AMC 10 · 2024 · #9
Grade 8 arithmeticProblem
Real numbers a,b, and c have arithmetic mean 0. The arithmetic mean of a2,b2, and c2 is 10. What is the arithmetic mean of ab,ac, and bc?
Pick an answer.
AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
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
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.
💡 An average times the count equals the sum — Grade 6 "a measure of center summarizes the values with one number."
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.
💡 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.
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.
💡 Undo the addition of $30$, then undo the multiplication by $2$ — classic Tool #11 inverse-operation chain.
6.SP.A.3 Step 4 Divide by $3$ to turn the sum into the mean the problem asked for.
💡 Mean is the sum divided by the count — the same Grade 6 definition we started with.
6.SP.A.3 Translate the two "mean" statements into sums. Multiplying each mean by $3$ remo 8.EE.C.7 Recall the trinomial-square identity. Expanding $(a + b + c)^2$ produces every s 8.EE.C.7 Substitute the known sums. The left side becomes $0^2 = 0$; the first term on th 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.3Recognize 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.7Solve 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!
More like this
Same archetype — closest grade level first.