Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2021 AIME I Problems/Problem 8

Revision as of 22:21, 12 March 2021 by MRENTHUSIASM (talk | contribs) (→‎Solution 3 (Piecewise Functions: Analysis and Graphs): 57 -> 057.)

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 1

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. Note: Be careful of counting at the end, you may mess up and get 59.

Solution 2 (also graphing)

Graph $y=|20|x|-x^2|$ (If you are having trouble, look at the description in the next two lines and/or the diagram in solution 3). 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 $\boxed{057}$.

Solution 3 (Piecewise Functions: Analysis and Graphs)

We take cases for the outermost absolute value, then rearrange: \[\left|20|x|-x^2\right|-c=\pm21.\] Let $f(x)=\left|20|x|-x^2\right|.$ We will rewrite $f(x)$ as a piecewise function without using any absolute value: \[f(x) = \begin{cases} \left|-20x-x^2\right| & \text{if} \ x \le 0 \begin{cases} 20x+x^2 & \text{if} \ x\le-20 \\ -20x-x^2 & \text{if} \ -20<x\leq0 \end{cases} \\ \left|20x-x^2\right| & \text{if} \ x > 0 \begin{cases} 20x-x^2 & \text{if} \ 0<x\leq20 \\ -20x+x^2 & \text{if} \ x>20 \end{cases} \end{cases}.\] We graph $f(x)$ as shown below, with some key points labeled.

File:2021 AIME I Problem 8.png

Graph in Desmos: https://www.desmos.com/calculator/g8jx63nvfh

Since $f(c)-c=\pm21$ has $12$ distinct real solutions, it is clear geometrically that each case has $6$ distinct real solutions. We need to shift the graph of $f(x)$ down by $c$ units:

  1. For $f(c)-c=21$ to have $6$ distinct real solutions, we get $0<c<79.$
  2. For $f(c)-c=-21$ to have $6$ distinct real solutions, we get $21<c<121.$

Taking the intersection of these two cases gives $21<c<79,$ from which there are $79-21-1=\boxed{057}$ such integers $c.$

~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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems/Problem_8&oldid=149344"