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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?

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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.

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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?

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

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