Difference between revisions of "2021 Fall AMC 10A Problems/Problem 16"
MRENTHUSIASM (talk | contribs) m (→Solution 2 (Piecewise Function)) |
Countmath1 (talk | contribs) m (→Solution 5 (Semi-Fakesolve)) |
||
(29 intermediate revisions by 7 users not shown) | |||
Line 5: | Line 5: | ||
\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 | + | 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 217: | Line 223: | ||
drawLabelAxes(); | drawLabelAxes(); | ||
− | draw((- | + | draw((-10,0)--(10,0),mediumblue+linewidth(1.25),"$y=|\lfloor x \rfloor|$"); |
for (int i = 0; i > -10; --i) { | for (int i = 0; i > -10; --i) { | ||
dot((i,-1),mediumblue+linewidth(4)); | dot((i,-1),mediumblue+linewidth(4)); | ||
Line 225: | Line 231: | ||
} | } | ||
for (int i = -9; i < 10; ++i) { | for (int i = -9; i < 10; ++i) { | ||
− | + | dot((i,0),mediumblue+linewidth(0.7),UnFill); | |
} | } | ||
</asy> | </asy> | ||
Line 233: | Line 239: | ||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
− | ==Solution | + | ==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 254: | Line 260: | ||
It follows that <math>f(x)=-x-(-x+1)=-1.</math><p> | It follows that <math>f(x)=-x-(-x+1)=-1.</math><p> | ||
<li><math>x=0</math></li><p> | <li><math>x=0</math></li><p> | ||
− | It follows that <math>f(x)0-1=-1.</math><p> | + | It follows that <math>f(x)=0-1=-1.</math><p> |
<li><math>x\in\mathbb{Z}^+</math></li><p> | <li><math>x\in\mathbb{Z}^+</math></li><p> | ||
It follows that <math>f(x)=x-(x-1)=1.</math><p> | It follows that <math>f(x)=x-(x-1)=1.</math><p> | ||
Line 272: | Line 278: | ||
so the graph of <math>f(x)</math> is symmetric about <math>\boxed{\textbf{(D) }\text{ the point }\left(\dfrac12, 0\right)}.</math> | so the graph of <math>f(x)</math> is symmetric about <math>\boxed{\textbf{(D) }\text{ the point }\left(\dfrac12, 0\right)}.</math> | ||
− | < | + | Alternatively, we can eliminate <math>\textbf{(A)}, \textbf{(B)}, \textbf{(C)},</math> and <math>\textbf{(E)}</math> once we finish with Case 3. This leaves us with <math>\textbf{(D)}.</math> |
+ | |||
+ | ~MRENTHUSIASM | ||
+ | |||
+ | == 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>. | ||
+ | Hence, <math>a</math> is the integer part of <math>x</math> and <math>b</math> is the decimal part of <math>x</math>. | ||
+ | |||
+ | '''Case 1''': <math>b = 0</math>. | ||
+ | |||
+ | We have | ||
+ | <cmath> | ||
+ | \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*} | ||
+ | </cmath> | ||
+ | |||
+ | '''Case 2''': <math>b \neq 0</math>. | ||
+ | |||
+ | We have | ||
+ | <cmath> | ||
+ | \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*} | ||
+ | </cmath> | ||
+ | |||
+ | Therefore, the graph of <math>f \left( x \right)</math> is symmetric through the point <math>\left( \frac{1}{2} , 0 \right)</math>. | ||
+ | |||
+ | Therefore, the answer is <math>\boxed{\textbf{(D) }\text{ the point }\left(\dfrac12, 0\right)}</math>. | ||
+ | |||
+ | ~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
Contents
Problem
The graph of is symmetric about which of the following? (Here
is the greatest integer not exceeding
.)
Solution 1 (Observations)
Note that
so
.
This means that the graph is symmetric about .
Solution 2 (Graphing)
Let and
Note that the graph of
is a reflection of the graph of
about the
-axis, followed by a translation
unit to the right.
The graph of is shown below:
The graph of
is shown below:
The graph of
is shown below:
Therefore, the graph of is symmetric about
~MRENTHUSIASM
Solution 3 (Casework)
For all and
note that:
and
We rewrite as
We apply casework to the value of
and
and
and
It follows that
It follows that
It follows that
It follows that
It follows that
It follows that
Together, we have
so the graph of
is symmetric about
Alternatively, we can eliminate and
once we finish with Case 3. This leaves us with
~MRENTHUSIASM
Solution 4 (Casework)
Denote , where
and
.
Hence,
is the integer part of
and
is the decimal part of
.
Case 1: .
We have
Case 2: .
We have
Therefore, the graph of is symmetric through the point
.
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
Solution 5 (Semi-Fakesolve)
Suppose making the equation equivalent to
We consider the cases when
If , we have
and
so
.
If or
, we trivially get
and
respectively.
If , we have
and
, giving
Since, for all ,
and
, we can conclude that it is symmetric across the coordinate pair
the midpoint of the "endpoints" of these line segments.
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 ", it becomes apparent that the graph of
is symmetric about
, or more specifically, the point
-Benedict T (countmath1)
Video Solution
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.