Printable

2005 AMC 10B Problem 4

Problem only Back to the problem

2005 AMC 10B Problem 4AMC 10 Step by Step

Name: Date: September 13, 2026

Functions · difficulty 1 of 5 · about 1 min · position P1-10

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

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

A)
$0$
B)
$\frac{17}{2}$
C)
$13$
D)
$13\sqrt{2}$
E)
$26$

Answer

Key insight

Squaring erases the signs, so both inner operations give sqrt(25+144) = 13, and 13 diamond 13 is sqrt(2 * 169) = 13 sqrt 2.

Solution

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

Why this works

A defined operation is just a function of two inputs; evaluate the innermost instances first and feed the results outward. Recognizing $a \diamond a = a\sqrt{2}$ (the hypotenuse of an isosceles right triangle with legs $a$ ) makes the final step instant.

The trap

Stopping at 13 after the inner steps, or letting the negative signs change the inner value.

Common mistakes

Stopping at 13 after the inner steps, or letting the negative signs change the inner value.
Computing $\sqrt{13 + 13} = \sqrt{26}$ or adding the inner results to get $26$ instead of applying the definition.

Techniques

Set up the equation/formula and compute; no special trick needed