Difference between revisions of "2021 Fall AMC 10A Problems/Problem 16"
MRENTHUSIASM (talk | contribs) |
MRENTHUSIASM (talk | contribs) (→Solution 2 (Graphing)) |
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path P[], Q[]; | path P[], Q[]; | ||
for (int i = 0; i < 9; ++i) { | for (int i = 0; i < 9; ++i) { | ||
| − | + | P[i] = (i,i)--(i+1,i); | |
Q[i] = (-i,i+1)--(-i-1,i+1); | Q[i] = (-i,i+1)--(-i-1,i+1); | ||
} | } | ||
draw(P^^Q,red+linewidth(1.25),"$y=|\lfloor x \rfloor|$"); | draw(P^^Q,red+linewidth(1.25),"$y=|\lfloor x \rfloor|$"); | ||
for (int i = 0; i < 9; ++i) { | for (int i = 0; i < 9; ++i) { | ||
| − | + | dot((i,i),red+linewidth(4)); | |
| − | dot((i+1,i),red,UnFill); | + | dot((i+1,i),red+linewidth(0.7),UnFill); |
| − | dot((-i-1,i+1),red); | + | dot((-i-1,i+1),red+linewidth(4)); |
| − | dot((-i,i+1),red,UnFill); | + | dot((-i,i+1),red+linewidth(0.7),UnFill); |
} | } | ||
</asy> | </asy> | ||
| Line 150: | Line 150: | ||
path P[], Q[]; | path P[], Q[]; | ||
for (int i = 0; i < 9; ++i) { | for (int i = 0; i < 9; ++i) { | ||
| − | + | P[i] = (i,i)--(i+1,i); | |
Q[i] = (-i,i+1)--(-i-1,i+1); | Q[i] = (-i,i+1)--(-i-1,i+1); | ||
} | } | ||
draw(P^^Q,heavygreen+linewidth(1.25),"$y=|\lfloor x \rfloor|$"); | draw(P^^Q,heavygreen+linewidth(1.25),"$y=|\lfloor x \rfloor|$"); | ||
for (int i = 0; i < 9; ++i) { | for (int i = 0; i < 9; ++i) { | ||
| − | + | dot((i,i),heavygreen+linewidth(0.7),UnFill); | |
| − | dot((i+1,i),heavygreen); | + | dot((i+1,i),heavygreen+linewidth(4)); |
| − | dot((-i-1,i+1),heavygreen,UnFill); | + | dot((-i-1,i+1),heavygreen+linewidth(0.7),UnFill); |
| − | dot((-i,i+1),heavygreen); | + | dot((-i,i+1),heavygreen+linewidth(4)); |
} | } | ||
</asy> | </asy> | ||
Revision as of 14:36, 25 November 2021
Problem
The graph of ![\[f(x) = |\lfloor x \rfloor| - |\lfloor 1 - x \rfloor|\]](http://latex.artofproblemsolving.com/0/8/5/0856042a755ebc88c0261897b7fad311209d9e99.png)
is symmetric about which of the following? (Here 
is the greatest integer not exceeding 
.)
Solution 1 (Piecewise Function)
IN PROGRESS AND WILL FINISH SOON. NO EDIT PLEASE. A MILLION THANKS.
~MRENTHUSIASM
Solution 2 (Graphing)
Let 
and 
The graph of 
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+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]](http://latex.artofproblemsolving.com/a/f/2/af2f7ebfaa5e9b7e27fa789adb3aee9dde14ab70.png)
Note that 
is a reflection of about the 
-axis, followed by a translation of 
unit right. The graph of 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+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]](http://latex.artofproblemsolving.com/7/c/3/7c306b7edcbdd8bc32b82c66f1abb2c6f3497367.png)
Taking the difference, we graph 
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 
~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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
