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A232329
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Integer areas A of the integer-sided triangles such that the product of the inradius and the circumradius is a square.
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1
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42, 168, 378, 672, 1050, 1512, 2058, 2088, 2688, 3000, 3402, 4200, 5082, 6048, 6960, 7098, 8232, 8352, 9450, 10752, 12000, 12138, 13608, 15162, 16800, 18522, 18792, 20328, 22218, 24192, 26250, 27000, 27840, 28392, 30618, 31416, 32928, 33408, 35322, 36000, 37800, 40362
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OFFSET
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1,1
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COMMENTS
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The areas of the primitive triangles of sides (a, b, c) and inradius, circumradius equals respectively to r and R are 42, 3000,... The sides of the nonprimitive triangles are of the form (a*k, b*k, c*k) with r’ = r*k and R’=R*k where r’, R’ are respectively the inradius and the circumradius of the nonprimitive triangles. The areas A’ of the nonprimitive triangles are A’ = A*k^2. The set {A016850} (numbers (5n)^2) is included in the set of the products r*R (see the table below).
The area A of a triangle whose sides have lengths a, b, and c is given by Heron's formula: A = sqrt(s*(s-a)*(s-b)*(s-c)), where s = (a+b+c)/2. The inradius r is given by r = A/s and the circumradius is given by R = abc/4A.
The product r*R is given by r*R = abc/2(a+b+c).
The following table gives the first values (A, a, b, c, r, R, r*R).
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| A | a | b | c | r | R | r*R |
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| 42 | 7 | 15 | 20 | 2 | 25/2 | 5^2 |
| 168 | 14 | 30 | 40 | 4 | 25 | 10^2 |
| 378 | 21 | 45 | 60 | 6 | 75/2 | 15^2 |
| 672 | 28 | 60 | 80 | 8 | 50 | 20^2 |
| 1050 | 35 | 75 | 100 | 10 | 125/2 | 25^2 |
| 1512 | 42 | 90 | 120 | 12 | 75 | 30^2 |
| 2058 | 49 | 105 | 140 | 14 | 175/2 | 35^2 |
| 2688 | 56 | 120 | 160 | 16 | 100 | 40^2 |
| 3000 | 80 | 85 | 85 | 24 | 289/6 | 34^2 |
| 3402 | 63 | 135 | 180 | 18 | 225/2 | 45^2 |
| 4200 | 70 | 150 | 200 | 20 | 125 | 50^2 |
| 5082 | 77 | 165 | 220 | 22 | 275/2 | 55^2 |
| 6048 | 84 | 180 | 240 | 24 | 150 | 60^2 |
| 6960 | 58 | 300 | 338 | 20 | 845/4 | 65^2 |
| 7098 | 91 | 195 | 260 | 26 | 325/2 | 65^2 |
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REFERENCES
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Mohammad K. Azarian, Circumradius and Inradius, Problem S125, Math Horizons, Vol. 15, Issue 4, April 2008, p. 32.
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LINKS
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EXAMPLE
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a(1) = 42 because, for (a,b,c) = (7, 15, 20):
the semiperimeter s = (7+15+20)/2 =21, and
A = sqrt(21*(21-7)*(21-15)*(21-20)) = 42
R = abc/4A = 7*15*20/(4*42) = 25/2
r = A/s = 42/21 = 2, hence r*R = 25 is a square.
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MATHEMATICA
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nn=800; lst={}; Do[s=(a+b+c)/2; rr=a*b*c/(2*(a+b+c))
; If[IntegerQ[s], area2=s(s-a)(s-b)(s-c); If[0<area2&&IntegerQ[Sqrt[area2]] &&IntegerQ[Sqrt[rr]], AppendTo[lst, Sqrt[area2]]]], {a, nn}, {b, a}, {c, b}]; Union[lst]
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PROG
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(PARI) lista(nn)=lst=[]; for (a = 1, nn, for (b=1, a, for (c=1, b, s=(a+b+c)/2; rr=a*b*c/(2*(a+b+c)); if ((type(s) == "t_INT") && (type(rr) == "t_INT"), area2=s*(s-a)*(s-b)*(s-c); if ((0<area2) && issquare(area2) && issquare(rr), lst = concat(lst, sqrtint(area2)); ); ); ); ); ); Set(lst); \\ after Mathematica; Michel Marcus, Jun 09 2015
(PARI) {for(a=20, 10000, forstep(b=a, 2, -1, forstep(c=min(b, a+b-1), a-b+1, -1, if((a+b+c)%2<1, s=(a+b+c)/2; if(issquare(s*(s-a)*(s-b)*(s-c), &A),
if((a*b*c)%(2*(a+b+c))<1&&if(issquare(a*b*c/(2*(a+b+c)), &d),
print([A, a, b, c, s, d]))))))))} \\ Faster version uesd for afile. Zak Seidov, Jun 06 2015
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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Missing term 33408 added by Zak Seidov, Jun 08 2015
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STATUS
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approved
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