Difference between revisions of "2021 AIME I Problems/Problem 8"
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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== | + | ==Solution 1 (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> | ||
+ | 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: | ||
+ | <cmath>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}.</cmath> | ||
+ | We graph <math>y=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>y=f(x)</math> is symmetric about the <math>y</math>-axis) should facilitate the process of graphing. | ||
+ | <asy> | ||
+ | /* Made by MRENTHUSIASM */ | ||
+ | size(1200,300); | ||
+ | |||
+ | real xMin = -65; | ||
+ | real xMax = 65; | ||
+ | real yMin = -50; | ||
+ | real yMax = 125; | ||
+ | |||
+ | draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); | ||
+ | draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); | ||
+ | label("$x$",(xMax,0),(2,0)); | ||
+ | label("$y$",(0,yMax),(0,2)); | ||
− | + | real f(real x) { return abs(20*abs(x)-x^2); } | |
+ | real g(real x) { return 21; } | ||
+ | real h(real x) { return -21; } | ||
+ | draw(graph(f,-25,25),red,"$y=\left|20|x|-x^2\right|$"); | ||
+ | draw(graph(g,-65,65),blue,"$y=\pm21$"); | ||
+ | draw(graph(h,-65,65),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); | ||
+ | |||
+ | add(legend(),point(E),40E,UnFill); | ||
+ | </asy> | ||
+ | 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 graphs of <math>y=\pm21</math> up <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 2 ( | + | ==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 | + | 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>. | 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>. | ||
Line 40: | Line 92: | ||
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>. | 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 ( | + | ==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> | |
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− | + | 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== | + | ==Solution 4 (Algebra)== |
Removing the absolute value bars from the equation successively, we get | Removing the absolute value bars from the equation successively, we get | ||
− | <cmath>\left||20|x|-x^2|-c\right|=21 | + | <cmath>\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*}</cmath> | ||
The discriminant of this equation is | The discriminant of this equation is | ||
− | <cmath>\sqrt{400-4(\pm c \pm 21)}</cmath> | + | <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. | 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>. | 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>. | ||
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~MRENTHUSIASM | ~MRENTHUSIASM | ||
− | ==See | + | ==Video Solution== |
+ | https://youtu.be/6k-uR71_jg0 | ||
+ | ~mathproblemsolvingskills.com | ||
+ | |||
+ | ==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}} | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 01:27, 24 December 2022
Contents
Problem
Find the number of integers such that the equation
has
distinct real solutions.
Solution 1 (Piecewise Function: Analysis and Graph)
We take cases for the outermost absolute value, then rearrange:
Let
We rewrite
as a piecewise function without using absolute values:
We graph
with all extremum points labeled, as shown below. The fact that
is an even function (
holds for all real numbers
so the graph of
is symmetric about the
-axis) should facilitate the process of graphing.
Since
has
distinct real solutions, it is clear that each case has
distinct real solutions geometrically. We shift the graphs of
up
units, where
- For
to have
distinct real solutions, we need
- For
to have
distinct real solutions, we need
Taking the intersection of these two cases gives from which there are
such integers
~MRENTHUSIASM
Solution 2 (Graphing)
Graph (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
and
.
We see that from left to right, the graph first dips from very positive to at
, then rebounds up to
at
, then falls back down to
at
.
The positive are symmetric, so the graph re-ascends to
at
, falls back to
at
, and rises to arbitrarily large values afterwards.
Now we analyze the (varied by
) values. At
, we will have no solutions, as the line
will have no intersections with our graph.
At , we will have exactly
solutions for the three zeroes.
At for any
strictly between
and
, we will have exactly
solutions.
At , we will have
solutions, because local maxima are reached at
.
At , we will have exactly
solutions.
To get distinct solutions for
, both
and
must produce
solutions.
Thus and
, so
is required.
It is easy to verify that all of these choices of produce
distinct solutions (none overlap), so our answer is
.
Solution 3 (Graphing)
Let Then the equation becomes
, or
. Note that since
,
is nonnegative, so we only care about nonnegative solutions in
. Notice that each positive solution in
gives two solutions in
(
), whereas if
is a solution, this only gives one solution in
,
. Since the total number of solutions in
is even,
must not be a solution. Hence, we require that
has exactly
positive solutions and is not solved by
If , then
is negative, and therefore cannot be the absolute value of
. This means the equation's only solutions are in
. There is no way for this equation to have
solutions, since the quadratic
can only take on each of the two values
at most twice, yielding at most
solutions. Hence,
.
also can't equal
, since this would mean
would solve the equation. Hence,
At this point, the equation will always have exactly
positive solutions, since
takes on each positive value exactly once when
is restricted to positive values (graph it to see this), and
are both positive. Therefore, we just need
to have the remaining
solutions exactly. This means the horizontal lines at
each intersect the parabola
in two places. This occurs when the two lines are above the parabola's vertex
. Hence we have
Hence, the integers
satisfying the conditions are those satisfying
There are
such integers.
Note: Be careful of counting at the end, you may mess up and get .
Solution 4 (Algebra)
Removing the absolute value bars from the equation successively, we get
The discriminant of this equation is
Equating the discriminant to
, we see that there will be two distinct solutions to each of the possible quadratics above only in the interval
. However, the number of zeros the equation
has is determined by where
and
intersect, namely at
. When
,
,
will have only
solutions, and when
,
, then there will be
real solutions, if they exist at all.
In order to have
solutions here, we thus need to ensure
, so that exactly
out of the
possible equations of the form
given above have y-intercepts below
and only
real solutions, while the remaining
equations have
solutions. This occurs when
, so our final bounds are
, giving us
valid values of
.
Remark
The graphs of and
are shown here in Desmos: https://www.desmos.com/calculator/i6l98lxwpp
Move the slider around for to observe how they intersect for
times.
~MRENTHUSIASM
Video Solution
https://youtu.be/6k-uR71_jg0 ~mathproblemsolvingskills.com
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.