Difference between revisions of "2024 AIME I Answer Key"
(→Note) |
|||
| (7 intermediate revisions by 2 users not shown) | |||
| Line 21: | Line 21: | ||
11. 371 | 11. 371 | ||
| − | 12. 384 ( | + | 12. <s>384</s> '''385''' (See below) |
13. 110 | 13. 110 | ||
| Line 29: | Line 29: | ||
15. 721 | 15. 721 | ||
| − | == | + | ==Problem 12== |
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 | 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 | ||
| + | |||
| + | Response to Furaken from Steven Chen (Professor Chen Education Palace, www.professorchenedu.com): | ||
| + | |||
| + | 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) = \left( 1 - 8.9 \cdot 10^{-6}, 1 - 2.2 \cdot 10^{-4} \right)</math>. | ||
| + | |||
| + | Fourth, you can see that the above solution is very close to <math>\left( 1 , 1 \right)</math>. This is why many people cannot realize its exitance, both using or without using any graphical calculator. | ||
| + | |||
| + | To further convince you my analysis above is correct, I plotted two functions near <math>\left( 1, 1 \right)</math>. As you can see, there are indeed two intersecting points. The non-<math>\left( 1,1 \right)</math> solution is consistent with my calculation above: | ||
| + | |||
| + | https://artofproblemsolving.com/wiki/index.php/File:2024_AIME_I_Problem_12,_two_solutions_near_(1,1).png | ||
| + | |||
| + | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
| + | |||
| + | |||
| + | |||
| + | When you draw the curves on a reasonable scale, the "width" of the gap between the two curves from (1,1) to that other intersection is less than the wavelength of visible light. No wonder most humans don’t see it —Furaken | ||
Latest revision as of 02:57, 4 February 2024
1. 204
2. 025
3. 809
4. 116
5. 104
6. 294
7. 540
8. 197
9. 480
10. 113
11. 371
12. 384 385 (See below)
13. 110
14. 104
15. 721
Problem 12
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 Solution page's answer has already been updated to 385. --Furaken
Response to Furaken from Steven Chen (Professor Chen Education Palace, www.professorchenedu.com):
You are correct. The number of intersecting points is 385, not 384. The controversy is whether there is one more solution near 
(beyond the solution at this point). The correct answer is YES, there is one more solution. This point is very very close to , but not the same as .
First, if you use any plotting tool and zoom in the region near , 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 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 
and 
. If such a second solution exists, then we should get a solution 
that are strictly positive and very close to 0. Since I restrict to small 
and 
, 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 and are sufficiently small, we can use approximations that 
and 
. 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 is 
.
Fourth, you can see that the above solution is very close to 
. This is why many people cannot realize its exitance, both using or without using any graphical calculator.
To further convince you my analysis above is correct, I plotted two functions near . As you can see, there are indeed two intersecting points. The non-
solution is consistent with my calculation above:
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
When you draw the curves on a reasonable scale, the "width" of the gap between the two curves from (1,1) to that other intersection is less than the wavelength of visible light. No wonder most humans don’t see it —Furaken
.png){kind=link}