Difference between revisions of "2020 AMC 12A Problems/Problem 25"
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− | ===Proof=== | + | ===Proof by Graph=== |
Clearly, the equation <math>f(x)=g(x)</math> has no negative solutions, and its positive solutions all satisfy <math>x>1.</math> Moreover, none of its solutions is an integer. | Clearly, the equation <math>f(x)=g(x)</math> has no negative solutions, and its positive solutions all satisfy <math>x>1.</math> Moreover, none of its solutions is an integer. | ||
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Now, we perform casework: | Now, we perform casework: | ||
− | <math> | + | <ol style="margin-left: 1.5em;"> |
− | + | <li><math>c=\frac{c}{c-1}>1\Longrightarrow c=2</math> (Trivial Case)</li><p> | |
− | It follows that the graphs of <math>g(x)</math> and <math>h(x)</math> only intersect at the point <math>(2,1),</math> which is not on the graph of <math>f(x).</math> So, the equation <math>f(x)=g(x)</math> has no solutions in this case, as the inequality <math>g(x)<h(x)</math> has no solutions. | + | It follows that the graphs of <math>g(x)</math> and <math>h(x)</math> only intersect at the point <math>(2,1),</math> which is not on the graph of <math>f(x).</math> So, the equation <math>f(x)=g(x)</math> has no solutions in this case, as the inequality <math>g(x)<h(x)</math> has no solutions.<p> |
− | + | <li><math>c>\frac{c}{c-1}>1\Longrightarrow c>2</math> and <math>1<\frac{c}{c-1}<2</math></li><p> | |
− | <math> | + | It follows that for <math>g(x)=h(x),</math> the smaller solution is <math>x=\frac{c}{c-1}\in(1,2),</math> and <math>g(x)<h(x)</math> holds for all <math>x\in\left(\frac{c}{c-1},c\right).</math><p> |
− | + | By the <b>Intermediate Value Theorem</b>, for each branch of <math>f(x)</math> (where <math>x\in\left[\lfloor t\rfloor,\lfloor t\rfloor+1\right)</math>), we have <math>g(x)</math> in between its left output and its right "output", namely <cmath>0=f\left(\lfloor t\rfloor\right)<g\left(\lfloor x\rfloor\right)<h\left(\lfloor t\rfloor+1\right)=\lfloor t\rfloor.</cmath> Therefore, for the equation <math>f(x)=g(x),</math> there is exactly one solution for each branch of <math>f(x),</math> where <math>x\in\left(\frac{c}{c-1},c\right).</math> Now, the proof of the bolded sentence of paragraph 2 is complete.<p> | |
− | It follows that for <math>g(x)=h(x),</math> the smaller solution is <math>x=\frac{c}{c-1}\in(1,2),</math> and <math>g(x)<h(x)</math> holds for all <math>x\in\left(\frac{c}{c-1},c\right).</math> | + | <li><math>\frac{c}{c-1}>c>1\Longrightarrow 1<c<2</math> and <math>\frac{c}{c-1}>1</math></li><p> |
− | + | This case uses the same argument as Case 2. The smaller solution is <math>x=c\in(1,2),</math> and for each branch of <math>f(x),</math> where <math>x\in\left(c,\frac{c}{c-1}\right),</math> the equation <math>f(x)=g(x)</math> has exactly one solution. | |
− | By the <b>Intermediate Value Theorem</b>, for each branch of <math>f(x)</math> (where <math>x\in\left[\lfloor t\rfloor,\lfloor t\rfloor+1\right)</math>), we have <math>g(x)</math> in between its left output and its right "output", namely <cmath>0=f\left(\lfloor t\rfloor\right)<g\left(\lfloor x\rfloor\right)<h\left(\lfloor t\rfloor+1\right)=\lfloor t\rfloor.</cmath> Therefore, for the equation <math>f(x)=g(x),</math> there is exactly one solution for each branch of <math>f(x),</math> where <math>x\in\left(\frac{c}{c-1},c\right).</math> Now, the proof of the bolded sentence of paragraph 2 is complete. | + | </ol> |
− | |||
− | <math> | ||
− | |||
− | This case uses the same argument as | ||
~MRENTHUSIASM | ~MRENTHUSIASM |
Revision as of 08:17, 28 March 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
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
Remarks of Solution 2 and Video Solution 3
Let and
Graph
We make the following table of values:
We graph by branches:
~MRENTHUSIASM (Graph by Desmos: https://www.desmos.com/calculator/ouvaiqjdzj)
Claim
For all positive integers the first nonzero solutions to are of the form where
Equivalently, for the intersections of the graphs of and occur in the consecutive branches of namely at
~MRENTHUSIASM
Proof by Graph
Clearly, the equation has no negative solutions, and its positive solutions all satisfy Moreover, none of its solutions is an integer.
Note that the upper bounds of the branches of are along the line (excluded). To prove the claim, we wish to show that for each branch of there is exactly one solution for (from the branch to the branch containing the larger solution of ). In 8:07-11:31 of Video Solution 3 (Art of Problem-Solving), Mr. Rusczyk questions whether two solutions of can be in the same branch of and he concludes that it is impossible in 16:25-16:43.
We analyze the upper bound of Let be one solution of It is clear that We substitute this point to find
We substitute this result back to find By the way, using the precondition that is a root of we can factor its left side easily by the Factor Theorem. Note that for all as quadratic functions always outgrow linear functions.
Now, we perform casework:
- (Trivial Case)
- and
- and
It follows that the graphs of and only intersect at the point which is not on the graph of So, the equation has no solutions in this case, as the inequality has no solutions.
It follows that for the smaller solution is and holds for all
By the Intermediate Value Theorem, for each branch of (where ), we have in between its left output and its right "output", namely Therefore, for the equation there is exactly one solution for each branch of where Now, the proof of the bolded sentence of paragraph 2 is complete.
This case uses the same argument as Case 2. The smaller solution is and for each branch of where the equation has exactly one solution.
~MRENTHUSIASM
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.