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A273752
Integer area of primitive bicentric quadrilateral with integer side, rational inradius and rational circumradius. Excluding right kites.
0
7140, 16380, 87780, 1543668, 1697892, 4444440, 5858580
OFFSET
1,1
COMMENTS
Bicentric quadrilaterals have the following properties:
1. a+c = b+d = s where s is the semiperimeter;
2. A+C = B+D = 180 degrees;
2. Area S = sqrt(a b c d);
3. Circumradius R = sqrt(a*b + c*d)*sqrt(a*c + b*d)*sqrt(a*d + b*c)/S;
4. Inradius r = S/s (it follows that r is always rational if sides and area are integers);
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.
There are only 7 primitive cases which are not right kites for S < 10^7.
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.)
Special cases of bicentric quadrilaterals are right kites and isosceles trapezium.
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).
Integer isosceles trapezium is impossible. Proof:
1. Let the sides of integer isosceles trapezium be (s-t,s,s+t,s);
2. S = s*sqrt(s^2 - t^2) and R = 2*s^2*sqrt(2s^2 - t^2)/S;
3. s^2 - t^2 and 2s^2 - t^2 are perfect squares;
4. Let u^2 = 2s^2 - t^2, v^2 = s^2 - t^2;
5. t^2,s^2,u^2 is an arithmetic progression with common difference = v^2;
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.
EXAMPLE
All examples with S < 10^7:
a, b, c, d, S, R, r
204, 140, 85, 21, 7140, 442, 476/15
315, 260, 91, 36, 16380, 650, 140/3
440, 399, 231, 190, 87780, 1885/2, 418/3
2397, 1564, 1316, 483, 1543668, 4810, 128639/240
4756, 3451, 1428, 123, 1697892, 15130, 348
2849, 2184, 2145, 1480, 4444440, 6290, 3080/3
5460, 5365, 1131, 1036, 5858580, 11050, 7215/8
MATHEMATICA
SMin=7140;
SMax=16380(*WARNING: runs very slow*);
dS=1(*assuming S mod 84 = 0, set to 84 to run faster*);
Do[
s=(a+b)/2+Sqrt[(a-b)^2/4+S^2/(a b)];
If[s//IntegerQ//Not, Continue[]];
If[GCD[a, b, s]>1, Continue[]];
R=(Sqrt[#1#2+#3#4]Sqrt[#1#3+#2#4]Sqrt[#1#4+#2#3])/S&[a, b, s-b, s-a];
If[R\[NotElement]Rationals, Continue[]];
S(*{a, b, s-b, s-a, S, R, S/s}*)//Sow;
, {S, Round[SMin, dS], SMax, dS}
, {a, S^2//Divisors//Select[#, S<#^2&&#<S&]&}
, {b, S^2/a//Divisors//Select[#, a/2<#<a&&1+a-#<=S^2/(a#)<=#(2#-a)&]&}
]//Reap//Last//Last(*//TableForm*)
{S, R, a, b, s}=.;
CROSSREFS
Sequence in context: A123292 A116180 A252303 * A213097 A063058 A214336
KEYWORD
nonn,more
AUTHOR
Albert Lau, May 29 2016
STATUS
approved