Let $ABC$ be a triangle with such conditions, such that $A=2B$. It's very well-known (or it can be deduced from the cosine law and the double-angle formula, or i think it's somewhere in the forum) that $a^2-b^2=bc$ Then, we just have to check the cases $(a,b,c)=(a,a-1,a-2), (a,b,c)=(a,a-2,a-1)$, and $(a,b,c)=(a,a-1,a+1)$. $(a,b,c)=(a,a-1,a-2)$ gives $a^2-5a+3=0$, but it has no integer roots. $(a,b,c)=(a,a-1,a+1)$ gives $a^2-2a=0 \Rightarrow a=2 \Rightarrow a+b=3=c \Rightarrow\Leftarrow$ finally, $(a,b,c)=(a,a-2,a-1)$ gives $a^2-7a+6=0$ and since $a=1 \Rightarrow c=0$, then we must have that $a=6$. then the triangle that satisfies such conditions is $(a,b,c)=(6,4,5)$ with $A=2B$

If <A=2<B, some simple manipulating with the sine rule gives \[ /a^2=b(b+c) /\]. Assuming the lengths of the sides to be \[/ (a-1,a,a+1)/\] and considering all possible cases we get the solutions in (a,b,c) as \[/(1,2,3) and (4,5,6)/\] and obviously all its permutations.

Campos missed the solution \[(1,2,3)\].

sayantanchakraborty wrote: Campos missed the solution \[(1,2,3)\]. That´s NOT a triangle, but a degenerated one! Best regards, sunken rock

Well technically a degenerate triangle is a triangle. The problem should be worded more clearly, or both solutions should be accepted as correct.

Bruh AMC 2024 #22