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<h4>
   Partial answer to Question arising from Problem 4 plus more questions.
  </h4>

<p><b>Problem:</b>
  Suppose Q is a (convex) quadliateral with sides 5, 7, 8 and 9.  Suppose also that the spread of the diagonals is 3/4.   Find the diagonals of the quadrilateral.  Generalize.
  </p>

<p>
  <b>Question:</b><em> Find the quadrilaterals with rational diagonal spread whose sides have integer length.</em>
<br>There are some easy examples -- Any kite with sides of integer length (in fact, <em>any kite </em>) has a spread of 1.  Also, <em>any</em> quadrilateral with integer sides can be flexed to an infinite number of positions so that the diagonal spread is rational, although in general it is difficult to see how to obtain even one position for a random quadrilateral.
<br><br><b>Integer quadrilaterals with rational spread using one Pythagorean triangle.</b>
  <br>We can form a quadrilateral from a single Pythagorean triangle (a right triangle whose sides have integer length $a,b$, and whose hypotenuse has integer length $c$) in several ways.  Here are some.
<br><br>Three ways involve taking two copies of the triangle:  1) Join them at the hypotenuse to form an $a$ by $b$ rectangle.  2) Also join them on side $a$ to form a $c$ by $b$ parallelogram or 3) on side $b$ to form a $c$ by $a$ parallelogram.

<br><br>Another three ways: 1) Take an integer $n\gt a$ and form a rectangle $b$ by $n$. Now cut off a triangular corner: $b$ by $a$, forming a trapezoid with sides of length $n$, $b$, $n-a$, and $c$.   2) Form a rectangle $2\,b$ by $3\,a$ and from one end cut off a $2\,b$ by $2\,a$ corner.  Then from the other end cut off two $a$ by $b$ corners.  What's left will be a trapezoid with sides $2\,a$, $c$, $c$, and $2\,c$.

Pythagorean triples ($a,\,b,\,c$) are generated  by picking random positive integers $x,y$ with $x\lt y$ and forming $a=2\,x\,y$, $b=y^2-x^2$, and $c=x^2+y^2$.  Below is a Sagelet which can be used to generate quadrilaterals in this class.
  </p>


<div id="mycell"><script type="text/x-sage">
def triple(a=3,b=5):
    return (2*a*b,b^2-a^2,a^2+b^2)
def spread(V1,V2):
    return det(matrix([V1,V2]))^2/((V1*V1)*(V2*V2))
@interact(layout={'top':[['a','b']]})
def quad1(a=input_box(1,width=6,label='x'),b=input_box(2,width=6,label='y'),auto_update=false):
    if not type(a)==Integer or not type(b)==Integer:
        print('x and y must be integers.')
        return
    if not a>0 or not a < b:
        print('Need 0 < x < y')
        return
    x,y,z=triple(a,b)
    A=(0,0);B=(2*x,0);C=(3*x,y) ;D=(2*x,2*y)
    sides=[2*x,z,z,2*z]
    A,B,C,D=map(vector,[A,B,C,D])
    R=[A,vector((3*x,0)),vector((3*x,2*y)),vector((0,2*y))]
    spd=spread(C-A,D-B)
    spd1=spread(R[2]-A,D-R[1])
    ang=(arcsin(sqrt(spd))*180/pi).n(digits=5)
    ang1=(arcsin(sqrt(spd1))*180/pi).n(digits=5)
    figure=polygon(R,alpha=.2)
    figure1=figure+polygon([A,B,C,D],color='red', alpha=.5,aspect_ratio=1,figsize=4,title="Sides = "+ str(sides)+"  Spread = "+str(spd)+"; Angle approx "+str(ang))
    figure1=figure1+line([A,C],color='black')+line([B,D],color='black')

    show(figure1)
    sides1=[3*x,2*y,x,2*z]
    figure2=figure+polygon([A,R[1],R[2],D],color='green', alpha=.5,aspect_ratio=1,figsize=4,title="Sides = "+ str(sides1)+"  Spread = "+str(spd1)+"; Angle approx "+str(ang1))
    figure2=figure2+line([A,R[2]],color='black')+line([R[1],D],color='black',aspect_ratio=1)

    show(figure2)

  </script>
</div>
<p>
  <b>Problem.</b>  Show that the first three ways can be accomplished in a manner similar to the way the second three ways proceed.
  </p>
<p>
  <b>Related Question.</b> <em>Is there a procedure for obtaining an arbitrary quadrilateral with integer sides and rational spread by cutting Pythagorean triangles off a rectangle with integer sides?</em>
  </p>
<p>
  <b>Related Question.</b> <em>Given a positive rational number $r\lt 1$, find a quadrilateral with integer sides and diagonal spread $r$.</em>

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Collaborative Calculation and Data Science

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