CEMC Euclid, 2014

Let $a=x-2013$ and let $b=y-2014$.

The given equation becomes $\dfrac{ab}{a^2+b^2} = -\dfrac{1}{2}$, which is equivalent to $2ab = -a^2-b^2$ and $a^2+2ab+b^2=0$.

This is equivalent to $(a+b)^2 = 0$ which is equivalent to $a+b=0$.

Since $a=x-2013$ and $b=y-2014$, then $x-2013+y-2014=0$ or $x+y=4027$.
Let $a = \log_{10} x$.

Then $(\log_{10}x)^{\log_{10}(\log_{10} x)} = 10\,000$ becomes $a^{\log_{10}a} = 10^4$.

Taking the base 10 logarithm of both sides and using the fact that $\log_{10}(a^b) = b\log_{10}a$, we obtain $(\log_{10} a)(\log_{10} a) = 4$ or $(\log_{10} a)^2 = 4$.

Therefore, $\log_{10} a = \pm 2$ and so $\log_{10}(\log_{10} x) = \pm 2$.

If $\log_{10}(\log_{10} x) = 2$, then $\log_{10}x = 10^2 = 100$ and so $x = 10^{100}$.

If $\log_{10}(\log_{10} x) = -2$, then $\log_{10}x = 10^{-2} = \frac{1}{100}$ and so $x = 10^{1/100}$.

Therefore, $x = 10^{100}$ or $x = 10^{1/100}$.

We check these answers in the original equation.

If $x = 10^{100}$, then $\log_{10}x = 100$.

Thus, $(\log_{10}x)^{\log_{10}(\log_{10} x)} = 100^{\log_{10} 100} = 100^2 = 10\,000$.

If $x = 10^{1/100}$, then $\log_{10}x = 1/100 = 10^{-2}$.

Thus, $(\log_{10}x)^{\log_{10}(\log_{10} x)} = (10^{-2})^{\log_{10} (10^{-2})} = (10^{-2})^{-2} = 10^4 = 10\,000$.