2021 Fall AMC 10A Problems/Problem 16

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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+linewidth(1.25),"$y=|\lfloor x \rfloor|$"); 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: [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),"$y=|\lfloor x \rfloor|$"); for (int i = 0; i < 9; ++i) { 	dot((i,i),heavygreen,UnFill);     dot((i+1,i),heavygreen);     dot((-i-1,i+1),heavygreen,UnFill);     dot((-i,i+1),heavygreen); } [/asy] Taking the difference, we graph $f(x) = y_1 - y_2,$ 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 (ProblemsAnswer KeyResources)
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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