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

Revision as of 10:49, 20 June 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

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: Analyses 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: $f(x) = \begin{cases} \left|-20x-x^2\right| & \mathrm{if} \ x \le 0 \begin{cases} 20x+x^2 & \mathrm{if} \ x\le-20 \\ -20x-x^2 & \mathrm{if} \ -20<x\leq0 \end{cases} \\ \left|20x-x^2\right| & \mathrm{if} \ x > 0 \begin{cases} 20x-x^2 & \mathrm{if} \ 0<x\leq20 \\ -20x+x^2 & \mathrm{if} \ x>20 \end{cases} \end{cases}.$ We graph $f(x)$ 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 $f(x)$ is symmetric about the $y$ -axis) should facilitate the process of graphing.

File:2021 AIME I Problem 8.png

Graph in Desmos: https://www.desmos.com/calculator/tdbbceyb9f

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 graph of $f(x)$ down $c$ 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 4

Removing the absolute value bars from the equation successively, we get $\left||20|x|-x^2|-c\right|=21$ $|20|x|-x^2|= c \pm21$ $20|x|-x^2 = \pm c \pm 21$ $x^2 \pm 20x \pm c \pm21 = 0$

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

@@ Line 1: / Line 1: @@
 ==Problem==
-These problems will not be available until the 2021 AIME I is released on Wednesday, March 10, 2021.
+Find the number of integers <math>c</math> such that the equation <cmath>\left||20|x|-x^2|-c\right|=21</cmath>has <math>12</math> distinct real solutions.
-==Solution==
+==Solution 1==
-==See also==
-{{AIME box|year=2021|n=I|num-b=6|num-a=8}}
+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>
+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>
+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>
+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.
+Note: Be careful of counting at the end, you may mess up and get 59.
+==Solution 2 (Also Graphing)==
+Graph <math>y=|20|x|-x^2|</math> (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 <math>c-21</math> and <math>c+21</math>.
+We see that from left to right, the graph first dips from very positive to <math>0</math> at <math>x=-20</math>, then rebounds up to <math>100</math> at <math>x=-10</math>, then falls back down to <math>0</math> at <math>x=0</math>.
+The positive <math>x</math> are symmetric, so the graph re-ascends to <math>100</math> at <math>x=10</math>, falls back to <math>0</math> at <math>x=10</math>, and rises to arbitrarily large values afterwards.
+Now we analyze the <math>y</math> (varied by <math>c</math>) values. At <math>y=k<0</math>, we will have no solutions, as the line <math>y=k</math> will have no intersections with our graph.
+At <math>y=0</math>, we will have exactly <math>3</math> solutions for the three zeroes.
+At <math>y=n</math> for any <math>n</math> strictly between <math>0</math> and <math>100</math>, we will have exactly <math>6</math> solutions.
+At <math>y=100</math>, we will have <math>4</math> solutions, because local maxima are reached at <math>x= \pm 10</math>.
+At <math>y=m>100</math>, we will have exactly <math>2</math> solutions.
+To get <math>12</math> distinct solutions for <math>y=|20|x|-x^2|=c \pm 21</math>, both <math>c +21</math> and <math>c-21</math> must produce <math>6</math> solutions.
+Thus <math>0<c-21</math> and <math>c+21<100</math>, so <math>c \in \{ 22, 23, \dots , 77, 78 \}</math> is required.
+It is easy to verify that all of these choices of <math>c</math> produce <math>12</math> distinct solutions (none overlap), so our answer is <math>\boxed{057}</math>.
+==Solution 3 (Piecewise Functions: Analyses and Graphs)==
+We take cases for the outermost absolute value, then rearrange: <cmath>\left|20|x|-x^2\right|-c=\pm21.</cmath>
+Let <math>f(x)=\left|20|x|-x^2\right|.</math> We will rewrite <math>f(x)</math> as a piecewise function without using any absolute value:
+<cmath>f(x) = \begin{cases}
+\left|-20x-x^2\right| & \mathrm{if} \ x \le 0
+\begin{cases}
+x+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}
+x-x^2 & \mathrm{if} \ 0<x\leq20 \\
+-20x+x^2 & \mathrm{if} \ x>20
+\end{cases}
+\end{cases}.</cmath>
+We graph <math>f(x)</math> with all extremum points labeled, as shown below. The fact that <math>f(x)</math> is an even function (<math>f(x)=f(-x)</math> holds for all real numbers <math>x,</math> so the graph of <math>f(x)</math> is symmetric about the <math>y</math>-axis) should facilitate the process of graphing.
+[[File:2021 AIME I Problem 8.png|center]]
+Graph in Desmos: https://www.desmos.com/calculator/tdbbceyb9f
+Since <math>f(x)-c=\pm21</math> has <math>12</math> distinct real solutions, it is clear that each case has <math>6</math> distinct real solutions geometrically. We shift the graph of <math>f(x)</math> down <math>c</math> units, where <math>c\geq0:</math>
+* For <math>f(x)-c=21</math> to have <math>6</math> distinct real solutions, we need <math>0\leq c<79.</math>
+* For <math>f(x)-c=-21</math> to have <math>6</math> distinct real solutions, we need <math>21<c<121.</math>
+Taking the intersection of these two cases gives <math>21<c<79,</math> from which there are <math>79-21-1=\boxed{057}</math> such integers <math>c.</math>
+~MRENTHUSIASM
+==Solution 4==
+Removing the absolute value bars from the equation successively, we get
+<cmath>\left||20|x|-x^2|-c\right|=21</cmath>
+<cmath>|20|x|-x^2|= c \pm21</cmath>
+<cmath>20|x|-x^2 = \pm c \pm 21</cmath>
+<cmath>x^2 \pm 20x \pm c \pm21 = 0</cmath>
+The discriminant of this equation is
+<cmath>\sqrt{400-4(\pm c \pm 21)}</cmath>
+Equating the discriminant to <math>0</math>, we see that there will be two distinct solutions to each of the possible quadratics above only in the interval <math>-79 < c < 79</math>. However, the number of zeros the equation <math>ax^2+b|x|+k</math> has is determined by where <math>ax^2+bx+k</math> and <math>ax^2-bx+k</math> intersect, namely at <math>(0,k)</math>. When <math>k<0</math>, <math>a>0</math>, <math>ax^2+b|x|+k</math> will have only <math>2</math> solutions, and when <math>k>0</math>, <math>a>0</math>, then there will be <math>4</math> real solutions, if they exist at all.
+In order to have <math>12</math> solutions here, we thus need to ensure <math>-c+21<0</math>, so that exactly <math>2</math> out of the <math>4</math> possible equations of the form <math>ax^2+b|x|+k</math> given above have y-intercepts below <math>0</math> and only <math>2</math> real solutions, while the remaining <math>2</math> equations have <math>4</math> solutions. This occurs when <math>c>21</math>, so our final bounds are <math>21<c<79</math>, giving us <math>\boxed{057}</math> valid values of <math>c</math>.
+==Remark==
+The graphs of <math>F(x)=\left||20|x|-x^2|-c\right|</math> and <math>G(x)=21</math> are shown here in Desmos: https://www.desmos.com/calculator/i6l98lxwpp
+Move the slider around for <math>21<c<79</math> to observe how they intersect for <math>12</math> times.
+~MRENTHUSIASM
+==See Also==
+{{AIME box|year=2021|n=I|num-b=7|num-a=9}}
+[[Category:Intermediate Algebra Problems]]
 {{MAA Notice}}

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