# 2020 AMC 12A Problems/Problem 25

## Problem

The number $a=\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers, has the property that the sum of all real numbers $x$ satisfying $$\lfloor x \rfloor \cdot \{x\} = a \cdot x^2$$ is $420$, where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$ and $\{x\}=x- \lfloor x \rfloor$ denotes the fractional part of $x$. What is $p+q$?

$\textbf{(A) } 245 \qquad \textbf{(B) } 593 \qquad \textbf{(C) } 929 \qquad \textbf{(D) } 1331 \qquad \textbf{(E) } 1332$

## Solutions

### Solution 1

Let $x$ be a root in the interval $(n,n+1)$. In this interval, $\lfloor x \rfloor = n$ and $\{x\}=x-n$, so we must have $ax^2 = nx-n^2$, i.e. $$x = \frac{n}{2a}\left(1 \pm \sqrt{1 -4a} \right) = \frac{2n}{1\mp \sqrt{1-4a}}$$Notice that only one of the expressions gives a root in $(n,n+1)$: since $1- \sqrt{1-4a} < 1$, so $2n/(1- \sqrt{1-4a})>2n>n+1$. Thus $$x = \frac{2n}{1+ \sqrt{1-4a}} =n\zeta.$$Suppose $N$ is the largest integer for which there is such a root; starting from $n we get \begin{align*} n < x &= n\zeta < n+1 \\ \Rightarrow \quad \tfrac 12 N(N+1) < 420 &= \tfrac 12 N(N+1)\zeta < \tfrac 12 N(N+3) \end{align*} From the right inequality we get $27< N$ and from the left one we get $N<29$. Thus $N=28$. Using this in the middle equality we get $\zeta = \frac{30}{29}$, i.e. $$a = \frac 14 \left[1-\left(\frac{2}{\zeta}-1\right)^2\right] = \frac 14 \left[1-\left(\frac{14}{15}\right)^2\right]= \frac{29}{900}.$$ The answer is $29+900=\boxed{\textbf{(C) } 929}.$

### Solution 2

Let $w=\lfloor x \rfloor$ and $f=\{x\}$ denote the whole part and the fractional part of $x,$ respectively, for which $0\leq f<1$ and $x=w+f.$

We rewrite the given equation as $$w\cdot f=a\cdot(w+f)^2. \hspace{38.75mm}(1)$$ Since $a\cdot(w+f)^2\geq0,$ it follows that $w\cdot f\geq0,$ from which $w\geq0.$

We expand and rearrange $(1)$ as $$af^2+(2a-1)wf+aw^2=0, \hspace{23mm}(2)$$ which is a quadratic with either $f$ or $w.$

For simplicity purposes, we will treat $w$ as some fixed nonnegative integer so that $(2)$ is a quadratic with $f.$ By the Quadratic Formula, we have \begin{align*} f&=\frac{(1-2a)w\pm\sqrt{(2a-1)^2w^2-4a^2w^2}}{2a} \\ &=w\Biggl(\frac{1-2a\pm\sqrt{(2a-1)^2-4a^2}}{2a}\Biggr) \\ &=w\Biggl(\frac{1-2a\pm\sqrt{1-4a}}{2a}\Biggr). \hspace{25mm}(3) \end{align*} If $w=0,$ then $f=0.$ We get $x=w+f=0,$ which does not affect the sum of the solutions. Therefore, we consider the case for $w\geq1:$

Recall that $0\leq f<1,$ so $\frac{1-2a\pm\sqrt{1-4a}}{2a}\geq0.$ From the discriminant, we require that $1-4a\geq0,$ or $a\leq\frac14.$ Combining this with the precondition $a>0,$ we need $$0 We consider each part of $0\leq f<1$ separately: 1. $f\geq0$ 2. From $(2),$ note that $a>0, (2a-1)w<0,$ and $aw^2>0.$ By Descartes' Rule of Signs, we deduce that $(2)$ must have two positive roots, so $f\geq0$ is always valid. Alternatively, from $(3)$ and $(4),$ note that all values of $a$ for which $0 satisfy $1-2a>\sqrt{1-4a}.$ We deduce that both roots in $(3)$ must be positive, so $f\geq0$ is always valid. 3. $f<1$ 4. We rewrite $(3)$ as $\[f=w\Biggl(\frac{1}{2a}-1\pm\frac{\sqrt{1-4a}}{2a}\Biggr).$$ From $(4),$ it follows that $\frac{1}{2a}\geq\frac{1}{1/2}=2.$ The larger root is $$f\geq w\left(2-1+2\sqrt{1-4a}\right) \geq 1\Biggl(2-1+2\sqrt{1-4\cdot\frac14}\Biggr) = 1,$$ which contradicts $f<1.$ So, we take the smaller root, from which $$f=w\Biggl(\frac{1}{2a}-1-\frac{\sqrt{1-4a}}{2a}\Biggr)$$ for some constant $k=\frac{1}{2a}-1-\frac{\sqrt{1-4a}}{2a}>0.$ We rewrite $f$ as $$f=wk,$$ in which $f<1$ is valid as long as $k<\frac 1w.$ Note that the solutions of $x$ are generated at $$w=1,2,3,\ldots,W,$$ up to some value $w=W$ such that $\frac{1}{W+1}\leq k<\frac1W.$

