2024 AMC 12A Problems/Problem 17

Problem

Let $f(x)$ be a nonzero continuous function such that $f\left(\sqrt{x^{2} + y^{2}}\right) = f(x)f(y)$ for all real numbers $x$ and $y$ . If $f(2) \leq 2024$ , then how many integers in the set $\{-20, -19, \cdots, 20\}$ could be the value of $f(1)$ ?

$\textbf{(A)}~6\qquad\textbf{(B)}~7\qquad\textbf{(C)}~12\qquad\textbf{(D)}~13\qquad\textbf{(E)}~16$

Solution

First, we observe that $f(x)$ must be an even function because $f\left(\sqrt{x^{2} + y^{2}}\right) = f(-x)f(y) = f(x)f(y) and f(y) \neq 0$ . We see that $f(x) = f(\lvert x \rvert) = f\left(\sqrt{\tfrac{x^{2}}{2} + \tfrac{x^{2}}{2}}\right) = f\left(\tfrac{x}{\sqrt{2}}\right)^{2} \geq 0$ . But since $f$ is nonzero, $f(x) > 0$ .

Consider $(x, y) = (1, 1)$ , which gives $f\left(\sqrt{1^{2} + 1^{2}}\right) = f\left(\sqrt{2}\right) = f(1)^{2}$ . Combining this with $(x, y) = \left(\sqrt{2}, \sqrt{2}\right)$ , we have $f\left(\sqrt{\left(\sqrt{2}\right)^{2} + \left(\sqrt{2}\right)^{2}}\right) = f(2) = f\left(\sqrt{2}\right)^{2} = f(1)^{4}$ .

Therefore, $f(2) = f(1)^{4} \leq 2024$ , and the only possible values for $f(1)$ are the positive integers less than $\sqrt[4]{2024}$ . Note that $6^{4} = 1296 < 2024$ but $7^{4} = 2401 > 2024$ , so the answer is $\boxed{\textbf{(A)}~6}$ .

2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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2024 AMC 12A Problems/Problem 17

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