Difference between revisions of "2021 AIME I Problems/Problem 8"
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+ | ==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== | ==Remark== |
Revision as of 13:03, 14 March 2021
Contents
Problem
Find the number of integers such that the equation has distinct real solutions.
Solution 1
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 59.
Solution 2 (also graphing)
Graph (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 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 (Piecewise Functions: Analyses and Graphs)
We take cases for the outermost absolute value, then rearrange: Let We will rewrite as a piecewise function without using any absolute value: We graph as shown below, with some key points labeled. The fact that is an even function ( holds for all real numbers from which the graph of is symmetric about the axis) should facilitate the process of graphing.
Graph in Desmos: https://www.desmos.com/calculator/fwvhtltxjr
Since has distinct real solutions, it is clear that each case has distinct real solutions geometrically. We need to shift the graph of down by units:
- For to have distinct real solutions, we get
- For to have distinct real solutions, we get
Taking the intersection of these two cases gives from which there are such integers
~MRENTHUSIASM
Solution 4
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 graph of is shown here in Desmos: https://www.desmos.com/calculator/i6l98lxwpp
Move the slider around for to observe how the graph of intersects the line for times.
~MRENTHUSIASM
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.