How to solve equations in two variables when they're not so appealing?

Question

I was trying to bash IMO 2009/4 using trigonometry and I landed up with the equations :

$$3y - 4y^3 = \frac{d+2}{2\sqrt{2} d}$$ $$2y\sqrt{1-y}=\frac{1}{d}$$.

From here, I have to write an equation in $d$. Since there are two variables, I believe it's possible but is there any sane method to do so?

Answer 1

Eliminating $1/d$, you can write

$$3y-4y^3=\frac1{2\sqrt2}+\sqrt2y\sqrt{1-y}$$

and from this, the polynomial equation

$$\left(3y-4y^3-\frac1{2\sqrt2}\right)^2=2y^2(1-y)$$ with the constraint $y<1$.

After numerical resolution, there are six real roots (four positive and two negative) that fulfill the constraint. But we need to reject those that correspond to a negative value of $\sqrt{1-y}$ (magenta curve). From the plot (LHS vs RHS), we see that three of the roots are valid.

$d$ easily follows from $y$. Due to the degree of the problem, there is no better way.

Answer 2

After squaring the second equation, you can eliminate $y$, but you end up with a nasty polynomial in $d$ of degree $6$.