Let $ a,b,c$ - are the sides of triangle, $ p = \frac {a + b + c}{2}$. $ S = \sqrt {p(p - a)(p - b)(p - c)}$ We have for some positive integers $ x,y$: $ a = \frac {2xy}{x + y},b = \sqrt {xy},c = \frac {x + y}{2}$ We know, that: $ \frac {2xy}{x + y}\leq \sqrt {xy}\leq \frac {x + y}{2}$ and if $ a = b$ then $ a = b = c$ and so on... But in this case $ S = \frac {\sqrt {3}a^2}{4}$ and it is not an integer. So, here should be $ a < b < c$. We can represent Heron`s formula as: $ 16S^2 = 4a^2c^2 - (a^2 + c^2 - b^2)^2$ We know that $ ac = b^2$, so we can write: $ 16S^2 = 4a^2c^2 - (a^2 + c^2 - ac)^2$ $ 16S^2 = 3a^2c^2 - (a^2 + c^2)(c - a)^2$ Let $ c = \alpha a$ for some $ \alpha > 1$. We will have: $ 16S^2 = a^4(3{\alpha}^2 - ({\alpha}^2 + 1)(\alpha - 1)^2)$ or $ 4S = a^2\sqrt {3{\alpha}^2 - ({\alpha}^2 + 1)(\alpha - 1)^2}$ Because $ a$ is an integer $ \sqrt {3{\alpha}^2 - ({\alpha}^2 + 1)(\alpha - 1)^2}$ alsow should be integer. So $ f(\alpha) = 3{\alpha}^2 - ({\alpha}^2 + 1)(\alpha - 1)^2 = m^2$ for some integer $ m$. For $ 1 < \alpha\leq \frac {3 + \sqrt {5}}{2}$ we have $ 0\leq f(\alpha) < 8$. Alsow function $ f(\alpha)$ has only one maximum. So there exist only two squares $ m^2 = 4$ and $ m^2 = 1$ such that $ f(\alpha) = m^2$. In this case the area of the triangle should be: $ S = \frac {a^2}{2}$ or $ S = \frac {a^2}{4}$ But $ c = \alpha a$ should be integer. The equations $ f(\alpha) = 4$ and $ f(\alpha) = 1$ has only irrational roots. This means that does not exist such triangle.

Yuriy Solovyov wrote: We can represent Heron`s formula as: $ 16S^2 = 4a^2c^2 - (a^2 + c^2 - b^2)^2$ Can someone detail please!Thanks!

$ 16S^2 = (a + b + c)(a + b - c)(a + c - b)(b + c - a)$ $ 16S^2 = (a + c + b)(a + c - b)(b + c - a)(b - c + a)$ $ 16S^2 = ((a + c)^2 - b^2)(b^2 - (c - a)^2)$ $ 16S^2 = (a^2 + c^2 - b^2 + 2ac)(b^2 - c^2 - a^2 + 2ac)$ $ 16S^2 = (2ac + (a^2 + c^2 - b^2))(2ac - (c^2 + a^2 - b^2))$ $ \boxed{16S^2 = 4a^2c^2 - (a^2 + c^2 - b^2)^2}$ But my previous solution is wrong... If $ 16S^2 = a^2\sqrt {f(\alpha)}$ then $ f(\alpha)$ should be $ \left(\frac {p}{q}\right)^2$, but not $ m^2$ I will try to get more ideas about this problem...

Ok, I found a solution. The complete set of solutions for integer Heronian triangles (the three side lengths and area can be multiplied by their least common multiple to make them all integers) were found by Euler (Buchholz 1992; Dickson 2005, p. 193), and parametric versions were given by Brahmagupta and Carmichael (1952) as $ a = n(m^2 + k^2)$ $ b = m(n^2 + k^2)$ $ c = (m + n)(mn - k^2)$ In this case the area of the triangle will be $ S = kmn(m + n)(mn - k^2)$ This produces one member of each similarity class of Heronian triangles for any integers $ m$, $ n$, and $ k$ such that $ \gcd {(m,n,k)} = 1$, $ mn > k^2\geq\frac {m^2n}{2m + n}$, and $ m\geq n\geq 1$ (Buchholz 1992). Easy to show that $ c\leq b\leq a$. Return to our problem, we got that $ ac = b^2$. This means that: $ n(m^2 + k^2)(m + n)(mn - k^2) = m^2(n^2 + k^2)^2$ Easy to see, that left side is divisible by $ n$ and right - not, because $ \gcd {(m,n,k)} = 1$. We are done

Yuriy Solovyov wrote: Ok, I found a solution. The complete set of solutions for integer Heronian triangles (the three side lengths and area can be multiplied by their least common multiple to make them all integers) were found by Euler (Buchholz 1992; Dickson 2005, p. 193), and parametric versions were given by Brahmagupta and Carmichael (1952) as $ a = n(m^2 + k^2)$ $ b = m(n^2 + k^2)$ $ c = (m + n)(mn - k^2)$ In this case the area of the triangle will be $ S = kmn(m + n)(mn - k^2)$ This produces one member of each similarity class of Heronian triangles for any integers $ m$, $ n$, and $ k$ such that $ \gcd {(m,n,k)} = 1$, $ mn > k^2\geq\frac {m^2n}{2m + n}$, and $ m\geq n\geq 1$ (Buchholz 1992). Sorry,I am wondering if it is a complete set of integer Heronian triangles. How about this one,$ a=43,b=68,c=61$,its area is $ 1290$ But we found no $ (m,n,k)$ to verify it

Quote: Sorry,I am wondering if it is a complete set of integer Heronian triangles. How about this one,a=68,b=61,c=43. But we found no m,n,k to verify it. Try this one: m=25,n=18, k=15 and divide the results by 225 (with the given constraints we must have a>=b>=c, thus I switched between a,b,c) A nice one is this: a=55, b=53, c=20 (hint m,n,k<50). But in fact I think you are looking for a proof that the parametric equations give a complete set without doubles; me too. And as I didn't find one on the internet it must be rather complicated and/or long. Can anyone give a hint where to find it or give a sketch of it? kweenie