
4. What is the minimal area of a right-angled triangle whose inradius is 1 unit?


Solution: Let $A B C$ be the right-angled triangle with $\angle B=90^{\circ}$. Let $I$ be its incentre and $D$ be the point where the incircle touches $A B$. Then $s-b=A D=r=1$. We also know that $[A B C]=r s=r(a+b+c) / 2$ and $[A B C]=a c / 2$. Thus

$\frac{a c}{2}=\frac{a+b+c}{2}=(a+c)-\frac{a+c-b}{2}=(a+c)-1$.

![](https://cdn.mathpix.com/cropped/2024_06_05_3cd360997e5dbe23c5e6g-2.jpg?height=361&width=531&top_left_y=381&top_left_x=1361)

Using AM-GM inequality, we get

$$
\frac{a c}{2}=a+c-1 \geq 2 \sqrt{a c}-1
$$

Taking $\sqrt{a c}=x$, we get $x^{2}-4 x+2 \geq 0$. Hence

$$
x \geq \frac{4+2 \sqrt{2}}{2}=2+\sqrt{2}
$$

Finally,

$$
[A B C]=\frac{a c}{2} \geq \frac{(2+\sqrt{2})^{2}}{2}=3+2 \sqrt{2}
$$

Thus the least area of such a triangle is $3+2 \sqrt{2}$.
