# Lower bound on triangle area in terms of semiperimeter.

I am currently trying to figure out if it's possible to give a better lower bound to the area of a triangle in terms of its semi-perimeter (or perimeter).

If the sides of a triangle are $a$, $b$ and $c$, and the semiperimeter is $s$, then using Heron's formula, I get: $16T^2 = s(s - a)(s - b)(s - c)$

Since all of $(s-a)$, $(s-b)$, $(s-c)$ are smaller than $s$, by substituting $s$ I can give an upper bound of $16T^2 < s^4$, and from there $T^2 < s^4/16$ pretty easily. On the other hand, to get a lower bound, other than simply eliminating the terms I mentioned above to get $s < 16T^2$ and then $T^2 > s/16$, I don't see what else I can do...

Surely, this lower bound can be improved right?

EDIT: I am also assuming the triangles I am working with, all have integer sides and an angle of 60 degrees.

• There is no lower bound (unless you include degenerate triangles, in which case the lower bound is $0$). You can "flatten" a triangle of given perimeter arbitrarily "thin". s < 16T^2 This doesn't make sense dimensionally, you cannot compare a length to the square of an area. – dxiv May 1 '17 at 20:23
• @dxiv Would it change anything if there is a further constraint that all side lengths be necessarily integers and the triangle has a 60 degree angle? – BizarreCake May 1 '17 at 20:25
• @quasi That's very interesting! Can you please elaborate a bit? – BizarreCake May 1 '17 at 21:02
• @BizarreCake -- Actually, my previous comment (now deleted) was an error. Even with your new conditions, the ratio $T/S^2$ can be made arbitrarily close to zero. – quasi May 1 '17 at 21:39
• @BizarreCake Those additional premises do in fact change the problem significantly. Such triangles would only exist for certain values of the perimeter, and there will only be a finite (usually small) number of them. Lookup Integer triangles with a 60° angle and Eisenstein triples. – dxiv May 1 '17 at 21:48

## 1 Answer

The greatest lower bound of $T$, for a given $s$, is $0.$

Given $s>0,$ take $t \in (0,\pi /2).$ Let $U=s/(1+\cos t)$

Take isosceles triangle $ABC$ with $AB=AC=U$ and $\angle ABC=\angle ACB=t.$ Then $BC=2(AB)\cos \angle ABC=2U\cos t .$

So the semi-perimeter of $ABC$ is $U(1+\cos t)=s.$

The area of $ABC$ is $$T= \frac {1}{2}(BC)(AB\sin \angle ABC)=\frac {1}{2} (2U\cos t)(U\sin t)=$$ $$= U^2\cos t \sin t<U^2 \sin t<s^2\sin t$$ which can be arbitrarily small if $t$ is small enough

• OP's (edited) question is asking for integer sides and an angle of 60 degrees. – dxiv May 2 '17 at 0:12