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### A Hexagon Inscribed within a Circle

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Date: 2/29/96 at 18:3:21
From: Heather Huang
Subject: Hexagon in a circle

Hello -

My name is Heather Huang.  I was wondering if you could help me
solve a problem. Here goes:

A hexagon is inscribed within a circle.  Three consecutive sides
have a length of 3 and the other three consecutive sides have a
length of 5.  A chord is drawn within the circle which divides
the hexagon into two trapezoids.  One of the trapezoids has three
sides that measure 3 and the other trapezoid has three sides that
measure 5.  The length of the chord is m/n.  What is the value of
m + n?

If you could help me solve this problem I would be eternally
grateful!

Heather
```

```
Date: 5/29/96 at 2:50:10
From: Doctor Pete
Subject: Hexagon in a circle

Hello, Heather -

Let 2p be the central angle subtended by a side of length 3, and
let 2q be the central angle subtended by a side of length 5.
Then we have the following relations:

sin(p) = 3/(2R), [1] sin(3p)=x/(2R), [3]
sin(q) = 5/(2R), [2]  p+q = Pi/3,    [4]

where R is the radius of the circumscribed circle, and x is the
length we wish to find.  Then [3] gives

sin(3p) = sin(p)cos(2p)+sin(2p)cos(p)
= sin(p)(cos^2(p)-sin^2(p))+2*sin(p)cos^2(p)
= 3*sin(p)cos^2(p)-sin^3(p)
= 3*sin(p)(1-sin^2(p))-sin^3(p)
= 3*sin(p)-4*sin^3(p) = x/2R,

and substituting [1], we get

9/(2R)-4*27/(2R)^3 = x/2R
==> x = 9-27/R^2.

To find R, we substitute [4] into [2]:

sin(q) = sin(Pi/3-p)
= sin(Pi/3)cos(p)-sin(p)cos(Pi/3)
= (sqrt(3*(1-sin^2(p)))-sin(p))/2 = 5/(2R),

and we now substitute [1] into this to get

5/2R = (sqrt(3*(1-(3/2R)^2))-3/2R)/2
==>   5 = (sqrt(3*(4R^2-9))-3)/2
==>  13 = sqrt(3*(4R^2-9))
==> 169 = 3*(4R^2-9)
==> R^2 = (169/3+9)/4 = 196/12 = 49/3.

Therefore x = 9-27/(49/3) = 360/49, and the sum of the numerator
and denominator is 409.  (Hope that's right!)

-Doctor Pete,  The Math Forum

```
Associated Topics:
High School Conic Sections/Circles
High School Geometry
High School Triangles and Other Polygons

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