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Integer area of primitive cyclic quadrilaterals with integer sides and rational radius.
(history; published version)
#5 by N. J. A. Sloane at Sat May 28 07:29:54 EDT 2016
STATUS

editing

approved

#4 by N. J. A. Sloane at Sat May 28 07:29:25 EDT 2016
NAME

Integer area of primitive cylic quadrilateral cyclic quadrilaterals with integer side sides and rational radius.

COMMENTS

Primitive means a,b,c,d share no common factorsfactor.

The area S = sqrt[(s-a)(s-b)(s-c)(s-d)] where s=(a+b+c+d)/2 is the semiperimeter.

The length of the diagonal separating a-b and c-d is (4S R)/(a b+c d), the other diagonal can be obtain by swapping b,c or swapping b,d.

It follows that if side the sides and area are integers, then (any diagonal is rational) <=> (circumradius is rational) <=> (all diagonals are rational).

From empirical observation, the area seems to be a multiple of 6. (If so, the program can could be modified to run 6 times fasteras fast.)

STATUS

proposed

editing

Discussion
Sat May 28
07:29
N. J. A. Sloane: Many edits were needed to grammar - hope that is OK
#3 by Albert Lau at Sat May 28 04:30:29 EDT 2016
STATUS

editing

proposed

#2 by Albert Lau at Sat May 28 03:52:17 EDT 2016
NAME

allocated for Albert LauInteger area of primitive cylic quadrilateral with integer side and rational radius.

DATA

12, 60, 108, 120, 120, 168, 192, 192, 234, 240, 300, 360, 360, 420, 420, 420, 420, 420, 420, 432, 540, 540, 588, 600, 660, 660, 714, 768, 840, 924, 960, 960, 966, 1008, 1008, 1008, 1080, 1080, 1080, 1092, 1134, 1200

OFFSET

1,1

COMMENTS

Given 4 segments a,b,c,d, there is a unique circumcircle such that these segments can be placed inside to form cyclic quadrilaterals. There are 3 ways to place these segments: abcd,acbd,adbc.

Primitive means a,b,c,d share no common factors.

The area S=sqrt[(s-a)(s-b)(s-c)(s-d)] where s=(a+b+c+d)/2 is the semiperimeter.

The circumradius R=Sqrt[a b+c d]*Sqrt[a c+b d]*Sqrt[a d+b c]/(4S)

The length of diagonal separating a-b and c-d is (4S R)/(a b+c d), the other diagonal can be obtain by swapping b,c or swapping b,d.

It follows that if side and area are integers, then (any diagonal is rational) <=> (circumradius is rational) <=> (all diagonals are rational).

From empirical observation, the area seems to be multiple of 6. (If so, the program can be modified to run 6 times faster)

EXAMPLE

a, b, c, d, S, r

4, 4, 3, 3, 12, 5/2

12, 12, 5, 5, 60, 13/2

14, 13, 13, 4, 108, 65/8

15, 15, 8, 8, 120, 17/2

21, 10, 10, 9, 120, 85/8

24, 24, 7, 7, 168, 25/2

21, 13, 13, 11, 192, 65/6

25, 15, 15, 7, 192, 25/2

24, 20, 15, 7, 234, 25/2

MATHEMATICA

SMax=1200;

Do[

x=S^2/(u v w);

If[u+v+w+x//OddQ, Continue[]];

If[v+w+x<=u, Continue[]];

{a, b, c, d}=(u+v+w+x)/2-{x, w, v, u};

If[GCD[a, b, c, d]>1, Continue[]];

R=(Sqrt[v w+u x]Sqrt[u w+v x]Sqrt[u v+w x])/(4S);

If[R\[NotElement]Rationals, Continue[]];

S(*{a, b, c, d, "", S, R, "", (4S R)/(a d+b c), (4S R)/(a c+b d), (4S R)/(a b+c d)}*)//Sow;

, {S, 1(*6*), SMax, 1(*6*)}(*assuming S mod 6 = 0, set to 6 to run faster*)

, {u, S^2//Divisors//Select[#, S<=#^2&&#<=S&]&}

, {v, S^2/u//Divisors//Select[#, S^2<=u#^3&&u/3<#<=u&]&}

, {w, S^2/(u v)//Divisors//Select[#, S^2<=u v#^2&&(u-v)/2<#<=v&]&}

]//Reap//Last//Last(*//TableForm*)

{S, R, x, a, b, c, d}=.;

KEYWORD

allocated

nonn

AUTHOR

Albert Lau, May 28 2016

STATUS

approved

editing

#1 by Albert Lau at Sat May 28 03:52:17 EDT 2016
NAME

allocated for Albert Lau

KEYWORD

allocated

STATUS

approved