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Difference between revisions of "User:Cxsmi"

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==Visit Count==
 
==Visit Count==
  
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</font></div><center><font size="100px">3</font></center>
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</font></div><center><font size="100px">21</font></center>
  
 
==About Me==
 
==About Me==
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# [https://artofproblemsolving.com/wiki/index.php?title=1987_AIME_Problems/Problem_11#Solution_3 1987 AIME Problem 11 Solution 3 ⭐]
 
# [https://artofproblemsolving.com/wiki/index.php?title=1987_AIME_Problems/Problem_11#Solution_3 1987 AIME Problem 11 Solution 3 ⭐]
 
# [https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_2#Solution_3 2012 AIME I Problem 2 Solution 3]
 
# [https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_2#Solution_3 2012 AIME I Problem 2 Solution 3]
−
# [https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_6#Solution_5 2024 AIME I Problem 6 Solution 5 ⭐]
+
# [https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_6#Solution_5 2024 AIME I Problem 6 Solution 5]
 +
# [https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_1#Solution_4 2000 AIME II Problem 1 Solution 4]
 +
# [https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_12#Solution_9 1985 AIME Problem 12 Solution 9 ⭐]
 +
# [https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_8#Solution_7 2012 AIME I Problem 8 Solution 7]
 +
# [https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_14#Solution_4 1987 AIME Problem 14 Solution 4]
 +
# [https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_10#Solution_5 1988 AIME Problem 10 Solution 5]
  
 
===AMC 8===
 
===AMC 8===
Line 22: Line 27:
 
# [https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_17#Solution_3 2002 AMC 8 Problem 17 Solution 3]
 
# [https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_17#Solution_3 2002 AMC 8 Problem 17 Solution 3]
 
# [https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_20#Solution_8_.28Answer_Choices.29 2007 AMC 8 Problem 20 Solution 8]
 
# [https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_20#Solution_8_.28Answer_Choices.29 2007 AMC 8 Problem 20 Solution 8]
−
# [https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_23#Solution_5 2018 AMC 8 Problem 23 Solution 5 ⭐]
+
# [https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_23#Solution_5_.28Casework.29 2018 AMC 8 Problem 23 Solution 5 ⭐]
 
# [https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_13#Solution_3_.28Casework.29 2016 AMC 8 Problem 13 Solution 3]
 
# [https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_13#Solution_3_.28Casework.29 2016 AMC 8 Problem 13 Solution 3]
 
# [https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_9#Solution_2 2017 AMC 8 Problem 9 Solution 2]
 
# [https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_9#Solution_2 2017 AMC 8 Problem 9 Solution 2]
 
# [https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_20#Solution_7 2012 AMC 8 Problem 20 Solution 7]
 
# [https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_20#Solution_7 2012 AMC 8 Problem 20 Solution 7]
 
# [https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_25#Solution_3_.28Recursion.29 2010 AMC 8 Problem 25 Solution 3]
 
# [https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_25#Solution_3_.28Recursion.29 2010 AMC 8 Problem 25 Solution 3]
−
# [https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_11#Solution_3 2024 AMC 8 Problem 11 Solution 3]
+
# [https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_11#Solution_3 2024 AMC 8 Problem 11 Solution 3 ⭐]
 +
# [https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_18#Solution_4 2024 AMC 8 Problem 18 Solution 4]
 +
# [https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_11#Solution_4 2024 AMC 8 Problem 11 Solution 4]
 +
# [https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_15#Solution_2 2019 AMC 8 Problem 15 Solution 2 ⭐]
  
 
===AJHSME===
 
===AJHSME===
Line 43: Line 51:
 
# [https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_45#Solution_3 1950 AHSME Problem 45 Solution 3]
 
# [https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_45#Solution_3 1950 AHSME Problem 45 Solution 3]
 
# [https://artofproblemsolving.com/wiki/index.php/1956_AHSME_Problems/Problem_23#Solution 1956 AHSME Problem 23 Solution 1]
 
