1973 AMC 12 第 5 题

Write the averaging operation as $a*b=\frac{a+b}{2}$

Expand each proposed identity; one counterexample is enough to reject a property

Let $a*b=\frac{a+b}{2}.$ Commutativity is immediate. Also $$\begin{aligned} a+(b*c) &=\frac{2a+b+c}{2}\\ &=(a+b)*(a+c). \end{aligned}$$ so addition distributes over averaging.

Associativity fails because $$\begin{aligned} (a*b)*c&=\frac{a+b+2c}{4},\\ a*(b*c)&=\frac{2a+b+c}{4}. \end{aligned}$$ Averaging does not distribute over addition since generally $$a*(b+c)\ne(a*b)+(a*c).$$ Finally, an identity $e$ would require $\frac{e+a}{2}=a$ for every $a,$ or $e=a,$ which is impossible for a fixed $e.$ Thus only $\mathrm{II}$ and $\mathrm{IV}$ always hold.

Therefore, the correct answer is D.