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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>
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== Solution 2 ==
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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.
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Let angle <math>\theta</math> be opposite the <math>2\sqrt7</math> side. By the Law of Cosines,
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<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>
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So <math>\sin\theta=\sqrt{1-\cos^2\theta}=\sqrt{\frac{13}{15}}</math>. The area of the triangle is then
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<cmath>\frac{\sin\theta}{2}\cdot 2\sqrt5\cdot 2\sqrt6=\sqrt{\frac{26}{30}}\cdot 2\sqrt{30}=\sqrt{104}</cmath>
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So our answer is <math>\left(\sqrt{104}\right)^2=\boxed{104}</math>
  
 
== See also ==
 
== See also ==

Revision as of 11:54, 17 July 2018

Problem

Evaluate the product $(\sqrt 5+\sqrt6+\sqrt7)(-\sqrt 5+\sqrt6+\sqrt7)(\sqrt 5-\sqrt6+\sqrt7)(\sqrt 5+\sqrt6-\sqrt7)$.

Solution

Simplify by repeated application of the difference of squares.

$\left((\sqrt{6} + \sqrt{7})^2 - \sqrt{5}^2\right)\left(\sqrt{5}^2 - (\sqrt{6} - \sqrt{7})^2\right)$
$= (13 + 2\sqrt{42} - 5)(5 - (13 - 2\sqrt{42}))$
$= (2\sqrt{42} + 8)(2\sqrt{42} - 8)$
$= (2\sqrt{42})^2 - 8^2 =$ $\boxed{104}$

Solution 2

Notice that in a triangle with side lengths $2\sqrt5,2\sqrt6,$ and $2\sqrt7$, by Heron's formula, the area is the square root of what we are looking for. Let angle $\theta$ be opposite the $2\sqrt7$ side. By the Law of Cosines, \[\cos\theta=\frac{(2\sqrt5)^2+(2\sqrt{6})^2-(2\sqrt7)^2}{2\cdot 2\sqrt5\cdot2\sqrt6}=\frac{16}{8\sqrt{30}}=\sqrt{\frac{2}{15}}\]

So $\sin\theta=\sqrt{1-\cos^2\theta}=\sqrt{\frac{13}{15}}$. The area of the triangle is then \[\frac{\sin\theta}{2}\cdot 2\sqrt5\cdot 2\sqrt6=\sqrt{\frac{26}{30}}\cdot 2\sqrt{30}=\sqrt{104}\]

So our answer is $\left(\sqrt{104}\right)^2=\boxed{104}$

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

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