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Difference between revisions of "2021 Fall AMC 10A Problems/Problem 16"

m (→‎Solution 1 (Graphing))
m (→‎Solution 5 (Semi-Fakesolve))
 
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\textbf{(D) }\text{ the point }\left(\dfrac12, 0\right)\qquad \textbf{(E) }\text{the point }(1,0)</math>
 
\textbf{(D) }\text{ the point }\left(\dfrac12, 0\right)\qquad \textbf{(E) }\text{the point }(1,0)</math>
  
−
==Solution 1 (Graphing)==
+
==Solution 1 (Observations)==
−
Let <math>y_1=|\lfloor x \rfloor|</math> and <math>y_2=|\lfloor 1 - x \rfloor|=|\lfloor -(x-1) \rfloor|.</math> Note that the graph of <math>y_2</math> is a reflection of the graph of <math>y_1</math> about the <math>y</math>-axis, followed by a translation of <math>1</math> unit right.
+
Note that <cmath>f(1-x)=|\lfloor 1-x\rfloor|-|\lfloor x\rfloor|=-f(x),</cmath>
 +
so <math>f\left(\frac12+x\right)=-f\left(\frac12-x\right)</math>.
 +
 
 +
This means that the graph is symmetric about <math>\boxed{\textbf{(D) }\text{ the point }\left(\dfrac12, 0\right)}</math>.
 +
 
 +
==Solution 2 (Graphing)==
 +
Let <math>y_1=|\lfloor x \rfloor|</math> and <math>y_2=|\lfloor 1 - x \rfloor|=|\lfloor -(x-1) \rfloor|.</math> Note that the graph of <math>y_2</math> is a reflection of the graph of <math>y_1</math> about the <math>y</math>-axis, followed by a translation <math>1</math> unit to the right.
  
 
The graph of <math>y_1</math> is shown below:
 
The graph of <math>y_1</math> is shown below:
Line 233: Line 239:
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  
−
==Solution 2 (Casework)==
+
==Solution 3 (Casework)==
 
For all <math>x\in\mathbb{R}</math> and <math>n\in\mathbb{Z},</math> note that:
 
For all <math>x\in\mathbb{R}</math> and <math>n\in\mathbb{Z},</math> note that:
 
<ol style="margin-left: 1.5em;">
 
<ol style="margin-left: 1.5em;">
Line 276: Line 282:
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  
−
== Solution 3 (Casework) ==
+
== Solution 4 (Casework) ==
 
Denote <math>x = a + b</math>, where <math>a \in \Bbb Z</math> and <math>b \in \left[ 0 , 1 \right)</math>.
 
Denote <math>x = a + b</math>, where <math>a \in \Bbb Z</math> and <math>b \in \left[ 0 , 1 \right)</math>.
 
Hence, <math>a</math> is the integer part of <math>x</math> and <math>b</math> is the decimal part of <math>x</math>.
 
Hence, <math>a</math> is the integer part of <math>x</math> and <math>b</math> is the decimal part of <math>x</math>.
  
−
<math>\textbf{Case 1}</math>: <math>b = 0</math>.
+
'''Case 1''': <math>b = 0</math>.
  
 
We have
 
We have
Line 312: Line 318:
 
</cmath>
 
</cmath>
  
−
<math>\textbf{Case 2}</math>: <math>b \neq 0</math>.
+
'''Case 2''': <math>b \neq 0</math>.
  
 
We have
 
We have
Line 331: Line 337:
  
 
~Steven Chen (www.professorchenedu.com)
 
~Steven Chen (www.professorchenedu.com)
 +
 +
== Solution 5 (Semi-Fakesolve) ==
 +
 +
Suppose <math>x\in \mathbb{Z},</math> making the equation equivalent to <math>f(x) = |x|-|1-x|.</math> We consider the cases when <math>x\in (-\infty, 0), 0, 1, (1, \infty).</math>
 +
 +
 +
If <math>x\in (-\infty, 0)</math>, we have <math>|x| = -x</math> and <math>|1-x| = 1-x,</math> so <math>f(x) = -x - (1-x) = -1</math>.
 +
 +
 +
If <math>x = 0</math> or <math>x = 1</math>, we trivially get <math>f(x) = -1</math> and <math>1</math> respectively.
 +
 +
 +
If <math>x\in (1, \infty)</math>, we have <math>|x| = x</math> and <math>|1-x| = x - 1</math>, giving <math>f(x) = x-(x-1)= 1.</math>
 +
 +
 +
Since, for all <math>x\in \mathbb{Z} \leq 0</math>, <math>f(x)  =-1</math> and <math>x\in \mathbb{Z} \geq 1, f(x) = 1</math>, we can conclude that it is symmetric across the coordinate pair
 +
<cmath>\left(\frac{0 + 1}{2}, \frac{-1 + 1}{2}\right) = \boxed{\textbf{(D)}\ \left(\frac{1}{2}, 0\right)},</cmath>
 +
the midpoint of the "endpoints" of these line segments.
 +
 +
<asy>
 +
size(250); //Credit to MRENTHUSIASM
 +
 +
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();
 +
 +
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));
 +
}
 +
 +
</asy>
 +
 +
 +
 +
Just considering the integers is never a good idea when dealing with any function, especially one with floor functions. However, after dealing with the case "when <math>x\in \mathbb{Z}</math>", it becomes apparent that the graph of <math>f(x)</math> is symmetric about <math>x = \frac{1}{2}</math>, or more specifically, the point <math>\left(\frac{1}{2}, 0\right).</math>
 +
 +
-Benedict T (countmath1)
 +
 +
==Video Solution ==
 +
https://youtu.be/RpxlZJRiSjk
 +
 +
~Education, the Study of Everything
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2021 Fall|ab=A|num-b=15|num-a=17}}
 
