2005 AMC 10B Problem 4

P1-10~1 minPrint

Functions Exponents, Logarithms & Radicals Set up the equation/formula and compute; no special trick needed

For real numbers $a$ and $b$ , define $a \diamond b = \sqrt{a^2 + b^2}$ . What is the value of

$(5 \diamond 12) \diamond ((-12) \diamond (-5))$ ?

Show solution without answering

Key insight

A negative number squared is positive, so both inner parts equal 13. Then 13 diamond 13 is 13 times the square root of 2.

Solution

Work from the inside of the parentheses out.

1. First inner part. $5 \diamond 12 = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$ .
2. Second inner part. $(-12) \diamond (-5) = \sqrt{(-12)^2 + (-5)^2}$ . A negative number times itself is positive, so $(-12)^2 = 144$ and $(-5)^2 = 25$ . This is again $\sqrt{169} = 13$ .
3. Outer step. Now the problem is $13 \diamond 13$ :
$13 \diamond 13 = \sqrt{13^2 + 13^2} = \sqrt{2 \cdot 13^2} = 13\sqrt{2}.$
(Pull the $13^2$ out of the square root as $13$ , and the $2$ stays inside.)

The answer is $\boxed{\textbf{(D)}\ 13\sqrt{2}}$ .

Why this works

A made-up symbol like $\diamond$ is just a rule: put the two numbers into the formula. With nested parentheses, do the innermost ones first and use their results in the outer step. It also helps to notice that $a \diamond a$ is always $a\sqrt{2}$ , the diagonal of a square with side $a$ .

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. Since squaring removes signs, the two inner operations are the same computation:
$5 \diamond 12 = \sqrt{25 + 144} = \sqrt{169} = 13, \qquad (-12) \diamond (-5) = \sqrt{144 + 25} = 13.$

Now apply the operation once more:
$13 \diamond 13 = \sqrt{13^2 + 13^2} = \sqrt{2 \cdot 169} = 13\sqrt{2}.$

The answer is $\boxed{\textbf{(D)}\ 13\sqrt{2}}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 16: Functions
Introduction to Algebra, Chapter 1: Follow the Rules (§1.6, §1.7, §1.8)
Introduction to Algebra, Chapter 17: Graphing Functions (§17.2)
Introduction to Algebra, Chapter 11: Special Factorizations (§11.4)