User:Cxsmi
Contents
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
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
- 1987 AIME Problem 11 Solution 3 ⭐
- 2012 AIME I Problem 2 Solution 3
- 2024 AIME I Problem 6 Solution 5 ⭐
AMC 8
- 2012 AMC 8 Problem 19 Solution 6
- 2002 AMC 8 Problem 17 Solution 3
- 2007 AMC 8 Problem 20 Solution 8
- 2018 AMC 8 Problem 23 Solution 5 ⭐
- 2016 AMC 8 Problem 13 Solution 3
- 2017 AMC 8 Problem 9 Solution 2
- 2012 AMC 8 Problem 20 Solution 7
- 2010 AMC 8 Problem 25 Solution 3
- 2024 AMC 8 Problem 11 Solution 3
AJHSME
- 1997 AJHSME Problem 22 Solution 1
- 1985 AJHSME Problem 1 Solution 2
- 1985 AJHSME Problem 24 Solution 2 ⭐
- 1985 AJHSME Problem 2 Solution 5
AHSME
- 1950 AHSME Problem 40 Solution 2
- 1950 AHSME Problem 41 Solution 2
- 1972 AHSME Problem 16 Solution 2
- 1950 AHSME Problem 45 Solution 3
- 1956 AHSME Problem 23 Solution 1
AMC 10/12
- 2021 Fall AMC 12B Problem 12 Solution 6
- 2021 Fall AMC 10B Problem 1/AMC 12B Problem 1 Solution 4
- 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.
- 2017 AMC 10A Problem 19 - First AMC 10 Solution of Difficulty 2 or Higher
- 2007 AMC 8 Problem 25 - First AMC 8 Final Five Solution
- 1984 AIME Problem 1 - First AIME Solution
- 2005 AMC 12B Problem 16 - First AMC 12 Solution of Difficulty 2.5 or Higher
- 2016 AMC 10A Problem 21 - First AMC 10 Final Five Solution
- 1987 AIME Problem 11 - First AIME Solution of Difficulty 4 or Higher; First AIME Solution of Difficulty 5 or Higher
- 2017 AMC 12A Problem 23 - First AMC 12 Final Five Solution
- 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 and , just like in the AIME.
All problems have been moved to this page.
Solutions
These are solutions for the problems above.
Problem 1 Solutions
Solution 1
We split the condition into two separate conditions, as listed below.
Rearranging the conditions, we find that
Recalling that where represents the fractional part of , we rewrite once more.
We now gain some valuable insight. From , we find that must divide . From , we find that cannot divide both and . It is impossible for to divide only of and , as this would make false. It must be that divides neither nor . For both this and to be true simultaneously, we must have that if , then . By inspection, this occurs when .We now test the factors of to see if we can find a smaller value. As both and are congruent to mod , is not a valid solution. However, with , , while . Clearly, , so our final answer is .
Problem 2 Solutions
Solution 1 (Meta-Solving)
We can consider smaller cases. If there is object to match, Bob will always make exactly correct match, so he will expect to make match. If there are objects, Bob can either match both correctly for a total of matches or match both incorrectly for a total of matches. Each of these has a chance of happening, so again Bob will expect to make match. If there are objects, we can list out each combination. Suppose we are matching to itself, and the "solution" is the set of ordered pairs . The possibilities are listed below:
(3 correct matches)
(1 correct match)
(1 correct match)
(1 correct match)
(0 correct matches)
(0 correct matches)
The expected value is . The expected value has remained at 1 for the first three cases, so we can safely assume that it will remain at for any positive integer number of items to match. Then Bob can expect to make match, and the remainder when this is divided by is .
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.
We notice that most of the terms telescope if we subtract from . . By adding to both sides, we find that . From here, we can consider . We note that for all , the second part of our definition (the term) is equal to one. From here, we can list out a few definitions for using our formula.
It appears that on the interval , . ( is the upper bound because if we tried to calculate using our alternate definition, we'd get , and is undefined.) We attempt to maximize the value of , as this will make finding the prime factorization easier. This value is maximized when we choose the lower bound. We take ; then and . From our original definition, . Thus, . We prime factorize ; by trying divisibility tests until one works, we find that . Recalling that , we find that . Then the desired sum is , and the remainder when this is divided by is .
Problem 4 Solutions
Solution 1
We can use coordinate bashing. We may assume the legs of the triangle have side length for ease of computation, and we can multiply by on our final answer to get our result. We place at the origin; in this case, has coordinates , has coordinates , and has coordinates . Then line segment is on the line that has slope and -intercept , so the line's equation is . Using the distance formula on and or noting that is a triangle, we find that the length of is . The area of is , and its semiperimeter is . Since or , the inradius of is . Also, because the side length of the square circumscribed around is equal to the length of the diameter of the incircle of , the square has side length . From here, we can find that the vertices of the square are, starting at the bottom left and going clockwise, , , , and . We label the top-right vertex of the square as , the leftmost of the two intersection points of the square with the triangle as , and the other intersection point as . is directly above , so they must share an x-coordinate. Since lies on the line , we take to find that has coordinates or . Using similar logic, has coordinates . The distance from to is just the positive difference of their y-coordinates, which is . Similarly, the distance from to is the same. Since is an isosceles right triangle, the length of is times the length of one of the legs, which happens to be . The perimeter of the diamond is equal to the perimeter of the square minus the lengths of and plus the length of , which is . We multiply this by to scale the triangle to a side length of , resulting in a perimeter of . Thus, , and the remainder when this is divided by is .
Problem 5 Solutions
I have not written a solution yet, but the answer is .
Problem 6 Solutions
I have not written a solution yet, but the answer is .