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Difference between revisions of "User:Cxsmi"

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==Some Problems I Wrote==
 
==Some Problems I Wrote==
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All problems have been moved to [https://www.dropbox.com/scl/fi/4m0xzo6pyry15m1y5ustn/Problem-Collection.pdf?rlkey=d8o7r0lezc34ej11g1otioplp&dl=0 this page].
  
 
==Solutions==
 
==Solutions==

Revision as of 17:52, 9 February 2024

Welcome

Welcome to my user page! Here, I compile lists of solutions I write on this wiki and post original problems I come up with for the world to see. Please add one to the visit count below if this is your first time on my user page (I got the idea from MRENTHUSIASM). Thanks for visiting my user page, and enjoy your stay!

Visit Count

3

About Me

Hi! I'm just another guy who happens to enjoy math. I often pop onto the AOPS wiki and look for problems to solve, and I sometimes even write solutions for them! I've starred ⭐ a few of my favorite solutions below; please feel free to take a look at any of them.

Solutions

AIME

  1. 1987 AIME Problem 11 Solution 3 ⭐
  2. 2012 AIME I Problem 2 Solution 3
  3. 2024 AIME I Problem 6 Solution 5 ⭐

AMC 8

  1. 2012 AMC 8 Problem 19 Solution 6
  2. 2002 AMC 8 Problem 17 Solution 3
  3. 2007 AMC 8 Problem 20 Solution 8
  4. 2018 AMC 8 Problem 23 Solution 5 ⭐
  5. 2016 AMC 8 Problem 13 Solution 3
  6. 2017 AMC 8 Problem 9 Solution 2
  7. 2012 AMC 8 Problem 20 Solution 7
  8. 2010 AMC 8 Problem 25 Solution 3
  9. 2024 AMC 8 Problem 11 Solution 3

AJHSME

  1. 1997 AJHSME Problem 22 Solution 1
  2. 1985 AJHSME Problem 1 Solution 2
  3. 1985 AJHSME Problem 24 Solution 2 ⭐
  4. 1985 AJHSME Problem 2 Solution 5

AHSME

  1. 1950 AHSME Problem 40 Solution 2
  2. 1950 AHSME Problem 41 Solution 2
  3. 1972 AHSME Problem 16 Solution 2
  4. 1950 AHSME Problem 45 Solution 3
  5. 1956 AHSME Problem 23 Solution 1

AMC 10/12

  1. 2021 Fall AMC 12B Problem 12 Solution 6
  2. 2021 Fall AMC 10B Problem 1/AMC 12B Problem 1 Solution 4
  3. 2003 AMC 10A Problem 17 Solution 2

Significant Problems

Here are some problems that, to me, have been significant on my math journey. This section is mainly for myself, but please please feel free to look at the problems if you're interested. Any referenced "difficulties" are from this page.

  1. 2017 AMC 10A Problem 19 - First AMC 10 Solution of Difficulty 2 or Higher
  2. 2007 AMC 8 Problem 25 - First AMC 8 Final Five Solution
  3. 1984 AIME Problem 1 - First AIME Solution
  4. 2005 AMC 12B Problem 16 - First AMC 12 Solution of Difficulty 2.5 or Higher
  5. 2016 AMC 10A Problem 21 - First AMC 10 Final Five Solution
  6. 1987 AIME Problem 11 - First AIME Solution of Difficulty 4 or Higher; First AIME Solution of Difficulty 5 or Higher
  7. 2017 AMC 12A Problem 23 - First AMC 12 Final Five Solution
  8. 2001 AIME I Problem 14 - First AIME Solution of Difficulty 6 or Higher

Some Problems I Wrote

All problems have been moved to this page.

Solutions

These are solutions for the problems above. $\textbf{Scroll down at your own risk!}$

Problem 1 Solutions

Solution 1

We split the condition into two separate conditions, as listed below.


$\frac{20^{23}}{n} + \frac{24^{23}}{n} = \lfloor{\frac{20^{23}}{n}+\frac{24^{23}}{n}}\rfloor$

$\frac{20^{23}}{n} + \frac{24^{23}}{n} \neq \lfloor{\frac{20^{23}}{n}}\rfloor + \lfloor{\frac{24^{23}}{n}}\rfloor$


Rearranging the conditions, we find that


$\frac{20^{23}+24^{23}}{n} - \lfloor \frac{20^{23}+24^{23}}{n}\rfloor = 0$

$(\frac{20^{23}}{n} - \lfloor \frac{20^{23}}{n} \rfloor) + (\frac{24^{23}}{n} - \lfloor \frac{24^{23}}{n} \rfloor) \neq 0$


Recalling that $x - \lfloor {x} \rfloor = frac(x)$ where $frac(x)$ represents the fractional part of $x$, we rewrite once more.


