Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2021 Fall AMC 10A Problems/Problem 16"
m (→‎Solution 2 (Graphing))
Line 13: Line 13:
 
We graph <math>y=|\lfloor x \rfloor|</math> and <math>y=|\lfloor 1 - x \rfloor|,</math> as shown below:
 
We graph <math>y=|\lfloor x \rfloor|</math> and <math>y=|\lfloor 1 - x \rfloor|,</math> as shown below:
  
−
<b>DIAGRAM WILL BE READY VERY SOON</b>
+
<b>DIAGRAM WILL BE READY VERY SOON. APPRECIATE IT IF THERE'S NO EDIT AT THIS PAGE WITHIN THE NEXT HOUR.</b>
  
 
Taking the difference, we graph <math>f(x) = |\lfloor x \rfloor| - |\lfloor 1 - x \rfloor|,</math> as shown below:
 
Taking the difference, we graph <math>f(x) = |\lfloor x \rfloor| - |\lfloor 1 - x \rfloor|,</math> as shown below:
  
−
<b>DIAGRAM WILL BE READY VERY SOON</b>
+
<b>DIAGRAM WILL BE READY VERY SOON. APPRECIATE IT IF THERE'S NO EDIT AT THIS PAGE WITHIN THE NEXT HOUR.</b>
  
 
Therefore, the answer is <math>\boxed{\textbf{(D) }\text{ the point }\left(\dfrac12, 0\right)}.</math>
 
Therefore, the answer is <math>\boxed{\textbf{(D) }\text{ the point }\left(\dfrac12, 0\right)}.</math>

Revision as of 13:04, 25 November 2021

Problem

The graph of \[f(x) = |\lfloor x \rfloor| - |\lfloor 1 - x \rfloor|\] is symmetric about which of the following? (Here $\lfloor x \rfloor$ is the greatest integer not exceeding $x$.)

$\textbf{(A) }\text{the }y\text{-axis}\qquad \textbf{(B) }\text{the line }x = 1\qquad \textbf{(C) }\text{the origin}\qquad \textbf{(D) }\text{ the point }\left(\dfrac12, 0\right)\qquad \textbf{(E) }\text{the point }(1,0)$

Solution 1 (Piecewise Function)

IN PROGRESS AND WILL FINISH SOON. NO EDIT PLEASE. A MILLION THANKS.

~MRENTHUSIASM

Solution 2 (Graphing)

We graph $y=|\lfloor x \rfloor|$ and $y=|\lfloor 1 - x \rfloor|,$ as shown below:

DIAGRAM WILL BE READY VERY SOON. APPRECIATE IT IF THERE'S NO EDIT AT THIS PAGE WITHIN THE NEXT HOUR.

Taking the difference, we graph $f(x) = |\lfloor x \rfloor| - |\lfloor 1 - x \rfloor|,$ as shown below:

DIAGRAM WILL BE READY VERY SOON. APPRECIATE IT IF THERE'S NO EDIT AT THIS PAGE WITHIN THE NEXT HOUR.

Therefore, the answer is $\boxed{\textbf{(D) }\text{ the point }\left(\dfrac12, 0\right)}.$

~MRENTHUSIASM

See Also

2021 Fall AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_10A_Problems/Problem_16&oldid=166311"