Minimum of perimeter

This question is similar to this question .

Let the legs of the right triangle be $a$ and $b$ , so the area is $A = \frac{1}{2}ab = 1$ and the perimeter is $P = a + b + \sqrt{a^2 + b^2}$ .

From the first equation, $b = \frac{2}{a}$ , and substituting this into second equation and rearranging gives $2Pa^2 - (P^2 + 4)a + 4P = 0$ , which solves to $a = \frac{1}{4}(P^2 + 4 \pm \sqrt{P^4 - 24P^2 + 16})$ .

To be a real number, the discriminant $P^4 - 24P^2 + 16 \geq 0$ , which solves to $P^2 \leq 12 - 8\sqrt{2}$ or $P^2 \geq 12 + 8\sqrt{2}$ . However, $P^2 \leq 12 - 8\sqrt{2}$ or $P \leq 0.828$ would not make $P$ an integer, so $P^2 \geq 12 + 8\sqrt{2}$ or $P > 4.828$ , which means the smallest integer $P$ can be is $P = \boxed{5}$ .

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Substituting $P = 5$ into $a = \frac{1}{4}(P^2 + 4 \pm \sqrt{P^4 - 24P^2 + 16})$ gives $a = \frac{1}{20}(29 \pm \sqrt{41})$ , which means $b = \frac{1}{20}(29 \mp \sqrt{41})$ and the hypotenuse $c = P - a - b = \frac{21}{10}$ .

Let the side lengths of the right angle triangle be $a, b$ and $c$ where $a\le b\le c$ . Hence $a^2+b^2=c^2$ .

Since the area is 1, we have $ab=2$ .

It is well-known that $u^2+v^2 \ge 2uv$ for all real numbers $u$ and $v$ . (Simple proof: $u^2+v^2 =\left(u-v \right)^2+2uv\ge 2uv$ ) .

Then the perimeter is $a+b+c=a+b+\sqrt{a^2+b^2} \ge 2\sqrt{ab} + \sqrt{2ab} = 2\sqrt{2} + 2 \approx 4.828$ .

Suppose the perimeter is 5. Then $a+b=5-c$ . Square both sides and obtain $a^2+b^2+2ab = 25-10c+c^2$ . Since $a^2+b^2=c^2$ and $ab=2$ , we have $c=\frac{21}{10}$ .

Now $a+b=5-\frac{21}{10}=\frac{29}{10}$ and $ab=2$ implies that $a$ and $b$ are roots of $u^2-\frac{29}{10}u +2 = 0$ . Solve it and obtain $u=\frac{29 \pm \sqrt{41}}{20}$ .

This means that a right angle triangle with side lengths $a=\frac{29 - \sqrt{41}}{20}, a=\frac{29 + \sqrt{41}}{20}, c=\frac{21}{10}$ fulfill the conditions area is 1 and perimeter is 5. So the minimum value of the perimeter is $\boxed{5}$ .

Discussion can be found in this video .