Mock AIME I 2012 Problems/Problem 1

Problem

A circle of maximal area is inscribed in the region bounded by the graph of $y=-x^2 -7x + 12$ and the $x$ -axis. The radius of this circle is $\dfrac{\sqrt{p} + q}{r}$ , where $p$ , $q$ , and $r$ are integers and $q$ and $r$ are relatively prime. What is $p+q+r$ ?

Solution

Let $C$ be the circle of maximal area, and $P$ be the given parabola. By symmetry, the center of $C$ will be on the axis of $P$ , at $x=-7/2$ . Because $C$ is tangent to the $x$ -axis, the y-coordinate of its center will be at $y=r$ (where $r$ is the radius). So $C$ has equation $(x+\frac{7}{2})^2+(y-r)^2=r^2$ . Now suppose that $(a,b)$ is one of the two intersections of $C$ and $P$ . Then $(a+\frac{7}{2})^2+(b-r)^2=r^2$ $-a^2-7a+12=b$ Adding these two equations and simplifying gives $b^2-(2r+1)b+\frac{97}{4}=0$ . By symmetry, there should only be one solution for $b$ , so the discriminant of this quadratic in $b$ is zero: $(2r+1)^2-97=0\Longrightarrow r=\frac{\sqrt{97}-1}{2}$ . The answer is $97-1+2=\boxed{098}$ .

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Mock AIME I 2012 Problems/Problem 1

Problem

Solution