Difference between revisions of "1986 AIME Problems/Problem 2"
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:<math>= (2\sqrt{42} + 8)(2\sqrt{42} - 8)</math> | :<math>= (2\sqrt{42} + 8)(2\sqrt{42} - 8)</math> | ||
:<math>= (2\sqrt{42})^2 - 8^2 =</math> <math>\boxed{104}</math> | :<math>= (2\sqrt{42})^2 - 8^2 =</math> <math>\boxed{104}</math> | ||
+ | |||
+ | == Solution 2 == | ||
+ | Notice that in a triangle with side lengths <math>2\sqrt5,2\sqrt6,</math> and <math>2\sqrt7</math>, by Heron's formula, the area is the square root of what we are looking for. | ||
+ | Let angle <math>\theta</math> be opposite the <math>2\sqrt7</math> side. By the Law of Cosines, | ||
+ | <cmath>\cos\theta=\frac{(2\sqrt5)^2+(2\sqrt{6})^2-(2\sqrt7)^2}{2\cdot 2\sqrt5\cdot2\sqrt6}=\frac{16}{8\sqrt{30}}=\sqrt{\frac{2}{15}}</cmath> | ||
+ | |||
+ | So <math>\sin\theta=\sqrt{1-\cos^2\theta}=\sqrt{\frac{13}{15}}</math>. The area of the triangle is then | ||
+ | <cmath>\frac{\sin\theta}{2}\cdot 2\sqrt5\cdot 2\sqrt6=\sqrt{\frac{26}{30}}\cdot 2\sqrt{30}=\sqrt{104}</cmath> | ||
+ | |||
+ | So our answer is <math>\left(\sqrt{104}\right)^2=\boxed{104}</math> | ||
== See also == | == See also == |
Latest revision as of 12:54, 17 July 2018
Contents
Problem
Evaluate the product .
Solution
Simplify by repeated application of the difference of squares.
Solution 2
Notice that in a triangle with side lengths and , by Heron's formula, the area is the square root of what we are looking for. Let angle be opposite the side. By the Law of Cosines,
So . The area of the triangle is then
So our answer is
See also
1986 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
- AIME Problems and Solutions
- American Invitational Mathematics Examination
- Mathematics competition resources
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.