Difference between revisions of "2020 AMC 12A Problems/Problem 25"
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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> | 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> | + | 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 obtain <cmath>0<a\leq\frac14. \hspace{54mm}(4)</cmath> |
+ | |||
+ | We consider both parts of <math>0\leq f<1:</math> | ||
+ | <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> always holds.<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 holds.<p> | ||
+ | <li><math>f<1</math></li><p> | ||
+ | </ol> | ||
<b>SOLUTION IN PROGRESS. NO EDIT PLEASE.</b> | <b>SOLUTION IN PROGRESS. NO EDIT PLEASE.</b> |
Revision as of 20:36, 19 May 2021
Contents
[hide]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 get
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 obtain
We consider both parts of
From note that
and
By Descartes' rule of signs, we deduce that
must have two positive roots. So,
always holds.
Alternatively, from solving the inequality
produces
which checks
So,
always holds.
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.