2008 iTest Problems/Problem 86

Problem

Let $a, b, c$ , and $d$ be positive real numbers such that

$\begin{array}{c@{\hspace{3pt}}c@{\hspace{3pt}}c@{\hspace{3pt}}c@{\hspace{3pt}}c}a^2+b^2&=&c^2+d^2&=&2008,\\ ac&=&bd&=&1000.\end{array}$

If $S=a+b+c+d$ , compute the value of $\lfloor S\rfloor$ .

Solution

Note that $c = \tfrac{1000}{a}$ and $d = \tfrac{1000}{b}$ . Substituting $c$ and $d$ results in $\frac{1000000}{a^2} + \frac{1000000}{b^2} = \frac{1000000(a^2 + b^2)}{a^2 b^2} = 2008$ . Since $a^2 + b^2 = 2008$ , $a^2 b^2 = 1000000$ , so $ab = 1000$ . Thus, $a^2 + 2ab + b^2 = 4008$ , so $a+b = \sqrt{4008} = 2\sqrt{1002}$ .

Note that if we solve for $a$ and $b$ and substitute, we can use the same steps to show that $c+d = 2\sqrt{1002}$ . Thus, $S = 4\sqrt{1002} \approx 126.62$ , so $\lfloor S\rfloor = \boxed{126}$ .

2008 iTest (Problems)
Preceded by: Problem 85	Followed by: Problem 87
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2008 iTest Problems/Problem 86

Problem

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