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Hexagonal Dilemma (Posted on 2004-06-02) Difficulty: 4 of 5
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

No Solution Yet Submitted by SilverKnight    
Rating: 4.0000 (5 votes)

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Another way; still not elegant | Comment 11 of 18 |
Here's a different method, algebra and geometry, no trig.

Picture the top half circle of radius 7, centered at the origin.  It's equation is x^2 + y^2 =49
Picture half of the hexagon INscribed in it, starting from the left the sides are of length 11 then 2 then 7.  Construct 3 triangles so each side connects to the origin.  We know that the one farthest to the right is equilateral and it's vertices are at (0,0), (7,0), and (7/2, 7*sqrt(3)/2).  Point P is where the sides of length 11 and length 2 meet.   Point Q is where the length 2 sided meets the length 7 side

We know we can make a hexagon with the above lengths, but we don't know for sure that all the vertices will be on the circle.   So pretend we are in geometry class, draw a circle of radius 2 centered at (7/2, 7*sqrt(3)/2), which is point Q.   Then draw a circle of radius 11 centered at (-7, 0).  (only do it with equations instead of a compass and pencil)  They will meet at point P, where the 11 side meets the 2 side.  We hope this will be on the big circle of radius 7.

In summary, we'll get 2 equations, subtract one equation from the other to get one equation with 2 unknowns.  We'll solve for x in terms of y, and use that to plug into the equation of the big circle.  This gives a quadratic in one variable.  Solving this gives the coordinates of point P as (1.642857 ,  6.804485)    I crunched through the algebra, it was messy, not very enlightening, and I'd rather not type it all in, unless someone really wants to see it.

The fact that we started from the assumption that 7 was the radius, and found a point P which satisfies the equations of all 3 circles again shows that 7 is the answer, but it's still not an elegant proof.
  Posted by Larry on 2004-06-03 23:48:56
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