# Difference between revisions of "Talk:2020 AMC 12A Problems/Problem 25"

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− | So... guys, I | + | ==Discussion 1 (Question)== |

+ | So... guys, I don't seem to understand the first solution. What is k and b and c? | ||

+ | |||

+ | Perhaps follow Mr. Rusczyk's video solution [https://www.youtube.com/watch?v=7_mdreGBPvg&t=428s&ab_channel=ArtofProblemSolving here.] Once you recognize that all solutions are multiples of <math>\frac{1-\sqrt{1-4a}}{2a},</math> you can understand Solution 1 hopefully. And yes, I agree that Solution 1 is not intuitive for non-genius readers (like most of us) to follow. To be honest, I do struggle to follow Solution 1. Therefore, I recommend Mr. Rusczyk's video solution. ~MRENTHUSIASM | ||

+ | |||

+ | == Discussion 2 (Remarks of Solution 2 and Video Solution 3) == | ||

+ | Let <math>f(x)=\lfloor x \rfloor \cdot \{x\}</math> and <math>g(x)=a \cdot x^2.</math> | ||

+ | |||

+ | ===Claim=== | ||

+ | For all positive integers <math>n,</math> the first <math>n</math> <b>nonzero</b> solutions to <math>f(x)=g(x)</math> are of the form <cmath>x=m\left(\frac{1-\sqrt{1-4a}}{2a}\right),</cmath> where <math>m=1,2,3,\cdots,n.</math> | ||

+ | |||

+ | Equivalently, for <math>x>0,</math> the <math>n</math> intersections of the graphs of <math>f(x)</math> and <math>g(x)</math> occur in the consecutive branches of <math>f(x),</math> namely at <math>x\in[1,2),[2,3),[3,4),\cdots,[n,n+1).</math> | ||

+ | |||

+ | ~MRENTHUSIASM | ||

+ | |||

+ | ===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. | ||

+ | |||

+ | Note that the upper bounds of the branches of <math>f(x)</math> are along the line <math>h(x)=x-1</math> (excluded). <i><b>To prove the claim, we wish to show that for each branch of <math>\boldsymbol{f(x),}</math> there is exactly one solution for <math>\boldsymbol{f(x)=g(x)}</math> (from the branch <math>\boldsymbol{x\in[1,2)}</math> to the branch containing the larger solution of <math>\boldsymbol{g(x)=h(x)}</math>).</b></i> In 8:07-11:31 of [https://www.youtube.com/watch?v=7_mdreGBPvg&t=428s&ab_channel=ArtofProblemSolving Video Solution 3 (Art of Problem-Solving)], Mr. Rusczyk questions whether two solutions of <math>f(x)=g(x)</math> can be in the same branch of <math>f(x),</math> and he concludes that it is impossible in 16:25-16:43. | ||

+ | |||

+ | We analyze the upper bound of <math>f(x):</math> Let <math>(c,c-1)</math> be one solution of <math>g(x)=h(x).</math> It is clear that <math>c>1.</math> We substitute this point to find <math>a:</math> | ||

+ | <cmath>\begin{align*} | ||

+ | g(c)&=h(c) \\ | ||

+ | ac^2&=c-1 \\ | ||

+ | a&=\frac{c-1}{c^2}. | ||

+ | \end{align*}</cmath> | ||

+ | |||

+ | We substitute this result back to find <math>x:</math> | ||

+ | <cmath>\begin{align*} | ||

+ | g(x)&=h(x) \\ | ||

+ | \left(\frac{c-1}{c^2}\right)x^2&=x-1 \\ | ||

+ | \left(\frac{c-1}{c^2}\right)x^2-x+1&=0 \\ | ||

+ | x^2-\left(\frac{c^2}{c-1}\right)x+\frac{c^2}{c-1}&=0 \ \ \ \ \ \ \ \ \ \ \ (*) \\ | ||

+ | (x-c)\left(x-\frac{c}{c-1}\right)&=0 \\ | ||

+ | x&=c,\ \frac{c}{c-1}. | ||

+ | \end{align*}</cmath> | ||

+ | By the way, using the precondition that <math>x=c</math> is a root of <math>(*),</math> we can factor its left side easily by the <b>Factor Theorem</b>. Note that <math>g(x)>h(x)</math> for all <math>x>\max{\left\{c, \frac{c}{c-1}\right\}},</math> as quadratic functions always outgrow linear functions. | ||

+ | |||

+ | Now, we perform casework: | ||

+ | |||

+ | <ol style="margin-left: 1.5em;"> | ||

+ | <li><math>c=\frac{c}{c-1}>1\implies 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.<p> | ||

+ | <li><math>c>\frac{c}{c-1}>1\implies c>2</math> and <math>1<\frac{c}{c-1}<2</math></li><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><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> | ||

+ | <li><math>\frac{c}{c-1}>c>1\implies 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. | ||

+ | </ol> | ||

+ | |||

+ | ~MRENTHUSIASM |

## Latest revision as of 02:49, 17 April 2021

## Contents

## Discussion 1 (Question)

So... guys, I don't seem to understand the first solution. What is k and b and c?

Perhaps follow Mr. Rusczyk's video solution here. Once you recognize that all solutions are multiples of you can understand Solution 1 hopefully. And yes, I agree that Solution 1 is not intuitive for non-genius readers (like most of us) to follow. To be honest, I do struggle to follow Solution 1. Therefore, I recommend Mr. Rusczyk's video solution. ~MRENTHUSIASM

## Discussion 2 (Remarks of Solution 2 and Video Solution 3)

Let and

### 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