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2024 AMC 10B Problem 9

2024 AMC 10B Problem 9AMC 10 Step by Step

Name: Date: September 13, 2026

Algebraic Manipulation · difficulty 2 of 5 · about 2 min · position P1-10

Real numbers $a, b,$ and $c$ have arithmetic mean $0$ . The arithmetic mean of $a^2, b^2,$ and $c^2$ is $10$ . What is the arithmetic mean of $ab, ac,$ and $bc$ ?

A)
$-5$
B)
$-\dfrac{10}{3}$
C)
$-\dfrac{10}{9}$
D)
$0$
E)
$\dfrac{10}{9}$

Answer

Key insight

Square the sum: (a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca), so 0 = 30 + 2(ab+bc+ca) gives ab+bc+ca = -15, mean -5.

Solution

The two conditions say $a + b + c = 0$ and $a^2 + b^2 + c^2 = 30$ .

Expand the square of the sum:
$(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca).$
Substituting the known values: $0 = 30 + 2(ab + bc + ca)$ , so $ab + bc + ca = -15$ .

The requested mean is $\frac{-15}{3} = -5$ .

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

Why this works

The three symmetric quantities $a+b+c$ , $a^2+b^2+c^2$ and $ab+bc+ca$ are tied together by one identity, so any two determine the third. Whenever a problem gives a sum and a sum of squares, squaring the sum is the reflex. The answer is forced even though $a, b, c$ themselves are not.

Alternative approach

Pick a concrete triple: $a = 0$ , $b = \sqrt{15}$ , $c = -\sqrt{15}$ has mean $0$ and squares averaging $\frac{0 + 15 + 15}{3} = 10$ . Then $ab + ac + bc = 0 + 0 - 15 = -15$ , mean $-5$ .

The trap

Reporting ab + bc + ca = -15 itself, or forgetting the factor 2 in the expansion and getting -10 (which divides to -10/3, choice (B)).

Common mistakes

Reporting ab + bc + ca = -15 itself, or forgetting the factor 2 in the expansion and getting -10 (which divides to -10/3, choice (B)).
Using the means $0$ and $10$ directly in the identity instead of the sums $0$ and $30$ .

Techniques

Apply an identity: SFFT, sum of squares, difference of cubes, Vieta · Test small/specific values or special cases to find or verify the answer