Difference between revisions of "2017 AIME I Problems/Problem 4"
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==Problem 4== | ==Problem 4== | ||
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>. | ||
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==Solution== | ==Solution== | ||
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>. | 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>. | ||
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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, | 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, | ||
− | < | + | <cmath>\overline {OM} + \overline {OC} = \overline {CM} = 16</cmath> |
− | <math>d + \sqrt {d^2 + 144} = 16</ | + | Equation <math>(1)</math>: |
+ | <cmath>d + \sqrt {d^2 + 144} = 16</cmath> | ||
Squaring both sides, we have | Squaring both sides, we have | ||
− | < | + | <cmath>d^2 + 144 + 2d\sqrt {d^2+144} + d^2 = 256</cmath> |
− | < | + | <cmath>2d^2 + 2d\sqrt {d^2+144} = 112</cmath> |
− | < | + | <cmath>2d(d + \sqrt {d^2+144}) = 112</cmath> |
− | Substituting with equation (1): | + | Substituting with equation <math>(1)</math>: |
− | < | + | <cmath>2d(16) = 112</cmath> |
− | < | + | <cmath>d = 7/2</cmath> |
We now find that <math>\sqrt{d^2 + 144} = 25/2</math>. | We now find that <math>\sqrt{d^2 + 144} = 25/2</math>. | ||
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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): | 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): | ||
− | < | + | <cmath>25^2 = h^2 + (\sqrt {d^2 + 144})^2</cmath> |
− | < | + | <cmath>625 = h^2 + 625/4</cmath> |
− | < | + | <cmath>1875/4 = h^2</cmath> |
− | < | + | <cmath>25\sqrt {3} / 2 = h</cmath> |
Finally, by the formula for volume of a pyramid, | Finally, by the formula for volume of a pyramid, | ||
− | < | + | <cmath>V = Bh/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>. | ||
− | + | ==See Also== | |
+ | {{AIME box|year=2017|n=I|num-b=3|num-a=5}} | ||
+ | {{MAA Notice}} |
Revision as of 19:20, 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,
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 .
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 |
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