# 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 .