Problem 5: Page 1
Marcos Pertierra
10890
modularmarc101
USAMTS Year 21 Round 2
5/2/21
Let
. By Stewart’s Theorem,
, then
.
Solving the equation for AP, we have
By Power of a Point on P, we have
So it suffices to prove that the global maximum of
is less than
(around
). We can find the maximum by finding the roots of
and plugging in those values to
. This happens when
(rounded),
so the maximum is about
which is clearly less than
. Therefore,
for all
.
, so
Problem 5: Page 2
Marcos Pertierra
10890
modularmarc101
USAMTS Year 21 Round 2
Here is a graph of
maximum.
. Clearly, it has only one critical point, which is the
Problem 3: Page 1
Marcos Pertierra
10890
modularmarc101
USAMTS Year 21 Round 2
3/2/21
First, notice that since
, we know that
or
.
Since the quadratic residues mod 7 are 0, 1, 2, and 4, we find that
and
, so
and
. Knowing this, we can
write a and b as
and
, for some naturals and . So,
Same as before, we find that
process, we get that
and
(
and
and
. Repeating the same
and
for
).
The most reduced term that we can get is
and
,
by setting
to 1 (since we won’t be able to divide the mod equation by
anymore.
Hence,
, similarly,
So,
(ab is a multiple of
, and therefore
).
Problem 1: Page 1
Marcos Pertierra
10890
modularmarc101
USAMTS Year 21 Round 2
1/2/21
Part A: First, we build a table to show each turn.
n
|
Left side
|
Right side
5
22
25
1
23
25
1
24
25
1
25
25
We see that Jeremy can win the game this way in 4 turns.
Part B:
Case 1:
Call the number on the left side . If
, we can choose the number 1
times, and both sides will be equal. Since is a positive integer, we know that
so
. Therefore, the game can be won in
or fewer
terms.
Case 2:
If
, then we first chose some positive integer , such that
we can win the game as in the first case. So it suffices to prove that
. Then,
,
since we used the first turn for , and the rest as in case 1. First of all, we want
the smallest possible value for , because any other value will not let us win the
game by choosing only 1’s after .
Problem 1: Page 2
Marcos Pertierra
10890
modularmarc101
USAMTS Year 21 Round 2
So we solve the inequality
for
to get
Therefore, the minimum of k is
, where the brackets [ ] represent the
floor function. Now arrange the inequality
as
Since
, we know that
for some positive integer
less than
,
.
So we must prove that
Since
is a positive integer, we know that
, which completes this case.
Problem 1: Page 3
Marcos Pertierra
10890
modularmarc101
USAMTS Year 21 Round 2
Case 3:
If
, it is an automatic win; meaning, we do not have to make any turns (the
game is won in 0 turns).
This completes our proof for this problem.
Problem 2: Page 1
Marcos Pertierra
10890
modularmarc101
USAMTS Year 21 Round 2
2/2/21
For the sake of brevity, I will refer to Alice’s daughters as d’s, her granddaughters
as g’s, and her granddaughters as gg’s.
We shall divide the solution into 4 main cases: 0 d’s, 1 d, 2 d’s, and 3 d’s are in
the set.
Case 1: 0 d ‘ s
Alice can and can not be in the set, so there are 2 choices for her.
If there are g’s in the set, then there are
sets.
This is because from the 6 g’s, we choose of them, and from the gg’s that are
not daughters of the g’s we choose (6 - ) , can or cannot be in the set (each one
has 2 choices: either they’re in the set, or not).
Since can range from 0 to 6, we have there are
sets.
However, Alice has 2 choices, which we have not counted, so there are actually
sets.
Problem 2: Page 2
Marcos Pertierra
10890
modularmarc101
USAMTS Year 21 Round 2
Case 2: 1 d
There are 3 choices for choosing d.
If there are g’s in the set, then there are
sets.
This is because there are 4 g’s we can choose (6 minus the 2 daughters of the d
we choose) and from the gg’s that are not daughters of the g’s we choose (6- ) ,
can or cannot be in the set (each one has 2 choices: either they’re in the set, or
not).
In this case, ranges from 0 to 4, since both daughters of the d we choose can
not be in the set.
So there are
sets. As
stated in the beginning of this case, there are 3 choices for d, so there are
sets.
Case 3: 2 d ‘s
There are
for choosing the 2 d’s.
If there are g’s in the set, then there are
sets.
This is because there are 2 g’s we can choose (6 minus the 4 daughters of the
d ‘s we choose) and from the gg’s that are not daughters of the g’s we choose
(6- ) , can or cannot be in the set (each one has 2 choices: either they’re in the
set, or not. So ranges from 0 to 2.
Problem 2: Page 3
Marcos Pertierra
10890
modularmarc101
USAMTS Year 21 Round 2
So there are
sets, but there are 3 choices for
the d’s as mentioned before so there are
sets.
Case 4: 3 d ‘ s
No g’s can be in the set, but any of the 6 gg’s can, so there are
Now we must add up all the sets. In total, there are
=
sets.
sets.
Problem 4: Page 1
Marcos Pertierra
10890
modularmarc101
USAMTS Year 21 Round 2
4/2/21
Part A:
To get to the point
coordinate.
, we will first reach the y-coordinate, and then the x-
# of minutes | coordinates of point after minute
_______________________________
0
|
1
|
2
|
3
|
5
|
6
|
7
|
We see that it is indeed possible to reach this point in a finite amount of time (7
minutes, to be exact).
Problem 4: Page 2
Marcos Pertierra
10890
modularmarc101
USAMTS Year 21 Round 2