# [https://artofproblemsolving.com/wiki/index.php/1956_AHSME_Problems/Problem_23#Solution 1956 AHSME Problem 23 Solution 1]
 +
# [https://artofproblemsolving.com/wiki/index.php/1982_AHSME_Problems/Problem_13#Solution_2 1982 AHSME Problem 13 Solution 2]
 +
# [https://artofproblemsolving.com/wiki/index.php/1974_AHSME_Problems/Problem_4# 1974 AHSME Problem 4 Solution 2]
  
 
===AMC 10/12===
 
===AMC 10/12===
Line 49: Line 59:
 
#[https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_1#Solution_4_.28Brute_Force.29 2021 Fall AMC 10B Problem 1/AMC 12B Problem 1 Solution 4]
 
#[https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_1#Solution_4_.28Brute_Force.29 2021 Fall AMC 10B Problem 1/AMC 12B Problem 1 Solution 4]
 
#[https://artofproblemsolving.com/wiki/index.php/2003_AMC_10A_Problems/Problem_17#Solution_2 2003 AMC 10A Problem 17 Solution 2]
 
#[https://artofproblemsolving.com/wiki/index.php/2003_AMC_10A_Problems/Problem_17#Solution_2 2003 AMC 10A Problem 17 Solution 2]
 +
#[https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_8#Solution_3 2015 AMC 12B Problem 8 Solution 3]
 +
#[https://artofproblemsolving.com/wiki/index.php/2003_AMC_12B_Problems/Problem_17#Solution_4 2003 AMC 12B Problem 17 Solution 4]
 +
#[https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_18#Solution_6 2020 AMC 10A Problem 18 Solution 6]
 +
#[https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_13#Solution_4 2021 AMC 10B Problem 13 Solution 4]
 +
#[https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_23#Solution_4 2007 AMC 10B Problem 23 Solution 4]
 +
#[https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_24#Solution_3 2005 AMC 10B Problem 24 Solution 3 ⭐]
 +
#[https://artofproblemsolving.com/wiki/index.php/2008_AMC_12B_Problems/Problem_24#Solution_4_.28Pattern_Observation.29 2008 AMC 12B Problem 24 Solution 4 ⭐]
 +
#[https://artofproblemsolving.com/wiki/index.php/2023_AMC_10B_Problems/Problem_24#Solution_2 2023 AMC 10B Problem 24 Solution 2 ⭐]
 +
#[https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_9#Solution_5_.28Desperate.29 2023 AMC 10B Problem 13/AMC 12B Problem 9 Solution 5]
 +
#[https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_5 2007 AMC 10B Problem 5 Solution 2]
  
 
==Significant Problems==
 
==Significant Problems==
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# [https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_23 2017 AMC 12A Problem 23 - First AMC 12 Final Five Solution]
 
# [https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_23 2017 AMC 12A Problem 23 - First AMC 12 Final Five Solution]
 
# [https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_14 2001 AIME I Problem 14 - First AIME Solution of Difficulty 6 or Higher]
 
# [https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_14 2001 AIME I Problem 14 - First AIME Solution of Difficulty 6 or Higher]
 +
# [https://artofproblemsolving.com/wiki/index.php/1984_USAMO_Problems/Problem_1 1984 USAMO Problem 1 - First USAMO Solution]
 +
# [https://artofproblemsolving.com/wiki/index.php/2011_USAJMO_Problems/Problem_1 2011 USAJMO Problem 1 - First USAJMO Solution]
 +
 +
==Other Major Contributions==
 +
# [https://artofproblemsolving.com/wiki/index.php/Titu%27s_Lemma Titu's Lemma (all examples and problems)]
  
 
==Some Problems I Wrote==
 
==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 <math>000</math> and <math>999</math>, just like in the AIME.
+
All problems and solutions have been moved to [https://www.dropbox.com/scl/fi/v947aqcm4cvnrcqys8jzj/Problem_Collection-1.pdf?rlkey=7olfeq431g81rxoc89q51pavb&st=iscedhjf&dl=0 this page].
−
 
 
−
===Problem 1===
 
−
Find the least positive integer <math>n</math> that satisfies <cmath>\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</cmath>. The notation <math>\lfloor{x}\rfloor</math> represents the greatest integer less than or equal to <math>x</math>.
 
