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

Latest revision as of 16:13, 21 July 2022

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 (Solves for Floor(x))

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 $f=w\Biggl(\frac{1-2a\pm\sqrt{1-4a}}{2a}\Biggr). \hspace{25mm}(3)$ 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 $0\leq1-4a<1,$ or $0<a\leq\frac14. \hspace{54mm}(4)$

We consider each part of $0\leq f<1$ separately:

$f\geq0$

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 $0<a\leq\frac14$ satisfy $1-2a>\sqrt{1-4a}.$ We deduce that both roots in $(3)$ must be positive, so $f\geq0$ is always valid.

$f<1$

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 (Solves for x)

Let $x_n$ be a root in the interval $(n,n+1)$ . In this interval, $\lfloor x_n \rfloor = n$ and $\{x_n\}=x_n-n$ , so we must have $ax_n^2 = nx_n-n^2$ , i.e., $ax_n^2-nx_n+n^2=0$ . We can homogenize this equation by setting $x_n=n\zeta$ ; then $x_1=\zeta$ , and $\zeta$ is a root of $a\zeta^2-\zeta+1=0$ .

Suppose $N$ is the largest integer for which there is such a root; we have, for $n=1,2,\ldots , N$ , $n < x_n = n\zeta < n+1$ Summing over $n\in \{1,2,\ldots , N\}$ we get $\tfrac 12 N(N+1) < 420 = \tfrac 12 N(N+1)\zeta < \tfrac 12 N(N+3)$ 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 $\zeta = \tfrac{30}{29}$ . Since $\zeta$ satisfies $a\zeta^2-\zeta+1=0$ , we get $a = \zeta^{-2}(\zeta-1)= \tfrac{29^2}{30^2}\cdot \tfrac 1{29}= \tfrac{29}{900}.$ The answer is $29+900=\boxed{\textbf{(C) } 929}.$

~Shihan

Solution 3 (Solves for x)

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 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

