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Difference between revisions of "2021 Fall AMC 10A Problems/Problem 16"
(→‎Solution 2 (Piecewise Function))
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We apply casework to the value of <math>x:</math>
 
We apply casework to the value of <math>x:</math>
 
<ol style="margin-left: 1.5em;">
 
<ol style="margin-left: 1.5em;">
 +
  <li><math>x\in\mathbb{Z}^-</math></li><p>
 +
It follows that <math>f(x)=-x-(-x+1)=-1.</math><p>
 
   <li><math>x=0</math></li><p>
 
   <li><math>x=0</math></li><p>
 
It follows that <math>f(x)=-1.</math><p>
 
It follows that <math>f(x)=-1.</math><p>
 
   <li><math>x\in\mathbb{Z}^+</math></li><p>
 
   <li><math>x\in\mathbb{Z}^+</math></li><p>
 
It follows that <math>f(x)=x-(x-1)=1.</math><p>
 
It follows that <math>f(x)=x-(x-1)=1.</math><p>
−
  <li><math>x\in\mathbb{Z}^-</math></li><p>
 
−
It follows that <math>f(x)=-x-(-x+1)=-1.</math><p>
 
 
   <li><math>x\not\in\mathbb{Z}</math> and <math>x<0</math></li><p>
 
   <li><math>x\not\in\mathbb{Z}</math> and <math>x<0</math></li><p>
 
It follows that <math>f(x)=-\lfloor x \rfloor - (-\lceil x\rceil+1)=(\lceil x \rceil - \lfloor x \rfloor)-1=0.</math><p>
 
It follows that <math>f(x)=-\lfloor x \rfloor - (-\lceil x\rceil+1)=(\lceil x \rceil - \lfloor x \rfloor)-1=0.</math><p>
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It follows that <math>f(x)=\lfloor x \rfloor - (\lceil x\rceil-1)=(\lfloor x \rfloor - \lceil x \rceil)+1=0.</math><p>
 
It follows that <math>f(x)=\lfloor x \rfloor - (\lceil x\rceil-1)=(\lfloor x \rfloor - \lceil x \rceil)+1=0.</math><p>
 
</ol>
 
</ol>
−
Together, we have ..., so the graph of <math>f(x)</math> is symmetric about <math>\boxed{\textbf{(D) }\text{ the point }\left(\dfrac12, 0\right)}.</math>
+
Together, we have  
 +
<cmath>f(x)=\begin{cases}
 +
-1 & \mathrm{if} \ x\in\mathbb{Z}^{-}\cup\{0\} \\
 +
1 & \mathrm{if} \ x\in\mathbb{Z}^{+} \\
 +
0 & \mathrm{if} \ x\not\in\mathbb{Z}
 +
\end{cases},</cmath>
 +
so the graph of <math>f(x)</math> is symmetric about <math>\boxed{\textbf{(D) }\text{ the point }\left(\dfrac12, 0\right)}.</math>
 +
 
 +
<u><b>Remark</b></u>
  
−
<b>IN PROGRESS AND WILL FINISH SOON. NO EDIT PLEASE. A MILLION THANKS.</b>
+
Certainly, if we read the answer choices sooner, then we can stop after Case 3 and eliminate <math>\textbf{(A)}, \textbf{(B)}, \textbf{(C)},</math> and <math>\textbf{(E)}.</math> This leaves us with <math>\textbf{(D)}.</math>
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM

Revision as of 17:07, 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 (Graphing)

Let $y_1=|\lfloor x \rfloor|$ and $y_2=|\lfloor 1 - x \rfloor|=|\lfloor -(x-1) \rfloor|.$ Note that the graph of $y_2$ is a reflection of the graph of $y_1$ about the $y$-axis, followed by a translation of $1$ unit right.

