A Quadratic Game

All Brilli needs to do is choose 3 distinct rational numbers $a,b,c$ such that $a+b+c = 0$ . Let the polynomial be $f(x) = Ax^2 + Bx + C$ where $A,B,C$ are some permutations of $a,b,c$ .

Clearly no matter how Brian arranges it we will have $f(1) = A+B+C = a+b+c = 0$ So the polynomial has a root at $x = 1$ .

The other root is at $C/A$ which is also clearly rational and not equal to $1$ and hence the equation has distinct roots.

So $\textbf{Brilli}$ has a winning strategy.

I think a similar problem was once used in the USA IMO training camp.

Have a look at the discriminant.

$D=b^2-4ac$

Brian's goal is to make it zero or negative. Assuming all numbers are non-zero, this can happen only if

$4ac>0$

Brian is now limited to picking either two positive numbers for $a, c$ or two negative in order to maintain the product positive. Here comes the problem: if Brilli gives him only one negative, it has to go as $b$ . Now Brilli picks a stupidly large negative and two tiny positives and insta-wins.

Obviously Brilli can put a number that can not be arranged in any way to get a quadratic equation that has real roots

Building on Eddie The Head's solution,

Let f(x) be $Ax^2 + Bx + C$ ,

Since x = 1 is a factor of f(x),

f(x) = (x-1)[Ax+(A+B)] + (A+B+C)

Since A + B + C = 0 and A+B = - C,

f(x) = (x-1)(Ax-C)

As according to remainder theorem, when Ax - C = 0, f(x) = 0, therefore, x = $\frac{C}{A}$ is another distinct rational solution since C and A are both non-zero real solutions.

Thus, as long as A + B + C = 0, Brilli has the winning strategy.