# Difference between revisions of "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 $p+q$?

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

## Solution 1

Let $1 be the unique solution in this range. Note that $ck$ is also a solution as long as $ck < c+1$, hence all our solutions are $k, 2k, ..., bk$ for some $b$. This sum $420$ must be between $\frac{b(b+1)}{2}$ and $\frac{(b+1)(b+2)}{2}$, which gives $b=28$ and $k=\frac{420}{406}=\frac{30}{29}$. Plugging this back in gives $a=\frac{29 \cdot 1}{30^2} = \frac{29}{900} \implies \boxed{\textbf{C}}$.

## 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 breakdown $\lfloor x\rfloor\cdot \{x\}$ 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 $$\sum_{k=1}^{28}\frac{k}{2a}\left(1-\sqrt{1-4a}\right)=\frac{1-\sqrt{1-4a}}{2a}\cdot 406=420$$ $$\frac{1-\sqrt{1-4a}}{2a}=\frac{420}{406}=\frac{30}{29}$$ $$29-29\sqrt{1-4a}=60a$$ $$29\sqrt{1-4a}=29-60a$$ $$29^2-4\cdot 29^2a=29^2+3600a^2-120\cdot 29a$$ $$3600a^2=116a$$ $$a=\frac{116}{3600}=\frac{29}{900} \implies \boxed{\textbf{(C) }929}$$ ~ktong

## Solution 3 (Comprehensive)

Let $w=\lfloor x \rfloor$ and $f=\{x\}$ denote the whole part and the fractional part of $x,$ respectively, in 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 get \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 have $x=w+f=0,$ 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 conclude that $1-4a\geq0,$ or $a\leq\frac14.$ Combining this with the precondition $a>0,$ we have $$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 $(3),$ solving the inequality $1-2a>\sqrt{1-4a}$ produces $0 which checks $(4).$ 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 $(4),$ we deduce that $\frac{1}{2a}\geq\frac{1}{\left(\frac12\right)}=2.$ The larger root is \begin{align*} f&\geq w\left(2-1+2\sqrt{1-4a}\right) \\ &\geq 1\Biggl(2-1+2\sqrt{1-4\cdot\frac14}\Biggr) \\ &= 1, \\ \end{align*}

which contradicts to $f<1.$ So, we take the smaller root.

We get $$f=w\Biggl(\frac{1}{2a}-1-\frac{\sqrt{1-4a}}{2a}\Biggr)$$ for some constant $k=\frac{1}{2a}-1-\frac{\sqrt{1-4a}}{2a}.$ Note that $k>0$ for all $a$ such that $0

Now, we obtain $$f=wk,$$ in which $f<1$ is valid as long as $w<\frac 1k.$

SOLUTION IN PROGRESS. NO EDIT PLEASE.

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

## Remarks

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

We make the following table of values:

$$\begin{array}{c|c|c|clc} & & & & & \\ [-1.5ex] \boldsymbol{x} & \boldsymbol{\lfloor x \rfloor} & \boldsymbol{f(x)} & & \hspace{4mm}\textbf{Equation} & \\ [1.5ex] \hline & & & & & \\ [-1ex] [0,1) & 0 & 0 & & y=0 & \\ [1.5ex] [1,2) & 1 & [0,1) & & y=x-1 & \\ [1.5ex] [2,3) & 2 & [0,2) & & y=2x-4 & \\ [1.5ex] [3,4) & 3 & [0,3) & & y=3x-9 & \\ [1.5ex] [4,5) & 4 & [0,4) & & y=4x-16 & \\ [1.5ex] \cdots & \cdots & \cdots & & \ \ \ \ \ \ \ \cdots & \\ [1.5ex] [m,m+1) & m & [0,m) & & y=mx-m^2 & \\ [2ex] \end{array}$$

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