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Difference between revisions of "2024 AMC 12A Problems/Problem 17"

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==Problem==
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Let <math>f(x)</math> be a nonzero continuous function such that <cmath>f\left(\sqrt{x^{2} + y^{2}}\right) = f(x)f(y)</cmath> for all real numbers <math>x</math> and <math>y</math>. If <math>f(2) \leq 2024</math>, then how many integers in the set <math>\{-20, -19, \cdots, 20\}</math> could be the value of <math>f(1)</math>?
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<math>\textbf{(A)}~6\qquad\textbf{(B)}~7\qquad\textbf{(C)}~12\qquad\textbf{(D)}~13\qquad\textbf{(E)}~16</math>
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==Solution==
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First, we observe that <math>f(x)</math> must be an even function because <math>f\left(\sqrt{x^{2} + y^{2}}\right) = f(-x)f(y) = f(x)f(y) and f(y) \neq 0</math>. We see that <math>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</math>. But since <math>f</math> is nonzero, <math>f(x) > 0</math>.
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Consider <math>(x, y) = (1, 1)</math>, which gives <math>f\left(\sqrt{1^{2} + 1^{2}}\right) = f\left(\sqrt{2}\right) = f(1)^{2}</math>. Combining this with <math>(x, y) = \left(\sqrt{2}, \sqrt{2}\right)</math>, we have <math>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}</math>.
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Therefore, <math>f(2) = f(1)^{4} \leq 2024</math>, and the only possible values for <math>f(1)</math> are the positive integers less than <math>\sqrt[4]{2024}</math>. Note that <math>6^{4} = 1296 < 2024</math> but <math>7^{4} = 2401 > 2024</math>, so the answer is <math>\boxed{\textbf{(A)}~6}</math>.
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==See also==
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{{AMC12 box|year=2024|ab=A|num-b=16|num-a=18}}
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{{MAA Notice}}

Revision as of 21:43, 20 March 2025

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 $\boxed{\textbf{(A)}~6}$.

See also

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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