Difference between revisions of "2020 AMC 12A Problems/Problem 25"
(→Solution 2) |
MRENTHUSIASM (talk | contribs) m (→Graph) |
||
(42 intermediate revisions by 6 users not shown) | |||
Line 1: | Line 1: | ||
− | ==Problem | + | ==Problem== |
The number <math>a=\frac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers, has the property that the sum of all real numbers <math>x</math> satisfying | The number <math>a=\frac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers, has the property that the sum of all real numbers <math>x</math> satisfying | ||
<cmath> \lfloor x \rfloor \cdot \{x\} = a \cdot x^2</cmath> | <cmath> \lfloor x \rfloor \cdot \{x\} = a \cdot x^2</cmath> | ||
Line 13: | Line 13: | ||
Next, we breakdown <math>\lfloor x\rfloor\cdot \{x\}</math> down for each interval <math>[n,n+1)</math>, where <math>n</math> is a positive integer. Assume <math>\lfloor x\rfloor=n</math>, then <math>\{x\}=x-n</math>. This means that when <math>x\in [n,n+1)</math>, <math>\lfloor x\rfloor \cdot \{x\}=n(x-n)=nx-n^2</math>. Setting this equal to <math>ax^2</math> gives | Next, we breakdown <math>\lfloor x\rfloor\cdot \{x\}</math> down for each interval <math>[n,n+1)</math>, where <math>n</math> is a positive integer. Assume <math>\lfloor x\rfloor=n</math>, then <math>\{x\}=x-n</math>. This means that when <math>x\in [n,n+1)</math>, <math>\lfloor x\rfloor \cdot \{x\}=n(x-n)=nx-n^2</math>. Setting this equal to <math>ax^2</math> gives | ||
<cmath>nx-n^2=ax^2\implies ax^2-nx+n^2=0 \implies x=\frac{n\pm \sqrt{n^2-4an^2}}{2a}</cmath> | <cmath>nx-n^2=ax^2\implies ax^2-nx+n^2=0 \implies x=\frac{n\pm \sqrt{n^2-4an^2}}{2a}</cmath> | ||
− | We're looking at the solution with | + | We're looking at the solution with the positive <math>x</math>, which is <math>x=\frac{n-n\sqrt{1-4a}}{2a}=\frac{n}{2a}\left(1-\sqrt{1-4a}\right)</math>. Note that if <math>\lfloor x\rfloor=n</math> is the greatest <math>n</math> such that <math>\lfloor x\rfloor \cdot \{x\}=ax^2</math> has a solution, the sum of all these solutions is slightly over <math>\sum_{k=1}^{n}k=\frac{n(n+1)}{2}</math>, which is <math>406</math> when <math>n=28</math>, just under <math>420</math>. Checking this gives |
− | <cmath>\sum_{k=1}^{28}\frac{ | + | <cmath>\sum_{k=1}^{28}\frac{k}{2a}\left(1-\sqrt{1-4a}\right)=\frac{1-\sqrt{1-4a}}{2a}\cdot 406=420</cmath> |
<cmath>\frac{1-\sqrt{1-4a}}{2a}=\frac{420}{406}=\frac{30}{29}</cmath> | <cmath>\frac{1-\sqrt{1-4a}}{2a}=\frac{420}{406}=\frac{30}{29}</cmath> | ||
<cmath>29-29\sqrt{1-4a}=60a</cmath> | <cmath>29-29\sqrt{1-4a}=60a</cmath> | ||
Line 22: | Line 22: | ||
<cmath>a=\frac{116}{3600}=\frac{29}{900} \implies \boxed{\textbf{(C) }929}</cmath> | <cmath>a=\frac{116}{3600}=\frac{29}{900} \implies \boxed{\textbf{(C) }929}</cmath> | ||
~ktong | ~ktong | ||
+ | |||
+ | ==Remarks== | ||
+ | ===Graph=== | ||
+ | Let <math>f(x)=\lfloor x \rfloor \cdot \{x\}</math> and <math>g(x)=a \cdot x^2.</math> | ||
+ | |||
+ | We make the following table of values: | ||
+ | |||
+ | <cmath>\begin{array}{c|c|c|clc} | ||
+ | \boldsymbol{x} & \boldsymbol{\lfloor x \rfloor} & \boldsymbol{f(x)} & & \hspace{4mm}\textbf{Equation} & \\ [1.5ex] | ||
+ | \hline | ||
+ | & & & & & \\ [-1ex] | ||
+ | [0,1) & 0 & 0 & & y=0 & \\ [1.5ex] | ||
+ | [1,2) & 1 & [0,1) & & y=x-1 & \\ [1.5ex] | ||
+ | [2,3) & 2 & [0,2) & & y=2x-4 & \\ [1.5ex] | ||
+ | [3,4) & 3 & [0,3) & & y=3x-9 & \\ [1.5ex] | ||
+ | [4,5) & 4 & [0,4) & & y=4x-16 & \\ [1.5ex] | ||
+ | \cdots & \cdots & \cdots & & \ \ \ \ \ \ \ \cdots & \\ [1.5ex] | ||
+ | [m,m+1) & m & [0,m) & & y=mx-m^2 & | ||
+ | \end{array}</cmath> | ||
+ | |||
+ | 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]] | ||
+ | |||
+ | Graph in Desmos: https://www.desmos.com/calculator/ouvaiqjdzj | ||
+ | |||
+ | ~MRENTHUSIASM | ||
+ | |||
+ | ===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 | ||
+ | |||
+ | ==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.