Difference between revisions of "2020 AMC 12A Problems/Problem 25"
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<math>\textbf{Case (1): }\boldsymbol{c=\frac{c}{c-1}>1\Longrightarrow c=2} \textbf{ (Trivial Case)}</math> | <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 the point <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 | + | It follows that the graphs of <math>g(x)</math> and <math>h(x)</math> only intersect at the point <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 solutions in this case, as the inequality <math>g(x)<h(x)</math> has no solutions. |
<math>\textbf{Case (2): }\boldsymbol{c>\frac{c}{c-1}>1\Longrightarrow c>2} \textbf{ and } \boldsymbol{1<\frac{c}{c-1}<2}</math> | <math>\textbf{Case (2): }\boldsymbol{c>\frac{c}{c-1}>1\Longrightarrow c>2} \textbf{ and } \boldsymbol{1<\frac{c}{c-1}<2}</math> |
Revision as of 11:38, 8 March 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
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
Remarks of Solution 2 and Video Solution 3
Let and
Graph
We make the following table of values:
We graph by branches:
~MRENTHUSIASM (Graph by Desmos: https://www.desmos.com/calculator/ouvaiqjdzj)
Claim
For all positive integers the first
nonzero solutions to
are of the form
where
Equivalently, for the
intersections of the graphs of
and
occur in the consecutive branches of
namely at
~MRENTHUSIASM
Proof
Clearly, the equation has no negative solutions, and its positive solutions all satisfy
Moreover, none of its solutions is an integer.
Note that the upper bounds of the branches of are along the line
(excluded). To prove the claim, we wish to show that for each branch of
there is exactly one solution for
(from the branch
to the branch containing the larger solution of
). In 8:07-11:31 of Video Solution 3 (Art of Problem-Solving), Mr. Rusczyk questions whether two solutions of
can be in the same branch of
and he concludes that it is impossible in 16:25-16:43.
We analyze the upper bound of Let
be one solution of
It is clear that
We substitute this point to find
We substitute this result back to find
By the way, using the precondition that
is a root of
we can factor its left side easily by the Factor Theorem. Note that
for all
as quadratic functions always outgrow linear functions.
Now, we perform casework:
It follows that the graphs of and
only intersect at the point
which is not on the graph of
So, the equation
has no solutions in this case, as the inequality
has no solutions.
It follows that for the smaller solution is
and
holds for all
By the Intermediate Value Theorem, for each branch of (where
), we have
in between its left output and its right "output", namely
Therefore, for the equation
there is exactly one solution for each branch of
where
Now, the proof of the bolded sentence of paragraph 2 is complete.
This case uses the same argument as The smaller solution is
and for each branch of
where
the equation
has exactly one solution.
~MRENTHUSIASM
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.