Question 1209330
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        The solution in the post by @ElectricPavlov is incorrect.


        It is incorrect, since it uses 


                "the distance  CF = side length of the heptagon = 1"   in  n.8,


        which is  FATALLY  WRONG.



<pre>
The side of this regular heptagon is 1 (given).  

Let O be the center of the heptagon ABCDEFG.

Let the radius of the circumscribed circle around the heptagon be r


Its central angle is a = {{{360/7}}} = 51.4286 degrees.


For the radius r we have this equation

    r*sin(a/2) = 1/2,  which gives  r = {{{0.5/sin(25.7143)}}} = {{{0.5/0.43388}}} = 1.1524.


Now consider triangle OCF.  It is isosceles triangle.

Its lateral sides OC and OF have the length r, and they conclude the angle COF of 3a = {{{3*(360/7)}}} = 154.2857 degrees.


So, the length of CF is (use the cosine law for triangle OCF)

    |CF| = {{{sqrt(r^2+r^2-2*r*r*cos(154.2857^o))}}} = {{{r*sqrt(2-2*(-0.900968))}}} = {{{1.1524*sqrt(3.801936)}}} = 2.247.



Let O' be the center of the circle, which touches  CD at C  and  touches EF at F.

Let R be the radius of this circle, which the problem asks to determine.


The angle at O' between perpendiculars to CD at C  and  to EF at F is 2a.


Now apply the cosine law to triangle  O'CF

    R^2 + R^2 - 2R*R*cos(2a) = |CF|^2


and find

    R = {{{abs(CF)/sqrt(2-2*cos(2a)))}}} = {{{2.247/sqrt(2-2*cos(102.8571))}}} = {{{2.247/sqrt(2-2*(-0.2252))}}} = 1.4354.    <U>ANSWER</U>

</pre>

It is how the problem &nbsp;SHOULD  &nbsp;be solved, &nbsp;if to do it correctly.