(From the **MSU Problem Corner**: Read the
problem statement
and the
solution.)
The answers:

- For an equilateral triangle with side length 1, the
circle should have radius
*r*= 1/3. - For a square with side length 1, the circle should have
radius
*r*= (1/2)sec (π/8) = 1/(2cos (π/8)) ≈ 0.5412. (All angle measures in radians.) - For a regular
*n*-gon with side length 1, the radius should be*r*= (1/2) cot (π/*n*) sec (π/(2*n*)).

Explanation: The first two parts follow as special cases
(*n*=3 and *n*=4) of the last part.

The optimal circle must have equal arc length inside and
outside the *n*-gon. To see why, take a circle with
radius *r*, and define *f*(*r*) to be the
total arc length of the circle inside the *n*-gon, so
that 2π*r* - *f*(*r*) is the arc length
outside the circle. Now increase the radius by an
infinitesimal amount (from *r* to *r*+*dr*).
The figure on the right illustrates (for a triangle) how the
"inside-outside" area changes: The region shaded red is
added, while the green-shaded region is subtracted.

The areas of these red- and green-shaded regions are
approximated by their width (*dr*) times their (arc)
lengths, so the resulting change in the area will be

(It is easily confirmed that if *dr* is
negative, so that we are actually decreasing the radius of
the circle, the expression given for *dA* is still
correct.) Therefore, *dA/dr* = 0 when
*f*(*r*) = π*r* — that is, when half of
the circle's circumference is inside, and half outside.

It remains to confirm the given expression for *r*,
which achieves the desired property
*f*(*r*)=π*r*. This is easiest to see with
the figure on the right.

In the figure, θ = 2π/*n*, since it is the central
angle of a regular *n*-gon. Therefore
the center-to-side distance is

*x* = (1/2) cot (θ/2) = (1/2) cot (π/*n*).

Since the arc length for a circular arc is determined by
the central angle of the arc (and the circle's radius), it is
clear that in order for the "inside" arc length to be the
same as the "outside" arc length, the circle must intersect
the *n*-gon so that the smaller angle shown is θ/4 =
π/(2*n*). This makes the radius

*r* = *x * / cos (θ/4)
= (1/2) cot (π/*n*) sec (π/(2*n*)).

Generalizations: If we fix the radius of the circle (at, say, 1) and vary the size of the triangle, the minimum area is achieved when the total "outside" side length of the triangle equals the total "inside" side length. This happens when the triangle's side length is 4*√(3/7) ≈2.619.

If we have a sphere and a regular tetrahedron, we can draw
similar conclusions (but with more complicated
computations!) by equating the inside and outside **surface
areas** of the sphere (with the tetrahedron's size held
constant) — or the tetrahedron (with the sphere's radius
held constant). Similarly, we can work with a cube and a
sphere, etc.