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

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<math>\textbf{(A) } 245 \qquad \textbf{(B) } 593 \qquad \textbf{(C) } 929 \qquad \textbf{(D) } 1331 \qquad \textbf{(E) } 1332</math>
 
<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 (Comprehensive Solution in Algebra)==
 +
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>
 +
 
 +
We rewrite the given equation as <cmath>w\cdot f=a\cdot(w+f)^2. \hspace{38.75mm}(1)</cmath>
 +
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 have
 +
<cmath>\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*}</cmath>
 +
 
 +
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>
 +
 
 +
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>1-4a\geq0,</math> or <math>a\leq\frac14.</math> Combining this with the precondition <math>a>0,</math> we need <cmath>0<a\leq\frac14. \hspace{54mm}(4)</cmath>
 +
 
 +
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>\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*}</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,\cdots,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>\begin{align*}
 +
x&=w+f \\
 +
&=w+wk \\
 +
&=w(1+k).
 +
\end{align*}</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 back into <math>(\bigstar)</math> produces <math>k=\frac{1}{29},</math> which satisfies <math>\frac{1}{W+1}\leq k<\frac1W,</math> as required.
 +
 
 +
Finally, we solve for <math>a</math> in <math>k=\frac{1}{2a}-1-\frac{\sqrt{1-4a}}{2a}:</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>
 +
 
 +
~MRENTHUSIASM (inspired by Math Jams's <b>2020 AMC 10/12A Discussion</b>)
 +
 
 +
==Solution 2 (Condensed Version of Solution 1)==
 
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>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>.
  
==Solution 2==
+
==Solution 3==
 
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).  
 
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 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  
 
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  
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~ktong
 
~ktong
  
==Video Solution 1 (Geometry)==
+
==Remark==
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 <math>f(x)=\lfloor x \rfloor \cdot \{x\}</math> and <math>g(x)=a \cdot x^2.</math>
 
Let <math>f(x)=\lfloor x \rfloor \cdot \{x\}</math> and <math>g(x)=a \cdot x^2.</math>
  
===Graph===
 
 
We make the following table of values:
 
We make the following table of values:
  
 
<cmath>\begin{array}{c|c|c|clc}
 
<cmath>\begin{array}{c|c|c|clc}
\boldsymbol{x} & \boldsymbol{\lfloor x \rfloor} & \boldsymbol{f(x)} & & \textbf{\ \ Equation} & \\ [1.5ex]
+
& & & & & \\ [-1.5ex]
 +
\boldsymbol{x} & \boldsymbol{\lfloor x \rfloor} & \boldsymbol{f(x)} & & \hspace{4mm}\textbf{Equation} & \\ [1.5ex]
 
\hline
 
\hline
 
& & & & & \\ [-1ex]
 
& & & & & \\ [-1ex]
Line 50: Line 95:
 
[4,5) & 4 & [0,4) & & y=4x-16 & \\ [1.5ex]
 
[4,5) & 4 & [0,4) & & y=4x-16 & \\ [1.5ex]
 
\cdots & \cdots & \cdots & & \ \ \ \ \ \ \ \cdots & \\ [1.5ex]
 
\cdots & \cdots & \cdots & & \ \ \ \ \ \ \ \cdots & \\ [1.5ex]
[m,m+1) & m & [0,m) & & y=mx-m^2 &
+
[m,m+1) & m & [0,m) & & y=mx-m^2 & \\ [2ex]
 
\end{array}</cmath>
 
\end{array}</cmath>
  
We graph <math>f(x)</math> by branches:
+
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.
  
 
[[File:2020 AMC 12A Problem 25.png|center]]
 
[[File:2020 AMC 12A Problem 25.png|center]]
  
~MRENTHUSIASM (Graph by Desmos: https://www.desmos.com/calculator/ouvaiqjdzj)
+
Graph in Desmos: https://www.desmos.com/calculator/ouvaiqjdzj
 
 
===Claim===
 
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>
 
 
 
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>
 
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  
===Proof===
+
==Video Solution 1 (Geometry)==
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.
+
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
  
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.
+
==Video Solution 2==
 +
https://www.youtube.com/watch?v=xex8TBSzKNE ~ MathEx
  
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>
+
==Video Solution 3 (by Art of Problem Solving)==
<cmath>\begin{align*}
+
https://www.youtube.com/watch?v=7_mdreGBPvg&t=428s&ab_channel=ArtofProblemSolving
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>
+
Created by Richard Rusczyk
<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:
 
 
 
<math>\textbf{Case (1): }\boldsymbol{c=\frac{c}{c-1}>1\Longrightarrow c=2} \textbf{ (Trivial Case)}</math>
 
 
 
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.
 
 
 
<math>\textbf{Case (2): }\boldsymbol{c>\frac{c}{c-1}>1\Longrightarrow c>2} \textbf{ and } \boldsymbol{1<\frac{c}{c-1}<2}</math>
 
 
 
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>
 
 
 
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.
 
 
 
<math>\textbf{Case (3): }\boldsymbol{\frac{c}{c-1}>c>1\Longrightarrow 1<c<2} \textbf{ and } \boldsymbol{\frac{c}{c-1}>1}.</math>
 
 
 
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==
 
==See Also==
 
{{AMC12 box|year=2020|ab=A|num-b=24|after=Last Problem}}
 
{{AMC12 box|year=2020|ab=A|num-b=24|after=Last Problem}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 07:38, 14 June 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 (Comprehensive Solution in Algebra)

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 \[0<a\leq\frac14. \hspace{54mm}(4)\]

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

  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 \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 $f<1.$ 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 \[w=1,2,3,\cdots,W,\] up to some value $w=W$ such that $\frac{1}{W+1}\leq k<\frac1W.$

Now, we express $x$ in terms of $w$ and $k:$ \begin{align*} x&=w+f \\ &=w+wk \\ &=w(1+k). \end{align*} 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 back 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 (Condensed Version of 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 3

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

Remark

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

We make the following table of values:

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

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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