Video Solution 4

https://youtu.be/i5b5P9RPuas

~MathProblemSolvingSkills

@@ Line 6: / Line 6: @@
 <math>\textbf{(A) } 245 \qquad \textbf{(B) } 593 \qquad \textbf{(C) } 929 \qquad \textbf{(D) } 1331 \qquad \textbf{(E) } 1332</math>
-==Solution 1==
+==Solution 1 (Solves for Floor(x))==
-Let <math>1<k<2</math> be the unique solution in this range. Note that <math>ck</math> is also a solution as long as <math>ck < c+1</math>, hence all our solutions are <math>k, 2k, ..., bk</math> for some <math>b</math>. This sum <math>420</math> must be between <math>\frac{b(b+1)}{2}</math> and <math>\frac{(b+1)(b+2)}{2}</math>, which gives <math>b=28</math> and <math>k=\frac{420}{406}=\frac{30}{29}</math>. Plugging this back in gives <math>a=\frac{29 \cdot 1}{30^2} = \frac{29}{900} \implies \boxed{\textbf{C}}</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, for which <math>0\leq f<1</math> and <math>x=w+f.</math>
-==Solution 2==
+We rewrite the given equation as <cmath>w\cdot f=a\cdot(w+f)^2. \hspace{38.75mm}(1)</cmath>
-First note that <math>\lfloor x\rfloor \cdot \{x\}<0</math> when <math>x<0</math> while <math>ax^2\ge 0\forall x\in \mathbb{R}</math>. Thus we only need to look at positive solutions (<math>x=0</math> doesn't affect the sum of the solutions).
+Since <math>a\cdot(w+f)^2\geq0,</math> it follows that <math>w\cdot f\geq0,</math> from which <math>w\geq0.</math>
-Next, we breakdown <math>\lfloor x\rfloor\cdot \{x\}</math> down for each interval <math>[n,n+1)</math>, where <math>n</math> is a positive integer. Assume <math>\lfloor x\rfloor=n</math>, then <math>\{x\}=x-n</math>. This means that when <math>x\in [n,n+1)</math>, <math>\lfloor x\rfloor \cdot \{x\}=n(x-n)=nx-n^2</math>. Setting this equal to <math>ax^2</math> gives
-<cmath>nx-n^2=ax^2\implies ax^2-nx+n^2=0 \implies x=\frac{n\pm \sqrt{n^2-4an^2}}{2a}</cmath>
-We're looking at the solution with the positive <math>x</math>, which is <math>x=\frac{n-n\sqrt{1-4a}}{2a}=\frac{n}{2a}\left(1-\sqrt{1-4a}\right)</math>. Note that if <math>\lfloor x\rfloor=n</math> is the greatest <math>n</math> such that <math>\lfloor x\rfloor \cdot \{x\}=ax^2</math> has a solution, the sum of all these solutions is slightly over <math>\sum_{k=1}^{n}k=\frac{n(n+1)}{2}</math>, which is <math>406</math> when <math>n=28</math>, just under <math>420</math>. Checking this gives
-<cmath>\sum_{k=1}^{28}\frac{k}{2a}\left(1-\sqrt{1-4a}\right)=\frac{1-\sqrt{1-4a}}{2a}\cdot 406=420</cmath>
-<cmath>\frac{1-\sqrt{1-4a}}{2a}=\frac{420}{406}=\frac{30}{29}</cmath>
-<cmath>29-29\sqrt{1-4a}=60a</cmath>
-<cmath>29\sqrt{1-4a}=29-60a</cmath>
-<cmath>29^2-4\cdot 29^2a=29^2+3600a^2-120\cdot 29a</cmath>
-<cmath>3600a^2=116a</cmath>
-<cmath>a=\frac{116}{3600}=\frac{29}{900} \implies \boxed{\textbf{(C) }929}</cmath>
-~ktong
-==Video Solution 1 (Geometry)==
+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>
-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==
+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 have <cmath>f=w\Biggl(\frac{1-2a\pm\sqrt{1-4a}}{2a}\Biggr). \hspace{25mm}(3)</cmath>
-https://www.youtube.com/watch?v=xex8TBSzKNE ~ MathEx
+If <math>w=0,</math> then <math>f=0.</math> We get <math>x=w+f=0,</math> which does not affect the sum of the solutions. Therefore, we consider the case for <math>w\geq1:</math>
-==Video Solution 3 (by Art of Problem-Solving)==
+Recall that <math>0\leq f<1,</math> so <math>\frac{1-2a\pm\sqrt{1-4a}}{2a}\geq0.</math> From the discriminant, we require that <math>0\leq1-4a<1,</math> or <cmath>0<a\leq\frac14. \hspace{54mm}(4)</cmath>
-https://www.youtube.com/watch?v=7_mdreGBPvg&t=428s&ab_channel=ArtofProblemSolving
