Difference between revisions of "2020 AMC 12A Problems/Problem 25"
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~ktong | ~ktong | ||
− | == | + | ==Remarks== |
− | + | ===Graph=== | |
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Let <math>f(x)=\lfloor x \rfloor \cdot \{x\}</math> and <math>g(x)=a \cdot x^2.</math> | Let <math>f(x)=\lfloor x \rfloor \cdot \{x\}</math> and <math>g(x)=a \cdot x^2.</math> | ||
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We make the following table of values: | We make the following table of values: | ||
<cmath>\begin{array}{c|c|c|clc} | <cmath>\begin{array}{c|c|c|clc} | ||
− | \boldsymbol{x} & \boldsymbol{\lfloor x \rfloor} & \boldsymbol{f(x)} & & \textbf{ | + | \boldsymbol{x} & \boldsymbol{\lfloor x \rfloor} & \boldsymbol{f(x)} & & \hspace{4mm}\textbf{Equation} & \\ [1.5ex] |
\hline | \hline | ||
& & & & & \\ [-1ex] | & & & & & \\ [-1ex] | ||
Line 53: | Line 42: | ||
\end{array}</cmath> | \end{array}</cmath> | ||
− | We graph <math>f(x)</math> by branches | + | We graph <math>f(x)</math> (in red, by branches) and <math>g(x)</math> (in blue, for <math>a=\frac{29}{900}</math>) as shown below. |
[[File:2020 AMC 12A Problem 25.png|center]] | [[File:2020 AMC 12A Problem 25.png|center]] | ||
− | + | Graph in Desmos: https://www.desmos.com/calculator/ouvaiqjdzj | |
− | === | + | ~MRENTHUSIASM |
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+ | ===Extension=== | ||
+ | Visit the [https://artofproblemsolving.com/wiki/index.php/Talk:2020_AMC_12A_Problems/Problem_25 Discussion Page] for the underlying arguments and additional questions. | ||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
− | == | + | ==Video Solution 1 (Geometry)== |
− | + | This video shows how things like The Pythagorean Theorem and The Law of Sines work together to solve this seemingly algebraic problem: https://www.youtube.com/watch?v=6IJ7Jxa98zw&feature=youtu.be | |
− | + | ==Video Solution 2== | |
+ | https://www.youtube.com/watch?v=xex8TBSzKNE ~ MathEx | ||
− | + | ==Video Solution 3 (by Art of Problem Solving)== | |
+ | https://www.youtube.com/watch?v=7_mdreGBPvg&t=428s&ab_channel=ArtofProblemSolving | ||
− | + | Created by Richard Rusczyk | |
==See Also== | ==See Also== | ||
{{AMC12 box|year=2020|ab=A|num-b=24|after=Last Problem}} | {{AMC12 box|year=2020|ab=A|num-b=24|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 22:03, 3 May 2021
Contents
Problem
The number , where and are relatively prime positive integers, has the property that the sum of all real numbers satisfying is , where denotes the greatest integer less than or equal to and denotes the fractional part of . What is ?
Solution 1
Let be the unique solution in this range. Note that is also a solution as long as , hence all our solutions are for some . This sum must be between and , which gives and . Plugging this back in gives .
Solution 2
First note that when while . Thus we only need to look at positive solutions ( doesn't affect the sum of the solutions). Next, we breakdown down for each interval , where is a positive integer. Assume , then . This means that when , . Setting this equal to gives We're looking at the solution with the positive , which is . Note that if is the greatest such that has a solution, the sum of all these solutions is slightly over , which is when , just under . Checking this gives ~ktong
Remarks
Graph
Let and
We make the following table of values:
We graph (in red, by branches) and (in blue, for ) as shown below.
Graph in Desmos: https://www.desmos.com/calculator/ouvaiqjdzj
~MRENTHUSIASM
Extension
Visit the Discussion Page for the underlying arguments and additional questions.
~MRENTHUSIASM
Video Solution 1 (Geometry)
This video shows how things like The Pythagorean Theorem and The Law of Sines work together to solve this seemingly algebraic problem: https://www.youtube.com/watch?v=6IJ7Jxa98zw&feature=youtu.be
Video Solution 2
https://www.youtube.com/watch?v=xex8TBSzKNE ~ MathEx
Video Solution 3 (by Art of Problem Solving)
https://www.youtube.com/watch?v=7_mdreGBPvg&t=428s&ab_channel=ArtofProblemSolving
Created by Richard Rusczyk
See Also
2020 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Problem |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.