AMC 10 Step by Step

2011 AMC 10A Problem 3

P1-10~2 minPrint

Suppose [a b][a\ b] denotes the average of aa and bb , and {a b c}\{a\ b\ c\} denotes the average of a,ba, b , and cc . What is {{1 1 0} [0 1] 0}\{\{1\ 1\ 0\}\ [0\ 1]\ 0\} ?

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Key insight

Work from the inside out. The inner brackets are 2/3 and 1/2; then average those two numbers together with 0.

Solution

The square brackets mean "average of two numbers" and the curly brackets mean "average of three numbers". Like ordinary parentheses, the innermost ones get worked out first.

1. The inner curly bracket. {1 1 0}\{1\ 1\ 0\} is the average of 11, 11 and 00:
1+1+03=23.\frac{1 + 1 + 0}{3} = \frac{2}{3}.
2. The inner square bracket. [0 1][0\ 1] is the average of 00 and 11:
0+12=12.\frac{0 + 1}{2} = \frac{1}{2}.
3. The outer curly bracket. Now the expression is {23 12 0}\{\tfrac{2}{3}\ \tfrac{1}{2}\ 0\}, the average of three numbers. Add them using the common denominator 66:
23+12+0=46+36=76.\frac{2}{3} + \frac{1}{2} + 0 = \frac{4}{6} + \frac{3}{6} = \frac{7}{6}.
4. There are three numbers, so divide by 33:
76÷3=718.\frac{7}{6} \div 3 = \frac{7}{18}.

The answer is (D) 718\boxed{\textbf{(D)}\ \frac{7}{18}}.

Why this works

A made-up symbol is just a rule for combining numbers. When symbols are nested, you handle them like nested parentheses: innermost first. The only thing to keep track of is how many numbers each average has, two or three, because that is what you divide by.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Work from the inside out.

The inner triple: {1 1 0}=1+1+03=23\{1\ 1\ 0\} = \dfrac{1 + 1 + 0}{3} = \dfrac{2}{3}.

The inner pair: [0 1]=0+12=12[0\ 1] = \dfrac{0 + 1}{2} = \dfrac{1}{2}.

The outer expression is the average of 23\frac{2}{3}, 12\frac{1}{2} and 00:
23+12+03=763=718.\frac{\frac{2}{3} + \frac{1}{2} + 0}{3} = \frac{\frac{7}{6}}{3} = \frac{7}{18}.

The answer is (D) 718\boxed{\textbf{(D)}\ \frac{7}{18}}.

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.