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

As suggested, *if* it matters, you may assume that the sides listed are given in order

The sides {7, 2, 11} may be inscribed in a semicircle of the same radius as the original circle.

Let AB = d be the diameter of the semicircle, BC = 11, CD = 2, and AD = 7.  Draw AC of length x, and BD of length y.

Note that angles ADB and ACB are right angles.  Hence, by Pythagoras:
x² = d² - 121
y² = d² - 49

Ptolemy's Theorem states that in a cyclic quadrilateral the sum of the products of the two pairs of opposite sides equals the product of its two diagonals.
Hence 2d + 77 = xy.

Therefore (2d + 77)² = (d² - 121)(d² - 49).

Simplifying, and dividing by d, we obtain d³ - 174d - 308 = 0.
This factorizes, giving (d - 14)(d² + 14d + 22) = 0.
The quadratic factor has negative real roots, so d = 14 is the only positive real root.

Therefore the radius of the circle in which the hexagon is inscribed is 7.

Posted by Nick Hobson on 2005-06-13 14:13:25