{{AMC10 box|year=2021 Fall|ab=A|num-b=15|num-a=17}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 15:56, 7 March 2024

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 (Observations)

Note that \[f(1-x)=|\lfloor 1-x\rfloor|-|\lfloor x\rfloor|=-f(x),\] so $f\left(\frac12+x\right)=-f\left(\frac12-x\right)$.

This means that the graph is symmetric about $\boxed{\textbf{(D) }\text{ the point }\left(\dfrac12, 0\right)}$.

Solution 2 (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 $1$ unit to the 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((-10,0)--(10,0),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 3 (Casework)

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 $f(x)=0-1=-1.$

  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 $f(x)=0-0=0.$

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

Alternatively, we can eliminate $\textbf{(A)}, \textbf{(B)}, \textbf{(C)},$ and $\textbf{(E)}$ once we finish with Case 3. This leaves us with $\textbf{(D)}.$

~MRENTHUSIASM

Solution 4 (Casework)

Denote $x = a + b$, where $a \in \Bbb Z$ and $b \in \left[ 0 , 1 \right)$. Hence, $a$ is the integer part of $x$ and $b$ is the decimal part of $x$.

Case 1: $b = 0$.

We have \begin{align*} f \left( x \right) & = \left| \lfloor x \rfloor \right| - \left| \lfloor 1 - x \rfloor \right| \\ & = | a | - | 1 - a | \\ & = \left\{ \begin{array}{ll} a - \left( a - 1 \right) & \mbox{ if } a \in \Bbb Z \mbox{ and } a \geq 1 \\ -1 & \mbox{ if } a = 0 \\ - a - \left( 1 - a \right) & \mbox{ if } a \in \Bbb Z \mbox{ and } a \leq -1 \end{array} \right. \\ & = \left\{ \begin{array}{ll} 1 & \mbox{ if } a \in \Bbb Z \mbox{ and } a \geq 1 \\ -1 & \mbox{ if } a = 0 \\ -1 & \mbox{ if } a \in \Bbb Z \mbox{ and } a \leq -1 \end{array} \right. \\ & = \left\{ \begin{array}{ll} 1 & \mbox{ if } a \in \Bbb Z \mbox{ and } a \geq 1 \\ -1 & \mbox{ if } a \in \Bbb Z \mbox{ and } a \leq 0 \end{array} \right. \end{align*}

Case 2: $b \neq 0$.

We have \begin{align*} f \left( x \right) & = \left| \lfloor x \rfloor \right| - \left| \lfloor 1 - x \rfloor \right| \\ & = | a | - | \lfloor 1 - a - b \rfloor | \\ & = | a | - | \lfloor - a + \left( 1 - b \right) \rfloor | \\ & = | a | - | - a | \\ & = 0 . \end{align*}

Therefore, the graph of $f \left( x \right)$ is symmetric through the point $\left( \frac{1}{2} , 0 \right)$.

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

~Steven Chen (www.professorchenedu.com)

Solution 5 (Semi-Fakesolve)

Suppose $x\in \mathbb{Z},$ making the equation equivalent to $f(x) = |x|-|1-x|.$ We consider the cases when $x\in (-\infty, 0), 0, 1, (1, \infty).$


If $x\in (-\infty, 0)$, we have $|x| = -x$ and $|1-x| = 1-x,$ so $f(x) = -x - (1-x) = -1$.


If $x = 0$ or $x = 1$, we trivially get $f(x) = -1$ and $1$ respectively.


If $x\in (1, \infty)$, we have $|x| = x$ and $|1-x| = x - 1$, giving $f(x) = x-(x-1)= 1.$


Since, for all $x\in \mathbb{Z} \leq 0$, $f(x) =-1$ and $x\in \mathbb{Z} \geq 1, f(x) = 1$, we can conclude that it is symmetric across the coordinate pair \[\left(\frac{0 + 1}{2}, \frac{-1 + 1}{2}\right) = \boxed{\textbf{(D)}\ \left(\frac{1}{2}, 0\right)},\] the midpoint of the "endpoints" of these line segments.

[asy] size(250); //Credit to MRENTHUSIASM 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(); 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)); } [/asy]


Just considering the integers is never a good idea when dealing with any function, especially one with floor functions. However, after dealing with the case "when $x\in \mathbb{Z}$", it becomes apparent that the graph of $f(x)$ is symmetric about $x = \frac{1}{2}$, or more specifically, the point $\left(\frac{1}{2}, 0\right).$

-Benedict T (countmath1)

Video Solution

https://youtu.be/RpxlZJRiSjk

~Education, the Study of Everything

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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