$frac(\frac{20^{23}+24^{23}}{n}) = 0$ $\textbf{(1)}$

$frac(\frac{20^{23}}{n}) + frac(\frac{24^{23}}{n}) \neq 0$ $\textbf{(2)}$


We now gain some valuable insight. From $\textbf{(1)}$, we find that $n$ must divide $20^{23} + 24^{23}$. From $\textbf{(2)}$, we find that $n$ cannot divide both $20^{23}$ and $24^{23}$. It is impossible for $n$ to divide only $1$ of $20^{23}$ and $24^{23}$, as this would make $\textbf{(1)}$ false. It must be that $n$ divides neither $20^{23}$ nor $24^{23}$. For both this and $\textbf{(1)}$ to be true simultaneously, we must have that if $20^{23} \equiv a \bmod n$, then $24^{23} \equiv -a \bmod n$. By inspection, this occurs when $n = 22$.We now test the factors of $22$ to see if we can find a smaller value. As both $20^{23}$ and $24^{23}$ are congruent to $0$ mod $2$, $n = 2$ is not a valid solution. However, with $n = 11$, $20^{23} \equiv (-2)^{23} \bmod 11$, while $24^{23} \equiv 2^{23} \bmod 11$. Clearly, $-(-2)^{23} = 2^{23}$, so our final answer is $\boxed{011}$.

Problem 2 Solutions

Solution 1

First, we note the statements below.


$f(n) = f(n-1) + f(n-2) + f(n-3) + ... + f(n - 2023) + f(n-2024)$ $\textbf{(1)}$ $f(n-1) = f(n-2) + f(n-3) + f(n-4) + ... + f(n - 2024) + f(n - 2025)$ $\textbf{(2)}$


We notice that most of the terms telescope if we subtract $\textbf{(2)}$ from $\textbf{(1)}$. \[f(n) - f(n-1) = f(n-1) - f(n-2025) \implies f(n)=2f(n-1) - f(n-2025)\]. By adding $f(n-1)$ to both sides, we find that $f(n) = 2f(n-1) - f(n-2025)$. From here, we can consider $f(4049)$. We note that for all $2026 \leq n \leq 4049$, the second part of our definition (the $f(n-2025)$ term) is equal to one. From here, we can list out a few definitions for $f(4049)$ using our formula.


$f(4049) = 2f(4048)-1 = 2^1f(4048) - (2^1-1)$

$= 2(2f(4047)-1)-1 = 4f(4047) - 3 = 2^2f(4047) - (2^2 - 1)$

$= 2(2(2f(4046)-1)-1)-1 = 8f(4046) - 7 = 2^3f(4046) - (2^3 - 1)$

$= 2(2(2(2f(4045)-1)-1)-1)-1 = 16f(4045) - 15 = 2^4f(4045) - (2^4-1)$


It appears that on the interval $2025 \leq n \leq 4049$, $f(4049) = 2^{4049-n}f(n) - (2^{4049-n} - 1) = 2^{4049-n}(f(n)-1) + 1$. ($2025$ is the upper bound because if we tried to calculate $f(2025)$ using our alternate definition, we'd get $2f(2024) - f(0)$, and $f(0)$ is undefined.) We attempt to maximize the value of $2^{4049-n}$, as this will make finding the prime factorization easier. This value is maximized when we choose the lower bound. We take $n = 2025$; then \[f(4049) = 2^{2024}(f(2025)-1)+1\] and \[f(4049) - 1 = 2^{2024}(f(2025)-1)\]. From our original definition, $f(2025) = 2024$. Thus, \[f(4049)-1=2023 \cdot 2^{2024}\]. We prime factorize $2023$; by trying divisibility tests until one works, we find that $2023 = 7 \cdot 289$. Recalling that $289 = 17^2$, we find that \[f(4049)-1 = 2^{2024} \cdot 7 \cdot 17^{2}\]. Then the desired sum is \[2 \cdot 2024 + 7 \cdot 1 + 17 \cdot 2 = 4089\], and the remainder when this is divided by $1000$ is $\boxed{089}$.