−
 
 
−
===Problem 2===
 
−
Bob has received a test. He must match a list of <math>2024</math> words to a list of <math>2024</math> 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 <math>n</math> matches. Find the remainder when <math>n</math> is divided by <math>1000</math>.
 
−
 
 
−
===Problem 3===
 
−
A sequence <math>f(n)</math> is defined recursively as <math>f(1) = f(2) = f(3) = ... = f(2024) = 1</math> and
 
−
 
 
−
<math>f(n) = \sum_{k=1} ^{2024} f(n - k)</math>
 
−
 
 
−
for <math>n \geq 2025</math>. <math>f(4049)-1</math> has prime factorization <math>p_1^{e_1} p_2^{e_2} \ldots p_n^{e_n}</math>. Find the remainder when <math>\sum_{i=1}^{n} p_i^{e_i}</math> is divided by <math>1000</math>.
 
−
 
 
−
===Problem 4===
 
−
A diamond is created by connecting the points at which a square circumscribed around the incircle of an isosceles right triangle <math>\triangle ABC</math> intersects <math>\triangle ABC</math> itself. <math>\triangle ABC</math> has leg length <math>2024</math>. The perimeter of this diamond is expressible as <math>a\sqrt{b}-c</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are integers, and <math>c</math> is not divisible by the square of any prime. What is the remainder when <math>a + b + c</math> is divided by <math>1000</math>?
 
−
 
 
−
<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));
 
−
</asy>
 
−
 
 
−
===Problem 5===
 
−
Compute the last three digits of <math>\underbrace{2024^{2024^{\ldots^{2024}}}}_\text{2024 2024s}</math>.
 
−
 
 
−
===Problem 6===
 
−
Let <math>P</math> be the set of polynomials of degree <math>2024</math> whose coefficients are all elements of the set <math>\{0, 1, 2, ..., 2023, 2024\}</math>. Let <math>H</math> be the set of all polynomials in <math>P</math> for which the harmonic mean of the polynomial's roots is not rational. The cardinality of <math>H</math> is expressible as <math>a \cdot 2025^{b}</math> such that <math>a</math> has no factors of <math>2025</math>. What is the remainder when <math>a + b</math> is divided by <math>1000</math>?
 
−
 
 
−
==Solutions==
 
−
These are solutions for the problems above. <math>\textbf{Scroll down at your own risk!}</math>
 
−
 
 
−
===Problem 1 Solutions===
 
−
====Solution 1====
 
−
We split the condition into two separate conditions, as listed below.
 
−
 
 
−
 
 
−
<math>\frac{20^{23}}{n} + \frac{24^{23}}{n} = \lfloor{\frac{20^{23}}{n}+\frac{24^{23}}{n}}\rfloor</math>
 
−
 
 
−
<math>\frac{20^{23}}{n} + \frac{24^{23}}{n} \neq \lfloor{\frac{20^{23}}{n}}\rfloor + \lfloor{\frac{24^{23}}{n}}\rfloor</math>
 
−
 
 
−
 
 
−
Rearranging the conditions, we find that
 
−
 
 
−
 
 
−
<math>\frac{20^{23}+24^{23}}{n} - \lfloor \frac{20^{23}+24^{23}}{n}\rfloor = 0</math>
 
−
 
 
−
<math>(\frac{20^{23}}{n} - \lfloor \frac{20^{23}}{n} \rfloor) + (\frac{24^{23}}{n} - \lfloor \frac{24^{23}}{n} \rfloor) \neq 0</math>
 
−
 
 
−
 
 
−
Recalling that <math>x - \lfloor {x} \rfloor = frac(x)</math> where <math>frac(x)</math> represents the fractional part of <math>x</math>, we rewrite once more.
 
