Difference between revisions of "2020 AMC 12A Problems/Problem 25"
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\end{align*}</cmath> | \end{align*}</cmath> | ||
− | If <math>w=0,</math> then <math>f=0.</math> We | + | If <math>w=0,</math> then <math>f=0.</math> We have <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 conclude that <math>1-4a\geq0,</math> or <math>a\leq\frac14.</math> Combining this with the precondition <math>a>0,</math> we | + | Recall that <math>0\leq f<1,</math> so <math>\frac{1-2a\pm\sqrt{1-4a}}{2a}\geq0.</math> From the discriminant, we conclude that <math>1-4a\geq0,</math> or <math>a\leq\frac14.</math> Combining this with the precondition <math>a>0,</math> we have <cmath>0<a\leq\frac14. \hspace{54mm}(4)</cmath> |
We consider each part of <math>0\leq f<1</math> separately: | We consider each part of <math>0\leq f<1</math> separately: | ||
<ol style="margin-left: 1.5em;"> | <ol style="margin-left: 1.5em;"> | ||
<li><math>f\geq0</math></li><p> | <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> always | + | 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> solving the inequality <math>1-2a>\sqrt{1-4a}</math> produces <math>0<a\leq\frac14,</math> which checks <math>(4).</math> So, <math>f\geq0</math> always | + | Alternatively, from <math>(3),</math> solving the inequality <math>1-2a>\sqrt{1-4a}</math> produces <math>0<a\leq\frac14,</math> which checks <math>(4).</math> So, <math>f\geq0</math> is always valid.<p> |
<li><math>f<1</math></li><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> we deduce that <math>\frac{1}{2a}\geq\frac{1}{\left(\frac12\right)}=2.</math> The larger root is | 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> we deduce that <math>\frac{1}{2a}\geq\frac{1}{\left(\frac12\right)}=2.</math> The larger root is | ||
Line 54: | Line 54: | ||
&= 1, \\ | &= 1, \\ | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | which contradicts to <math>f<1.</math> So, we take the smaller root. | + | which contradicts to <math>f<1.</math> So, we take the smaller root.<p> |
+ | We get <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}.</math> Note that <math>k>0</math> for all <math>a</math> such that <math>0<a\leq\frac14.</math> <p> | ||
+ | Now, we obtain <cmath>f=wk,</cmath> in which <math>f<1</math> is valid as long as <math>w<\frac 1k.</math> | ||
</ol> | </ol> | ||
Revision as of 23:22, 19 May 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
Let be the unique solution in this range. Note that is also a solution as long as , hence all our solutions are for some . This sum must be between and , which gives and . Plugging this back in gives .
Solution 2
First note that when while . Thus we only need to look at positive solutions ( doesn't affect the sum of the solutions). Next, we breakdown 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
Solution 3 (Comprehensive)
Let and denote the whole part and the fractional part of respectively, in 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 get
If then We have which does not affect the sum of the solutions. Therefore, we consider the case for
Recall that so From the discriminant, we conclude that or Combining this with the precondition we have
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 solving the inequality produces which checks So, is always valid.
We rewrite as From we deduce that The larger root is
which contradicts to So, we take the smaller root.We get for some constant Note that for all such that
Now, we obtain in which is valid as long as
SOLUTION IN PROGRESS. NO EDIT PLEASE.
~MRENTHUSIASM (inspired by Math Jams's 2020 AMC 10/12A Discussion)
Remarks
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.