Difference between revisions of "2020 AMC 12A Problems/Problem 25"
MRENTHUSIASM (talk | contribs) (→Remarks) |
MRENTHUSIASM (talk | contribs) (Added in Sol 3.) |
||
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 | ||
+ | |||
+ | ==Solution 3 (Comprehensive)== | ||
+ | Let <math>w=\lfloor x \rfloor</math> and <math>f=\{x\}</math> denote the whole part and the fractional part of <math>x,</math> respectively, in which <math>0\leq f<1</math> and <math>x=w+f.</math> | ||
+ | |||
+ | We rewrite the given equation as <cmath>w\cdot f=a\cdot(w+f)^2.</cmath> | ||
+ | We expand and rearrange as <cmath>af^2+(2a-1)wf+aw^2=0, \hspace{23mm}(1)</cmath> which is a quadratic with either <math>f</math> or <math>w.</math> For simplicity purposes, we will treat <math>w</math> as some fixed integer so that <math>(1)</math> is a quadratic with <math>f.</math> | ||
+ | |||
+ | By the quadratic formula, we get | ||
+ | <cmath>\begin{align*} | ||
+ | f&=\frac{(1-2a)w\pm\sqrt{(2a-1)^2w^2-4a^2w^2}}{2a} \\ | ||
+ | &=w\left(\frac{1-2a\pm\sqrt{(2a-1)^2-4a^2}}{2a}\right) \\ | ||
+ | &=w\Biggl(\frac{1-2a\pm\sqrt{1-4a}}{2a}\Biggr). \hspace{25mm}(2) | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | <b>SOLUTION IN PROGRESS. NO EDIT PLEASE.</b> | ||
+ | |||
+ | ~MRENTHUSIASM (inspired by Math Jams's 2020 AMC 10/12A Discussion) | ||
==Remarks== | ==Remarks== |
Revision as of 19:00, 19 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
Solution 3 (Comprehensive)
Let and denote the whole part and the fractional part of respectively, in which and
We rewrite the given equation as We expand and rearrange as which is a quadratic with either or For simplicity purposes, we will treat as some fixed integer so that is a quadratic with
By the quadratic formula, we get
SOLUTION IN PROGRESS. NO EDIT PLEASE.
~MRENTHUSIASM (inspired by Math Jams's 2020 AMC 10/12A Discussion)
Remarks
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
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.