** Solution by the organizers **. This solution does not require the concept
of curvature .

We first find the largest circle that touches the parabola at (0, 0). The center

of such disk must be located at (0, r) where r is the radius of the circle. Then
the

equation of this circle is

To find the largest possible r we require that this circle intersects the
parabola y = x^{2}

at a single point. Solving the system y = x^{2} and x^{2} + (y − r)^{2} = r^{2} we get

Thus we get two solution , y = 0 and y = 2r −1. However, y needs to be
non- negative

since y = x^{2}, thus to have only one solution we need 2r − 1 ≤ 0. That is r ≤
1/2.

Thus r = 1/2 is the largest radius of a circle touching the parabola at (0, 0).

Now, let
be a point on the parabola. By symmetry, the largest

circle C on the concave part of the parabola touching the parabola at
must

also touch the parabola at
,
and thus have a center on the y-axis. Suppose

that the center of such circle is (0, a). The slope of the tangent line to the
parabola ,

at
,
equals the derivative y' = 2x evaluated at x_{0}. Thus the slope of that line

is 2x_{0}. Since the segment joining (0, a) and
is perpendicular to the tangent

line at
, we have that

Thus
.
Let r be the radius of C, then we have that
.

Plugging the value of a we get
.
Therefore r > 1/2. Thus the

circle of radius 1/2 will freely rotate on the parabola, and 1/2 is the radius
of the

largest such possible circle.

y = x^{2}