|
|
| (8 intermediate revisions by 4 users not shown) |
| Line 1: |
Line 1: |
| − | 1. 204
| + | Return to [[2024 AIME I]] ([[2024 AIME I Problems]]) |
| | | | |
| − | 2. 025
| + | # 204 |
| | + | # 025 |
| | + | # 809 |
| | + | # 116 |
| | + | # 104 |
| | + | # 294 |
| | + | # 540 |
| | + | # 197 |
| | + | # 480 |
| | + | # 113 |
| | + | # 371 |
| | + | # 385 |
| | + | # 110 |
| | + | # 104 |
| | + | # 721 |
| | | | |
| − | 3. 809
| |
| | | | |
| − | 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]]}} |
| − | | |
| − | 5. 104
| |
| − | | |
| − | 6. 294
| |
| − | | |
| − | 7. 540
| |
| − | | |
| − | 8. 197
| |
| − | | |
| − | 9. 480
| |
| − | | |
| − | 10. 113
| |
| − | | |
| − | 11. 371
| |
| − | | |
| − | 12. 384 (385?)
| |
| − | | |
| − | 13. 110
| |
| − | | |
| − | 14. 104
| |
| − | | |
| − | 15. 721
| |
| − | | |
| − | ==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
| |
| − | | |
| − | \textbf{Response to Furaken: }
| |
| − | | |
| − | 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>.
| |
| − | | |
| − | 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.
| |
| − | | |
| − | 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.
| |
| − | | |
| − | 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} , 1 - \frac{1}{8 \cdot 18 \pi^3} \right)</math>.
| |
