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2021 Fall AMC 10A Problem 16

Problem 16 of 25IntermediateAlgebra

Consider the function f(x)=∣⌊x⌋∣−∣⌊1−x⌋∣f(x) = |\lfloor x \rfloor| - |\lfloor 1 - x \rfloor| Its graph is symmetric about which of the following? (Here ⌊x⌋\lfloor x \rfloor is the greatest integer not exceeding x.x.)

Answer choices

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Solution

For every real x,x, f(1−x)=∣⌊1−x⌋∣−∣⌊x⌋∣=−f(x). \begin{aligned} &f(1-x)=|\lfloor 1-x\rfloor| \\ &\quad {}-|\lfloor x\rfloor|=-f(x). \end{aligned} Equivalently, replacing xx by 12+t\frac{1}{2}+t gives f ⁣(12−t)=−f ⁣(12+t).f\!\left(\frac{1}{2}-t\right)=-f\!\left(\frac{1}{2}+t\right). This is point symmetry about (12,0).\left(\frac{1}{2},0\right). Thus, D is the correct answer.
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Tagged: floor and ceiling functions · symmetry (algebra)

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