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Bicentric Trapezoids (Posted on 2010-07-24) Difficulty: 3 of 5
A certain trapezoid happens to be a bicentric quadrilateral. The trapezoid's circumradius is twice the inradius. What are the measures of the angles of the trapezoid?

If the circumradius is five times the inradius, then what are the measures of the angles?

What is the ratio of the circumradius and inradius when the longer base of the trapezoid is a diameter of the circumcircle?

See The Solution Submitted by Brian Smith    
Rating: 3.0000 (1 votes)

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Solution Solution Comment 1 of 1

Let r, R, and s be the lengths of the
inradius, circumradius, and the distance
betweeen the incenter and circumcenter
respectively. The relation between these
three is
   (R^2 - s^2)^2 = 2*r^2*(R^2 + s^2).  (*)
Solving for s^2 gives
   s^2 = [r^2 + R^2] - r*sqrt{r^2 + 4*R^2]     
Let the shorter of the two parallel lines
have length 2a and the longer 2b. Clearly,
the slant sides have an equal length of
a + b.
Using the Pathagorean theorem we have
   a^2 = R^2 - (r + s)^2
   b^2 = R^2 - (r - s)^2
   r^2 = a*b
The four angles of the trapezoid have 
two values that are supplementary. The
smaller of these two values is
   arccos([b - a]/[b + a]) 
     = arccos([b^2 - a^2]/[b + a]^2)
     = arccos(2*r*s/[R^2 - s^2])
With R = kr,
   s^2 = r^2*[(k^2 + 1) - sqrt(4*k^2 + 1)]
   2*r*s = 2*r^2*(sqrt[4*k^2 + 1] + 1)*
           sqrt[(k^2 + 1) - sqrt(4*k^2 + 1)]
   R^2 - s^2 = r^2*[sqrt(4*k^2 + 1) - 1]
   arccos(2*r*s/[R^2 - s^2]) =
     arccos(sqrt[([2*k^2 - 1] -
       sqrt[4*k^2 + 1])/(2*k^2)]
Problem#1 (k = 2)
   arccos(sqrt[(7 - sqrt[17])/8]) 
          ~= 53.1533 deg.
Problem#2 (k = 5)
   arccos(sqrt[(49 - sqrt[101])/50]) 
          ~= 28.0410 deg.
Problem#3 (2*R = 2*b  <==>  s = r)
   Solving equation (*) for R,
     (R^2 - r^2)^2 = 2*r^2*(R^2 + r^2)
                   or
     R^2 = r^2*(2 + sqrt[5])
   Therefore,
     R/r = sqrt(2 + sqrt[5]) ~= 2.0582

 

  Posted by Bractals on 2010-07-24 12:45:46
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