A cyclic hexagon has side lengths of 2, 2, 7, 7, 11, 11, in that order. Find the length of its circumradius.

Not sure if there is a theorem or formula for this, but I tried dividing it into 30°, 60°, 90° triangles. Is that a possible way to approach the problem? If there is a theorem that can be used for this I would love to know. Any help is greatly appreciated.

  • $\begingroup$ Perhaps you could connect opposite vertices and using cyclic quadrilateral theorems to find the circumradius? $\endgroup$ – Pakquebchsoflwty Oct 26 '14 at 3:24
  • $\begingroup$ I've tried, but it's not working out, or perhaps I'm forgetting something $\endgroup$ – Lulu Uy Oct 26 '14 at 3:35
  • $\begingroup$ Check this out, see if it helps-ftp.jssac.org/Editor/Suushiki/V18/No1/V18N1_102.pdf $\endgroup$ – Pakquebchsoflwty Oct 26 '14 at 3:40
  • $\begingroup$ I didn't check my working below but I think I found the trick. =) $\endgroup$ – user21820 Oct 26 '14 at 3:43
  • $\begingroup$ It makes sense thanks a million. The only problem is the answer key says its 49π so perhaps they meant to ask for the area of the circumscribed circle? Im not sure $\endgroup$ – Lulu Uy Oct 26 '14 at 3:47

Adjacent edges in a cyclic polygon can be swapped without changing the circumradius. Thus all the edges can be permuted without changing the circumradius. Thus we can rearrange the side lengths of the given hexagon to be (2,7,11,2,7,11). Now opposite points are diameters of the circumcircle, and so it reduces to a cyclic quadrilateral with one side being a diameter and the others being (2,7,11). Let the diameter be $d$ and the diagonals of the quadrilateral be $x,y$. We get:

$x y = 2 \times 11 + 7 d$ [by Ptolemy's theorem]

$x^2 + 11^2 = d^2$ [by Pythagoras' theorem]

$y^2 + 2^2 = d^2$ [by Pythagoras' theorem]

Then we get:

$(d^2-121)(d^2-4) = x^2 y^2 = (7d+22)^2$

$d^4-125d^2+484 = 49d^2+308d+484$

$d^3-125d = 49d+308$ [because $d>0$]

$d^3 - 174d - 308 = 0$

$(d-14)(d^2+14d+22) = 0$

$(d-14)((d+7)^2-27) = 0$

$d=14$ [because $d>0$]

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In general would be the solution for

$4 R^3 -(a^2+b^2+c^2) R - a b c = 0$

$P = \sqrt[3]{ \sqrt{4(-12)^3(a^2+b^2+c^2)^3 + 2^8 3^6 a^2 b^2 c^2} + 2^4 3^3 a b c }$

$ R = \frac{P}{12 \sqrt[3]{2}} + \frac{\sqrt[3]{2} (a^{2}+b^{2}+c^{2})}{P} $


$a=2, b=7, c=11$

$4 R^3 - 174 R - 154 = 0$

$R = 7$

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  • $\begingroup$ It's impossible to follow you, right from the beginning... $\endgroup$ – Jean Marie Sep 20 '17 at 13:55

Let us look at a general development involving a hexagon $ABCDEF$ with sides $a,a,b,b,c,c$. To avoid degenerate cases it is assumed these lengths are distinct, and wlog for definiteness $c$ is assumed to be the maximum of the three. Following @user21820 we can reorder the sides as desired, and I choose $|AB|=|FA|=a,|BC|=|EF|=b,|CD|=|DE|=c$. This gives a cyclic hexagon with a mirror line $DA$, thus the hexagon is reduced to quadrilateral $ABCD$ with $DA$ as a diameter of the circumcircle, thus $|DA|=2r$.

Draw diagonal $AC$. With $DA$ a diameter, $\angle ACD$ is a right angle and then $\triangle ACD$ gives:

$|AC|^2=4r^2-c^2$ Eq. 1

$\cos \angle CDA=(c/2r)$ Eq. 2

The Law of Cosines is also applied to $\triangle ABC$:

$\cos \angle ABC=(a^2+b^2+c^2-4r^2)/(2ab)$ Eq. 3

Here, Eq. 1 was used to render the length of $AC$.

In a convex cyclic quadrilateral, opposite angles are supplementary and thus the cosines in Eqs. 2 and 3 must be negatives of each other. Forming that equation, clearing fractions and collecting terms gives a cubic equation for $r$:

$P_3(r)=4r^3-(a^2+b^2+c^2)r-abc=0$ Eq. 4

Eq. 4 is in general irreducible but we can render its roots with sufficient precision to identify its key properties. First off, Descartes' Rule of Signs identifies exactly one positive root. This will be the root corresponding to the convex quadrilateral, and thus to the convex hexagon formed from the quadrilateral and it's mirror image, provided that the (nondegenerate) existence condition

$c/2 < r < (a+b+c)/2$

is satisfied. This says the diameter has to be longer than the longest given side but shorter than the sum of all three other sides. To check this, substitute the bounding values of $r$ into $P_3(r)$ from Eq. 4:



By continuity the root must lie between these bounds guaranteeing the geometric existence of a convex solution.

We then throw out the remaining, negative roots because they do not correspond to real objects ... or do they? It is possible to render $ABCD$ not as a convex quadrilateral but as a "crossed" quadrilateral in which two "opposite" sides, such as $AB$ and $CD$, are crossed over each other. The hexagon formed by combining the quadrilateral with its mirror image across $DA$ will also be crossed.

Crossing the sides like this effectively reverses the orientation of one side, leading to a reversed sign; but we require $a,b,c>0$. So the diameter must be "reversed" and thus a crossed quadrilateral will show up as negative root for $r$. The radius of the circumcircle in this case will be the absolute value of $r$ which must satisfy existence conditions analogous to the convex case:

$-(a+b+c)/2 < r < -c/2$

So the values of $P_3(r)$ at the bounding values may be checked as before:



Thus one negative root corresponds to the existence of crossed polygons. The other does not correspond to real geometric objects, but it still has its influence: since this "bad" root is negative (it must be so by Descartes' Rule of Signs) and all three roots of Eq. 4 must sum to zero, the "good" crossed polygon root must correspond to a smaller circumcircle than the positive, convex root.

For the specific case at hand, Eq. 4 divided by a common factor of $2$ gives:


And the roots may be interpreted as follows:

$r=+7$: convex quadrilaterals/hexagons inscribed in a circle of radius $7$

$r=-(7+3\sqrt{3})/2$: crossed polygons inscribed in a circle of radius approximately $6.1$

$r=-(7-3\sqrt{3})/2$: No geometric solutions.

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