WMC 2018 Senior Problem 36

Multiplying the two equations, we get:

$\frac { 1 }{ x^{ 2 }y^{ 2 } } =\left( \frac { 1 }{ 16 } -\frac { 1 }{ xy } \right) \left( \frac { 1 }{ 2 } -\frac { 1 }{ xy } \right)$

$\left( \frac { 1}{xy } \right) ^2=\left( \frac { 1 }{ 16 } -\frac { 1 }{ xy } \right) \left( \frac { 1 }{ 2 } -\frac { 1 }{ xy } \right)$

Let $u=\frac { 1 }{ xy }$

The result can be then rewritten as:

$u^{ 2 }=\left( \frac { 1 }{ 16 } -u \right) \left( \frac { 1 }{ 2 } -u \right)$

Solving for u:

$u^{ 2 }=\frac { 1 }{ 32 } -\frac { 9 }{ 16 } u+u^2$

$0=\frac { 1 }{ 32 } -\frac { 9 }{ 16 } u$

$0=1-18u$

$u=\frac { 1 }{ 18 }$

Substituting back the definition of u:

$\frac { 1 }{ xy } =\frac { 1 }{ 18 }$

Taking the reciprocal of both sides gives us our final answer:

$\boxed { xy=18 }$

We have $16y=x^2y-16x,2x=xy^2-2y$

$16(x+y)=x^2y=8xy^2,x=8y$

$\dfrac1{x^2}=\dfrac1{8xy}=\dfrac1{16}-\dfrac1{xy}$

Let $\dfrac1{xy}=z$ , then $\dfrac18z=\dfrac1{16}-z,z=\dfrac1{18},xy=18$