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

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== Solution 5 (Semi-Fakesolve) ==
 
== Solution 5 (Semi-Fakesolve) ==
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Suppose <math>x\in \mathbb{Z},</math> making 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>
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If, for all <math>x\in \mathbb{R}</math>, <math>f(x)</math> is symmetric about a line (or a point), we know that for all <math>x\in \mathbb{Z}</math>, <math>f(x)</math> is symmetric about the same point/line because <math>\mathbb{Z}\subset \mathbb{R}.</math>
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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>
  
  
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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  
 
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  
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<cmath>\left(\frac{0 + 1}{2}, \frac{-1 + 1}{2}\right) = \boxed{\textbf{(D)}\ \left(\frac{1}{2}, 0\right)}.</cmath>
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<cmath>\left(\frac{0 + 1}{2}, \frac{-1 + 1}{2}\right) = \boxed{\textbf{(D)}\ \left(\frac{1}{2}, 0\right)},</cmath>
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the midpoint of the "endpoints" of these line segments.
 
-Benedict T (countmath1)
 
-Benedict T (countmath1)
  

Revision as of 11:33, 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)

If, for all $x\in \mathbb{R}$, $f(x)$ is symmetric about a line (or a point), we know that for all $x\in \mathbb{Z}$, $f(x)$ is symmetric about the same point/line because $\mathbb{Z}\subset \mathbb{R}.$


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. -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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