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?

 

HERON abc a+b+c 1

 

(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
This entry was posted in Number Puzzles and tagged . Bookmark the permalink.

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.

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