Difference between revisions of "2021 Fall AMC 10A Problems/Problem 16"
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Countmath1 (talk | contribs) m (→Solution 5 (Semi-Fakesolve)) |
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== Solution 5 (Semi-Fakesolve) == | == Solution 5 (Semi-Fakesolve) == | ||
− | 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> | + | |
+ | 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> | ||
Line 352: | Line 353: | ||
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 | ||
− | <cmath>\left(\frac{0 + 1}{2}, \frac{-1 + 1}{2}\right) = \boxed{\textbf{(D)}\ \left(\frac{1}{2}, 0\right)}.</ | + | <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) | -Benedict T (countmath1) | ||
Latest revision as of 14:56, 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.