Lesson A problem on a regular heptagon

A problem on a regular heptagon

Problem 1

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 =  = 51.4286 degrees.


For the radius r we have this equation

    r*sin(a/2) = 1/2,  which gives  r =  =  = 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 =  = 154.2857 degrees.


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

    |CF| =  =  =  = 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 =  =  =  = 1.4354.    ANSWER

Let's continue side DC from C to D and further,
and let's continue side EF from F to E and further until intersection at point H.


Also, draw a perpendicular to EF at point F and perpendicular to DC at point C 
till intersection at point P.


Then we have four figures/shapes that we will consider:

    - triangle CFH,

    - trapezoid CFED,

    - triangle CFP,

    - quadrilateral PCHF. 


Trapezoid CFED is isosceles due to symmetry.  Two its angles E and D are interior
angles of the regular heptagon ABCDEFG. Therefore, the measure of every of these two angles
is  = .


Let 'a' be the measure of any of two angles, ∠CFE and/or ∠FCD at the base of the trapezoid CFED.

The sum of all interior angles of any trapezoid is , so

    a + a + m(∠D) + m(∠E) = ,  

or

     +  = .


It gives  a =  -  = .



OK.  So, we know now that the angles  ∠HFC  and  ∠HCF  at the base of triangle CFH are    each.

Hence, angle CHF is   -  -  = .


Now we can easy calculate the base CF of the trapezoid CFED: it is

    CF = ED +  +  =  = 1 + 2*0.62360675659 = 2.247213513.



Now let's consider triangle CFP.  It is isosceles, CP = FP = R,  the radius of the circle,
which touches DC at C and EF at F.


We want to find its angle  ∠CPF  at  P.

For it, consider quadrilateral PCHF.  It has two right angles ∠PFH and ∠PCH.

We also know its angle  ∠CHF  at vertex H: it is  .

So, we can write the formula for the sum of its angles

    m(∠CPF) +  +  +  = .


From it, we find  

    ∠CPF =  -  -  -  = ^2 =  +  - 

or

     =  = ,

     = ,

    R =  =  = 1.436943888.


At this point, the solution is complete.


The radius of the circle, which is tangent to DC at C and to EF at F is about 1.4369 units, 
if to round with 4 decimal places.