Difference between revisions of "2021 AIME I Problems/Problem 8"
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+ | We graph <math>f(x)</math> as shown below, with some key points labelled. | ||
+ | [[File:2021 AIME I Problem 8.png|center]] | ||
+ | Graph in Desmos: https://www.desmos.com/calculator/g8jx63nvfh | ||
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<b>I WILL RETURN AND FINISH THIS SOLUTION SOON. IN THE MEANWHILE PLEASE DO NOT EDIT THIS SOLUTION. THANKS. :)</b> | <b>I WILL RETURN AND FINISH THIS SOLUTION SOON. IN THE MEANWHILE PLEASE DO NOT EDIT THIS SOLUTION. THANKS. :)</b> |
Revision as of 21:48, 12 March 2021
Contents
[hide]Problem
Find the number of integers such that the equation
has
distinct real solutions.
Solution 1
Let Then the equation becomes
, or
. Note that since
,
is nonnegative, so we only care about nonnegative solutions in
. Notice that each positive solution in
gives two solutions in
(
), whereas if
is a solution, this only gives one solution in
,
. Since the total number of solutions in
is even,
must not be a solution. Hence, we require that
has exactly
positive solutions and is not solved by
If , then
is negative, and therefore cannot be the absolute value of
. This means the equation's only solutions are in
. There is no way for this equation to have
solutions, since the quadratic
can only take on each of the two values
at most twice, yielding at most
solutions. Hence,
.
also can't equal
, since this would mean
would solve the equation. Hence,
At this point, the equation will always have exactly
positive solutions, since
takes on each positive value exactly once when
is restricted to positive values (graph it to see this), and
are both positive. Therefore, we just need
to have the remaining
solutions exactly. This means the horizontal lines at
each intersect the parabola
in two places. This occurs when the two lines are above the parabola's vertex
. Hence we have:
Hence, the integers satisfying the conditions are those satisfying
There are
such integers.
Note: Be careful of counting at the end, you may mess up and get 59.
Solution 2 (also graphing)
Graph . Notice that we want this to be equal to
and
.
We see that from left to right, the graph first dips from very positive to at
, then rebounds up to
at
, then falls back down to
at
.
The positive are symmetric, so the graph re-ascends to
at
, falls back to
at
, and rises to arbitrarily large values afterwards.
Now we analyze the (varied by
) values. At
, we will have no solutions, as the line
will have no intersections with our graph.
At , we will have exactly
solutions for the three zeroes.
At for any
strictly between
and
, we will have exactly
solutions.
At , we will have
solutions, because local maxima are reached at
.
At , we will have exactly
solutions.
To get distinct solutions for
, both
and
must produce
solutions.
Thus and
, so
is required.
It is easy to verify that all of these choices of produce
distinct solutions (none overlap), so our answer is
.
Solution 3 (Piecewise Functions: Analysis and Graphs)
We take cases for the outermost absolute value, then rearrange:
Let
We will rewrite
as a piecewise function without using any absolute value:
We graph
as shown below, with some key points labelled.
Graph in Desmos: https://www.desmos.com/calculator/g8jx63nvfh
I WILL RETURN AND FINISH THIS SOLUTION SOON. IN THE MEANWHILE PLEASE DO NOT EDIT THIS SOLUTION. THANKS. :)
~MRENTHUSIASM
See also
2021 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.