# Pair of real number satisfying $56x+33y=-\frac{y}{x^2+y^2}$ and $33x+56y=\frac{x}{x^2+y^2}$.

Let (x,y) be a pair of real number satisfying $$56x+33y=-\frac{y}{x^2+y^2}$$ and $$33x+56y=\frac{x}{x^2+y^2}$$. If $$|x|+|y|=\frac{p}{q}$$ (where p and q are relatively prime), then find the value (6p – q).

I used the concept $$\frac{x}{y}=t$$ while solving i get $$56t+33=-\frac{1}{t^2+1}$$ and $$33t+56=\frac{t}{t^2+1}$$ on dividing the two equation i end up getting quadratic equation but that is not helping me

• have you tried polar cordinates. May 20, 2020 at 19:03
• Have you tried searching the problem up? hmmt-archive.s3.amazonaws.com/tournaments/2008/feb/guts/… page 8 problem 29 has your answer (and it's not "no solutions"!). Sep 26, 2022 at 20:31

Let $$z = x - yi$$
Then, subtract the first equation times $$i$$ from the second equation. \begin{align*} &33x - 56y - 56xi - 33yi = \frac{x + yi}{x^2 + y^2} \\ \implies &33(x-yi) - 56(y + xi) = \frac{1}{x - yi} \\ &\text{you can see that y + xi = i \cdot (x - yi) , so} \\ \implies & z(33 - 56i) = \frac{1}{z} \implies z^2 = \frac{1}{33-56i} \end{align*} $$33 - 56i = (7 - 4i)^2$$, so $$z = \pm \frac{1}{7 - 4i} = \pm(\frac{7}{65} + \frac{4}{65} i) \implies (x, y) = (\pm \frac{7}{65}, \mp \frac{4}{65})$$. Thus, $$|x| + |y| = \boxed{\frac{11}{65}} \hspace{.5em} \blacksquare$$
Let $$z =x +iy \\ \bar{z}=x-iy \\ now \ subtract \ 2 \ equations \ by multiplying \ first \ with \ i$$ $$\bar{z}(33-56i) = \frac{1}{z}$$ Now you can proceed
• @SamarImamZaidi based on this solution also on RHS you have $\frac{1}{|z|^2}$ and on the LHS you have $33-56i$. So RHS is a purely real number, whereas the LHS is non-real. Hence no solution. May 21, 2020 at 7:49
Equation (1) $$\times x$$ + Equation (2) $$\times y$$ when added together gives \begin{align*} 56x^2+66xy+56y^2&=0\\ 28x^2+33xy+28y^2&=0\\ 28\left(x+\frac{33}{56}y\right)^2+\frac{2047}{112}y^2&=0. \end{align*} Since left side is always non-negative (for $$x,y \in \mathbb{R}$$), this can only happen when both \begin{align*} x+\frac{33}{56}y&=0\\ y^2&=0 \end{align*} This gives $$x=y=0$$ as the only solution. But this does not satisfy the original equation (denominator will become $$0$$). Thus no such real $$x$$ and $$y$$ can exist.