Does there exist a right triangle with area 7 and perimeter 12?

This question is really trivial. I can prove that there is no right triangle with area 7 and perimeter 12, but what I do is solve the following system: if $a$, $b$ and $c$ are, respectively, the two legs and hypotenuse of such a triangle, then

$$a^2 + b^2 = c^2,$$ $$a + b +c = 12,$$ $$ab = 14$$

It is easy (although a bit boring and long) to see that there are no real solutions to this system.

But I feel that there is a simple answer to this question - perhaps using the triangle inequality - but I just cannot see it.

Solution 1:

I'm not sure if this counts as a "simple answer", but let $x$ be the length of one leg of a right triangle of area $7$; then the other leg is $\frac{14}x$, and the hypotenuse is $\sqrt{x^2 + \left(\frac{14}x\right)^2}$, so the perimeter is given by the function $$P(x) = x + \frac{14}x + \sqrt{x^2 + \left(\frac{14}x\right)^2}$$

Asymptotically, we have $\lim_{x\to 0} P(x) = +\infty$ and $\lim_{x\to\infty} P(x) = +\infty$, and intuitively it seems clear that the absolute minimum occurs when the two legs of the triangle are equal in length, i.e. when $x=\sqrt{14}$. At this value, we have $$P(\sqrt{14}) = 2\sqrt{14} +2\sqrt{7}$$ This is the smallest possible perimeter for a right triangle of area $7$. It remains only to convince yourself that this number is greater than $12$.

Solution 2:

If the perimeter is $12$, the triangle of maximum area is the equilateral triangle of side $4$ which has area $2\times 2\sqrt 3\lt 7$ (square both sides $48\lt 49$)

Now it is a question of how you show that the triangle of maximum area is equilateral. First fix a base AB. The third point C lies on an ellipse, and symmetry/convexity show that the maximum area is if the two remaining sides AC and BC are equal. So if the sides are not equal, you will get a greater area by averaging the shortest and longest. For maximum area (no further gains) all sides must be equal - hence equilateral.

In summary, a more formal way of showing that the maximal area of a triangle of fixed perimeter is reached when the triangle is equilateral is to go via Heron's formula (the area of a triangle as a function of its sides) and AM/GM. With the perimeter $2s=a+b+c$ fixed, note that $(-a+b+c)+(a-b+c)+(a+b-c)=2s$ and the product of the three terms in brackets is therefore greatest when they are equal.

Heron's formula is area =$\sqrt {s(s-a)(s-b)(s-c)}$