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

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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.
 
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.
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==Solution 2 (also graphing)==
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Graph <math>y=|20|x|-x^2|</math>. Notice that we want this to be equal to <math>c-21</math> and <math>c+21</math>.
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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>.
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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.
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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.
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At <math>y=0</math>, we will have exactly <math>3</math> solutions for the three zeroes.
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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.
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At <math>y=100</math>, we will have <math>4</math> solutions, because local maxima are reached at <math>x= \pm 10</math>.
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At <math>y=m>100</math>, we will have exactly <math>2</math> solutions.
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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.
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Thus <math>0<c-21</math> and <math>c+21<100</math>, so <math>c \in \{ 22, 23, \dots , 77, 78 \}</math> is required.
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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>.
  
 
==See also==
 
==See also==
 
{{AIME box|year=2021|n=I|num-b=7|num-a=9}}
 
{{AIME box|year=2021|n=I|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 13:12, 12 March 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

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 $\boxed{057}$.

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

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