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

(Added in Sol 3.)
m (Solution 3 (Comprehensive))
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Let <math>w=\lfloor x \rfloor</math> and <math>f=\{x\}</math> denote the whole part and the fractional part of <math>x,</math> respectively, in which <math>0\leq f<1</math> and <math>x=w+f.</math>
 
Let <math>w=\lfloor x \rfloor</math> and <math>f=\{x\}</math> denote the whole part and the fractional part of <math>x,</math> respectively, in which <math>0\leq f<1</math> and <math>x=w+f.</math>
  
We rewrite the given equation as <cmath>w\cdot f=a\cdot(w+f)^2.</cmath>
+
We rewrite the given equation as <cmath>w\cdot f=a\cdot(w+f)^2. \hspace{38.75mm}(1)</cmath>
We expand and rearrange as <cmath>af^2+(2a-1)wf+aw^2=0, \hspace{23mm}(1)</cmath> which is a quadratic with either <math>f</math> or <math>w.</math> For simplicity purposes, we will treat <math>w</math> as some fixed integer so that <math>(1)</math> is a quadratic with <math>f.</math>
+
Since <math>a\cdot(w+f)^2\geq0,</math> it follows that <math>w\cdot f\geq0,</math> from which <math>w\geq0.</math>
 +
 
 +
We expand and rearrange <math>(1)</math> as <cmath>af^2+(2a-1)wf+aw^2=0, \hspace{23mm}(2)</cmath> which is a quadratic with either <math>f</math> or <math>w.</math> For simplicity purposes, we will treat <math>w</math> as some fixed nonnegative integer so that <math>(2)</math> is a quadratic with <math>f.</math>
  
 
By the quadratic formula, we get
 
By the quadratic formula, we get
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f&=\frac{(1-2a)w\pm\sqrt{(2a-1)^2w^2-4a^2w^2}}{2a} \\
 
f&=\frac{(1-2a)w\pm\sqrt{(2a-1)^2w^2-4a^2w^2}}{2a} \\
 
&=w\left(\frac{1-2a\pm\sqrt{(2a-1)^2-4a^2}}{2a}\right) \\
 
&=w\left(\frac{1-2a\pm\sqrt{(2a-1)^2-4a^2}}{2a}\right) \\
&=w\Biggl(\frac{1-2a\pm\sqrt{1-4a}}{2a}\Biggr). \hspace{25mm}(2)
+
&=w\Biggl(\frac{1-2a\pm\sqrt{1-4a}}{2a}\Biggr). \hspace{25mm}(3)
 
\end{align*}</cmath>
 
\end{align*}</cmath>
  
 
<b>SOLUTION IN PROGRESS. NO EDIT PLEASE.</b>
 
<b>SOLUTION IN PROGRESS. NO EDIT PLEASE.</b>
  
~MRENTHUSIASM (inspired by Math Jams's 2020 AMC 10/12A Discussion)
+
~MRENTHUSIASM (inspired by Math Jams's <b>2020 AMC 10/12A Discussion</b>)
  
 
==Remarks==
 
==Remarks==

Revision as of 19:41, 19 May 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

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\left(\frac{1-2a\pm\sqrt{(2a-1)^2-4a^2}}{2a}\right) \\ &=w\Biggl(\frac{1-2a\pm\sqrt{1-4a}}{2a}\Biggr). \hspace{25mm}(3) \end{align*}

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.

2020 AMC 12A Problem 25.png

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 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

See Also

2020 AMC 12A (ProblemsAnswer KeyResources)
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Problem 24
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