New Year Fun !!!!!

Let the shortest side length of the triangle be $x$ and the smallest angle be $\theta$ .

Then using Sine Rule, we have:

$\quad \dfrac {x}{\sin {\theta}} = \dfrac {x+2}{\sin {2\theta}} \quad \Rightarrow x\sin {2\theta} = (x+2) \sin {\theta}$

$\quad \Rightarrow 2x\sin {\theta}\cos {\theta} = (x+2)\sin {\theta} \quad \Rightarrow 2x\cos {\theta} = x+2$

$\quad \Rightarrow \cos {\theta} = \dfrac {x+2}{2x}$

Using Cosine Rule, we have:

$\quad x^2=(x+1)^2+(x+2)^2-2(x+1)(x+2)\cos {\theta}$

$\quad \Rightarrow x^2 = (x+1)^2 + x^2+4x+4 - \dfrac {2(x+1)(x+2)^2} {2x}$

$\quad \Rightarrow 0 = (x+1)^2 +4x+4 - \dfrac {(x+1)(x+2)^2} {x}$

$\quad \Rightarrow (x+1)^2 +4(x+1) = \dfrac {(x+1)(x+2)^2} {x}$

$\quad \Rightarrow (x+1) +4 = \dfrac {(x+2)^2} {x} \quad \Rightarrow x(x+5) = (x+2)^2$

$\quad \Rightarrow x^2+5x = x^2+4x+4 \quad \Rightarrow x = 4$

Therefore the sum of sides $=x+(x+1)+(x+2) = 4+5+6 = \boxed{15}$

Let the side opposite to (1) angle A (i.e 2a) be x+1 (2) angle C (i.e a) be x-1 (3) angle B be x Construct the internal bisector of angle A and let it intersect BC at D. Now, since angle CAD = angle ACB => AD=DC Now, in triangle ABC and DBA (1) angle DAB = angle ACB (2) angle ABC is common => triangle ABC is similar to triangle DBA.

$\rightarrow \frac{AC}{AD}=\frac{AB}{BD}=\frac{BC}{AB}$ $or$ $\frac{a}{DC}=\frac{a-1}{BD}=\frac{a+1}{a-1}$ $=>DC+BD=\frac{(a)(a-1)+(a-1)^{2}}{a+1}$ $=> a^{2} - a + a^{2} +1 -2a = a^{2} + 1 + 2a$ $=> a^{2} - 5a = 0$ $=> a=5$ Therefore sum of the sides of the triangle = 5 + (5-1) + (5+1) = 15

Actually i have already posted this problem - see click on this . The only thing that is modified is that this question asks for perimeter and mine asks for side.

Let the triangle be ABC with largest angle A and smallest as C

Let the sides be x,x+1,x+2

The side BC=x+2, AB=x and AC=x+1

Let angle C =y

=> angle A=2y

Now,

construct a line CD such that it forms angle 'y' with AC and meets BA extended at D. By the exterior angle property in triangle CAD we have angle ADC=y

Now,

Apply similarity in triangles ABC and CBD and take the ratios excluding CD Form a Quadratic Equation and then SOLVE!!!!!!!!

nice solution: