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We need to find all cyclic quadrilaterals (or formulas that gives its sides), which have integer sides $a,b,c,d$. The constrain is that its area must be an integer multiple of its perimeter. We can find its area by Brahmagupta's formula: $$s=(a+b+c+d)/2 \implies A=[(s-a)(s-b)(s-c)(s-d)]^{1/2}$$ And as you know its perimeter is $P=a+b+c+d=2s$. For all integer $k$, we need to find $(a,b,c,d)$ integer quadruples that satisfy $A=kP$.

  1. aabc
  2. abcd
aslan95
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  • The case of integer multiple $k = A/P = 1$ has been completely solved previously (see this Question and links to earlier posts). It seems to me that simply scaling up the nine cases (which include five degenerate quadrilaterals) by a factor $k$ will produce solutions to this Question. However it might not be true that all solutions to this Question are produced in such fashion (and my intuition is that it would not). – hardmath Oct 18 '14 at 03:44

2 Answers2

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To begin, write the formula of the solution of the following equation:

$$S_4=\sqrt{(a+b+c)(a+b+d)(a+c+d)(b+c+d)}$$

Formulas of the solutions can be written.

$$a=((t^2+k^2)s-tkp)pn$$

$$b=((t^2+k^2)s-tkp)pj$$

$$c=(tkp^2-t^2ps+tks^2)n+(tkp^2-(t^2-k^2)ps-tks^2)j$$

$$d=(tkp^2+(t^2-k^2)ps-tks^2)n+(tkp^2-k^2ps+tks^2)j$$

The square will be equal to:

$$S_4=pstk((2pk-ts)j+(pk+ts)n)((pt+sk)j+(2pt-sk)n)$$

And the perimeter.

$$P_4=a+b+c+d=p((ktp+sk^2)j+t(pk+ts)n)$$

$p,k,t,s,j,n $ - Integers asked us.

And now about the idea of solving the equation:

$$S_4=FP_4$$

$F$ - the integer specified by the problem statement. To do this, put this number on the multiplier. And we will solve the linear equation in the unknown $j,n$ .

$$st((2pk-ts)j+(pk+ts)n)=F$$

The number $p,k$ determine from the equation.

$$ptk=s(t^2-k^2)$$

individ
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  • Does this really answer the Question? Are you able to solve the case $k=1$ with this approach? Posting a partial Answer to a difficult Question is acceptable, provided the post makes clear what limited claims are being advanced. – hardmath Oct 18 '14 at 15:20
  • @hardmath You want I showed all arithmetic? I formula recorded and then everything is simple. – individ Oct 18 '14 at 15:45
  • What I'm trying to ask is not for showing "all arithmetic" but simply an indication of the extent to which you think your Answer tells, given integer $k$, how "to find $(a,b,c,d)$ integer quadruples that satisfy $A = kP$." For example, can your method tell us how many solutions exist for (say) $k=5$? – hardmath Oct 19 '14 at 01:04
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I became entangled in the problem and decided another equation. It is necessary to solve this equation:

$$S_4=\frac{1}{4}\sqrt{(a+b+c-d)(a+b-c+d)(a-b+c+d)(b-a+c+d)}$$

Solutions will be:

$$a=(pt+ks)(ts-pk)n$$

$$b=(pt+ks)(ts-pk)j$$

$$c=tk(p^2+s^2)j-ps(t^2+k^2)n$$

$$d=ps(t^2+k^2)j-tk(p^2+s^2)n$$

$p,t,k,s,j,n - $ integers asked us. Then the area of the quadrilateral are equal.

$$S_4=pstk(ts-pk)(pt+sk)(j-n)(j+n)$$

And its perimeter is equal to:

$$P_4=2(pt+sk)(tsj-pkn)$$

To solve the equation:

$$S_4=FP_4$$

When the number of $F$ is set for the problem. Come to the need to solve the following equation:

$$pstk(ts-pk)(j-n)(j+n)=2F(tsj-pkn)$$

Left to think. What is the easiest way to solve this equation.

individ
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