## Primitive Heron triangles(a,b,c) and abc(a+b+c)=x^2, (b+c,c+a,a+b) is also Heron

Here are some primitive Heron triangles $(a, \; b, \; c)$   for which

$a \, b \, c \, (a + b + c)$   is a perfect square.

then,   $(b+c, \; c+a, \; a+b)$   is also a Heron.

Can you find other such primitive Heron triangles?

(5,5,8)   ……………   isosceles obtuse triangle
(13,13,10)   …………   isosceles acute triangle

(32,65,65)   …………   isosceles acute triangle
(130,97,97)   ………..   isosceles acute triangle

(72,85,85)   …………   isosceles acute triangle
(170,157,157)   ………   isosceles acute triangle

(72,325,325)   ……….   isosceles acute triangle
(650,397,397)   ………   isosceles obtuse triangle

(128,1025,1025)   …….   isosceles acute triangle
(2050,1153,1153)   ……   isosceles obtuse triangle

(200,629,629)   ………   isosceles acute triangle
(1258,829,829)   ……..   isosceles obtuse triangle

(200,2501,2501)   …….   isosceles acute triangle
(5002,2701,2701)   ……   isosceles obtuse triangle

(288,145,145)   ………   isosceles obtuse triangle
(290,433,433)   ………   isosceles acute triangle

(288,5185,5185)   …….   isosceles acute triangle
(10370,5473,5473)   …..   isosceles obtuse triangle

(392,2405,2405)   …….   isosceles acute triangle
(4810,2797,2797)   ……   isosceles obtuse triangle

Claim :

When the Heron   $(a, \; b, \; c)$   is isosceles and   $a \, b \, c \, (a + b + c)$   is a square, then the perimeter of the triangle   $(b+c, \; c+a, \; a+b)$   is a square number.

Can you explain why?

**************************************************

Paul found examples of primitive Heron   (non-isosceles)   for which the perimeter of the triangle   $(b+c, \; c+a, \; a+b)$   is not a square number:

(1)

$(29, \; 174, \; 175)$
area   =   2520
perimeter   =   378

$abc(a+b+c) \; = \; 29\times 174\times 175\times(29+174+175) = 333792900 = 18270^2$

$(b+c, \; c+a, \; a+b) \; = \; (349, \; 204, \; 203)$
area   =   18270
perimeter   =   756   =   $2^2\times 3^3\times 7$

(2)

$(39, \; 221, \; 250)$
area   =   3060
perimeter   =   510

$abc(a+b+c) = 39\times 221\times 250\times(39+221+250) = 1098922500 = 33150^2$

$(b+c, \; c+a, \; a+b) \; = \; (471, \; 289, \; 260)$
area   =   33150
perimeter   =   1020   =   $2^2\times 3\times 5\times 17$

(3)

$(150, \; 169, \; 275)$
area   =   11088
perimeter   =   594

$abc(a+b+c) = 150\times 169\times 275\times(150+169+275) = 4140922500 = 64350^2$

$(b+c, \; c+a, \; a+b) = (444, \; 425, \; 319)$
area   =   64350
perimeter   =   1188   =   $2^2\times 3^3\times 11$

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## About benvitalis

math grad - Interest: Number theory
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### 4 Responses to Primitive Heron triangles(a,b,c) and abc(a+b+c)=x^2, (b+c,c+a,a+b) is also Heron

1. paul says:

There is another one that doesn’t have a square perimeter (last column). Format follows as above

{{5,5,8},12,18,60,{13,13,10},60,6}

{{29,174,175},2520,378,18270,{349,204,203},18270,6 Sqrt[21]}

{{32,65,65},1008,162,4680,{130,97,97},4680,18}
{{72,85,85},2772,242,11220,{170,157,157},11220,22}.

Paul.

• paul says:

a couple more the same

{{39,221,250},3060,510,33150,{471,289,260},33150,2 Sqrt[255]}
{{150,169,275},11088,594,64350,{444,425,319},64350,6 Sqrt[33]}

P.

• benvitalis says:

Good catch! My statement is not accurate. It works for Heron isosceles

• benvitalis says:

I’ve modified my claim.