math>12\sqrt{32(8)}=192</math>. |
− | <math>\ | + | Let the fourth vertex of the tetrahedron be <math>P</math>, and let the midpoint of <math>\overline {AB}</math> be <math>M</math>. Since <math>P</math> is equidistant from <math>A</math>, <math>B</math>, and <math>C</math>, the line through <math>P</math> perpendicular to the plane of <math>\triangle ABC</math> will pass through the circumcenter of <math>\triangle ABC</math>, which we will call <math>O</math>. Note that <math>O</math> is equidistant from each of <math>A</math>, <math>B</math>, and <math>C</math>. We find that <math>\overline {CM} = 16</math>. Then, |
− | + | <math>\overline {OM} + \overline {OC} = \overline {CM} = 16</math> | |
+ | |||
+ | <math>d + \sqrt {d^2 + 144} = 16</math> (1) | ||
+ | |||
+ | Squaring both sides, we have | ||
+ | |||
+ | <math>d^2 + 144 + 2d\sqrt {d^2+144} + d^2 = 256</math> | ||
+ | |||
+ | <math>2d^2 + 2d\sqrt {d^2+144} = 112</math> | ||
+ | |||
+ | <math>2d(d + \sqrt {d^2+144}) = 112</math> | ||
+ | |||
+ | Substituting with equation (1): | ||
+ | |||
+ | <math>2d(16) = 112</math> | ||
+ | |||
+ | <math>d = 7/2</math>. | ||
+ | |||
+ | We now find that <math>\sqrt{d^2 + 144} = 25/2</math>. | ||
+ | |||
+ | Let the distance <math>\overline {OP} = h</math>. Using the Pythagorean Theorem on triangle <math>AOP</math>, <math>BOP</math>, or <math>COP</math> (all three are congruent by SSS): | ||
+ | |||
+ | <math>25^2 = h^2 + (\sqrt {d^2 + 144})^2</math> | ||
+ | |||
+ | <math>625 = h^2 + 625/4</math> | ||
+ | |||
+ | <math>1875/4 = h^2</math> | ||
+ | |||
+ | <math>25\sqrt {3} / 2 = h</math>. | ||
+ | |||
+ | |||
+ | Finally, by the formula for volume of a pyramid, | ||
+ | |||
+ | <math>V = Bh/3</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 16:40, 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
.