What is the smallest possible area for a triangle of integral side lengths whose perimeter is 15cm?

Question

What is the smallest possible area for a triangle of integral side lengths whose perimeter is 15cm? Round your decimal answer to the nearest hundredth.

How do you find this?

Answer 1

If the triangle has (integer) sides $a,b,c$ then as given $a+b+c = 15$. Now let's use imranfat's suggestion of the triangle inequality: $$a+b > c$$ $$a+c > b$$ $$b+c >a.$$

Since we want to minimize area and all our data is in terms of side lengths I think Heron's Formula is our best bet: let $s = \frac{a+b+c}{2} = 15/2 =7.5$ then the area $A$ of our triangle is $$A = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{7.5(7.5-a)(7.5-b)(7.5-c)}.$$

From looking at the above one can see to make the area small we essentially want some integer, say $a$, closest to $7.5$, in particular $7$ is good because it leaves $0.5$ which scales down our area. Any integer small leaves a whole number multiply which increases the area and we can't go bigger than $7$ by the triangle inequalities and the bound on perimeter. All of which should motivate the choice of $(7,7,1)$ as our triple for $(a,b,c)$ which gives minimal area (for intuition this is a very skinny isosceles triangle). Doing the calculation and rounding gives $A \approx 3.49$.

[Note this approach isn't the most elegant, and we could be more exact by using the tools of calculus, but from a purely high-school geometry point of view it's a workable way of thinking.]