-Created by Richard Rusczyk
+We consider each part of <math>0\leq f<1</math> separately:
+<ol style="margin-left: 1.5em;">
+  <li><math>f\geq0</math></li><p>
+From <math>(2),</math> note that <math>a>0, (2a-1)w<0,</math> and <math>aw^2>0.</math> By Descartes' Rule of Signs, we deduce that <math>(2)</math> must have two positive roots, so <math>f\geq0</math> is always valid.<p>
+Alternatively, from <math>(3)</math> and <math>(4),</math> note that all values of <math>a</math> for which <math>0<a\leq\frac14</math> satisfy <math>1-2a>\sqrt{1-4a}.</math> We deduce that both roots in <math>(3)</math> must be positive, so <math>f\geq0</math> is always valid.<p>
+  <li><math>f<1</math></li><p>
+We rewrite <math>(3)</math> as <cmath>f=w\Biggl(\frac{1}{2a}-1\pm\frac{\sqrt{1-4a}}{2a}\Biggr).</cmath> From <math>(4),</math> it follows that <math>\frac{1}{2a}\geq\frac{1}{1/2}=2.</math> The larger root is <cmath>f\geq w\left(2-1+2\sqrt{1-4a}\right) \geq 1\Biggl(2-1+2\sqrt{1-4\cdot\frac14}\Biggr) = 1,</cmath>
+which contradicts <math>f<1.</math> So, we take the smaller root, from which <cmath>f=w\Biggl(\frac{1}{2a}-1-\frac{\sqrt{1-4a}}{2a}\Biggr)</cmath> for some constant <math>k=\frac{1}{2a}-1-\frac{\sqrt{1-4a}}{2a}>0.</math> We rewrite <math>f</math> as <cmath>f=wk,</cmath> in which <math>f<1</math> is valid as long as <math>k<\frac 1w.</math> Note that the solutions of <math>x</math> are generated at <cmath>w=1,2,3,\ldots,W,</cmath> up to some value <math>w=W</math> such that <math>\frac{1}{W+1}\leq k<\frac1W.</math>
+</ol>
+Now, we express <math>x</math> in terms of <math>w</math> and <math>k:</math> <cmath>x=w+f=w+wk=w(1+k).</cmath>
+The sum of all solutions to the original equation is <cmath>\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)</cmath>
+As <math>1+k<1+\frac1W,</math> we conclude that <math>1+k</math> is slightly above <math>1</math> so that <math>\frac{W(W+1)}{2}</math> is slightly below <math>420,</math> or <math>W(W+1)</math> is slightly below <math>840.</math> By observations, we get <math>W=28.</math> Substituting this into <math>(\bigstar)</math> produces <math>k=\frac{1}{29},</math> which satisfies <math>\frac{1}{W+1}\leq k<\frac1W,</math> as required.
-==Remarks of Solution 2 and Video Solution 3==
+Finally, we solve for <math>a</math> in <math>k=\frac{1}{2a}-1-\frac{\sqrt{1-4a}}{2a}:</math>
-Let <math>f(x)=\lfloor x \rfloor \cdot \{x\}</math> and <math>g(x)=a \cdot x^2.</math>
+<cmath>\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*}</cmath>
+Since <math>a>0,</math> we obtain <math>\frac{60^2}{29^2}a-\frac{4}{29}=0,</math> from which <cmath>a=\frac{4}{29}\cdot\frac{29^2}{60^2}=\frac{29}{900}.</cmath>
+The answer is <math>29+900=\boxed{\textbf{(C) } 929}.</math>
-===Graph===
+~MRENTHUSIASM (inspired by Math Jams's <b>2020 AMC 10/12A Discussion</b>)
-We make the following table of values:
-<cmath>\begin{array}{c|c|c|clc}
+==Solution 2 (Solves for x)==
-\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}</cmath>
-We graph <math>f(x)</math> by branches:
+Let <math>x_n</math> be a root in the interval <math>(n,n+1)</math>. In this interval, <math>\lfloor x_n \rfloor = n</math> and <math>\{x_n\}=x_n-n</math>, so we must have <math>ax_n^2 = nx_n-n^2</math>, i.e., <math>ax_n^2-nx_n+n^2=0</math>. We can homogenize this equation by setting <math>x_n=n\zeta</math>; then <math>x_1=\zeta</math>, and <math>\zeta</math> is a root of <math>a\zeta^2-\zeta+1=0</math>.
-[[File:2020 AMC 12A Problem 25.png|center]]
