AMC 10 · 2005 · #4
Grade 8 arithmeticPick an answer.
AMC 10 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The expression nests one diamond inside another, so Tool #7 (Identify Subproblems) is the natural fit: compute each inner diamond first, then feed the two results into the outer diamond. Tool #3 (Eliminate Possibilities) catches the built-in traps — reading ◆ as plain addition gives 13+13=26 (choice E), and forgetting the final square root of the sum leaves a bare 13 (choice C), so knowing the real rule steers past both.
Apply the rule to 5 ♦ 12
Plug a=5, b=12 into √(a²+b²): 25+144 = 169, so 5 ◆ 12 = 13.
The diamond is just a recipe: drop the two numbers into √(a²+b²) and turn the crank.
6.EE.A.2Identify SubproblemsApply the rule to (-12) ♦ (-5)
Same rule with a=-12, b=-5: squaring erases the signs, so 144+25 = 169 and (-12) ◆ (-5) = 13 again.
Squaring erases minus signs, so the negatives give the very same 169 as the positives.
Squaring erases minus signs, so the negatives give the very same value as the positives.
▸ Why?
A number and its opposite have the same square, so the sign cannot survive.
▸ Why?
The rule adds two squares under a root, exactly as a right triangle's hypotenuse is built.
Feed both results into the outer diamond
Both inner values are 13, so the outer one is 13 ◆ 13 = √(169+169) = √(338).
Two equal inputs mean the sum inside the root is just double one square: 169+169.
6.EE.A.1Identify SubproblemsSimplify the square root
Since 338 = 169·2 and 169 = 13² is a perfect square, √(338) = 13√(2) — choice (D).
Pull the perfect square 169 out of the root, leaving the irrational √(2) behind.
8.EE.A.2Eliminate PossibilitiesWork a nested custom operation from the inside out, and remember squaring wipes out minus signs — so the negatives here changed nothing.
- Apply the rule to 5 ♦ 12
- Apply the rule to (-12) ♦ (-5)
- Feed both results into the outer diamond
- Simplify the square root