## Is it Possible to Construct a Triangle With an Angle of 60 Degrees With Sides in an Arithmetic Progression?

It is possible to constrict a Right Angled Triangle With Sides in Arithmetic Progression. One such triangle has sides 3, 4 and 5 units respectively, and this triangle can be scaled up to give an infinite number of triangles with sides in an Arithmetic Sequence.
Is it possible to construct a triangle with a 60 degree angle with sides in an arithmetic progression? If so, then the smallest side must be opposite the smallest angle, and the largest side opposite the largest angle. We obtain the triangle below.

Applying the Cosine Rule:
$a^2 = b^2+c^2-2 \times b \times c \times cos A$

$(x+y)^2 = x^2+(x+2y)^2-2 \times x \times (x+2y) \times cos 60$

$(x+y)^2 = x^2+(x+2y)^2-2x(x+2y)cos 60$

$x^2+2xy+y^2 = x^2+x^2+4xy+4y^2-2x(x+2y) \times \frac{1}{2}$

$x^2+2xy+y^2 = x^2+x^2+4xy+4y^2-x^2-2xy$

$0 = 3y^2$

Hence
$y=0$
.
All the sides are length
$x$
units and the triangle is isosceles, so no such triangle exists.