AMC10 2024 B

AMC10 2024 B · Q9

AMC10 2024 B · Q9. It mainly tests Linear equations, Manipulating equations.

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 $bc$?

实数 $a, b,$ 和 $c$ 的算术平均数为 $0$。$a^2, b^2,$ 和 $c^2$ 的算术平均数为 $10$。求 $ab, ac,$ 和 $bc$ 的算术平均数？

(A) -5 -5

(B) -\dfrac{10}{3} -\dfrac{10}{3}

(D) 0 0

(E) \dfrac{10}{9} \dfrac{10}{9}

Answer

Correct choice: (A)

正确答案：（A）

Solution

If $\frac{a+b+c}{3} = 0$, that means $a+b+c=0$, and $(a+b+c)^2=0$. Expanding that gives \[(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc\] If $\frac{a^2+b^2+c^2}{3} = 10$, then $a^2+b^2+c^2=30$. Thus, we have \[30 + 2ab + 2ac + 2bc = 0\] Arithmetic will give you that $ab + bc + ac = -15$. To find the arithmetic mean, divide that by 3, so $\frac{ab + bc + ac}{3} = \boxed{\textbf{(A) }-5}$

若 $\frac{a+b+c}{3} = 0$，则 $a+b+c=0$，$(a+b+c)^2=0$。展开得： \[(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc\] 若 $\frac{a^2+b^2+c^2}{3} = 10$，则 $a^2+b^2+c^2=30$。因此， \[30 + 2ab + 2ac + 2bc = 0\] 算术运算得 $ab + bc + ac = -15$。算术平均数除以3，即 $\frac{ab + bc + ac}{3} = \boxed{\textbf{(A) }-5}$

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