The length of the hypotenuse of a right triangle is 3.

a) The Absent-Minded Scholar calculated the variance of the side lengths of this triangle and found it to be 2. Could he have made a mistake in his calculations?

b) What is the smallest standard deviation of the sides that such a right triangle can have? What are the legs of the triangle in this case?
a) Let $a$ and $b$ be the legs of a triangle. Then $a^{2}+b^{2}=9$, and $D=1 / 3\left(a^{2}+b^{2}+9\right)-\& \operatorname{frac} 19 ;(a+b+3)^{2}=18 / 3-1 / 9(a$ $+b+3)^{2}=6-1 / 9(a+b+3)^{2}$. From the triangle inequality, it follows that $a+b>3$. Therefore, $D<6-1 / 9 \cdot 6^{2}=2$.

b) From the inequality between the arithmetic mean and the quadratic mean, it follows that

$a+b \leq 2 \sqrt{\frac{a^{2}+b^{2}}{2}}=2 \frac{3}{\sqrt{2}}=3 \sqrt{2} \cdot$ Therefore,

$D \geq 6-\left(\frac{3 \sqrt{2}+3}{3}\right)^{2}=6-(\sqrt{2}+1)^{2}=3-2 \sqrt{2}=(\sqrt{2}-1)^{2}$.

Thus, the standard deviation $S=\sqrt{D}$ of the lengths of the three sides does not exceed $\sqrt{2}-1=0.41421356 \ldots$ This minimum value is achieved only if the variance of the legs is zero, that is, if $a=b=\frac{3 \sqrt{2}}{2}$.

## Answer

a) Incorrect; b) $\sqrt{2}-1 ; a=b=\frac{3 \sqrt{2}}{2}$.