Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.0 AIME, harder Find the answer

Example 6. The vertex of the parabola $y=x^{2}+b x+c$ is located inside or on the square with vertices at $(0,0) 、(1,0) 、(1,1) 、(0,1)$ . Find the range of values for $b$ and $c$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

Let the vertex be $(p, q)$ , then $p, q$ satisfy the conditions $0 \leqslant p \leqslant 1,0 \leqslant 1 \leqslant 1$ .
The vertex of the parabola $y=x^{2}+b x+c$ is $\left(-\frac{b}{2}, \frac{4 c-b^{2}}{4}\right)$ .
$\therefore 0 \geqslant-\frac{b}{2}-\leqslant 1,0 \leqslant \frac{4 c-b^{2}}{4} \quad 1$ .
So $-2 \leqslant b \leqslant 0$ .
And $0<c-\frac{b^{2}}{4} \quad 1$ .
That is $\frac{b^{2}}{4} \leqslant c \leqslant \frac{b^{2}}{4}+1$ .
From (1) and (2), we get the relationship between $\mathrm{b}$ and $\mathrm{c}$ , as shown in the figure below:

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.