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Difference between revisions of "2021 Fall AMC 10A Problems/Problem 16"
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==Solution 2 (Graphing)==
 
==Solution 2 (Graphing)==
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Let <math>y_1=|\lfloor x \rfloor|</math> and <math>y_2=|\lfloor 1 - x \rfloor|=|\lfloor -(x-1) \rfloor|.</math> Note that <math>y_2</math> is a reflection of <math>y_1</math> about the <math>y</math>-axis, followed by a translation of <math>1</math> unit right.
+
Let <math>y_1=|\lfloor x \rfloor|</math> and <math>y_2=|\lfloor 1 - x \rfloor|=|\lfloor -(x-1) \rfloor|.</math>
  
 
The graph of <math>y_1</math> is shown below:
 
The graph of <math>y_1</math> is shown below:
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}
 
}
 
</asy>
 
</asy>
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The graph of <math>y_2</math> is shown below:
+
Note that <math>y_2</math> is a reflection of <math>y_1</math> about the <math>y</math>-axis, followed by a translation of <math>1</math> unit right. The graph of <math>y_2</math> is shown below:
  
 
<b>DIAGRAM WILL BE READY VERY SOON. APPRECIATE IT IF THERE'S NO EDIT AT THIS PAGE WITHIN THE NEXT HOUR.</b>
 
<b>DIAGRAM WILL BE READY VERY SOON. APPRECIATE IT IF THERE'S NO EDIT AT THIS PAGE WITHIN THE NEXT HOUR.</b>

Revision as of 14:19, 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)

Let $y_1=|\lfloor x \rfloor|$ and $y_2=|\lfloor 1 - x \rfloor|=|\lfloor -(x-1) \rfloor|.$

The graph of $y_1$ is shown below: [asy] /* Made by MRENTHUSIASM */ size(250); int xMin = -10; int xMax = 10; int yMin = -10; int yMax = 10; //Draws the horizontal gridlines void horizontalLines() { for (int i = yMin+1; i < yMax; ++i) { draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4)); } } //Draws the vertical gridlines void verticalLines() { for (int i = xMin+1; i < xMax; ++i) { draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4)); } } //Draws the horizontal ticks void horizontalTicks() { for (int i = yMin+1; i < yMax; ++i) { draw((-3/16,i)--(3/16,i), black+linewidth(1)); } } //Draws the vertical ticks void verticalTicks() { for (int i = xMin+1; i < xMax; ++i) { draw((i,-3/16)--(i,3/16), black+linewidth(1)); } } //Draws and labels coordinate axes void drawLabelAxes() { draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("$x$",(xMax,0),(2,0)); label("$y$",(0,yMax),(0,2)); } horizontalLines(); verticalLines(); horizontalTicks(); verticalTicks(); drawLabelAxes(); path P[], Q[]; for (int i = 0; i < 9; ++i) { P[i] = (i,i)--(i+1,i); Q[i] = (-i,i+1)--(-i-1,i+1); } draw(P^^Q,red,"$y=|\lfloor x \rfloor|$"); for (int i = 0; i < 9; ++i) { dot((i,i),red); dot((i+1,i),red,UnFill); dot((-i-1,i+1),red); dot((-i,i+1),red,UnFill); } [/asy] Note that $y_2$ is a reflection of $y_1$ about the $y$-axis, followed by a translation of $1$ unit right. The graph of $y_2$ is 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

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