One Large Triangle

Just like Euclid's formula for Pythagorean triples, there is a formula to generate integer triangles with an angle of $120^\circ$ ; if $n<m$ are integers, then the triangles with sides $2mn+n^2,\;\;\; m^2-n^2,\;\;\; m^2+mn+n^2$ has an angle of $120^\circ$ opposite the longest side. To ensure the sides are coprime, we need $\gcd(m,n)=1$ and also $m-n$ cannot be a multiple of $3$ .

There are two ways to assign the shorter sides, so two cases to check. For neatness, write $k=1500$ :

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Case $a=2mn+n^2$ , $b=m^2-n^2$ , $c=m^2+mn+n^2$ :

$\begin{aligned} \frac{a-b-2c}{a+2b-2c} &= \frac{2mn+n^2-m^2+n^2-2m^2-2mn-2n^2}{2mn+n^2+2m^2-2n^2-2m^2-2mn-2n^2} \\ &=\frac{-3m^2}{-3n^2} \\ &=\frac{m^2}{n^2} \end{aligned}$

so that $m=kn$ . The only coprime pair satisfying this is $(m,n)=(1,k)$ giving sides $a=2k+1$ , $b=k^2-1$ , $c=k^2+k+1$ and perimeter $P=2k^2+3k+1=(k+1)(2k+1)$ . In this case, the perimeter is $P=1501\times 3001=\boxed{4504501}$ .

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Case $a=m^2-n^2$ , $b=2mn+n^2$ , $c=m^2+mn+n^2$ :

Switching the values of $a$ and $b$ , $\begin{aligned} \frac{a-b-2c}{a+2b-2c} &= \frac{m^2-n^2-2mn-n^2-2m^2-2mn-2n^2}{m^2-n^2+4mn+2n^2-2m^2-2mn-2n^2} \\ &=\frac{-m^2-4mn-4n^2}{-m^2+2mn-n^2} \\ &=\frac{(m+2n)^2}{(m-n)^2} \end{aligned}$

so now we need $m+2n=k(m-n)$ . Rearranging, this leads to $(k-1)m=(k+2)n$

In this case, $\gcd(k-1,k+2)=1$ so the only coprime pair that works is $(m,n)=(1502,1499$ . But these don't satisfy the condition that $m-n$ isn't a multiple of $3$ , so the sides $a,b,c$ in this case are not coprime. In fact, this triangle is just a scaled version of the one found in the first case.

So the unique answer is $P=\boxed{4504501}$ .

Since the diameter of the circumcircle of $\triangle ABC$ , $D=\dfrac a{\sin A} = \dfrac b{\sin B} = \dfrac c{\sin C}$ , then

$\begin{aligned} \sqrt{\frac {a-b-2c}{a+2b-2c}} & = 1500 \\ \sqrt{\frac {\sin A - \sin B - 2\sin C}{\sin A + 2\sin B-2\sin C}} & = 1500 \\ \sqrt{\frac {\sin A - \sin (60^\circ - A) - 2\sin 120^\circ}{\sin A + 2\sin (60^\circ - A) -2\sin 120^\circ}} & = 1500 \\ \sqrt{\frac {\sqrt 3 \sin A - \cos A - 2}{2\cos A - 2}} & = 1500 & \small \blue{\text{Let }t = \tan \frac A2} \\ \sqrt{\frac {3 - 2\sqrt 3 t + t^2}{6t^2}} & = 1500 \\ \frac {\sqrt 3-t}{2t} & = 1500 & \small \blue{\text{Since }A < 60^\circ} \\ \implies t & = \frac {\sqrt 3}{3001} \end{aligned}$

Then we have:

$\begin{aligned} a:b:c & = \sin A : \sin (60^\circ - A) : \sin 120^\circ \\ & = \frac {2t}{1+t^2} : \frac {\sqrt 3(1-t^2)-2t}{2(1+t^2)}: \frac {\sqrt 3}2 \\ & = 4t : \sqrt 3(1-t^2)-2t : \sqrt 3 (1+t^2) \\ & = \frac 4{3001} : 1 - \frac 3{3001^2} - \frac 2{3001} : 1 + \frac 3{3001^2} \\ & = 12004 : 8999996 : 9006004 \\ & = 3001 : 2249999 : 2251501 \end{aligned}$

Therefore $a+b+c = 3001 + 2249999 + 2251501 = \boxed{4504501}$ .