−
 
 
−
 
 
−
<math>frac(\frac{20^{23}+24^{23}}{n}) = 0</math> <math>\textbf{(1)}</math>
 
−
 
 
−
<math>frac(\frac{20^{23}}{n}) + frac(\frac{24^{23}}{n}) \neq 0</math> <math>\textbf{(2)}</math>
 
−
 
−
 
 
−
We now gain some valuable insight. From <math>\textbf{(1)}</math>, we find that <math>n</math> must divide <math>20^{23} + 24^{23}</math>. From <math>\textbf{(2)}</math>, we find that <math>n</math> cannot divide both <math>20^{23}</math> and <math>24^{23}</math>. It is impossible for <math>n</math> to divide only <math>1</math> of <math>20^{23}</math> and <math>24^{23}</math>, as this would make <math>\textbf{(1)}</math> false. It must be that <math>n</math> divides neither <math>20^{23}</math> nor <math>24^{23}</math>. For both this and <math>\textbf{(1)}</math> to be true simultaneously, we must have that if <math>20^{23} \equiv a \bmod n</math>, then <math>24^{23} \equiv -a \bmod n</math>. By inspection, this occurs when <math>n = 22</math>.We now test the factors of <math>22</math> to see if we can find a smaller value. As both <math>20^{23}</math> and <math>24^{23}</math> are congruent to <math>0</math> mod <math>2</math>, <math>n = 2</math> is not a valid solution. However, with <math>n = 11</math>, <math>20^{23} \equiv (-2)^{23} \bmod 11</math>, while <math>24^{23} \equiv 2^{23} \bmod 11</math>. Clearly, <math>-(-2)^{23} = 2^{23}</math>, so our final answer is <math>\boxed{011}</math>.
 
−
 
 
−
===Problem 2 Solutions===
 
−
====Solution 1 (Meta-Solving)====
 
−
We can consider smaller cases. If there is <math>1</math> object to match, Bob will always make exactly <math>1</math> correct match, so he will expect to make <math>1</math> match. If there are <math>2</math> objects, Bob can either match both correctly for a total of <math>2</math> matches or match both incorrectly for a total of <math>0</math> matches. Each of these has a <math>\frac{1}{2}</math> chance of happening, so again Bob will expect to make <math>1</math> match. If there are <math>3</math> objects, we can list out each combination. Suppose we are matching <math>{1, 2, 3}</math> to itself, and the "solution" is the set of ordered pairs <math>{(1,1),(2,2), (3,3)}</math>. The possibilities are listed below:
 
−
 
 
−
 
 
−
<math>{(1,1), (2,2), (3,3)}</math> (3 correct matches)
 
−
 
 
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<math>{(1,2), (2,1), (3,3)}</math> (1 correct match)
 
−
 
 
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<math>{(1,3), (2,2), (3,1)}</math> (1 correct match)
 
−
 
 
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<math>{(1,1), (2,3), (3,2)}</math> (1 correct match)
 
−
 
 
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<math>{(1,2), (2,3), (3,1)}</math> (0 correct matches)
 
−
 
 
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<math>{(1,3), (2,1), (3,2)}</math> (0 correct matches)
 
−
 
 
−
 
 
−
The expected value is <math>\frac{1}{6} \cdot (3+1+1+1) = 1</math>. The expected value has remained at 1 for the first three cases, so we can safely assume that it will remain at <math>1</math> for any positive integer number of items to match. Then Bob can expect to make <math>1</math> match, and the remainder when this is divided by <math>1000</math> is <math>\boxed{001}</math>.
 
−
 
 
−
 
 
−
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.
 
−
 
 
−
 
 
−
<math>f(n) = f(n-1) + f(n-2) + f(n-3) + ... + f(n - 2023) + f(n-2024)</math> <math>\textbf{(1)}</math>
 
−
<math>f(n-1) = f(n-2) + f(n-3) + f(n-4) + ... + f(n - 2024) + f(n - 2025)</math> <math>\textbf{(2)}</math>
 
−
 
 
−
 
 
−
We notice that most of the terms telescope if we subtract <math>\textbf{(2)}</math> from <math>\textbf{(1)}</math>.
 
