
A3 Let the real parameter $p$ be such that the system

$$
\left\{\begin{array}{l}
p\left(x^{2}-y^{2}\right)=\left(p^{2}-1\right) x y \\
|x-1|+|y|=1
\end{array}\right.
$$

has at least three different real solutions. Find $p$ and solve the system for that $p$.


## Solution

The second equation is invariant when $y$ is replaced by $-y$, so let us assume $y \geq 0$. It is also invariant when $x-1$ is replaced by $-(x-1)$, so let us assume $x \geq 1$. Under these conditions the equation becomes $x+y=2$, which defines a line on the coordinate plane. The set of points on it that satisfy the inequalities is a segment with endpoints $(1,1)$ and $(2,0)$. Now taking into account the invariance under the mentioned replacements, we conclude that the set of points satisfying the second equation is the square $\diamond$ with vertices $(1,1),(2,0),(1,-1)$ and $(0,0)$.

The first equation is equivalent to

$p x^{2}-p^{2} x y+x y-p y^{2}=0$

$p x(x-p y)+y(x-p y)=0$

$(p x+y)(x-p y)=0$.

Thus $y=-p x$ or $x=p y$. These are equations of two perpendicular lines passing through the origin, which is also a vertex of $\diamond$. If one of them passes through an interior point of the square, the other cannot have any common points with $\diamond$ other than $(0,0)$, so the system has two solutions. Since we have at least three different real solutions, the lines must contain some sides of $\diamond$, i.e. the slopes of the lines have to be 1 and -1 . This happens if $p=1$ or $p=-1$. In either case $x^{2}=y^{2},|x|=|y|$, so the second equation becomes $|1-x|+|x|=1$. It is true exactly when $0 \leq x \leq 1$ and $y= \pm x$.