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+linewidth(1.25)); for (int i = 0; i < 9; ++i) { dot((i,i),red+linewidth(4)); dot((i+1,i),red+linewidth(0.7),UnFill); dot((-i-1,i+1),red+linewidth(4)); dot((-i,i+1),red+linewidth(0.7),UnFill); } [/asy] The graph of $y_2$ 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,heavygreen+linewidth(1.25)); for (int i = 0; i < 9; ++i) { dot((i,i),heavygreen+linewidth(0.7),UnFill); dot((i+1,i),heavygreen+linewidth(4)); dot((-i-1,i+1),heavygreen+linewidth(0.7),UnFill); dot((-i,i+1),heavygreen+linewidth(4)); } [/asy] The graph of $f(x)=y_1-y_2$ 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(); draw((-9,0)--(9,0),mediumblue+linewidth(1.25),"$y=|\lfloor x \rfloor|$"); for (int i = 0; i > -10; --i) { dot((i,-1),mediumblue+linewidth(4)); } for (int i = 1; i < 10; ++i) { dot((i,1),mediumblue+linewidth(4)); } for (int i = -9; i < 10; ++i) { dot((i,0),mediumblue+linewidth(0.7),UnFill); } [/asy]

Therefore, the graph of $f(x)$ is symmetric about $\boxed{\textbf{(D) }\text{ the point }\left(\dfrac12, 0\right)}.$

~MRENTHUSIASM

Solution 2 (Piecewise Function)

For all $x\in\mathbb{R}$ and $n\in\mathbb{Z},$ note that:

  1. $\lfloor x+n \rfloor = \lfloor x \rfloor + n$ and $\lceil x+n \rceil = \lceil x \rceil + n$

  2. $\lfloor -x \rfloor = -\lceil x \rceil$

  3. $\lceil x \rceil - \lfloor x \rfloor = \begin{cases} 0 & \mathrm{if} \ x\in\mathbb{Z} \\ 1 & \mathrm{if} \ x\not\in\mathbb{Z} \end{cases}$

We rewrite $f(x)$ as \begin{align*} f(x) &= |\lfloor x \rfloor| - |\lfloor 1 - x \rfloor| \\ &= |\lfloor x \rfloor| - |-\lceil x - 1 \rceil| \\ &= |\lfloor x \rfloor| - |-\lceil x \rceil + 1|. \end{align*} We apply casework to the value of $x:$

  1. $x\in\mathbb{Z}^-$

    2. It follows that $f(x)=-x-(-x+1)=-1.$

  2. $x=0$

    3. It follows that $f(x)=-1.$

  3. $x\in\mathbb{Z}^+$

    4. It follows that $f(x)=x-(x-1)=1.$

  4. $x\not\in\mathbb{Z}$ and $x<0$

    5. It follows that $f(x)=-\lfloor x \rfloor - (-\lceil x\rceil+1)=(\lceil x \rceil - \lfloor x \rfloor)-1=0.$

  5. $x\not\in\mathbb{Z}$ and $0<x<1$

    6. It follows that $f(x)=0.$

  6. $x\not\in\mathbb{Z}$ and $x>1$

    7. It follows that $f(x)=\lfloor x \rfloor - (\lceil x\rceil-1)=(\lfloor x \rfloor - \lceil x \rceil)+1=0.$

Together, we have \[f(x)=\begin{cases} -1 & \mathrm{if} \ x\in\mathbb{Z}^{-}\cup\{0\} \\ 1 & \mathrm{if} \ x\in\mathbb{Z}^{+} \\ 0 & \mathrm{if} \ x\not\in\mathbb{Z} \end{cases},\] so the graph of $f(x)$ is symmetric about $\boxed{\textbf{(D) }\text{ the point }\left(\dfrac12, 0\right)}.$

Remark

Certainly, if we read the answer choices sooner, then we can stop after Case 3 and eliminate $\textbf{(A)}, \textbf{(B)}, \textbf{(C)},$ and $\textbf{(E)}.$ This leaves us with $\textbf{(D)}.$

~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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