9. Right Triangle (7-11). The length of the hypotenuse of a right triangle is 3.

a) (1 pt.) 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) (2 pts.) What is the smallest standard deviation of the side lengths that such a right triangle can have? What are the lengths of its legs in this case?
Solution. a) Let $a$ and $b$ be the legs of the triangle. Then $a^{2}+b^{2}=9$. We will find the variance of the set of side lengths. For this, we will use the formula $S^{2}=\overline{x^{2}}-\bar{x}^{2}$ - the mean of the squares minus the square of the mean:

$$
S^{2}=\frac{a^{2}+b^{2}+9}{3}-\left(\frac{a+b+3}{2}\right)^{2}=\frac{18}{3}-\left(\frac{a+b+3}{2}\right)^{2}=6-\left(\frac{a+b+3}{3}\right)^{2}.
$$

From the triangle inequality, it follows that $a+b>3$. Therefore,

$$
S^{2}<6-\left(\frac{6}{3}\right)^{2}=6-4=2
$$

Answer: incorrect.
b) The variance of any set is non-negative. We will calculate the variance of the two legs separately: $S_{\text {legs }}^{2}=\frac{a^{2}+b^{2}}{2}-\left(\frac{a+b}{2}\right)^{2}$. Since the variance is non-negative, we have

$$
\frac{a^{2}+b^{2}}{2}-\left(\frac{a+b}{2}\right)^{2} \geq 0, \text { hence } \frac{a+b}{2} \leq \sqrt{\frac{a^{2}+b^{2}}{2}}
$$

The obtained inequality (a special case of the Cauchy-Bunyakovsky inequality) is one of the inequalities about means: "The arithmetic mean of any numbers is not less than their quadratic mean, and equality is achieved only if all the numbers are equal to each other." This inequality could have been proven in various ways; we chose an unusual method - through variance.

From (2) it follows: $a+b \leq 2 \sqrt{\frac{a^{2}+b^{2}}{2}}=2 \frac{3}{\sqrt{2}}=3 \sqrt{2}$. Substituting this into (1), we get a lower bound for $S^{2}$:

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

Therefore, the standard deviation $S$ 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 $a=b$. Then $a=b=\frac{3 \sqrt{2}}{2}$.

Answer: the minimum standard deviation $\sqrt{2}-1$ is achieved in an isosceles right triangle with legs $\frac{3 \sqrt{2}}{2}$.