+Suppose <math>N</math> is the largest integer for which there is such a root; we have, for <math>n=1,2,\ldots , N</math>, <cmath>n < x_n = n\zeta < n+1</cmath> Summing over <math>n\in \{1,2,\ldots , N\}</math> we get <cmath>\tfrac 12 N(N+1) < 420 = \tfrac 12 N(N+1)\zeta < \tfrac 12 N(N+3)</cmath> From the right inequality we get <math>27< N</math> and from the left one we get  <math>N<29</math>. Thus <math>N=28</math>. Using this in the middle equality we get <math>\zeta = \tfrac{30}{29}</math>. Since <math>\zeta</math> satisfies <math>a\zeta^2-\zeta+1=0</math>, we get
+<cmath>a = \zeta^{-2}(\zeta-1)= \tfrac{29^2}{30^2}\cdot \tfrac 1{29}= \tfrac{29}{900}.</cmath>
+The answer is <math>29+900=\boxed{\textbf{(C) } 929}.</math>
-~MRENTHUSIASM (Graph by Desmos: https://www.desmos.com/calculator/ouvaiqjdzj)
+~Shihan
-===Claim===
+==Solution 3 (Solves for x)==
-For all positive integers <math>n,</math> the first <math>n</math> <b>nonzero</b> solutions to <math>f(x)=g(x)</math> are of the form <cmath>x=m\left(\frac{1-\sqrt{1-4a}}{2a}\right),</cmath> where <math>m=1,2,3,\cdots,n.</math>
+First note that <math>\lfloor x\rfloor \cdot \{x\}<0</math> when <math>x<0</math> while <math>ax^2\ge 0\forall x\in \mathbb{R}</math>. Thus we only need to look at positive solutions (<math>x=0</math> doesn't affect the sum of the solutions).
+Next, we break <math>\lfloor x\rfloor\cdot \{x\}</math> down for each interval <math>[n,n+1)</math>, where <math>n</math> is a positive integer. Assume <math>\lfloor x\rfloor=n</math>, then <math>\{x\}=x-n</math>. This means that when <math>x\in [n,n+1)</math>, <math>\lfloor x\rfloor \cdot \{x\}=n(x-n)=nx-n^2</math>. Setting this equal to <math>ax^2</math> gives
+<cmath>nx-n^2=ax^2\implies ax^2-nx+n^2=0 \implies x=\frac{n\pm \sqrt{n^2-4an^2}}{2a}</cmath>
+We're looking at the solution with the positive <math>x</math>, which is <math>x=\frac{n-n\sqrt{1-4a}}{2a}=\frac{n}{2a}\left(1-\sqrt{1-4a}\right)</math>. Note that if <math>\lfloor x\rfloor=n</math> is the greatest <math>n</math> such that <math>\lfloor x\rfloor \cdot \{x\}=ax^2</math> has a solution, the sum of all these solutions is slightly over <math>\sum_{k=1}^{n}k=\frac{n(n+1)}{2}</math>, which is <math>406</math> when <math>n=28</math>, just under <math>420</math>. Checking this gives
+<cmath>\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\sqrt{1-4a}&=60a \\
+\sqrt{1-4a}&=29-60a \\
+^2-4\cdot 29^2a&=29^2+3600a^2-120\cdot 29a \\
+a^2&=116a \\
+a&=\frac{116}{3600}=\frac{29}{900} \implies \boxed{\textbf{(C) } 929}.
+\end{align*}</cmath>
+~ktong
-Equivalently, for <math>x>0,</math> the <math>n</math> intersections of the graphs of <math>f(x)</math> and <math>g(x)</math> occur in the consecutive branches of <math>f(x),</math> namely at <math>x\in[1,2),[2,3),[3,4),\cdots,[n,n+1).</math>
+==Remark==
+Let <math>f(x)=\lfloor x \rfloor \cdot \{x\}</math> and <math>g(x)=a \cdot x^2.</math>
-~MRENTHUSIASM
+We make the following table of values:
-===Proof===
+<cmath>\begin{array}{c|c|c|l}
-Clearly, the equation <math>f(x)=g(x)</math> has no negative solutions, and its positive solutions all satisfy <math>x>1.</math>Moreover, none of its solutions is an integer.
+& & & \\ [-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}</cmath>
+We graph <math>f(x)</math> (in red, by branches) and <math>g(x)</math> (in blue, for <math>a=\frac{29}{900}</math>) as shown below.
-Note that the upper bounds of the branches of <math>f(x)</math> are along the line <math>h(x)=x-1</math> (excluded). <i><b>To prove the claim, we wish to show that for each branch of <math>\boldsymbol{f(x),}</math> there is exactly one solution for <math>\boldsymbol{f(x)=g(x)}</math> (from the branch <math>\boldsymbol{x\in(1,2)}</math> to the branch containing the larger solution of <math>\boldsymbol{g(x)=h(x)}</math>).</b></i> In 8:07-11:31 of [https://www.youtube.com/watch?v=7_mdreGBPvg&t=428s&ab_channel=ArtofProblemSolving Video Solution 3 (Art of Problem-Solving)], Mr. Rusczyk questions whether two solutions of <math>f(x)=g(x)</math> can be in the same branch of <math>f(x),</math> and he concludes that it is impossible in 16:25-16:43.
