AMC 10 · 2005 · #4
Grade 8 arithmeticFor real numbers a and b, define a⋄b=a2+b2. What is the value of
(5⋄12)⋄((−12)⋄(−5))?
Pick an answer.
AMC 10 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A new operation is defined by $a \diamond b = \sqrt{a^2 + b^2}$ for real numbers $a$ and $b$. Using this rule, find the value of $(5 \diamond 12) \diamond ((-12) \diamond (-5))$.
Givens: The rule $a \diamond b = \sqrt{a^2 + b^2}$: square both inputs, add them, take the square root; The expression to evaluate is $(5 \diamond 12) \diamond ((-12) \diamond (-5))$; Answer choices: (A) $0$, (B) $\dfrac{17}{2}$, (C) $13$, (D) $13\sqrt{2}$, (E) $26$
Unknowns: The single number that $(5 \diamond 12) \diamond ((-12) \diamond (-5))$ equals
Understand
Restated: A new operation is defined by $a \diamond b = \sqrt{a^2 + b^2}$ for real numbers $a$ and $b$. Using this rule, find the value of $(5 \diamond 12) \diamond ((-12) \diamond (-5))$.
Givens: The rule $a \diamond b = \sqrt{a^2 + b^2}$: square both inputs, add them, take the square root; The expression to evaluate is $(5 \diamond 12) \diamond ((-12) \diamond (-5))$; Answer choices: (A) $0$, (B) $\dfrac{17}{2}$, (C) $13$, (D) $13\sqrt{2}$, (E) $26$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #3 Eliminate Possibilities
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 $\diamond$ 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.
Execute — Answer: D
6.EE.A.2 Step 1 Apply the rule to 5 ♦ 12
- Plug $a=5$ and $b=12$ into the rule $\sqrt{a^2+b^2}$.
- Square each: $5^2=25$ and $12^2=144$.
- Add them: $25+144=169$.
- Then take the square root: $\sqrt{169}=13$.
- So $5 \diamond 12 = 13$.
💡 The diamond is just a recipe: drop the two numbers into $\sqrt{a^2+b^2}$ and turn the crank.
7.NS.A.2 Step 2 Apply the rule to (-12) ♦ (-5)
- Now plug $a=-12$ and $b=-5$ into the same rule.
- Squaring a negative gives a positive: $(-12)^2 = 144$ and $(-5)^2 = 25$.
- Add them: $144+25=169$, the same sum as before.
- So $(-12) \diamond (-5) = \sqrt{169} = 13$ — the minus signs made no difference.
💡 Squaring erases minus signs, so the negatives give the very same $169$ as the positives.
6.EE.A.1 Step 3 Feed both results into the outer diamond
- Both inner diamonds equal $13$, so the outer expression is $13 \diamond 13$.
- Apply the rule once more: square each $13$ to get $169$, and add: $13^2+13^2 = 169+169 = 338$.
- The value is $\sqrt{338}$ — not yet in simplest form.
💡 Two equal inputs mean the sum inside the root is just double one square: $169+169$.
8.EE.A.2 Step 4 Simplify the square root
- Factor $338 = 169 \cdot 2$, and $169=13^2$ is a perfect square.
- Pull it out of the root: $\sqrt{169\cdot 2} = \sqrt{169}\cdot\sqrt{2} = 13\sqrt{2}$.
- That is choice (D).
- The trap $26$ (E) comes from adding $13+13$ as if $\diamond$ were ordinary addition, and the bare $13$ (C) from skipping the final square root — both are ruled out.
💡 Pull the perfect square $169$ out of the root, leaving the irrational $\sqrt{2}$ behind.
6.EE.A.2 Plug $a=5$ and $b=12$ into the rule $\sqrt{a^2+b^2}$. Square each: $5^2=25$ and 7.NS.A.2 Now plug $a=-12$ and $b=-5$ into the same rule. Squaring a negative gives a posi 6.EE.A.1 Both inner diamonds equal $13$, so the outer expression is $13 \diamond 13$. App 8.EE.A.2 Factor $338 = 169 \cdot 2$, and $169=13^2$ is a perfect square. Pull it out of t Review
Reasonableness: Each inner diamond is the classic $5\text{-}12\text{-}13$ right-triangle relationship, so both collapse to $13$ — a clean whole number, a good sign the setup is right. The outer diamond then combines two equal $13$s, and $13\sqrt{2} \approx 13(1.414) \approx 18.4$ sits sensibly between $13$ and $26$: bigger than either input's raw value but smaller than their plain sum. Choices $0$ and $\tfrac{17}{2}$ are far too small, and $26$ would require $\diamond$ to be addition, which it is not.
Alternative: Use the general fact $x \diamond x = \sqrt{x^2+x^2} = \sqrt{2x^2} = x\sqrt{2}$ for any positive $x$. Once both inner diamonds are seen to equal $13$, the outer diamond is $13 \diamond 13 = 13\sqrt{2}$ in a single step, with no need to compute $338$ explicitly.
CCSS standards used (min grade 8)
6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Substituting the given numbers for $a$ and $b$ in the defined rule $\sqrt{a^2+b^2}$ to evaluate each diamond.)7.NS.A.2Apply and extend understanding of multiplication and division of rational numbers (Recognizing that squaring the negatives $(-12)^2$ and $(-5)^2$ gives positive $144$ and $25$.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Evaluating the squares $5^2,\,12^2,\,13^2$ and summing them inside the square roots.)8.EE.A.2Use square root and cube root symbols to represent solutions (Evaluating $\sqrt{169}=13$ and simplifying $\sqrt{338}=13\sqrt{2}$ by extracting the perfect square.)
⭐ Work a nested custom operation from the inside out, and remember squaring wipes out minus signs — so the negatives here changed nothing.
⭐ Work a nested custom operation from the inside out, and remember squaring wipes out minus signs — so the negatives here changed nothing.
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