# Talk:2020 AMC 12A Problems/Problem 25

## 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) 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) 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) 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 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