Find out if the (3,4,5) is the only possible right triangle whose semiperimeter equals its area?

Question

Please list some other such right triangle if possible. Also, I'd really appreciate a proper proof if (3,4,5) is the only such right triangle
The sides of the triangle must be positive integers.

Answer 1

Let triangle have sides 'a' and 'b' with hypotenuse 'c'.
By Pythagoras, c^2 = a^2 + b^2.
Semiperimeter = (a + b + c) / 2
Area = ab / 2

If semiperimeter = area, then :
(a + b + c) / 2 = ab / 2
Multiplying through by 2 gives :
a + b + c = ab
or, c = ab - a - b

Therefore, c^2 = (ab - a - b)^2,
but, as above, c^2 = a^2 + b^2.

Therefore, (ab - a - b)^2 = a^2 + b^2

Expanding gives :
a^2b^2 + a^2 + b^2 - 2a^2b - 2ab^2 + 2ab = a^2 + b^2

Subtracting a^2 + b^2 from both sides gives :
a^2b^2 - 2a^2b - 2ab^2 + 2ab = 0

We can divide through by ab because neither are zero.
This gives :
ab - 2a - 2b + 2 = 0

Factor out the 'a' and add 2b - 2 to both sides :
a(b - 2) = 2b - 2

Dividing by b - 2 gives :
a = (2b - 2) / (b - 2)

Dividing this using synthetic division gives :
a = 2 + 2 / (b - 2)

Now 'a' and '2' are integers, so 2 / (b - 2) must be an integer.
This can only be true if b - 2 = 1 or if b - 2 = 2.
That is, if b = 3 or b = 4.

If b = 3, then a = 4
and if b = 4, then a = 3.

In both cases, c = 5, so there is only one possible triangle.

Answer 2

I shall repeat the part by the 1st person and we get
The numerical value of the area of a right triangle is x y / 2, where x and y are the numerical values of the catheti. The numerical value of the semiperimeter of this triangle is (x + y + Sqrt(x^2 + y^2)) / 2, because the hypothenuse has a numerical value Sqrt(x^2 + y^2). Thus, your question reduces to the following equation:

xy / 2 = (x + y + Sqrt(x^2 + y^2)) / 2 <=>

xy = x + y + Sqrt(x^2 + y^2) <=>

xy - x - y = Sqrt(x^2 + y^2) /()^2 (be careful, the lhs has to be positive)

x^2 y^2 + x^2 + y^2 - 2 x^2 y - 2 x y^2 + 2 x y = x^2 + y^2 <=>
x^2 y^2 - 2 x^2 y - 2 x y^2 + 2 xy = 0 /: xy (this is always different from zero)

xy - 2x - 2y + 2 = 0

x (y - 2) = 2 (y - 1)
from the x = 2(y-1)/(y-2)

clearly y-1 and y-2 are coprimes unless y-2 = 1 which gives y = 3 and x = 4

so y-2 must be a factor of 2 and must be integer
so y-2 = 1 or 2
y = 3 or 4
y =3 has been considered
y =4 gives x = 3
so (3,4) or (4,3) both being same (3,4) are the 2 legs and (3,4,5) are the 3 sides
and there is no other solution

Answer 3

The numerical value of the area of a right triangle is x y / 2, where x and y are the numerical values of the catheti. The numerical value of the semiperimeter of this triangle is (x + y + Sqrt(x^2 + y^2)) / 2, because the hypothenuse has a numerical value Sqrt(x^2 + y^2). Thus, your question reduces to the following equation:

xy / 2 = (x + y + Sqrt(x^2 + y^2)) / 2 <=>

xy = x + y + Sqrt(x^2 + y^2) <=>

xy - x - y = Sqrt(x^2 + y^2) /()^2 (be careful, the lhs has to be positive)

x^2 y^2 + x^2 + y^2 - 2 x^2 y - 2 x y^2 + 2 x y = x^2 + y^2 <=>
x^2 y^2 - 2 x^2 y - 2 x y^2 + 2 xy = 0 /: xy (this is always different from zero)

xy - 2x - 2y + 2 = 0

x (y - 2) = 2 (y - 1)

This equation is with two unknowns, so I'm guessing you're asking for integer solutions (x, y). Then we have a Diophantine equation. The only non-trivial solutions of this equation are (3,4) and (4,3), which is the same!

Answer 4

All integer right triangles have the form (for suitable integers k, m, n):
k(n²-m²), 2kmn, k(n²+m²)

So you want to know what for k, n, m:
(k(m²-n²) + 2kmn + k(n²+m²))/2 = k(n²-m²) * 2kmn / 2
n² + nm = k(n²-m²)nm
n+m = k(n+m)(n-m)m
1 = k(n-m)m

This can happen only if k=1, m=1, n-m=1
so that n=2
Thus (a, b, c) = (3, 4, 5) is the only right triangle.


PS. If you ask how many right triangles have their area equal to their perimiter, solve it the same way to get
2 = k(n-m)m
leading to triangles
(5, 12, 13) and (6, 8, 10)

Answer 5

{3,4,5} * (2^2) or (3^2) or "(any higher positive whole number^2)" depicts same area perimeter relations you have stated.

Further we will find said endless possibilities in decimal values. 'In between a-zero and a-one', which will start from {0,3, 0.4, 0.5} *1/1^2 or 1/2^2 or 1/3^2 or 1/ any higher (whole number-denominator)^2 .

However decimal values are not easy to plot and verify on computers!

(1) in between 0 and 0.99999999999... and (2) in between 0 and 99999999999..., same principle applies!

Answer 6

consider that a,b,c - sides of the triangle.
Solve the system:
{
F(a,b,c) = a+b+c ,
a+b>c
a>0, b>0, c>0
}
where F(a,b,c) = S (formula to calc area)
---------------------------------------
Choose from result those a,b,c that
a,b,c E N

Answer 7

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