Difference between revisions of "2021 AIME I Problems/Problem 8"
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+ | Let <math>y = |x|.</math> Then the equation becomes <math>\left|\left|20y-y^2\right|-c\right| = 21</math>, or <math>\left|y^2-20y\right| = c \pm 21</math>. Note that since <math>y = |x|</math>, <math>y</math> is nonnegative, so we only care about nonnegative solutions in <math>y</math>. Notice that each positive solution in <math>y</math> gives two solutions in <math>x</math> (<math>x = \pm y</math>), whereas if <math>y = 0</math> is a solution, this only gives one solution in <math>x</math>, <math>x = 0</math>. Since the total number of solutions in <math>x</math> is even, <math>y = 0</math> must not be a solution. Hence, we require that <math>\left|y^2-20y\right| = c \pm 21</math> has exactly <math>6</math> positive solutions and is not solved by <math>y = 0.</math> | ||
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+ | If <math>c < 21</math>, then <math>c - 21</math> is negative, and therefore cannot be the absolute value of <math>y^2 - 20y</math>. This means the equation's only solutions are in <math>\left|y^2-20y\right| = c + 21</math>. There is no way for this equation to have <math>6</math> solutions, since the quadratic <math>y^2-20y</math> can only take on each of the two values <math>\pm(c + 21)</math> at most twice, yielding at most <math>4</math> solutions. Hence, <math>c \ge 21</math>. <math>c</math> also can't equal <math>21</math>, since this would mean <math>y = 0</math> would solve the equation. Hence, <math>c > 21.</math> | ||
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+ | At this point, the equation <math>y^2-20y = c \pm 21</math> will always have exactly <math>2</math> positive solutions, since <math>y^2-20y</math> takes on each positive value exactly once when <math>y</math> is restricted to positive values (graph it to see this), and <math>c \pm 21</math> are both positive. Therefore, we just need <math>y^2-20y = -(c \pm 21)</math> to have the remaining <math>4</math> solutions exactly. This means the horizontal lines at <math>-(c \pm 21)</math> each intersect the parabola <math>y^2 - 20y</math> in two places. This occurs when the two lines are above the parabola's vertex <math>(10,-100)</math>. Hence we have: | ||
+ | <cmath>-(c + 21) > -100</cmath> | ||
+ | <cmath>c + 21 < 100</cmath> | ||
+ | <cmath>c < 79</cmath> | ||
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+ | Hence, the integers <math>c</math> satisfying the conditions are those satisfying <math>21 < c < 79.</math> There are <math>\boxed{057}</math> such integers. | ||
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==See also== | ==See also== | ||
{{AIME box|year=2021|n=I|num-b=7|num-a=9}} | {{AIME box|year=2021|n=I|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 16:46, 11 March 2021
Problem
Find the number of integers such that the equation
has
distinct real solutions.
Solution
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.
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.