Areteem: AIME Sprint Problems

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AIME Sprint Problems

Here is the full solution.

Let the coordinates of $P$ be $(x_0,y_0)$. Let $A$ be $(a,0)$, and $B$ be $(b,0)$, and assume $a < b$ without loss of generality. We try to minimize the area of $\triangle PAB$, which is $\dfrac{1}{2}(b-a)\cdot y_0$.

The equation of line $\overline{PB}$ is \[ \frac{y}{y_0} = \frac{x-b}{x_0-b}, \] which is \[ y_0x - (x_0-b)y - y_0b = 0. \] Since the center of the given circle $(0,1)$ has distance $1$ to $\overline{PB}$, we have \[ \frac{|(x_0-b) - y_0b|}{\sqrt{y_0^2 + (x_0-b)^2}} = 1, \] so \[ |(x_0-b) - y_0b| = \sqrt{y_0^2 + (x_0-b)^2}, \] and squaring, \[ (x_0-b)^2 + 2y_0b(x_0-b) + y_0^2b^2 = y_0^2 + (x_0-b)^2, \] Clearly $y_0 > 2$ (otherwise either $A$ or $B$ cannot exist), and the above equation becomes \[ (y_0-2)b^2 + 2x_0b - y_0 = 0. \] Similarly, \[ (y_0-2)a^2 + 2x_0a - y_0 = 0. \] Thus, $a$ and $b$ are the two distinct real roots of the quadratic equation \[ (y_0-2)t^2 + 2x_0t - y_0 = 0, \] and by Vieta's Formulas, \[ a+b = -\frac{2x_0}{y_0-2}, \quad ab = -\frac{y_0}{y_0-2}. \] so \[ (b-a)^2 = (a+b)^2 - 4ab = \frac{4x_0^2}{(y_0-2)^2} + \frac{4y_0}{y_0-2} = \frac{4x_0^2 + 4y_0^2 - 8y_0}{(y_0-2)^2}. \] Since $(x_0,y_0)$ is on the parabola, $y_0 = x_0^2/2$, so \[ (b-a)^2 = \frac{4y_0^2}{(y_0-2)^2}, \] hence \[ b-a = \frac{2y_0}{y_0-2}, \] and the area of $\triangle PAB$ is \[ \frac{1}{2}(b-a)\cdot y_0 = \frac{y_0}{y_0-2}\cdot y_0 = (y_0-2) + \frac{4}{y_0-2} + 4 \geq 2\sqrt{4}+4=8. \] The last step uses AM-GM Inequality.

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