Consider an octagon and a heptagon, where each is circumscribed by a
circle. If the area of the heptagon and the octagon are equal, and the
radius of the circle containing the octagon is assigned the value of
one, what is the radius of the circle containing the heptagon?

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Request for Question Clarification by mathtalk-ga on 06 Aug 2005 10:22 PDT

Hi, g22-ga:

As ticbol-ga points out, the Question begs for an additional
assumption, presumably that the polygons are regular.  If not, a
circumscribing circle for each figure would exists, but then the
radius would not depend only on the polygon's area.

With regular polygons the center of the circumscribing circle (we
would conventionally refer to the polygons as "inscribed" in the
circle) is the same as the center of the polygon.

You can then follow ticbol-ga's analysis, dividing the polygons into
isoceles triangles with apex at this common center.  The apex angle is
given by dividing the whole rotation (360 degrees or 2pi radians) into
an equal number of parts for the number of sides of the regular
polygon.  In both cases (octagon and heptagon) the area of the polygon
is the sum of the (equal) areas of these isoceles triangles (8 copies
vs. 7 copies, resp.).


regards, mathtalk-ga

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Clarification of Question by g22-ga on 07 Aug 2005 09:47 PDT

See comment.

I assume you mean the heptagon and octagon are regular polygons. A
regular polygon is equilateral and equiangular.

One way to find the area of a regular n-gon is to subdivide it into n
numbers of congruent isosceles triangles. Draw line segments from the
vertices to the center of the polygon. The area of the n-gon is equal
to n times the area of one of these subdivisions.

Each of these isosceles triangle subdivisions has: 

>>>base = side of the n-gon.

>>>two equal sides = each is a line from a vertex to the center of the
n-gon. If the n-gon is circumscribed by a circle, then each of the two
equal sides is a radius of the cotaining circle.
For the octagon, radius = 1
For the heptagon, radius = r

>>>two base angles = each is half of an interior angle of the n-gon.
So, the sum of the two base angles is equal in measure to an interior
angle of the n-gon.
Hence, the apex angle = 180 minus the measure of an interior angle of
the n-gon, in degrees.
Let us call this apex angle, angle A.
An interior angle of the n-gon is (n-2)(180deg)/n

For the octagon,
A1 = 180 -[(8-2)(180)/8] = 180 -(3/4)180 = (1/4)180 = 45 deg.

For the heptagon, 
A2 = 180 -[(7-2)(180)/7] = 180 -(5/7)180 = (2/7)180 = 360/7 deg.

---------
One way to get the area of a triangle is by
area = (1/2)(side)(another side)*sin(included angle).
In the isosceles triangle above, the apex angle is included by the two equal sides.

For the octagon,
area1 = (1/2)(1)(1)sin(45deg).
area1 = (1/2)sin(45deg)


For the heptagon,
area2 = (1/2)(r)(r)sin(360/7 deg)
area2 = (1/2)(r^2)sin(360/7 deg)

The two areas are equal, so,
(1/2)(r^2)sin(360/7 deg) = (1/2)sin(45deg)
(r^2)sin(360/7 deg) = sin(45deg)
r^2 = [sin(45deg)]/[sin(360/7 deg)]
r^2 = 0.90442352
r = sqrt(0.90442352)
r = 0.951  -------------the radius of the circle containing the heptagon.

Oopps, sorry, Let me correct that one.

Area of octagon = 8[(1/2)sin(45deg)]
Area of heptagon = 7[(1/2)(r^2)sin(360/7 deg)]

7[(1/2)(r^2)sin(360/7 deg)] = 8[(1/2)sin(45deg)]
r^2 = [4sin(45 deg)]/[(7/2)sin(360/7 deg)]
r^2 = 2.828427 / 2.736410
r^2 = 1.033627
r = 1.0167   --------the radius containing the heptagon. (answer)

Thank you very much ticbol and mathtalk. Yes I was referring to
regular polygons. I had forgotten that irregular polygons can still be
circumscribed by circles, even if all corners are not in contact with
the circle.

My understanding is a polygon, whether regular or not, is said to be
inscribed in a circle if all its vertices are on the circumference of
the circumscribing circle.
Another term for these inscribed polygons is cyclic polygons.

If one or more of the corners/vertices of the polygon does or do not
touch the circle, can that polygon still be considered as
circumscribed by the circle?
I am not sure about that.

Good terminology question.  If one says that a polygon is inscribed in
a circle, that clearly conveys that all the vertices are lie on the
circle.  Hence the characterization of those quadrilaterals that can
be inscribed in a circle (opposite angles are supplementary) strictly
limits these to a proper subset of all quadrilaterals (implying in
particular they are convex).

For this reason it might be best to avoid referring to a circle of
minimum radius that contains a given polygon as the circumscribing
circle.  An alternative name for this situation is a (minimal)
"bounding" circle.

regards, mathtalk-ga