−
 
 
−
 
 
−
<math>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))</math>
 
−
 
 
−
<math>= f(n-1) + (f(n-2) - f(n-2)) + (f(n-3) - f(n-3)) + ... + (f(n-2024) - f(n-2024)) - f(n - 2025)</math>
 
−
 
 
−
<math>= f(n-1) - f(n-2025)</math>
 
−
 
 
−
 
 
−
By adding <math>f(n-1)</math> to both sides, we find that <math>f(n) = 2f(n-1) - f(n-2025)</math>. From here, we can consider <math>f(4049)</math>. We note that for all <math>2026 \leq n \leq 4049</math>, the second part of our definition (the <math>f(n-2025)</math> term) is equal to one. From here, we can list out a few definitions for <math>f(4049)</math> using our formula.
 
−
 
 
−
 
 
−
<math>f(4049) = 2f(4048)-1 = 2^1f(4048) - (2^1-1)</math>
 
−
 
 
−
<math>= 2(2f(4047)-1)-1 = 4f(4047) - 3 = 2^2f(4047) - (2^2 - 1)</math>
 
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<math>= 2(2(2f(4046)-1)-1)-1 = 8f(4046) - 7 = 2^3f(4046) - (2^3 - 1)</math>
 
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<math>= 2(2(2(2f(4045)-1)-1)-1)-1 = 16f(4045) - 15 = 2^4f(4045) - (2^4-1)</math>
 
−
 
 
−
It appears that on the interval <math>2025 \leq n \leq 4049</math>, <math>f(4049) = 2^{4049-n}f(n) - (2^{4049-n} - 1) = 2^{4049-n}(f(n)-1) + 1</math>. (<math>2025</math> is the upper bound because if we tried to calculate <math>f(2025)</math> using our alternate definition, we'd get <math>2f(2024) - f(0)</math>, and <math>f(0)</math> is undefined.) Clearly, to maximize <math>4049 - n</math> (to maximize <math>b</math> in the problem), we choose the lower bound. Then we get <math>f(4049) = 2^{2024}(f(2025)-1) + 1 = 2^{2024}(2024-1) + 1 = 2^{2024}(2023) + 1</math>, so <math>a + b = 4047</math>. The remainder when this is divided by <math>1000</math> is <math>\boxed{047}</math>.
 
−
 
 
−
===Problem 4 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));
 
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draw((0,0)--(0,1.172),red);
 
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draw((0,1.172)--(1.172,1.172));
 
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draw((1.172,1.172)--(1.172,0));
 
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draw((1.172,0)--(0,0),red);
 
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draw((0,1.172)--(0.828,1.172),red);
 
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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>
 
