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 (Solves for Floor(x))== |
− | Let <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> |
+ | 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>f=w\Biggl(\frac{1-2a\pm\sqrt{1-4a}}{2a}\Biggr). \hspace{25mm}(3)</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>0\leq1-4a<1,</math> or <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>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. | ||
+ | |||
+ | 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 (Solves for x)== | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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> | ||
+ | |||
+ | ~Shihan | ||
+ | |||
+ | ==Solution 3 (Solves for x)== | ||
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 | + | 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> | <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 | 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>\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*}</cmath> | ||
~ktong | ~ktong | ||
− | ==Solution | + | ==Remark== |
+ | Let <math>f(x)=\lfloor x \rfloor \cdot \{x\}</math> and <math>g(x)=a \cdot x^2.</math> | ||
+ | |||
+ | We make the following table of values: | ||
+ | |||
+ | <cmath>\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}</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. | ||
+ | |||
+ | [[File:2020 AMC 12A Problem 25.png|center]] | ||
+ | |||
+ | 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 | 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== | + | ==Video Solution 2== |
https://www.youtube.com/watch?v=xex8TBSzKNE ~ MathEx | 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== | ==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}} |
Latest revision as of 19:21, 4 November 2021
Contents
Problem
The number , where and are relatively prime positive integers, has the property that the sum of all real numbers satisfying is , where denotes the greatest integer less than or equal to and denotes the fractional part of . What is ?
Solution 1 (Solves for Floor(x))
Let and denote the whole part and the fractional part of respectively, for which and
We rewrite the given equation as Since it follows that from which
We expand and rearrange as which is a quadratic with either or
For simplicity purposes, we will treat as some fixed nonnegative integer so that is a quadratic with By the Quadratic Formula, we have If then We get which does not affect the sum of the solutions. Therefore, we consider the case for
Recall that so From the discriminant, we require that or
We consider each part of separately:
From note that and By Descartes' Rule of Signs, we deduce that must have two positive roots, so is always valid.
Alternatively, from and note that all values of for which satisfy We deduce that both roots in must be positive, so is always valid.
We rewrite as From it follows that The larger root is which contradicts So, we take the smaller root, from which for some constant We rewrite as in which is valid as long as Note that the solutions of are generated at up to some value such that
Now, we express in terms of and The sum of all solutions to the original equation is As we conclude that is slightly above so that is slightly below or is slightly below By observations, we get Substituting this into produces which satisfies as required.
Finally, we solve for in Since we obtain from which The answer is
~MRENTHUSIASM (inspired by Math Jams's 2020 AMC 10/12A Discussion)
Solution 2 (Solves for x)
Let be a root in the interval . In this interval, and , so we must have , i.e., . We can homogenize this equation by setting ; then , and is a root of .
Suppose is the largest integer for which there is such a root; we have, for , Summing over we get From the right inequality we get and from the left one we get . Thus . Using this in the middle equality we get . Since satisfies , we get The answer is
~Shihan
Solution 3 (Solves for x)
First note that when while . Thus we only need to look at positive solutions ( doesn't affect the sum of the solutions). Next, we break down for each interval , where is a positive integer. Assume , then . This means that when , . Setting this equal to gives We're looking at the solution with the positive , which is . Note that if is the greatest such that has a solution, the sum of all these solutions is slightly over , which is when , just under . Checking this gives ~ktong
Remark
Let and
We make the following table of values:
We graph (in red, by branches) and (in blue, for ) 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
See Also
2020 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Problem |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.