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

(Solution 3 (Piecewise Function: Analysis and Graph): Shifting the lines might be slighter easier than shifting the curve. Made the minor changes so it is more consistent with Sol 1 and 2.)
(Switched the order of solutions, and prioritized solution with graphs. Please PM me if you are disagree with this change.)
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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.
 
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 1 (Graphing)==
+
==Solution 1 (Piecewise Function: Analysis and Graph)==
 
 
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>\begin{align*}
 
-(c + 21) &> -100 \\
 
c + 21 &< 100 \\
 
c &< 79.
 
\end{align*}</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 <math>59</math>.
 
 
 
==Solution 2 (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 Function: Analysis and Graph)==
 
 
We take cases for the outermost absolute value, then rearrange: <cmath>\left|20|x|-x^2\right|=c\pm21.</cmath>
 
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 rewrite <math>f(x)</math> as a piecewise function without using absolute values:
 
Let <math>f(x)=\left|20|x|-x^2\right|.</math> We rewrite <math>f(x)</math> as a piecewise function without using absolute values:
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~MRENTHUSIASM
 
~MRENTHUSIASM
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 +
==Solution 2 (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 1). 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 (Graphing)==
 +
 +
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>\begin{align*}
 +
-(c + 21) &> -100 \\
 +
c + 21 &< 100 \\
 +
c &< 79.
 +
\end{align*}</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 <math>59</math>.
  
 
==Solution 4 (Algebra)==
 
==Solution 4 (Algebra)==

Revision as of 03:01, 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],(-1.5,0),red); label("$(0,0)$",A[2],(0,-1.5),red,UnFill); label("$(10,100)$",A[3],(1.5,0),red); label("$(20,0)$",A[4],(1.5,-1.5),red,UnFill); label("$y=f(x)$",(45,130),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

See Also

2021 AIME I (ProblemsAnswer KeyResources)
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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