+[[File:2020 AMC 12A Problem 25.png|center]]
-We analyze the upper bound of <math>f(x):</math> Let <math>(c,c-1)</math> be one solution of <math>g(x)=h(x).</math> It is clear that <math>c>1.</math> We substitute this point to find <math>a:</math>
+Graph in Desmos: https://www.desmos.com/calculator/ouvaiqjdzj
-<cmath>\begin{align*}
-g(c)&=h(c) \\
-ac^2&=c-1 \\
-a&=\frac{c-1}{c^2}.
-\end{align*}</cmath>
-We substitute this result back to find <math>x:</math>
+~MRENTHUSIASM
-<cmath>\begin{align*}
-g(x)&=h(x) \\
-\left(\frac{c-1}{c^2}\right)x^2&=x-1 \\
-\left(\frac{c-1}{c^2}\right)x^2-x+1&=0 \\
-x^2-\left(\frac{c^2}{c-1}\right)x+\frac{c^2}{c-1}&=0 \ \ \ \ \ \ \ \ \ \ \ (*) \\
-(x-c)\left(x-\frac{c}{c-1}\right)&=0 \\
-x&=c,\ \frac{c}{c-1}.
-\end{align*}</cmath>
-By the way, using the precondition that <math>x=c</math> is a root of <math>(*),</math> we can factor its left side easily by the <b>Factor Theorem</b>. Note that <math>g(x)>h(x)</math> for all <math>x>\max{\left(c, \frac{c}{c-1}\right)},</math> as quadratic functions always outgrow linear functions.
-Now, we perform casework:
+==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
-<math>\textbf{Case (1): }\boldsymbol{c=\frac{c}{c-1}>1\Longrightarrow c=2} \textbf{ (Trivial Case)}</math>
+==Video Solution 2==
+https://www.youtube.com/watch?v=xex8TBSzKNE
-It follows that the graphs of <math>g(x)</math> and <math>h(x)</math> only intersect at <math>(2,1),</math> which is not on the graph of <math>f(x).</math> So, the equation <math>f(x)=g(x)</math> has no solution in this case, as the inequality <math>g(x)<h(x)</math> has no solution.
+~MathEx
-<math>\textbf{Case (2): }\boldsymbol{c>\frac{c}{c-1}>1\Longrightarrow c>2} \textbf{ and } \boldsymbol{1<\frac{c}{c-1}<2}</math>
+==Video Solution 3 (by Art of Problem Solving)==
+https://www.youtube.com/watch?v=7_mdreGBPvg&t=428s&ab_channel=ArtofProblemSolving
-It follows that for <math>g(x)=h(x),</math> the smaller solution is <math>x=\frac{c}{c-1}\in(1,2),</math> and <math>g(x)<h(x)</math> holds for all <math>x\in\left(\frac{c}{c-1},c\right).</math>
+Created by Richard Rusczyk
-By the <b>Intermediate Value Theorem</b>, for each branch of <math>f(x)</math> (where <math>x\in\left[c',c'+1\right)</math>), we have <math>g(x)</math> in between its left output and its right "output", namely <math>0=f\left(c'\right)<g\left(c'\right)<c'.</math> Therefore, for the equation <math>f(x)=g(x),</math> there is exactly one solution for each branch of <math>f(x),</math> where <math>x\in\left(\frac{c}{c-1},c\right).</math> Now, the proof of the bolded sentence of paragraph 2 is complete.
+==Video Solution 4==
+https://youtu.be/i5b5P9RPuas
-<math>\textbf{Case (3): }\boldsymbol{\frac{c}{c-1}>c>1\Longrightarrow 1<c<2} \textbf{ and } \boldsymbol{\frac{c}{c-1}>1}.</math>
+~MathProblemSolvingSkills
-This case uses the same argument as <math>\text{Case (2)}.</math> The smaller solution is <math>x=c\in(1,2),</math> and for each branch of <math>f(x),</math> where <math>x\in\left(c,\frac{c}{c-1}\right),</math> the equation <math>f(x)=g(x)</math> has exactly one solution.
-~MRENTHUSIASM
 ==See Also==
 {{AMC12 box|year=2020|ab=A|num-b=24|after=Last Problem}}
 {{MAA Notice}}

2020 AMC 12A (Problems • Answer Key • Resources)
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