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

Revision as of 02:06, 13 September 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 1 (Piecewise Function: Analysis and Graph)

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 rewrite $f(x)$ as a piecewise function without using absolute values: $f(x) = \begin{cases} \left|-20x-x^2\right| & \mathrm{if} \ x \le 0 \begin{cases} 20x+x^2 & \mathrm{if} \ x\le-20 \\ -20x-x^2 & \mathrm{if} \ -20<x\leq0 \end{cases} \\ \left|20x-x^2\right| & \mathrm{if} \ x > 0 \begin{cases} 20x-x^2 & \mathrm{if} \ 0<x\leq20 \\ -20x+x^2 & \mathrm{if} \ x>20 \end{cases} \end{cases}.$ We graph $y=f(x)$ with all extremum points labeled, as shown below. The fact that $f(x)$ is an even function ( $f(x)=f(-x)$ holds for all real numbers $x,$ so the graph of $y=f(x)$ is symmetric about the $y$ -axis) should facilitate the process of graphing. $[asy] /* Made by MRENTHUSIASM */ size(300); import TrigMacros; draw((-65,0)--(65,0),Arrows); draw((0,-50)--(0,150),Arrows); label("$x$",(75,0)); label("$y$",(0,160)); real f(real x) { return 20x+x^2; } real g(real x) { return -20x-x^2; } real h(real x) { return 20x-x^2; } real j(real x) { return -20x+x^2; } draw(graph(f,-25.8,-20),red); draw(graph(g,-20,0),red); draw(graph(h,0,20),red); draw(graph(j,20,25.8),red); draw((-65,-21)--(65,-21),blue); draw((-65,21)--(65,21),blue); pair A[]; A[0] = (-20,0); A[1] = (-10,100); A[2] = (0,0); A[3] = (10,100); A[4] = (20,0); for(int i = 0; i <= 4; ++i) { dot(A[i],red+linewidth(4.5)); } label("$(-20,0)$",A[0],(-1.5,-1.5),red,UnFill); label("$(-10,100)$",A[1],(-1.5,1.5),red); label("$(0,0)$",A[2],(0,-1.5),red,UnFill); label("$(10,100)$",A[3],(1.5,1.5),red); label("$(20,0)$",A[4],(1.5,-1.5),red,UnFill); label("$y=f(x)$",(45,135),red); label("$y=21$",(45,27),blue); label("$y=-21$",(45,-27),blue); [/asy]$ Since $f(x)=c\pm21$ has $12$ distinct real solutions, it is clear that each case has $6$ distinct real solutions geometrically. We shift the graphs of $y=\pm21$ up $c$ units, where $c\geq0:$

For $f(x)=c+21$ to have $6$ distinct real solutions, we need $0\leq c<79.$

For $f(x)=c-21$ to have $6$ distinct real solutions, we need $21<c<121.$

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

~MRENTHUSIASM

Solution 2 (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 1). 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 (Graphing)

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 $\begin{align*} -(c + 21) &> -100 \\ c + 21 &< 100 \\ c &< 79. \end{align*}$ 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 4 (Algebra)

Removing the absolute value bars from the equation successively, we get $\begin{align*} \left|\left|20|x|-x^2\right|-c\right|&=21 \\ \left|20|x|-x^2\right|&= c \pm21 \\ 20|x|-x^2 &= \pm c \pm 21 \\ x^2 \pm 20x \pm c \pm21 &= 0. \end{align*}$ The discriminant of this equation is $\sqrt{400-4(\pm c \pm 21)}.$ Equating the discriminant to $0$ , we see that there will be two distinct solutions to each of the possible quadratics above only in the interval $-79 < c < 79$ . However, the number of zeros the equation $ax^2+b|x|+k$ has is determined by where $ax^2+bx+k$ and $ax^2-bx+k$ intersect, namely at $(0,k)$ . When $k<0$ , $a>0$ , $ax^2+b|x|+k$ will have only $2$ solutions, and when $k>0$ , $a>0$ , then there will be $4$ real solutions, if they exist at all. In order to have $12$ solutions here, we thus need to ensure $-c+21<0$ , so that exactly $2$ out of the $4$ possible equations of the form $ax^2+b|x|+k$ given above have y-intercepts below $0$ and only $2$ real solutions, while the remaining $2$ equations have $4$ solutions. This occurs when $c>21$ , so our final bounds are $21<c<79$ , giving us $\boxed{057}$ valid values of $c$ .

Remark

The graphs of $F(x)=\left||20|x|-x^2|-c\right|$ and $G(x)=21$ are shown here in Desmos: https://www.desmos.com/calculator/i6l98lxwpp

Move the slider around for $21<c<79$ to observe how they intersect for $12$ times.

~MRENTHUSIASM

2021 AIME I (Problems • Answer Key • Resources)
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