AMC10 2005 B

AMC10 2005 B · Q4

AMC10 2005 B · Q4. It mainly tests Exponents & radicals, Pythagorean theorem.

For real numbers $a$ and $b$, define $a \diamond b = \sqrt{a^2 + b^2}$. What is the value of $(5 \diamond 12) \diamond ((-12) \diamond (-5))$?

对于实数$a$和$b$，定义$a \diamond b = \sqrt{a^2 + b^2}$。求$(5 \diamond 12) \diamond ((-12) \diamond (-5))$的值。

(A) 0 0

(B) $\frac{17}{2}$ $\frac{17}{2}$

(D) $13\sqrt{2}$ $13\sqrt{2}$

(E) 26 26

Answer

Correct choice: (D)

正确答案：（D）

Solution

(D) It follows from the definition that $$(5\diamond 12)\diamond((-12)\diamond(-5))=\sqrt{5^2+12^2}\diamond\sqrt{(-12)^2+(-5)^2}$$ $$=13\diamond 13=\sqrt{13^2+13^2}=13\sqrt{2}.$$

（D）由定义可知 $$(5\diamond 12)\diamond((-12)\diamond(-5))=\sqrt{5^2+12^2}\diamond\sqrt{(-12)^2+(-5)^2}$$ $$=13\diamond 13=\sqrt{13^2+13^2}=13\sqrt{2}。$$

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