AMC12 2020 A

Answer (C): Note that $x>0$, and the only solution with $[x]=0$ is $x=0$, so it follows that $x\ge 1$. The graph of $y=[x]\{x\}$, for $x\ge 1$, consists of line segments $\ell_k$ from $(k,0)$ to $(k+1,k)$ for $k=1,2,\ldots$ with the right endpoint of each segment excluded. If $[x]\{x\}=ax^2$ has no solutions, then the graph of $y=ax^2$ must intersect all the segments $\ell_1,\ell_2,\ldots,\ell_n$, but not $\ell_{n+1}$. Then $$1+2+\cdots+n\le 420<2+3+\cdots+(n+1),$$ which forces $n=28$. Also, note that it must be that $a<\frac14$ for there to be solutions other than $x=0$. The following figure gives an idea of what is happening. Writing $x=[x]+\{x\}$, it follows that $[x]\{x\}=a([x]+\{x\})^2$, so $$0=a\{x\}^2+(2a-1)[x]\{x\}+a[x]^2.$$ Then by the quadratic formula, $$\{x\}=\frac{-(2a-1)[x]\pm\sqrt{(2a-1)^2[x]^2-4a^2[x]^2}}{2a}$$ $$=\left(\frac{(1-2a)\pm\sqrt{1-4a}}{2a}\right)[x].$$ The minus sign must be chosen, because for $x\ge 1$, $$\{x\}=\left(\frac{(1-2a)+\sqrt{1-4a}}{2a}\right)[x]$$ $$\ge \frac{1-2a}{2a}$$ $$=\frac{1}{2a}-1$$ $$>\frac{1}{2\cdot\frac14}-1$$ $$=1,$$ which is impossible. Therefore the solutions are $$x=[x]+\{x\}=\left(\frac{1-\sqrt{1-4a}}{2a}\right)[x]$$ for $1\le [x]\le 28$. Note that this equation implies that for each positive integer $k$, there is at most one solution with $k\le x<k+1$. It follows that the sum of the solutions is $$420=\left(\frac{1-\sqrt{1-4a}}{2a}\right)(1+2+\cdots+28)=\left(\frac{1-\sqrt{1-4a}}{2a}\right)\cdot 14\cdot 29.$$ Thus $$\frac{1-\sqrt{1-4a}}{2a}=\frac{30}{29}.$$ Writing this in the form $29\sqrt{1-4a}=29-60a$, then squaring and simplifying gives $3600a^2-116a=0,$ so