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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A273752 Integer area of primitive bicentric quadrilateral with integer side, rational inradius and rational circumradius. Excluding right kites.

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%I A273752 #13 Aug 17 2018 19:47:27
%S A273752 7140,16380,87780,1543668,1697892,4444440,5858580
%N A273752 Integer area of primitive bicentric quadrilateral with integer side, rational inradius and rational circumradius. Excluding right kites.
%C A273752 Bicentric quadrilaterals have the following properties:
%C A273752 1. a+c = b+d = s where s is the semiperimeter;
%C A273752 2. A+C = B+D = 180 degrees;
%C A273752 2. Area S = sqrt(a b c d);
%C A273752 3. Circumradius R = sqrt(a*b + c*d)*sqrt(a*c + b*d)*sqrt(a*d + b*c)/S;
%C A273752 4. Inradius r = S/s (it follows that r is always rational if sides and area are integers);
%C A273752 5. Length of the diagonal separating a-b and c-d is (4S*R)/(a*b + c*d), the other diagonal can be obtained by swapping b,c or swapping b,d. It follows that all diagonals are rational iff a,b,c,d,R,S are rationals.
%C A273752 There are only 7 primitive cases which are not right kites for S < 10^7.
%C A273752 From empirical observation, the area seems to be a multiple of 84. (If proven, the program could be modified to run 84 times as fast.)
%C A273752 Special cases of bicentric quadrilaterals are right kites and isosceles trapezium.
%C A273752 Integer right kites can be generated by joining two (a,b,c) Pythagorean triangles, which gives S=a b/2, R=c/2, r=ab/(a+b+c).
%C A273752 Integer isosceles trapezium is impossible. Proof:
%C A273752 1. Let the sides of integer isosceles trapezium be (s-t,s,s+t,s);
%C A273752 2. S = s*sqrt(s^2 - t^2) and R = 2*s^2*sqrt(2s^2 - t^2)/S;
%C A273752 3. s^2 - t^2 and 2s^2 - t^2 are perfect squares;
%C A273752 4. Let u^2 = 2s^2 - t^2, v^2 = s^2 - t^2;
%C A273752 5. t^2,s^2,u^2 is an arithmetic progression with common difference = v^2;
%C A273752 6. Fermat's right triangle theorem states that no integer solution exists, except v=0 which corresponds to (0,s,2s,s), a degenerate quadrilateral. QED.
%H A273752 Wikipedia, <a href="https://en.wikipedia.org/wiki/Bicentric_quadrilateral">Bicentric quadrilateral</a>.
%H A273752 Wikipedia, <a href="https://en.wikipedia.org/wiki/Fermat%27s_right_triangle_theorem">Fermat's right triangle theorem</a>.
%e A273752 All examples with S < 10^7:
%e A273752 a,    b,    c,    d,    S,       R,      r
%e A273752 204,  140,  85,   21,   7140,    442,    476/15
%e A273752 315,  260,  91,   36,   16380,   650,    140/3
%e A273752 440,  399,  231,  190,  87780,   1885/2, 418/3
%e A273752 2397, 1564, 1316, 483,  1543668, 4810,   128639/240
%e A273752 4756, 3451, 1428, 123,  1697892, 15130,  348
%e A273752 2849, 2184, 2145, 1480, 4444440, 6290,   3080/3
%e A273752 5460, 5365, 1131, 1036, 5858580, 11050,  7215/8
%t A273752 SMin=7140;
%t A273752 SMax=16380(*WARNING: runs very slow*);
%t A273752 dS=1(*assuming S mod 84 = 0, set to 84 to run faster*);
%t A273752 Do[
%t A273752   s=(a+b)/2+Sqrt[(a-b)^2/4+S^2/(a b)];
%t A273752   If[s//IntegerQ//Not,Continue[]];
%t A273752   If[GCD[a,b,s]>1,Continue[]];
%t A273752   R=(Sqrt[#1#2+#3#4]Sqrt[#1#3+#2#4]Sqrt[#1#4+#2#3])/S&[a,b,s-b,s-a];
%t A273752   If[R\[NotElement]Rationals,Continue[]];
%t A273752   S(*{a,b,s-b,s-a,S,R,S/s}*)//Sow;
%t A273752   ,{S,Round[SMin,dS],SMax,dS}
%t A273752   ,{a,S^2//Divisors//Select[#,S<#^2&&#<S&]&}
%t A273752   ,{b,S^2/a//Divisors//Select[#,a/2<#<a&&1+a-#<=S^2/(a#)<=#(2#-a)&]&}
%t A273752 ]//Reap//Last//Last(*//TableForm*)
%t A273752 {S,R,a,b,s}=.;
%K A273752 nonn,more
%O A273752 1,1
%A A273752 _Albert Lau_, May 29 2016