Question

Suppose that a parabola has vertex $\left(\frac{1}{4}, -\frac{9}{8}\right)$ and equation $y = ax^2 + bx + c,$ where $a \gt 0$ and $a + b + c$ is an integer. The minimum possible value of $a$ can be written in the form $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p + q.$

答案

Solution:

In vertex form the parabola is $y = a\left(x - \frac{1}{4}\right)^2 - \frac{9}{8}.$ Since $a + b + c$ equals the value of $y$ at $x = 1,$ $a + b + c = a\left(\frac{3}{4}\right)^2 - \frac{9}{8} = \frac{9(a - 2)}{16}.$

If this equals the integer $n,$ then $a = 2 + \frac{16n}{9}.$ The condition $a \gt 0$ requires $16n \gt -18,$ that is $n \ge -1,$ and $a$ is smallest when $n = -1,$ giving $a = 2 - \frac{16}{9} = \frac{2}{9}.$

Thus $p + q = 2 + 9 = 11.$

2011 AIME I Problem 6

Solution:

Problem 6 in Other Years