2021 AIME I Problems/Problem 8

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.

Solution 2 (also graphing)

Graph $y=|20|x|-x^2|$ . Notice that we want this to be equal to $c-21$ and $c+21$ .

We see that from left to right, the graph first dips from very positive to $0$ at $x=-20$ , then rebounds up to $100$ at $x=-10$ , then falls back down to $0$ at $x=0$ .

The positive $x$ are symmetric, so the graph re-ascends to $100$ at $x=10$ , falls back to $0$ at $x=10$ , and rises to arbitrarily large values afterwards.

Now we analyze the $y$ (varied by $c$ ) values. At $y=k<0$ , we will have no solutions, as the line $y=k$ will have no intersections with our graph.

At $y=0$ , we will have exactly $3$ solutions for the three zeroes.

At $y=n$ for any $n$ strictly between $0$ and $100$ , we will have exactly $6$ solutions.

At $y=100$ , we will have $4$ solutions, because local maxima are reached at $x= \pm 10$ .

At $y=m>100$ , we will have exactly $2$ solutions.

To get $12$ distinct solutions for $y=|20|x|-x^2|=c \pm 21$ , both $c +21$ and $c-21$ must produce $6$ solutions.

Thus $0<c-21$ and $c+21<100$ , so $c \in \{ 22, 23, \dots , 77, 78 \}$ is required.

It is easy to verify that all of these choices of $c$ produce $12$ distinct solutions (none overlap), so our answer is $\boxed{057}$ .

2021 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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2021 AIME I Problems/Problem 8

Contents

Problem

Solution

Solution 2 (also graphing)

See also