应该还是得用海伦公式.
设三边长分别为a,b,c,则三角形面积平方 = (a+b+c)(-a+b+c)(a-b+c)(a+b-c)/16.
由条件得4(a+b+c) = (-a+b+c)(a-b+c)(a+b-c).
设x = -a+b+c,y = a-b+c,z = a+b-c,则x,y,z为同奇同偶的正整数,且x+y+z = a+b+c.
于是4(x+y+z) = xyz,即1/(xy)+1/(yz)+1/(zx) = 1/4.
不妨设x ≥ y ≥ z,则3/(yz) ≥ 1/(xy)+1/(yz)+1/(zx) = 1/4,即yz ≤ 12.进一步z² ≤ yz ≤ 12,故z ≤ 3.
若z = 1,代入得4(x+y+1) = xy,即(x-4)(y-4) = 20.
满足x ≥ y ≥ z = 1的正整数解(x,y,z)有(24,5,1),(14,6,1),(9,8,1).
若z = 2,代入得2(x+y+2) = xy,即(x-2)(y-2) = 8.
满足x ≥ y ≥ z = 2的正整数解(x,y,z)有(10,3,2),(6,4,2).
若z = 3,代入得4(x+y+3) = 3xy,即(3x-4)(3y-4) = 52.
不存在满足x ≥ y ≥ z = 3的正整数解.
综上,同奇同偶的正整数解只有(6,4,2)一组.
a = (y+z)/2 = 3,b = (x+z)/2 = 4,c = (x+y)/2 = 5.
即问题的解只有边长3,4,5的三角形.
再问： 为什么设x ≥ y ≥ z，3/(yz) ≥ 1/(xy)+1/(yz)+1/(zx)成立？
再答： ∵x ≥ y ≥ z ≥ 0, ∴xy ≥ xz ≥ yz ≥ 0, ∴1/(yz) ≥ 1/(xz) ≥ 1/(xy). 于是3/(yz) ≥ 1/(xy)+1/(yz)+1/(zx).
