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

(Created page with "So... guys, I dont seem to understand the first solution. What is k and b and c?")
 
m (Discussion 1 (Question))
 
(7 intermediate revisions by the same user not shown)
Line 1: Line 1:
So... guys, I dont seem to understand the first solution. What is k and b and c?
+
==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

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 $\frac{1-\sqrt{1-4a}}{2a},$ 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 $f(x)=\lfloor x \rfloor \cdot \{x\}$ and $g(x)=a \cdot x^2.$

Claim

For all positive integers $n,$ the first $n$ nonzero solutions to $f(x)=g(x)$ are of the form \[x=m\left(\frac{1-\sqrt{1-4a}}{2a}\right),\] where $m=1,2,3,\cdots,n.$

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

~MRENTHUSIASM

Proof by Graph

Clearly, the equation $f(x)=g(x)$ has no negative solutions, and its positive solutions all satisfy $x>1.$ Moreover, none of its solutions is an integer.

Note that the upper bounds of the branches of $f(x)$ are along the line $h(x)=x-1$ (excluded). To prove the claim, we wish to show that for each branch of $\boldsymbol{f(x),}$ there is exactly one solution for $\boldsymbol{f(x)=g(x)}$ (from the branch $\boldsymbol{x\in[1,2)}$ to the branch containing the larger solution of $\boldsymbol{g(x)=h(x)}$). In 8:07-11:31 of Video Solution 3 (Art of Problem-Solving), Mr. Rusczyk questions whether two solutions of $f(x)=g(x)$ can be in the same branch of $f(x),$ and he concludes that it is impossible in 16:25-16:43.

We analyze the upper bound of $f(x):$ Let $(c,c-1)$ be one solution of $g(x)=h(x).$ It is clear that $c>1.$ We substitute this point to find $a:$ \begin{align*} g(c)&=h(c) \\ ac^2&=c-1 \\ a&=\frac{c-1}{c^2}. \end{align*}

We substitute this result back to find $x:$ \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*} By the way, using the precondition that $x=c$ is a root of $(*),$ we can factor its left side easily by the Factor Theorem. Note that $g(x)>h(x)$ for all $x>\max{\left\{c, \frac{c}{c-1}\right\}},$ as quadratic functions always outgrow linear functions.

Now, we perform casework:

  1. $c=\frac{c}{c-1}>1\implies c=2$ (Trivial Case)
  2. It follows that the graphs of $g(x)$ and $h(x)$ only intersect at the point $(2,1),$ which is not on the graph of $f(x).$ So, the equation $f(x)=g(x)$ has no solutions in this case, as the inequality $g(x)<h(x)$ has no solutions.

  3. $c>\frac{c}{c-1}>1\implies c>2$ and $1<\frac{c}{c-1}<2$
  4. It follows that for $g(x)=h(x),$ the smaller solution is $x=\frac{c}{c-1}\in(1,2),$ and $g(x)<h(x)$ holds for all $x\in\left(\frac{c}{c-1},c\right).$

    By the Intermediate Value Theorem, for each branch of $f(x)$ (where $x\in\left[\lfloor t\rfloor,\lfloor t\rfloor+1\right)$), we have $g(x)$ in between its left output and its right "output", namely \[0=f\left(\lfloor t\rfloor\right)<g\left(\lfloor x\rfloor\right)<h\left(\lfloor t\rfloor+1\right)=\lfloor t\rfloor.\] Therefore, for the equation $f(x)=g(x),$ there is exactly one solution for each branch of $f(x),$ where $x\in\left(\frac{c}{c-1},c\right).$ Now, the proof of the bolded sentence of paragraph 2 is complete.

  5. $\frac{c}{c-1}>c>1\implies 1<c<2$ and $\frac{c}{c-1}>1$
  6. This case uses the same argument as Case 2. The smaller solution is $x=c\in(1,2),$ and for each branch of $f(x),$ where $x\in\left(c,\frac{c}{c-1}\right),$ the equation $f(x)=g(x)$ has exactly one solution.

~MRENTHUSIASM

Invalid username
Login to AoPS