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User:Cxsmi

Revision as of 22:39, 28 January 2024 by Cxsmi (talk | contribs) (→‎Problem 4)

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. Thanks for visiting my user page, and enjoy your stay!

Solutions

AIME

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

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

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

AMC 12

  1. 2021 AMC 12B Problem 12 Solution 6

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. 2009 HMMT February Algebra Problem 1 - First February HMMT Solution
  9. 2001 AIME I Problem 14 - First AIME Solution of Difficulty 6 or Higher

Some Problems I Wrote

I enjoy writing problems when I see concepts that interest me. I've written a few below; please feel free to solve them! Also, feel free to add solutions or change any mistakes you see. The questions should all be roughly low to mid-AIME difficulty; however, I am not a very good judge of difficulty, so they may be easier or harder than they should be. All problems have integer answers with values between $001$ and $999$, just like in the AIME.

Problem 1

Find the least positive integer $n$ that satisfies the following. The notation $\lfloor{x}\rfloor$ represents the greatest integer less than or equal to $x$.

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

Problem 2

Bob has received a test. He must match a list of $2024$ words to a list of $2024$ definitions such that each word matches to exactly one definition and vice versa. He does not know these words, so he will guess randomly. If he does so, he will expect to get $n$ matches. Find the remainder when $n$ is divided by $1000$.

Problem 3

A sequence $f(n)$ is defined recursively as $f(1) = f(2) = f(3) = ... = f(2024) = 1$ and

$f(n) = \sum_{k=1} ^{2024} f(n - k)$

for $n \geq 2025$. The value of $f(4049)$ can be written in the form $a(2)^b+1$ for positive integers $a$ and $b$ such that $b$ is maximized. What is the remainder when $a + b$ is divided by $1000$?

Problem 4

A simple pentagon is created by connecting the points at which a square circumscribed around the incircle of an isosceles right triangle $\triangle ABC$ with leg length $2024$ intersects $\triangle ABC$ itself. The perimeter of this pentagon is expressible as $a\sqrt{b}-c$, where $a$, $b$, and $c$ are integers, and $c$ is not divisible by the square of any prime. What is the remainder when $a + b + c$ is divided by $1000$? Note that a simple polygon is a non-self-intersecting polygon.

[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),blue); draw((0,1.172)--(1.172,1.172)); draw((1.172,1.172)--(1.172,0)); draw((1.172,0)--(0,0),blue); draw((0,1.172)--(0.828,1.172),blue); draw((0.828,1.172)--(1.172,0.828),blue); draw((1.172,0.828)--(1.172,0),blue); draw((0,0.1)--(0.1,0.1)); draw((0.1,0.1)--(0.1,0)); label("$A$",(0,2.2)); label("$B$",(0,-0.2)); label("$C$",(2,-0.2)); label("$2024$",(-0.2,1)); label("$2024$",(1,-0.2)); [/asy]

Hints

These hints will suggest topics that are relevant to my solutions to these problems. If you get stuck, try thinking along the routes of these concepts.

Problem 1

Remainders

Problem 2

Recursion

Problem 3

Telescoping

Problem 4

Coordinate Bashing

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 (Meta-Solving)

We can consider smaller cases. If there is $1$ object to match, Bob will always make exactly $1$ correct match, so he will expect to make $1$ match. If there are $2$ objects, Bob can either match both correctly for a total of $2$ matches or match both incorrectly for a total of $0$ matches. Each of these has a $\frac{1}{2}$ chance of happening, so again Bob will expect to make $1$ match. If there are $3$ objects, we can list out each combination. Suppose we are matching ${1, 2, 3}$ to itself, and the "solution" is the set of ordered pairs ${(1,1),(2,2), (3,3)}$. The possibilities are listed below:


${(1,1), (2,2), (3,3)}$ (3 correct matches)

${(1,2), (2,1), (3,3)}$ (1 correct match)

${(1,3), (2,2), (3,1)}$ (1 correct match)

${(1,1), (2,3), (3,2)}$ (1 correct match)

${(1,2), (2,3), (3,1)}$ (0 correct matches)

${(1,3), (2,1), (3,2)}$ (0 correct matches)


The expected value is $\frac{1}{6} \cdot (3+1+1+1) = 1$. The expected value has remained at 1 for the first three cases, so we can safely assume that it will remain at $1$ for any positive integer number of items to match. Then Bob can expect to make $1$ match, and the remainder when this is divided by $1000$ is $\boxed{001}$.


Note: This is indeed the correct answer, and it can be found more rigorously. Maybe that solution will come in the future :)

Problem 3 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 $\textbf{(1)}$.


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

$= f(n-1) + (f(n-2) - f(n-2)) + (f(n-3) - f(n-3)) + ... + (f(n-2024) - f(n-2024)) - f(n - 2025)$

$= f(n-1) - f(n-2025)$


By adding $f(n-1)$ to both sides, we find that $f(n) = 2f(n-1) - f(n-2025)$. We can verify by finding $f(2026)$; from the original definition, we find that $f(2025) = 2024$, and $f(2026) = 4047$. From our definition, we find that $f(2026) = 2 \cdot f(2025) - f(1) = 2 \cdot 2024 - 1 = 4047$. Of course, this doesn't guarantee that our definition is indeed correct, but it gives us additional verification to our algebraic method. 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 lower 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.) Clearly, to maximize $4049 - n$ (to maximize $b$ in the problem), we choose the lower bound. Then we get $f(4049) = 2^{2024}(f(2025)-1) + 1 = 2^{2024}(2024-1) + 1 = 2^{2024}(2023) + 1$, so $a + b = 4047$. The remainder when this is divided by $1000$ is $\boxed{047}$.

Problem 4 Solutions

[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),blue); draw((0,1.172)--(1.172,1.172)); draw((1.172,1.172)--(1.172,0)); draw((1.172,0)--(0,0),blue); draw((0,1.172)--(0.828,1.172),blue); draw((0.828,1.172)--(1.172,0.828),blue); draw((1.172,0.828)--(1.172,0),blue); draw((0,0.1)--(0.1,0.1)); draw((0.1,0.1)--(0.1,0)); label("$A$",(0,2.2)); label("$B$",(0,-0.2)); label("$C$",(2.2,-0.2)); label("$2024$",(-0.2,1)); label("$2024$",(1,-0.2)); [/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 formula for the coordinates of the incenter of a triangle, we find that the incenter of this triangle is at $(\frac{4}{2+\sqrt{2}},\frac{4}{2+\sqrt{2}})$ or $(2-\sqrt{2},2-\sqrt{2})$. 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 in 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 pentagon 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 $6072\sqrt{2} - 4048$. Since $a = 6072$, $b = 2$, and $c = 4048$, $a + b + c = 10122$, and the remainder when this is divided by $1000$ is $\boxed{122}$.

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