Problem 3 Solutions

Solution 1

[asy] unitsize(1inch); draw((0,0)--(0,2)); draw((0,2)--(2,0)); draw((2,0)--(0,0)); draw(circle((0.586,0.586),0.586)); draw((0,0)--(0,1.172),red); draw((0,1.172)--(1.172,1.172)); draw((1.172,1.172)--(1.172,0)); draw((1.172,0)--(0,0),red); draw((0,1.172)--(0.828,1.172),red); draw((0.828,1.172)--(1.172,0.828),red); draw((1.172,0.828)--(1.172,0),red); draw((0,0.1)--(0.1,0.1)); draw((0.1,0.1)--(0.1,0)); label("$A$",(0,2.1)); label("$B$",(0,-0.1)); label("$C$",(2,-0.1)); label("$2024$",(-0.2,1)); label("$2024$",(1,-0.2)); label("$D$",(1.272, 1.272)); label("$E$",(0.928,1.272)); label("$F$",(1.272,0.928)); [/asy]

We can use coordinate bashing. We may assume the legs of the triangle have side length $2$ for ease of computation, and we can multiply by $1012$ on our final answer to get our result. We place $B$ at the origin; in this case, $A$ has coordinates $(0,2)$, $B$ has coordinates $(0, 0)$, and $C$ has coordinates $(2, 0)$. Then line segment $\overline{\rm AC}$ is on the line that has slope $\frac{0-2}{2-0} = -1$ and $y$-intercept $2$, so the line's equation is $y = 2 - x$. Using the distance formula on $A$ and $C$ or noting that $\triangle ABC$ is a $45-45-90$ triangle, we find that the length of $\overline{\rm AC}$ is $2\sqrt{2}$. The area of $\triangle ABC$ is $\frac{1}{2}(2)(2) = 2$, and its semiperimeter is $\frac{2+2+2\sqrt{2}}{2} = 2 + \sqrt{2}$. Since $rs = A$ or $r = \frac{A}{s}$, the inradius of $\triangle ABC$ is $\frac{2}{2+\sqrt{2}} = 2-\sqrt{2}$. Also, because the side length of the square circumscribed around $\triangle ABC$ is equal to the length of the diameter of the incircle of $\triangle ABC$, the square has side length $4 - 2\sqrt{2}$. From here, we can find that the vertices of the square are, starting at the bottom left and going clockwise, $(0,0)$, $(0, 4 - 2\sqrt{2})$, $(4 - 2\sqrt{2}, 4 - 2\sqrt{2})$, and $(4 - 2\sqrt{2}, 0)$. We label the top-right vertex of the square as $D$, the leftmost of the two intersection points of the square with the triangle as $E$, and the other intersection point as $F$. $D$ is directly above $F$, so they must share an x-coordinate. Since $F$ lies on the line $y = 2 - x$, we take $x = 4 - 2\sqrt{2}$ to find that $F$ has coordinates $(4 - 2\sqrt{2}, 2 - (4 - 2\sqrt{2}))$ or $(4 - 2\sqrt{2}, 2\sqrt{2} - 2)$. Using similar logic, $E$ has coordinates $(2\sqrt{2} - 2, 4 - 2\sqrt{2})$. The distance from $D$ to $F$ is just the positive difference of their y-coordinates, which is $4 - 2\sqrt{2} - (2\sqrt{2} - 2) = 6 - 4\sqrt{2}$. Similarly, the distance from $D$ to $E$ is the same. Since $\triangle DEF$ is an isosceles right triangle, the length of $\overline{\rm EF}$ is $\sqrt{2}$ times the length of one of the legs, which happens to be $6\sqrt{2} - 8$. The perimeter of the diamond is equal to the perimeter of the square minus the lengths of $\overline{\rm ED}$ and $\overline{\rm DF}$ plus the length of $\overline{\rm EF}$, which is $4(4 - 2\sqrt{2}) - 2(6 - 4\sqrt{2}) + (6\sqrt{2} - 8) = 6\sqrt{2} - 4$. We multiply this by $1012$ to scale the triangle to a side length of $2024$, resulting in a perimeter of $6072\sqrt{2} - 4048$. Thus, $a + b + c = 10122$, and the remainder when this is divided by $1000$ is $\boxed{122}$.

Problem 4 Solutions

I have not written a solution yet, but the answer is $\boxed{976}$.

Problem 5 Solutions

I have not written a solution yet, but the answer is $\boxed{198}$.

Problem 6 Solutions

I have not written a solution yet.

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