Now, we express $x$ in terms of $w$ and $k:$ $$x=w+f=w+wk=w(1+k).$$ The sum of all solutions to the original equation is $$\sum_{w=1}^{W}w(1+k)=(1+k)\cdot\sum_{w=1}^{W}w=(1+k)\cdot\frac{W(W+1)}{2}=420. \hspace{10mm}(\bigstar)$$ As $1+k<1+\frac1W,$ we conclude that $1+k$ is slightly above $1$ so that $\frac{W(W+1)}{2}$ is slightly below $420,$ or $W(W+1)$ is slightly below $840.$ By observations, we get $W=28.$ Substituting this into $(\bigstar)$ produces $k=\frac{1}{29},$ which satisfies $\frac{1}{W+1}\leq k<\frac1W,$ as required.

Finally, we solve for $a$ in $k=\frac{1}{2a}-1-\frac{\sqrt{1-4a}}{2a}:$ \begin{align*} \frac{1}{29}&=\frac{1}{2a}-1-\frac{\sqrt{1-4a}}{2a} \\ \frac{2}{29}a&=1-2a-\sqrt{1-4a} \\ \frac{60}{29}a-1&=-\sqrt{1-4a} \\ \frac{60^2}{29^2}a^2-\frac{120}{29}a+1&=1-4a \\ \frac{60^2}{29^2}a^2-\frac{4}{29}a&=0 \\ a\left(\frac{60^2}{29^2}a-\frac{4}{29}\right)&=0. \end{align*} Since $a>0,$ we obtain $\frac{60^2}{29^2}a-\frac{4}{29}=0,$ from which $$a=\frac{4}{29}\cdot\frac{29^2}{60^2}=\frac{29}{900}.$$ The answer is $29+900=\boxed{\textbf{(C) } 929}.$

~MRENTHUSIASM (inspired by Math Jams's 2020 AMC 10/12A Discussion)

## Solution 2

First note that $\lfloor x\rfloor \cdot \{x\}<0$ when $x<0$ while $ax^2\ge 0\forall x\in \mathbb{R}$. Thus we only need to look at positive solutions ($x=0$ doesn't affect the sum of the solutions). Next, we break $\lfloor x\rfloor\cdot \{x\}$ down for each interval $[n,n+1)$, where $n$ is a positive integer. Assume $\lfloor x\rfloor=n$, then $\{x\}=x-n$. This means that when $x\in [n,n+1)$, $\lfloor x\rfloor \cdot \{x\}=n(x-n)=nx-n^2$. Setting this equal to $ax^2$ gives $$nx-n^2=ax^2\implies ax^2-nx+n^2=0 \implies x=\frac{n\pm \sqrt{n^2-4an^2}}{2a}$$ We're looking at the solution with the positive $x$, which is $x=\frac{n-n\sqrt{1-4a}}{2a}=\frac{n}{2a}\left(1-\sqrt{1-4a}\right)$. Note that if $\lfloor x\rfloor=n$ is the greatest $n$ such that $\lfloor x\rfloor \cdot \{x\}=ax^2$ has a solution, the sum of all these solutions is slightly over $\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$, which is $406$ when $n=28$, just under $420$. Checking this gives \begin{align*} \sum_{k=1}^{28}\frac{k}{2a}\left(1-\sqrt{1-4a}\right)&=\frac{1-\sqrt{1-4a}}{2a}\cdot 406=420 \\ \frac{1-\sqrt{1-4a}}{2a}&=\frac{420}{406}=\frac{30}{29} \\ 29-29\sqrt{1-4a}&=60a \\ 29\sqrt{1-4a}&=29-60a \\ 29^2-4\cdot 29^2a&=29^2+3600a^2-120\cdot 29a \\ 3600a^2&=116a \\ a&=\frac{116}{3600}=\frac{29}{900} \implies \boxed{\textbf{(C) } 929}. \end{align*} ~ktong

## Remark

Let $f(x)=\lfloor x \rfloor \cdot \{x\}$ and $g(x)=a \cdot x^2.$

We make the following table of values:

$$\begin{array}{c|c|c|l} & & & \\ [-2ex] \boldsymbol{x} & \boldsymbol{\lfloor x \rfloor} & \boldsymbol{f(x)} & \multicolumn{1}{c}{\textbf{Equation}} \\ [1.5ex] \hline & & & \\ [-1ex] [0,1) & 0 & 0 & y=0 \\ [1.5ex] [1,2) & 1 & [0,1) & y=x-1 \\ [1.5ex] [2,3) & 2 & [0,2) & y=2x-4 \\ [1.5ex] [3,4) & 3 & [0,3) & y=3x-9 \\ [1.5ex] [4,5) & 4 & [0,4) & y=4x-16 \\ [1.5ex] \cdots & \cdots & \cdots & \cdots \\ [1.5ex] [m,m+1) & m & [0,m) & y=mx-m^2 \\ [1.5ex] \end{array}$$ We graph $f(x)$ (in red, by branches) and $g(x)$ (in blue, for $a=\frac{29}{900}$) as shown below.

Graph in Desmos: https://www.desmos.com/calculator/ouvaiqjdzj

~MRENTHUSIASM

## 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 3 (by Art of Problem Solving)

Created by Richard Rusczyk