Triangle perimeter and area

Question

If you take the square root of the perimeter of a triangle and double it, you get the area of the triangle. The sides have integer lengths which are mutually different from each other. What are the lengths of its sides?

Answer 1

If we label the sides of the triangle $a,b,c$ then we know that the perimeter of the triangle is $$P=a+b+c$$

We can use Heron's Formula to calculate the area of a triangle when we know all three side lengths:

$$A=\sqrt{s(s-a)(s-b)(s-c)}$$ where $$s=\frac{a+b+c}2$$

Can you continue from here?

Answer 2

Let the side lengths of the triangle be $a,b,c$, and let $P = a+b+c$.

By Heron's formula, $$A = \sqrt{\frac P2\left(\frac P2-a\right)\left(\frac P2-b\right)\left(\frac P2-c\right)}$$

From the given information,

$$A = 2\sqrt{P}$$

Equating the two and check that $P\ne 0$,

$$\begin{align*} \sqrt{\frac 12\left(\frac P2-a\right)\left(\frac P2-b\right)\left(\frac P2-c\right)} &= 2\\ \left(\frac P2-a\right)\left(\frac P2-b\right)\left(\frac P2-c\right) &= 8\\ (P-2a)(P-2b)(P-2c) &= 64\\ (b+c-a)(c+a-b)(a+b-c) &= 64 \end{align*}$$

$a, b,c$ are all positive integers and are all different, so $P-2a, P-2b, P-2c$ are different positive integral factors of $64$.

First, all decomposed factor triplet with a $1$ is rejected. (Why?)

Then the only decomposition with different integers is $2\times4\times8$. Without loss of generality, let

$$\begin{align*} P-2a &= 2\\ P-2b &= 4\\ P-2c &= 8\\ 3P - 2(a+b+c) &= 2+4+8\\ P &= 14\\ a &= 6\\ b &= 5\\ c &= 3 \end{align*}$$

So the triangle is a $6-5-3$ triangle.

Answer 3

Without the restriction that the integers are different, there are two other solutions: 2, 9, 9 and 4, 4, 4.