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Irregularly Inscribing a Circle ... of What Radius?

Date: 12/24/2011 at 09:42:14
From: huey
Subject: find the radius of a circle given a inscribed hexagon

A hexagon with consecutive sides of lengths 2, 2, 7, 7, 11, and 11 is
inscribed in a circle. Find the radius of the circle.

No angles are given, and it is a irregular hexagon, so I don't know how to
start.



Date: 12/26/2011 at 19:38:57
From: Doctor Greenie
Subject: Re: find the radius of a circle given a inscribed hexagon

Hi, Huey --

I haven't found an analytical solution yet, but a numerical one using a
graphing calculator was not too difficult.

If we draw the radii to each vertex of the hexagon and the perpendiculars
from the center of the circle to the midpoint of each side of the hexagon,
then we end up with 12 right triangles with a common vertex at the center
of the circle, and all with hypotenuse r, where r is the radius of the
circle. Four of these triangles have opposite side length 2/2 = 1, four
have opposite side length 7/2 = 3.5, and four have opposite side length
11/2 = 5.5.

Then, since the central angles of all these 12 triangles add up to 360
degrees, we have (in degree measure)

   arcsin(1/r) + arcsin(3.5/r) + arcsin(5.5/r) = 90

A graphing calculator easily determines that the solution to this equation
is r = 7, so the radius of the circle is 7.

If you need an analytical solution, write back. I'm still playing with
possible approaches to an analytical solution, and I will ask other
volunteers here if they have any ideas.

- Doctor Greenie, The Math Forum
  http://mathforum.org/dr.math/ 



Date: 12/27/2011 at 09:55:53
From: huey
Subject: find the radius of a circle given a inscribed hexagon

Hi,

I do really need analytical solutions. 

Thank you.

Huey



Date: 04/11/2012 at 14:55:17
From: Doctor Floor
Subject: Re: find the radius of a circle given a inscribed hexagon

Hi, Huey,

I've had your question on my watchlist. This is probably of no help any
more, but still I was eager to find a solution -- and just succeeded. So
here we go:

We had an inscribed hexagon with side lengths 2, 2, 7, 7, 11 and 11. Note
that we may change the order of the sides without changing the radius of
the circumcircle.

Now if we change into the order 2, 7, 11, 11, 7, 2 we get a line symmetric
hexagon. The line of symmetry of course is a diameter. So we may cut the
hexagon in half and find two cyclic quadrilaterals with sides 2, 7, 11,
and x, where x denotes the diameter of the circle.

More precisely, we may consider one of the quadrilaterals. Name the
vertices ABCD, such that AB = x, BC = 11, CD = 7 and DA = 2. Of course,
triangles ABC and ABD are right, having AB as hypotenuse. 

Hence we have:

   AC = sqrt(x^2 - 121)
   BD = sqrt(x^2 - 4)

Now we may apply Ptolemy's theorem:

   AB*DC + BC*AD = AC*BD

From the Dr. Math Archives, see

  http://mathforum.org/library/drmath/view/54956.html 

This gives:

   7x + 22 = sqrt(x^2 - 121)*sqrt(x^2 - 4)

Squaring both sides:

   49x^2 + 308x + 484 = x^4 - 125x^2 + 484
                    0 = x^4 - 174x^2 - 308x
                    0 = x(x^3 - 174x - 308)

We may check for rational solutions of x^3 - 174x - 308 = 0 by the
rational root theorem:

  http://mathforum.org/library/drmath/view/56425.html 

As 308 = 2*2*7*11, the number of possibilities is limited. We succeed with
x = 14. This means that 14 is a possible diameter length for our circle.

Now factor:

   x^3 - 174x - 308 = (x - 14)(x^2 + 14x + 22)

Note that the second factor has two negative roots. So x = 14 is THE
solution we are looking for.

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Conic Sections/Circles
High School Triangles and Other Polygons

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