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

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We make the following table of values:
 
We make the following table of values:
  
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<cmath>\begin{array}{c|c|c|l}
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<cmath>\begin{array}{c|c|c|clc}
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\boldsymbol{x} & \boldsymbol{\lfloor x \rfloor} & \boldsymbol{f(x)} & \textbf{\ \ \ Equation} \\ [1.5ex]
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\boldsymbol{x} & \boldsymbol{\lfloor x \rfloor} & \boldsymbol{f(x)} & & \textbf{\ \ Equation} & \\ [1.5ex]
 
\hline
 
\hline
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& & & \\ [-1ex]
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& & & & & \\ [-1ex]
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[0,1) & 0 & 0 & y=0 \\ [1.5ex]
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[0,1) & 0 & 0 & & y=0 & \\ [1.5ex]
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[1,2) & 1 & [0,1) & y=x-1 \\ [1.5ex]
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[1,2) & 1 & [0,1) & & y=x-1 & \\ [1.5ex]
−
[2,3) & 2 & [0,2) & y=2x-4 \\ [1.5ex]
+
[2,3) & 2 & [0,2) & & y=2x-4 & \\ [1.5ex]
−
[3,4) & 3 & [0,3) & y=3x-9 \\ [1.5ex]
+
[3,4) & 3 & [0,3) & & y=3x-9 & \\ [1.5ex]
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[4,5) & 4 & [0,4) & y=4x-16 \\ [1.5ex]
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[4,5) & 4 & [0,4) & & y=4x-16 & \\ [1.5ex]
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\cdots & \cdots & \cdots & \ \ \ \ \ \ \ \cdots \\ [1.5ex]
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\cdots & \cdots & \cdots & & \ \ \ \ \ \ \ \cdots & \\ [1.5ex]
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[m,m+1) & m & [0,m) & y=mx-m^2
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[m,m+1) & m & [0,m) & & y=mx-m^2 &
 
\end{array}</cmath>
 
\end{array}</cmath>
  

Revision as of 14:36, 7 March 2021

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 $1<k<2$ 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

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 2

https://www.youtube.com/watch?v=xex8TBSzKNE ~ MathEx

Video Solution 3 (by Art of Problem-Solving)

https://www.youtube.com/watch?v=7_mdreGBPvg&t=428s&ab_channel=ArtofProblemSolving

Created by Richard Rusczyk

Remarks of Solution 2 and Video Solution 3

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

Graph

We make the following table of values:

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

We graph $f(x)$ by branches:

2020 AMC 12A Problem 25.png

~MRENTHUSIASM (Graph by Desmos: https://www.desmos.com/calculator/ouvaiqjdzj)

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 $m=1,2,3,\cdots,n.$

~MRENTHUSIASM

See Also

2020 AMC 12A (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

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