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| + | Return to [[2024 AIME I]] ([[2024 AIME I Problems]]) |
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| − | 2. 025
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| | + | # 385 |
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| − | 3. 809
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| − | 4. 116
| + | {{AIME box|year=2024|n=I|before=[[2023 AIME II Answer Key|2023 AIME II]]|after=[[2025 AIME I Answer Key|2025 AIME I]]}} |
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| − | 7. 540
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| − | 12. 384 (385?)
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| − | ==Note== | |
| − | Both me and another person on AoPS got 385 for problem 12. It seems that 385 is the correct answer, and someone verified it by Desmos. The [[2024_AIME_I_Problems/Problem_12|Solution page]]'s answer has already been updated to 385. --Furaken
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| − | Response to Furaken:
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| − | You are correct. The number of intersecting points is 385, not 384. The controversy is whether there is one more solution near <math>\left( 1, 1 \right)</math> (beyond the solution at this point). The correct answer is YES, there is one more solution. This point is very very close to <math>\left( 1, 1 \right)</math>, but not the same as <math>\left( 1, 1 \right)</math>.
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| − | First, if you use any plotting tool and zoom in the region near <math>\left( 1, 1 \right)</math>, you can see two distinct solutions.
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| − | Second, a more realistic thing is how could we find this second solution in the contest since we were not allowed to use any graphing calculator. On the page [[2024_AIME_I_Problems/Problem_12|Solution page]], I provided my solution (both my text solution and video solution) to answer this question. The key idea is as follows. I denote <math>x' = 1 - x</math> and <math>y' = 1 - y</math>. If such a second solution exists, then we should get a solution <math>\left( x', y' \right)</math> that are strictly positive and very close to 0. Since I restrict to small <math>x'</math> and <math>y'</math>, I can get closed forms without any absolution signs in the two given functions. After this step, we still need to solve a system of two non-trivial equations. Again, because <math>x'</math> and <math>y'</math> are sufficiently small, we can use approximations that <math>\sin \theta \approx \theta</math> and <math>\cos \theta \approx 1 - \frac{\theta^2}{2}</math>. This reduces two complicated equations to one linear and one quadratic equation. I can then easily find a non-zero solution and even get the closed form.
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| − | Third, based on my above analysis, the closed-form (up to the second order approximation) of the second solution near <math>\left( 1, 1 \right)</math> is <math>\left( 1 - \frac{1}{8^2 \cdot 18 \pi^4} , \frac{1}{8 \cdot 18 \pi^3} \right)</math>.
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