Question 1208064
<pre>

Maybe Ikleyn or Greenestamps can simplify my solution but here it is at last.

Since quadrilaterals are normally lettered counter-clockwise, and PS is a
diameter, the quadrilateral must be inscribed in a semi-circle. I will need
some right triangles so I will bisect everything including the diameter. The
green line segments are the perpendicular bisectors of the upper 3 sides and
the angles as well since the triangles are isosceles. So the radius of the
circle is u/2, and will be the hypotenuse of all 6 right triangles in this figure:

{{{drawing(400,800/3,-1.5,1.5,-.5,1.5,

arc(0,0,2,-2,0,180), line(-1,0,1,0), 

line(0,0,-.3420201433,.9396926208),
line(0,0,.5735764364,.8191520443),

line(1,0,.5735764364,.8191520443), line(.5735764364,.8191520443,-.3420201433,.9396926208),
line(-.3420201433,.9396926208,-1,0),

locate(1,0,P), locate(.57,.95,Q),locate(-.35,1.08,R), locate(-1,0,S), 

green(line(0,0,.7867882182,.4095760221), line(0,0,.1157781465,.8794223325),

line(0,0,-.6710100717,.4698463104 )),
locate(0,0,O),
red(locate(.25,.15,alpha),
locate(.17,.25,alpha),
locate(.05,.28,alpha),
locate(-.065,.28,alpha),
locate(-.16,.23,beta),
locate(-.22,.13,beta)),

locate(.8,.3,v/2),
locate(.6,.64,v/2),
locate(.25,.89,v/2),
locate(-.15,.91,v/2),

locate(-.5,.73,w/2),

locate(-.77,.36,w/2),

locate(.5,0,u/2),
locate(-.5,0,u/2)



)}}}

[eqs. A]  {{{sin(alpha)="v/2"/"u/2"= v/u}}} and {{{sin(beta)="w/2"/"u/2"= w/u}}}

The sum of all 6 angles is 180<sup>o</sup>

{{{4alpha+2beta=180^o}}}
{{{2alpha+beta=90^o}}}
2&alpha; and &beta; are complementary,
{{{alpha+(alpha+beta)=90^o}}}
&alpha; and &alpha;+&beta; are also complementary

So {{{sin(alpha)=cos(alpha+beta)= cos(alpha)cos(beta)-sin(alpha)sin(beta)}}}

[eq. B]   {{{cos(alpha)cos(beta)=sin(alpha)+sin(alpha)sin(beta)}}}

{{{2alpha+beta=90^o}}}. Taking sines of both sides:

{{{sin(2alpha+beta)=1}}}

{{{sin(2alpha)cos(beta)+cos(2alpha)sin(beta)=1}}}
{{{(2sin(alpha)cos(alpha)^"")cos(beta)+cos(2alpha)sin(beta)=1}}}
Using the associative law and since 2&alpha; and &beta; are complementary,
{{{2sin(alpha)(cos(alpha)cos(beta)^"")+sin(beta)sin(beta)=1}}}
Substituting from [eq. B] above

{{{2sin(alpha)(sin(alpha)+sin(alpha)sin(beta)^"")+sin^2(beta)=1}}}

{{{2sin^2(alpha)+2sin^2(alpha)sin(beta)+sin^2(beta)=1}}}

From [eqs. A]

{{{2(v/u)^2+2(v/u)^2(w/u)+(w/u)^2=1}}}

{{{2v^2/u^2+(2v^2w)/u^3+w^2/u^2=1}}}

Multiply through by u<sup>3</sup>

{{{2uv^2+2v^2w+uw^2=u^3}}}

{{{(u)w^2+(2v^2)w+(2uv^2-u^3)=0}}}

Since {{{""+-u}}} are factors of the last term, we try both and
find that w=-u is a solution and so we factor it by synthetic 
division

-u | u      2v<sup>2</sup>   2uv<sup>2</sup>-u<sup>3</sup>
   |<u>        -u<sup>2</sup>  -2uv<sup>2</sup>+u<sup>3</sup></u>
     u   2v<sup>2</sup>-u<sup>2</sup>         0

{{{(w+u^""^"")(u*w+(2v^2-u^2)^"")=0}}}

{{{w+u=0}}};  {{{u*w+(2v^2-u^2)=0}}}
{{{w=-u}}};    {{{u*w=u^2-2v^2}}}
              {{{w=(u^2-2v^2)/u^""}}}
              {{{w=u-2v^2/u^""}}}
              {{{2v^2/u^""=u-w}}} 
              
So the term {{{2v^2/u^""}}} must be an integer, since the terms 
on the right are integers.

For contradiction, assume u is a prime number.

If u=2, then the radius is 1, and the semicircle is &pi;.

So {{{2v+w<pi}}}, that would make v=1, w=2. then triangle
SRO would have sides 1,1,2 which violates the triangle
inequality.  So u is not the prime number 2.

Thus u must divide evenly into v<sup>2</sup>.   But if u is a prime 
then u must divide evenly into v as well. This cannot be true 
because v < u, the diameter.

Thus we have a contradiction and u cannot be a prime number. 

Edwin</pre>