2011 AMC 10B Problem 15

Test the laws with real numbers first, since one failed example is enough to rule a law out. Take $x = 2$, $y = 0$, $z = 0$. Remember $a @ b$ is the average of $a$ and $b$.

1. Law I. Left: $2 @ (0 + 0) = 2 @ 0 = 1$. Right: $(2 @ 0) + (2 @ 0) = 1 + 1 = 2$. Since $1 \ne 2$, law I is false.
2. That already rules out (A) and (D). To choose among the rest, check II and III for all numbers by writing them out.
3. Law II. Left: $x + \frac{y + z}{2} = \frac{2x + y + z}{2}$. Right: the average of $x + y$ and $x + z$ is $\frac{(x + y) + (x + z)}{2} = \frac{2x + y + z}{2}$. The two sides match, so law II is true.
4. Law III. Left: $y @ z = \frac{y + z}{2}$, and averaging that with $x$ gives
$$\frac{x + \frac{y+z}{2}}{2} = \frac{2x + y + z}{4}.$$
Right: $x @ y = \frac{x + y}{2}$ and $x @ z = \frac{x + z}{2}$. Their average is
$$\frac{\frac{x+y}{2} + \frac{x+z}{2}}{2} = \frac{2x + y + z}{4}.$$
The two sides match, so law III is true.

The answer is $\boxed{\textbf{(E)}\ \text{II and III only}}$.

A made-up symbol becomes easy once you replace it with ordinary arithmetic. Law I fails because the right side takes half of $x$ twice, which adds up to a whole $x$ instead of half. To show a law is false, one example is enough; to show it is always true, you need the algebra.

Write out both sides of each law using $a @ b = \frac{a + b}{2}$.

I. Left: $x @ (y + z) = \frac{x + y + z}{2}$. Right: $\frac{x + y}{2} + \frac{x + z}{2} = \frac{2x + y + z}{2}$. These differ by $\frac{x}{2}$, so I fails (for example $x = 2$, $y = z = 0$ gives $1 \ne 2$).

II. Left: $x + \frac{y + z}{2} = \frac{2x + y + z}{2}$. Right: $\frac{(x + y) + (x + z)}{2} = \frac{2x + y + z}{2}$. Equal, so II holds.

III. Left: $x @ \frac{y + z}{2} = \frac{x + \frac{y+z}{2}}{2} = \frac{2x + y + z}{4}$. Right: $\frac{x+y}{2} @ \frac{x+z}{2} = \frac{\frac{x+y}{2} + \frac{x+z}{2}}{2} = \frac{2x + y + z}{4}$. Equal, so III holds.

The answer is $\boxed{\textbf{(E)}\ \text{II and III only}}$.