To infinity and beyond
If the radius of the largest circle which can be inscribed inside the region bounded by
and
can be expressed in the form
where a and b are integers and
is not divisible by the square of any prime, find
.
The answer is 55.
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Clearly the center of the circle must be at {x=0} due to symmetry. Let the equation of the circle be x 2 + ( y − k ) 2 = r 2 . The shortest distance from the center of the circle to the line must be the radius, which is k / 2 . Plugging in the value of the radius and the equation of the parabola into the equation of the circle, we get x 2 − k 2 / 2 + ( 8 − k − x 2 / 2 ) 2 = 0 . Now we want to find the value(s) of k for which this equation has only two solutions. The discriminant with respect to x is ( k 2 − 3 2 k + 1 2 8 ) ∗ ( k 2 + 4 k − 3 0 ) 2 / 2 . Setting the discriminant equal to {0} and solving for k , we get k = 8 ( 2 − 2 ) , k = 8 ( 2 + 2 ) , k = − 2 − 3 4 , k = − 2 + 3 4 , and the only value of k that satisfies the constraints of the problem is k = − 2 + 3 4 .