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Difference between revisions of "2024 AIME I Answer Key"

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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
  
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\textbf{Response to Furaken: }
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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>.  
 
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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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.  
 
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} , 1 - \frac{1}{8 \cdot 18 \pi^3} \right)</math>.
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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} , 1 - \frac{1}{8 \cdot 18 \pi^3} \right) = \left( 1 - 8.9 \cdot 10^{-6}, 1 - 2.2 \cdot 10^{-4} \right)</math>.
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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.
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Revision as of 00:49, 4 February 2024

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3. 809

4. 116

5. 104

6. 294

7. 540

8. 197

9. 480

10. 113

11. 371

12. 384 (385?)

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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 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 $\left( 1, 1 \right)$ (beyond the solution at this point). The correct answer is YES, there is one more solution. This point is very very close to $\left( 1, 1 \right)$, but not the same as $\left( 1, 1 \right)$.

First, if you use any plotting tool and zoom in the region near $\left( 1, 1 \right)$, 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 $x' = 1 - x$ and $y' = 1 - y$. If such a second solution exists, then we should get a solution $\left( x', y' \right)$ that are strictly positive and very close to 0. Since I restrict to small $x'$ and $y'$, 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 $x'$ and $y'$ are sufficiently small, we can use approximations that $\sin \theta \approx \theta$ and $\cos \theta \approx 1 - \frac{\theta^2}{2}$. 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 $\left( 1, 1 \right)$ is $\left( 1 - \frac{1}{8^2 \cdot 18 \pi^4} , 1 - \frac{1}{8 \cdot 18 \pi^3} \right) = \left( 1 - 8.9 \cdot 10^{-6}, 1 - 2.2 \cdot 10^{-4} \right)$.

Fourth, you can see that the above solution is very close to $\left( 1 , 1 \right)$. This is why many people cannot realize its exitance, both using or without using any graphical calculator.

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

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