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
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 ⭐
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
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
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
- 2009 HMMT February Algebra Problem 1 - First February HMMT 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.
Problem 1
Find the least positive integer that satisfies the following. The notation
represents the greatest integer less than or equal to
.
Problem 2
Bob has received a test. He must match a list of words to a list of
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
matches. Find the remainder when
is divided by
.
Problem 3
A sequence is defined recursively as
and
for . The value of
can be written in the form
for positive integers
and
such that
is maximized. What is the remainder when
is divided by
?
Problem 4
A diamond is created by connecting the points at which a square circumscribed around the incircle of an isosceles right triangle intersects
itself.
has leg length
. The perimeter of this diamond is expressible as
, where
,
, and
are integers, and
is not divisible by the square of any prime. What is the remainder when
is divided by
?
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. I have given each hint a certain number of stars ⭐ to represent how revealing it is, with a 1-star hint being minor or rather obvious and a 5-star hint revealing the main mechanics of the problem.
Problem 1 ⭐⭐⭐⭐
Remainders
Problem 2 ⭐⭐⭐
Recursion
Problem 3 ⭐
Telescoping
Problem 4 ⭐⭐⭐⭐⭐
Coordinate Bashing
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
. We can verify by finding
; from the original definition, we find that
, and
. From our definition, we find that
. 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
. 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.) Clearly, to maximize
(to maximize
in the problem), we choose the lower bound. Then we get
, so
. 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
. Since
,
, and
,
, and the remainder when this is divided by
is
.