Difference between revisions of "2020 AMC 12A Problems/Problem 25"
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==Video Solution 2== | ==Video Solution 2== | ||
− | https://www.youtube.com/watch?v=xex8TBSzKNE ~ MathEx | + | https://www.youtube.com/watch?v=xex8TBSzKNE |
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==Video Solution 3 (by Art of Problem Solving)== | ==Video Solution 3 (by Art of Problem Solving)== | ||
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Created by Richard Rusczyk | Created by Richard Rusczyk | ||
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+ | https://youtu.be/i5b5P9RPuas | ||
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+ | ~MathProblemSolvingSkills | ||
==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 17:13, 21 July 2022
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
Video Solution 4
~MathProblemSolvingSkills
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.