# Condition of existence of a triangle

It's easy to prove that the triangle inequality holds for any triangle with the lengths of sides $a$, $b$ and $c$. But how can one prove that if the triangle inequality holds for any given positives $a$, $b$ and $c$ then a triangle (geometric figure) with the lengths of the sides equal to $a,b$ and $c$ can necessarily be formed?

• Would a geometrical argument do? Draw one of the edges (say $a$) and then draw circles with radii $b$ and $c$ at each end Aug 7, 2017 at 16:01
• @Shuri2060 How do you know that the circles meet each other? Aug 7, 2017 at 16:08
• @MarkBennet Can it be taken intuitively that if the distance between the centres of two circles is less than the sum of their radii then they must coincide? If not, you could go for an algebraic approach instead as is suggested in an answer below. Aug 7, 2017 at 16:28
• @Shuri2060 Consider a circle of radius $1$ centred at the origin and a circle of radius $4$ centred at $(0,2)$ - you need all three triangle equalities to make it work. I think it is a comment of Coxeter on Euclid's first proposition (constructing an equilateral triangle) that he doesn't prove that the circles meet - I think that's in his "Introduction to Geometry", but I haven't the reference to hand. Aug 7, 2017 at 16:43
• @MarkBennet I didn't think it through fully. But yeah - using the other two inequalities, you'll also get that one circle can't be contained in another so they must intersect Aug 7, 2017 at 16:50

Let $B(0,0)$ and $C(a,0)$.

Hence, $BC=a$ and we need to prove that there exists $A(x,y)$ such that $AB=c$ and $AC=b$.

You can write equations of two circles and prove that there are intersection points.

For example $x^2+y^2=c^2$ and $(x-a)^2+y^2=b^2$.

Thus, $-2ax+a^2+c^2=b^2$ or $x=\frac{a^2+c^2-b^2}{2a}$ and $$y^2=c^2-\left(\frac{a^2+c^2-b^2}{2a}\right)^2$$ or $$y^2=\frac{(a+b+c)(a+b-c)(a+c-b)(b+c-a)}{4a^2},$$ which says that there are two intersection points.

• Note that to have $y^2\gt 0$ you need the various triangle inequalities to work. Aug 7, 2017 at 16:44
• @Mark Bennet It's given that $a+b-c>0$, $a+c-b>0$ and $b+c-a>0$. We need to prove that there exists a triangle with sides-lengths $a$, $b$ and $c$, which I did. Aug 7, 2017 at 16:47
• I know it is given - the other half of the point is that if the formula didn't reference the inequalities, it would [almost certainly] be wrong. Aug 7, 2017 at 16:48
• @Mark Bennet Yes, of course, but I thought that it's obvious. Thank you! Aug 7, 2017 at 16:51

It depends what you are able to assume. Take sides of lengths $a,b,c$ and choose two points $A, B$ distance $c$ apart. Construct a circle of radius $b$ centred at $A$ and a circle of radius $a$ centred at $B$.

The circle centred at $A$ crosses the segment $AB$ extended - at two points. In the direction towards $B$ we have $c-a\lt b\lt c+a$ by the triangle inequality, so the crossing point is within the circle centred at $A$.

For the point away from $B$ we have $c+b\gt a$ so the crossing point is outside the circle centred at $A$.

Therefore the circles must meet (this is where you heed to know what you can assume - in fact at two points). Choose a meeting point and call it $C$. $ABC$ is an example of the triangle you wanted to exist.

Note how all three of the "triangle inequalities" are invoked here.