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Difference between revisions of "2021 AIME I Problems/Problem 8"

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==Solution==
 
==Solution==
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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:
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<cmath>-(c + 21) > -100</cmath>
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<cmath>c + 21 < 100</cmath>
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<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 17:46, 11 March 2021

Problem

Find the number of integers $c$ such that the equation\[\left||20|x|-x^2|-c\right|=21\]has $12$ distinct real solutions.

Solution

Let $y = |x|.$ Then the equation becomes $\left|\left|20y-y^2\right|-c\right| = 21$, or $\left|y^2-20y\right| = c \pm 21$. Note that since $y = |x|$, $y$ is nonnegative, so we only care about nonnegative solutions in $y$. Notice that each positive solution in $y$ gives two solutions in $x$ ($x = \pm y$), whereas if $y = 0$ is a solution, this only gives one solution in $x$, $x = 0$. Since the total number of solutions in $x$ is even, $y = 0$ must not be a solution. Hence, we require that $\left|y^2-20y\right| = c \pm 21$ has exactly $6$ positive solutions and is not solved by $y = 0.$

If $c < 21$, then $c - 21$ is negative, and therefore cannot be the absolute value of $y^2 - 20y$. This means the equation's only solutions are in $\left|y^2-20y\right| = c + 21$. There is no way for this equation to have $6$ solutions, since the quadratic $y^2-20y$ can only take on each of the two values $\pm(c + 21)$ at most twice, yielding at most $4$ solutions. Hence, $c \ge 21$. $c$ also can't equal $21$, since this would mean $y = 0$ would solve the equation. Hence, $c > 21.$

At this point, the equation $y^2-20y = c \pm 21$ will always have exactly $2$ positive solutions, since $y^2-20y$ takes on each positive value exactly once when $y$ is restricted to positive values (graph it to see this), and $c \pm 21$ are both positive. Therefore, we just need $y^2-20y = -(c \pm 21)$ to have the remaining $4$ solutions exactly. This means the horizontal lines at $-(c \pm 21)$ each intersect the parabola $y^2 - 20y$ in two places. This occurs when the two lines are above the parabola's vertex $(10,-100)$. Hence we have: \[-(c + 21) > -100\] \[c + 21 < 100\] \[c < 79\]

Hence, the integers $c$ satisfying the conditions are those satisfying $21 < c < 79.$ There are $\boxed{057}$ 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

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