FTW Equation

• Ok, so I set the first member A and the second member B , such $A\times B = 2013$ .
• Solving for A, we have:
  $\sqrt{x^{2} + 2013} = (A + x)$ .
  $x^{2} + 2013 = A^{2} + 2Ax + x^{2}$ .
  $A^{2} + 2Ax = 2013$ .
  
  
  
• Using the same logic with y and B , we find that: $2013 = B^{2} +2By$ .
• With $A \times B = 2013$ and those two equations above, we state that: $A^{2} + 2Ax = A \times B = B^{2} + 2By$ .
• So, we can conclude that $A + 2x = B$ and $B + 2y = A$ .
• Replacing $B + 2y = A$ in $A + 2x = B$ , we have:
  $B + 2x + 2y = B => 2x + 2y = 0$ .

• With this, we can finally state that $x = -y$ .

• Now, for $\frac{2013x + y}{5x + y}$ , we have:
  $\frac{2013x - x}{5x - x} => \frac{2012x}{4x} = 503$ .

• So, this adventure ends with the answer being 503 .

(hope you enjoyed the solution).

Let

$x = \sqrt{2013}\tan{\theta}$

$y = \sqrt{2013}\tan{\phi}$

We can do this without loss of generality because the range of tangent is all real numbers. Then:

$(\sqrt{2013}\sec{\theta}-\sqrt{2013}\tan{\theta})(\sqrt{2013}\sec{\phi}-\sqrt{2013}\tan{\phi})=2013 \\ (\sec{\theta}-\tan{\theta})(\sec{\phi}-\tan{\phi})=1 \\$

Notice that:

$\frac{1}{\sec{\theta}+\tan{\theta}} \\ = \frac{1}{\sec{\theta}+\tan{\theta}}\frac{\sec{\theta}-\tan{\theta}}{\sec{\theta}-\tan{\theta}} \\ = \frac{\sec{\theta}-\tan{\theta}}{\sec^2{\theta}-\tan^2{\theta}} \\ = \sec{\theta}-\tan{\theta} \\$

Hence: $(\sec{\theta}-\tan{\theta})(\sec{\phi}-\tan{\phi})=1 \\ = \frac{\sec{\phi}-\tan{\phi}}{\sec{\theta}+\tan{\theta}}=1$

Notice that secant is even and tangent is odd. We can set the numerator and denominator equal if we let $\theta = -\phi$ . Referring back to the equations for $x$ and $y$ , this means $x = -y$ , and hence:

$\frac{2013x+y}{5x+y} =\frac{2012x}{4x} =\boxed{503}$

Let $y=-x \ne 0$ .

Then, $(\sqrt{x^2+2013}-x)(\sqrt{y^2+2013}-y)=(\sqrt{x^2+2013}-x)(\sqrt{x^2+2013}+x)$

$=(x^2+2013)-x^2=2013$

Now, $\frac{2013x+y}{5x+y}=\frac{2012x}{4x}=\boxed{503}$