AMC 10 · 2024 · #9

Grade 8 arithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Successive Multiplicative Steps

mean-median-mode-rangeperfect-squareslinear-equations-two-var convert-to-algebraidentify-subproblems ↑ Prerequisites: mean-median-mode-rangelinear-equations-one-var

📏 Short solution 💡 2 insights

Problem

Three real numbers a, b, c average to 0, and their squares a², b², c² average to 10. Find the average of the three pairwise products ab, ac, bc.

Pick an answer.

(A)

-5

(B)

$-\dfrac{10}{3}$

(C)

$-\dfrac{10}{9}$

(D)

(E)

$\dfrac{10}{9}$

AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Convert to Algebra

We do not need individual values of a, b, c — only the symmetric combination ab + ac + bc. That is the signature of Tool #13 (Convert to Algebra): write down what each "mean" statement says as a clean equation, then look for an identity that connects the three sums (a + b + c), (a² + b² + c²), and (ab + ac + bc). The trinomial-square identity (a + b + c)² = a² + b² + c² + 2(ab + ac + bc) links exactly those three. Tool #11 (Work Backwards) reaches the target by treating ab + ac + bc as the unknown and solving for it after substituting the known sums.

1STEP 1

Multiply each average by 3 to get the two sums a + b + c = 0 and a² + b² + c² = 30.

a + b + c = 0, a² + b² + c² = 30

2STEP 2

Expand the trinomial square (a + b + c)²: it gives every square plus twice every pairwise product — the bridge between our sums.

(a + b + c)² = a² + b² + c² + 2(ab + ac + bc)

3STEP 3

Substitute the known sums: 0 = 30 + 2(ab + ac + bc), then work backwards to isolate ab + ac + bc = -15.

0 = 30 + 2(ab + ac + bc) → ab + ac + bc =

\frac{0 - 30}{2}

= -15

4STEP 4

Divide the sum by 3: the mean of ab, ac, bc is - $\frac{15}{3}$ = -5, choice (A).

Mean =

\frac{ab + ac + bc}{3}

= -

\frac{15}{3}

= -5 → (A)

Answer

-5

Sign check: the average of the squares (10) is positive but the average of the values is 0, so a, b, c must have mixed signs — the pairwise products ab, ac, bc will then have a net negative contribution, matching the negative answer -5. Concrete check: take a = √(15), b = -√(15), c = 0. Mean is 0 ✓, mean of squares is

\frac{15 + 15 + 0}{3}

= 10 ✓, pairwise products: ab = -15, ac = 0, bc = 0, mean = -5 ✓. Answer (A) confirmed.

💡Key takeaway

This AMC 10 problem only needs Grade 8 equation-solving — squaring the sum links the three sums together, and one line of algebra extracts the pairwise-product average!

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