# Perimeter of a triangle with an angle of $120^\circ$ and sides in arithmetic progression

I need to find the perimeter of the following triangle in terms of $$\ell$$. The only extra info I have is the three sides form an arithmetic sequence. I've been tutoring maths on these topics for a couple years now and I am still struggling with this problem. I'm sure there's something rather simple I can't see. Any help is appreciated!

• I am pretty sure that there are not sufficient amount of information. We can also let the angle 120 to be a variable point, so I suppose the total perimeter wont be fixed. – sentheta Feb 2 at 0:52
• It seems to me that this triangle is similar to triangle with sides 3, 5, 7. – richrow Feb 5 at 12:52

It is reasonable to assume that the sides of the triangle, have a form $$a=3l-2d$$, $$b=3l-d$$ and $$c=3l$$ for some $$d>0$$. You can use the law of cosines as $$c^2=a^2+b^2-2ab\cos {2\pi\over 3}=a^2+b^2+ab$$and by substitution, first find $$d$$ then $$a+b+c$$.
• Because this can be the only case that the sides form an arithmetic sequence alongside the side $3l$ being the largest side of the triangle. – Mostafa Ayaz Feb 2 at 1:34
Let $$a$$, $$b$$ and $$3l$$ be sides-lengths of the triangle.
Thus, $$2b=a+3l,$$ or $$a=2b-3l$$ and $$(2b-3l)^2+b^2+b(2b-3l)=(3l)^2$$ or $$7b^2-15bl=0,$$ which gives $$(a,b)=\left(\frac{9l}{7},\frac{15l}{7}\right)$$ and $$a+b+c=\frac{45}{7}l.$$