A unique triangle

Let the shortest side length of the triangle be $a$ . Then the other two side lengths are $a+1$ and $a+2$ . Let the smallest angle be $\theta$ ; the largest angle is $2\theta$ . By sine rule , we have:

$\begin{aligned} \frac a{\sin \theta} & = \frac {a+2}{\sin 2\theta} \\ a\sin 2\theta & = (a+2)\sin \theta \\ 2a\sin \theta \cos \theta & = (a+2)\sin \theta & \small \color{#3D99F6} \text{Since }\sin \theta \ne 0 \\ 2a\cos \theta & = a+2 \\ \implies \cos \theta & = \frac {a+2}{2a} \end{aligned}$

By cosine rule ,

$\begin{aligned} a^2 & = (a+1)^2+(a+2)^2 - 2(a+1)(a+2) \color{#3D99F6} \cos \theta & \small \color{#3D99F6} \text{Substituting }\cos \theta = \frac {a+2}{2a} \\ a^2 & = 2a^2 + 6a+5 - \frac {(a+1)(a+2)^2}a & \small \color{#3D99F6} \text{Rearranging} \\ a^2 + 6a + 5 & = \frac {(a+1)(a+2)^2}a \\ a(a+1)(a+5) & = (a+1)(a+2)^2 & \small \color{#3D99F6} \text{Since }(a+1) \ne 0 \\ a^2 + 5a & = a^2 + 4a+4 \\ \implies a & = 4 \end{aligned}$

Therefore the perimeter is $4+5+6 = \boxed{15}$ .

Let $AC=x-1, CB=x$ and $BA=x+1$ . Also $\angle C=2\angle B$ as the largest side is opposite to the largest angle and similarly the smallest side is opposite to the smallest angle.

Extend side $AC$ to $D$ such that $CD=CB=x$ . This implies $\angle CDB =\angle CBD = \frac{\angle C}{2}=\angle B.$ It follows that $\triangle ABC\sim \triangle ADB \implies AB^2=AC\cdot AD\implies (x+1)^2=(x-1)(2x-1).$

Simplifying the above equation gives $x^2+2x+1=2x^2-3x+1\implies x^2=5x\implies x=5\implies x-1=4, x+1=6$ .

Therefore, perimeter of $\triangle ABC=AB+BC+CA=6+5+4=\boxed{15}$

Note: There is only one such triangle with these characteristics and it has side lengths of 4,5 and 6.

Let the sides of the triangle be $x - 1$ , $x$ , and $x + 1$ , and let the smallest angle be $\theta$ and the largest angle be $2\theta$ .

By the law of cosines, $\cos 2\theta = \frac{x^2 + (x - 1)^2 - (x + 1)^2}{2x(x - 1)}$ which simplifies to $\cos 2\theta = \frac{x - 4}{2(x - 1)}$ , and $\cos \theta = \frac{x^2 + (x + 1)^2 - (x - 1)^2}{2x(x + 1)}$ which simplifies to $\cos \theta = \frac{x + 4}{2(x + 1)}$ .

Since $\cos 2\theta = 2 \cos^2 \theta - 1$ , by substitution $\frac{x - 4}{2(x - 1)} = 2 \Big(\frac{x + 4}{2(x + 1)} \Big)^2 - 1$ , which solves to $x = 5$ for $x > 1$ .

Therefore, the three integer sides are $4$ , $5$ , and $6$ , for a perimeter of $4 + 5 + 6 = \boxed{15}$ .

We will use the cosine rule: $\cosφ = (a^2+b^2-c^2)/2ab$ Apply this to the smallest angle (opposite c, so a and b are the longest sides): $a=x+1, b=x+2, c=x, φ=α$

$\cos α =\frac{(x+1)^2+(x+2)^2-x^2} {2(x+1)(x+2)} = \frac{x^2+6x+5}{2x^2+6x+4}$

Also apply it to the largest angle (opposite c, so a and b are the shortest sides): $a=x, b=x+1, c=x+2, φ=2α$

$\cos 2α = \frac{x^2-2x-3}{2x(x+1)}$

Now using to the double angle formula $\cos 2α= 2\cos^2α-1$ we can relate these values to each other:

$\frac{x^2-2x-3}{2x(x+1)} = 2\frac{(x^2+6x+5)^2}{(2x^2+6x+4)^2}-1$

Working this out eventually leads to the polynomial $2x^6+5x^5-22x^4-88x^3-112x^2-61x-12=0$ which can be rewritten as $(x-4)(x+1)^3(x+3)(2x+1)=0$ with roots $x \in\{-3,-1,-\frac{1}{2},4\}$ .

We are looking for a positive integer solution, which is x=4. So the sides are 4, 5, and 6 and the perimeter is $4+5+6=\boxed{15}$ .

Let the triangle sides be n - 1, n, and n + 1. Let B be the angle opposite n - 1 and 2B the angle opposite n + 1. By the Law of Sines: (1) (n - 1)/sin(B) = (n + 1)/sin(2B). (2) 2(n - 1)cos(B) = n + 1 (3) cos(B) = (n + 1)/[2(n - 1)] By the law of cosines, (4) (n - 1)^2 = n^2 + n + 1)^2 - 2n(n + 1)(n + 1)/[2(n - 1)]. Simplifying, (5) n^2 - 5n = 0. So n = 5, n - 1 = 4, n + 1 = 6, perimeter = 15.