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Difference between revisions of "Mock AIME I 2012 Problems/Problem 1"

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Let <math>C</math> be the circle of maximal area, and <math>P</math> be the given parabola. By symmetry, the center of <math>C</math> will be on the axis of <math>P</math>, at <math>x=-7/2</math>. Because <math>C</math> is tangent to the <math>x</math>-axis, the y-coordinate of its center will be at <math>y=r</math> (where <math>r</math> is the radius). So <math>C</math> has equation <math>(x+\frac{7}{2})^2+(y-r)^2=r^2</math>. Now suppose that <math>(a,b)</math> is one of the two intersections of <math>C</math> and <math>P</math>. Then <cmath>(a+\frac{7}{2})^2+(b-r)^2=r^2</cmath> <cmath>-a^2-7a+12=b</cmath>
 
Let <math>C</math> be the circle of maximal area, and <math>P</math> be the given parabola. By symmetry, the center of <math>C</math> will be on the axis of <math>P</math>, at <math>x=-7/2</math>. Because <math>C</math> is tangent to the <math>x</math>-axis, the y-coordinate of its center will be at <math>y=r</math> (where <math>r</math> is the radius). So <math>C</math> has equation <math>(x+\frac{7}{2})^2+(y-r)^2=r^2</math>. Now suppose that <math>(a,b)</math> is one of the two intersections of <math>C</math> and <math>P</math>. Then <cmath>(a+\frac{7}{2})^2+(b-r)^2=r^2</cmath> <cmath>-a^2-7a+12=b</cmath>
 
Adding these two equations and simplifying gives <math>b^2-(2r+1)b+\frac{97}{4}=0</math>. By symmetry, there should only be one solution for <math>b</math>, so the discriminant of this quadratic in <math>b</math> is zero: <math>(2r+1)^2-97=0\Longrightarrow r=\frac{\sqrt{97}-1}{2}</math>. The answer is <math>97-1+2=\boxed{098}</math>.
 
Adding these two equations and simplifying gives <math>b^2-(2r+1)b+\frac{97}{4}=0</math>. By symmetry, there should only be one solution for <math>b</math>, so the discriminant of this quadratic in <math>b</math> is zero: <math>(2r+1)^2-97=0\Longrightarrow r=\frac{\sqrt{97}-1}{2}</math>. The answer is <math>97-1+2=\boxed{098}</math>.
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==See Also==
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*[[Mock AIME I 2012 Problems/Problem 2| Next Problem]]
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*<math>\textbf{First Problem}</math>
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*[[Mock AIME I 2012 Problems]]

Latest revision as of 09:54, 11 March 2025

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 $97-1+2=\boxed{098}$.

See Also

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