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

(→‎Solution 2 (Graphing))
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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 (Piecewise Function)==
+
==Solution 1 (Graphing)==
−
<b>IN PROGRESS AND WILL FINISH SOON. NO EDIT PLEASE. A MILLION THANKS.</b>
 
−
 
 
−
~MRENTHUSIASM
 
−
 
 
−
==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 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> 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.
  
Line 234: Line 229:
 
</asy>
 
</asy>
  
−
Therefore, the answer is <math>\boxed{\textbf{(D) }\text{ the point }\left(\dfrac12, 0\right)}.</math>
+
Therefore, the graph of <math>f(x)</math> is symmetric about <math>\boxed{\textbf{(D) }\text{ the point }\left(\dfrac12, 0\right)}.</math>
 +
 
 +
~MRENTHUSIASM
 +
 
 +
==Solution 2 (Piecewise Function)==
 +
For all real numbers <math>x</math> and integers <math>n,</math> note that:
 +
<ol style="margin-left: 1.5em;">
 +
  <li><math>\lfloor x+n \rfloor = \lfloor x \rfloor + n</math> and <math>\lceil x+n \rceil = \lceil x \rceil + n</math></li><p>
 +
  <li><math>\lfloor -x \rfloor = -\lceil x \rceil</math></li><p>
 +
  <li><math>\lceil x \rceil - \lfloor x \rfloor = \begin{cases}
 +
0 & \mathrm{if} \ x\text{ is an integer} \\
 +
1 & \mathrm{if} \ x\text{ is not an integer}
 +
\end{cases}</math></li>
 +
</ol>
 +
We rewrite <math>f(x)</math> as
 +
<cmath>\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*}</cmath>
 +
We apply casework to the value of <math>x:</math>
 +
<ol style="margin-left: 1.5em;">
 +
  <li></li><p>
 +
  <li></li><p>
 +
  <li></li><p>
 +
  <li></li><p>
 +
  <li></li><p>
 +
  <li></li><p>
 +
</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>
 +
 
 +
<b>IN PROGRESS AND WILL FINISH SOON. NO EDIT PLEASE. A MILLION THANKS.</b>
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM

Revision as of 16:36, 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 real numbers $x$ and integers $n,$ 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. $\lceil x \rceil - \lfloor x \rfloor = \begin{cases} 0 & \mathrm{if} \ x\text{ is an integer} \\ 1 & \mathrm{if} \ x\text{ is not an integer} \end{cases}$

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 $x:$

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

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

~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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