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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the midpoint of the "endpoints" of these line segments. | the midpoint of the "endpoints" of these line segments. | ||
+ | <asy> | ||
+ | size(250); //Credit to MRENTHUSIASM | ||
− | Our assumption was not a good one to make initially, but it works after we find out the behavior of <math>f</math> on the integers. There are many functions that behave wildly between integers, so be wary of taking this route in the future. | + | 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> | ||
+ | |||
+ | |||
+ | |||
+ | Our assumption was not a good one to make initially, but it works after we find out the behavior of <math>f</math> on the integers (or looking at the above graph of <math>f(x)</math> where <math>x\in \mathbb{Z}</math>). There are many functions that behave wildly between integers, so be wary of taking this route without double checking everything in the future. | ||
-Benedict T (countmath1) | -Benedict T (countmath1) |
Revision as of 14:52, 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)
We make the bold assumption that for all , if is symmetric about a line (or a point),then for all , is symmetric about the same point/line because
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.
Our assumption was not a good one to make initially, but it works after we find out the behavior of on the integers (or looking at the above graph of where ). There are many functions that behave wildly between integers, so be wary of taking this route without double checking everything in the future.
-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.