2024 AMC 10B Problem 9

1. Turn the means into sums. A mean of $0$ for three numbers means $a + b + c = 0$. A mean of $10$ for the squares means $a^2 + b^2 + c^2 = 30$.
2. We want $ab + ac + bc$. The one step that connects a sum to products is squaring the sum, so square $a + b + c$:
$$(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc.$$
(Each product shows up twice, for example $ab$ from $a \cdot b$ and from $b \cdot a$.)
3. Put in what we know. The left side is $0^2 = 0$, and the squares add to $30$:
$$0 = 30 + 2(ab + ac + bc).$$
4. So $2(ab + ac + bc) = -30$, and $ab + ac + bc = -15$.
5. The question asks for the mean of the three products, so divide by $3$:
$$\frac{-15}{3} = -5.$$

The answer is $\boxed{\textbf{(A)}\ -5}$.

The sum $a + b + c$, the sum of squares, and the sum of the products $ab + ac + bc$ are linked by squaring the sum. If you know any two of them, you can find the third. So when a problem hands you a sum and a sum of squares, squaring the sum is the first thing to try.

Since the answer must work for any numbers that fit, pick easy ones. Let $a = 0$ and $c = -b$, so the mean is $0$. The squares must add to $30$, so $b^2 + b^2 = 30$ and $b^2 = 15$. The products are $ab = 0$, $ac = 0$, $bc = -b^2 = -15$. Their mean is $\frac{-15}{3} = -5$.