  
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We can use coordinate bashing. We may assume the legs of the triangle have side length <math>2</math> for ease of computation, and we can multiply by <math>1012</math> on our final answer to get our result. We place <math>B</math> at the origin; in this case, <math>A</math> has coordinates <math>(0,2)</math>, <math>B</math> has coordinates <math>(0, 0)</math>, and <math>C</math> has coordinates <math>(2, 0)</math>. Then line segment <math>\overline{\rm AC}</math> is on the line that has slope <math>\frac{0-2}{2-0} = -1</math> and <math>y</math>-intercept <math>2</math>, so the line's equation is <math>y = 2 - x</math>. Using the distance formula on <math>A</math> and <math>C</math> or noting that <math>\triangle ABC</math> is a <math>45-45-90</math> triangle, we find that the length of <math>\overline{\rm AC}</math> is <math>2\sqrt{2}</math>. The area of <math>\triangle ABC</math> is <math>\frac{1}{2}(2)(2) = 2</math>, and its semiperimeter is <math>\frac{2+2+2\sqrt{2}}{2} = 2 + \sqrt{2}</math>. Since <math>rs = A</math> or <math>r = \frac{A}{s}</math>, the inradius of <math>\triangle ABC</math> is <math>\frac{2}{2+\sqrt{2}} = 2-\sqrt{2}</math>. Also, because the side length of the square circumscribed around <math>\triangle ABC</math> is equal to the length of the diameter of the incircle of <math>\triangle ABC</math>, the square has side length <math>4 - 2\sqrt{2}</math>. From here, we can find that the vertices of the square are, starting at the bottom left and going clockwise, <math>(0,0)</math>, <math>(0, 4 - 2\sqrt{2})</math>, <math>(4 - 2\sqrt{2}, 4 - 2\sqrt{2})</math>, and <math>(4 - 2\sqrt{2}, 0)</math>. We label the top-right vertex of the square as <math>D</math>, the leftmost of the two intersection points of the square with the triangle as <math>E</math>, and the other intersection point as <math>F</math>. <math>D</math> is directly above <math>F</math>, so they must share an x-coordinate. Since <math>F</math> lies on the line <math>y = 2 - x</math>, we take <math>x = 4 - 2\sqrt{2}</math> to find that <math>F</math> has coordinates <math>(4 - 2\sqrt{2}, 2 - (4 - 2\sqrt{2}))</math> or <math>(4 - 2\sqrt{2}, 2\sqrt{2} - 2)</math>. Using similar logic, <math>E</math> has coordinates <math>(2\sqrt{2} - 2, 4 - 2\sqrt{2})</math>. The distance from <math>D</math> to <math>F</math> is just the positive difference of their y-coordinates, which is <math>4 - 2\sqrt{2} - (2\sqrt{2} - 2) = 6 - 4\sqrt{2}</math>. Similarly, the distance from <math>D</math> to <math>E</math> is the same. Since <math>\triangle DEF</math> is an isosceles right triangle, the length of <math>\overline{\rm EF}</math> is <math>\sqrt{2}</math> times the length of one of the legs, which happens to be <math>6\sqrt{2} - 8</math>. The perimeter of the diamond is equal to the perimeter of the square minus the lengths of <math>\overline{\rm ED}</math> and <math>\overline{\rm DF}</math> plus the length of <math>\overline{\rm EF}</math>, which is <math>4(4 - 2\sqrt{2}) - 2(6 - 4\sqrt{2}) + (6\sqrt{2} - 8) = 6\sqrt{2} - 4</math>. We multiply this by <math>1012</math> to scale the triangle to a side length of <math>2024</math>, resulting in a perimeter of <math>6072\sqrt{2} - 4048</math>. Thus, <math>a + b + c = 10122</math>, and the remainder when this is divided by <math>1000</math> is <math>\boxed{122}</math>.
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Last Updated: May 25, 2024
  
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===Problem 5 Solutions===
 
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I have not written a solution yet, but the answer is <math>\boxed{976}</math>.
 
  
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===Problem 6 Solutions===
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If you find an issue or have an alternate solution, please put it below! I'll add it to the document.
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I have not written a solution yet, but the answer is <math>\boxed{198}</math>.
 

Latest revision as of 21:20, 14 November 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

21

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
  4. 2000 AIME II Problem 1 Solution 4
  5. 1985 AIME Problem 12 Solution 9 ⭐
  6. 2012 AIME I Problem 8 Solution 7
  7. 1987 AIME Problem 14 Solution 4
  8. 1988 AIME Problem 10 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 ⭐
  10. 2024 AMC 8 Problem 18 Solution 4
  11. 2024 AMC 8 Problem 11 Solution 4
  12. 2019 AMC 8 Problem 15 Solution 2 ⭐

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
  6. 1982 AHSME Problem 13 Solution 2
  7. 1974 AHSME Problem 4 Solution 2

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
  4. 2015 AMC 12B Problem 8 Solution 3
  5. 2003 AMC 12B Problem 17 Solution 4
  6. 2020 AMC 10A Problem 18 Solution 6
  7. 2021 AMC 10B Problem 13 Solution 4
  8. 2007 AMC 10B Problem 23 Solution 4
  9. 2005 AMC 10B Problem 24 Solution 3 ⭐
  10. 2008 AMC 12B Problem 24 Solution 4 ⭐
  11. 2023 AMC 10B Problem 24 Solution 2 ⭐
  12. 2023 AMC 10B Problem 13/AMC 12B Problem 9 Solution 5
  13. 2007 AMC 10B Problem 5 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
  9. 1984 USAMO Problem 1 - First USAMO Solution
  10. 2011 USAJMO Problem 1 - First USAJMO Solution

Other Major Contributions

  1. Titu's Lemma (all examples and problems)

Some Problems I Wrote

All problems and solutions have been moved to this page.

Last Updated: May 25, 2024


If you find an issue or have an alternate solution, please put it below! I'll add it to the document.

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