2021 AIME I Problems/Problem 8
Problem
Find the number of integers such that the equationhas distinct real solutions.
Solution
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.
Solution 2 (also graphing)
Graph . 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 .
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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