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AMC 10 · 2024 · #9

Grade 8 arithmetic
Archetype Arithmetic Expression Decomposition
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

Real numbers a,b,a, b, and cc have arithmetic mean 0. The arithmetic mean of a2,b2,a^2, b^2, and c2c^2 is 10. What is the arithmetic mean of ab,ac,ab, ac, and bcbc?

(A) −5(B) −103(C) −109(D) 0(E) 109\textbf{(A) } -5 \qquad\textbf{(B) } -\dfrac{10}{3} \qquad\textbf{(C) } -\dfrac{10}{9} \qquad\textbf{(D) } 0 \qquad\textbf{(E) } \dfrac{10}{9}

Pick an answer.

(A)
-5
(B)
$-\dfrac{10}{3}$
(C)
$-\dfrac{10}{9}$
(D)
0
(E)
$\dfrac{10}{9}$
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Toolkit + CCSS Solution

Understand

Restated: Three real numbers $a, b, c$ average to $0$, and their squares $a^2, b^2, c^2$ average to $10$. Find the average of the three pairwise products $ab, ac, bc$.

Givens: Mean of $a, b, c$ is $0$, so $a + b + c = 0$; Mean of $a^2, b^2, c^2$ is $10$, so $a^2 + b^2 + c^2 = 30$; Answer choices: (A) $-5$, (B) $-\tfrac{10}{3}$, (C) $-\tfrac{10}{9}$, (D) $0$, (E) $\tfrac{10}{9}$

Unknowns: The arithmetic mean $\tfrac{ab + ac + bc}{3}$

Understand

Restated: Three real numbers $a, b, c$ average to $0$, and their squares $a^2, b^2, c^2$ average to $10$. Find the average of the three pairwise products $ab, ac, bc$.

Givens: Mean of $a, b, c$ is $0$, so $a + b + c = 0$; Mean of $a^2, b^2, c^2$ is $10$, so $a^2 + b^2 + c^2 = 30$; Answer choices: (A) $-5$, (B) $-\tfrac{10}{3}$, (C) $-\tfrac{10}{9}$, (D) $0$, (E) $\tfrac{10}{9}$

Plan

Primary tool: #13 Convert to Algebra

Secondary: #11 Work Backwards

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 $(a + b + c)$, $(a^2 + b^2 + c^2)$, and $(ab + ac + bc)$. The trinomial-square identity $(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc)$ 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.

Execute — Answer: A

#13 Convert to Algebra 6.SP.A.3 Step 1
  • Translate the two "mean" statements into sums.
  • Multiplying each mean by $3$ removes the fraction and exposes the two sums we will work with.
$$a + b + c = 0, \quad a^2 + b^2 + c^2 = 30$$

💡 An average times the count equals the sum — Grade 6 "a measure of center summarizes the values with one number."

#13 Convert to Algebra 8.EE.C.7 Step 2
  • Recall the trinomial-square identity.
  • Expanding $(a + b + c)^2$ produces every square plus twice every pairwise product, which is exactly the bridge between our two known sums and the unknown sum.
$$(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc)$$

💡 This identity is the algebra version of "area of a big square split into three squares plus six identical rectangles" — a Grade 8 linear/quadratic equation tool.

#11 Work Backwards 8.EE.C.7 Step 3
  • Substitute the known sums.
  • The left side becomes $0^2 = 0$; the first term on the right is $30$.
  • Solve for the bracket — this is the Tool #11 (Work Backwards) step: the target $ab + ac + bc$ sits inside the identity, and we peel the equation back to isolate it.
$$0 = 30 + 2(ab + ac + bc) \;\Rightarrow\; ab + ac + bc = \dfrac{0 - 30}{2} = -15$$

💡 Undo the addition of $30$, then undo the multiplication by $2$ — classic Tool #11 inverse-operation chain.

#13 Convert to Algebra 6.SP.A.3 Step 4

Divide by $3$ to turn the sum into the mean the problem asked for.

$$\text{Mean} = \dfrac{ab + ac + bc}{3} = \dfrac{-15}{3} = -5 \;\Rightarrow\; \textbf{(A)}$$

💡 Mean is the sum divided by the count — the same Grade 6 definition we started with.

[1] #13 6.SP.A.3 Translate the two "mean" statements into sums. Multiplying each mean by $3$ remo
[2] #13 8.EE.C.7 Recall the trinomial-square identity. Expanding $(a + b + c)^2$ produces every s
[3] #11 8.EE.C.7 Substitute the known sums. The left side becomes $0^2 = 0$; the first term on th
[4] #13 6.SP.A.3 Divide by $3$ to turn the sum into the mean the problem asked for.

Review

Reasonableness: 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 $a = \sqrt{15}, b = -\sqrt{15}, c = 0$. Mean is $0\;\checkmark$, mean of squares is $\tfrac{15 + 15 + 0}{3} = 10\;\checkmark$, pairwise products: $ab = -15, ac = 0, bc = 0$, mean $= -5\;\checkmark$. Answer (A) confirmed.

Alternative: Tool #9 (Solve an Easier Related Problem): start with the simpler test triple above ($a = \sqrt{15}, b = -\sqrt{15}, c = 0$). It already gives mean $-5$. To trust the answer for *every* valid triple, the identity in Step 2 shows the value is forced — it depends only on the two given sums, not on which specific $a, b, c$ realize them. So $-5$ is the unique answer (A).

CCSS standards used (min grade 8)

  • 6.SP.A.3 Recognize that a measure of center summarizes all its values with a single number (Converting "mean of $a, b, c$ is $0$" into the sum $a + b + c = 0$, and recovering the requested mean by dividing the sum $ab + ac + bc = -15$ by $3$.)
  • 8.EE.C.7 Solve linear equations in one variable (Applying the trinomial-square identity $(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc)$ and solving the resulting linear equation in the single unknown $ab + ac + bc$.)

⭐ 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!

⭐ 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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