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Duke Math Meet (DMM)
2022 Duke Math Meet
2022 Duke Math Meet
Part of
Duke Math Meet (DMM)
Subcontests
(1)
3
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2022 DMM Tiebreaker Round - Duke Math Meet
p1. The sequence
{
x
n
}
\{x_n\}
{
x
n
}
is defined by
x
n
+
1
=
{
2
x
n
−
1
,
i
f
1
2
≤
x
n
<
1
2
x
n
,
i
f
0
≤
x
n
<
1
2
x_{n+1} = \begin{cases} 2x_n - 1, \,\, if \,\, \frac12 \le x_n < 1 \\ 2x_n, \,\, if \,\, 0 \le x_n < \frac12 \end{cases}
x
n
+
1
=
{
2
x
n
−
1
,
i
f
2
1
≤
x
n
<
1
2
x
n
,
i
f
0
≤
x
n
<
2
1
where
0
≤
x
0
<
1
0 \le x_0 < 1
0
≤
x
0
<
1
and
x
7
=
x
0
x_7 = x_0
x
7
=
x
0
. Find the number of sequences satisfying these conditions. p2. Let
M
=
{
1
,
.
.
.
,
2022
}
M = \{1, . . . , 2022\}
M
=
{
1
,
...
,
2022
}
. For any nonempty set
X
⊆
M
X \subseteq M
X
⊆
M
, let
a
X
a_X
a
X
be the sum of the maximum and the minimum number of
X
X
X
. Find the average value of
a
X
a_X
a
X
across all nonempty subsets
X
X
X
of
M
M
M
. PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here.
2022 DMM Team Round - Duke Math Meet
p1. The serpent of fire and the serpent of ice play a game. Since the serpent of ice loves the lucky number
6
6
6
, he will roll a fair
6
6
6
-sided die with faces numbered
1
1
1
through
6
6
6
. The serpent of fire will pay him
log
10
x
\log_{10} x
lo
g
10
x
, where
x
x
x
is the number he rolls. The serpent of ice rolls the die
6
6
6
times. His expected total amount of winnings across the
6
6
6
rounds is
k
k
k
. Find
1
0
k
10^k
1
0
k
. p2. Let
a
=
log
3
5
a = \log_3 5
a
=
lo
g
3
5
,
b
=
log
3
4
b = \log_3 4
b
=
lo
g
3
4
,
c
=
−
log
3
20
c = - \log_3 20
c
=
−
lo
g
3
20
, evaluate
a
2
+
b
2
a
2
+
b
2
+
a
b
+
b
2
+
c
2
b
2
+
c
2
+
b
c
+
c
2
+
a
2
c
2
+
a
2
+
c
a
\frac{a^2+b^2}{a^2+b^2+ab} +\frac{b^2+c^2}{b^2+c^2+bc} +\frac{c^2+a^2}{c^2+a^2+ca}
a
2
+
b
2
+
ab
a
2
+
b
2
+
b
2
+
c
2
+
b
c
b
2
+
c
2
+
c
2
+
a
2
+
c
a
c
2
+
a
2
. p3. Let
△
A
B
C
\vartriangle ABC
△
A
BC
be an isosceles obtuse triangle with
A
B
=
A
C
AB = AC
A
B
=
A
C
and circumcenter
O
O
O
. The circle with diameter
A
O
AO
A
O
meets
B
C
BC
BC
at points
X
,
Y
X, Y
X
,
Y
, where X is closer to
B
B
B
. Suppose
X
B
=
Y
C
=
4
XB = Y C = 4
XB
=
Y
C
=
4
,
X
Y
=
6
XY = 6
X
Y
=
6
, and the area of
△
A
B
C
\vartriangle ABC
△
A
BC
is
m
n
m\sqrt{n}
m
n
for positive integers
m
m
m
and
n
n
n
, where
n
n
n
does not contain any square factors. Find
m
+
n
m + n
m
+
n
. p4. Alice is not sure what to have for dinner, so she uses a fair
6
6
6
-sided die to decide. She keeps rolling, and if she gets all the even numbers (i.e. getting all of
2
,
4
,
6
2, 4, 6
2
,
4
,
6
) before getting any odd number, she will reward herself with McDonald’s. Find the probability that Alice could have McDonald’s for dinner. p5. How many distinct ways are there to split
50
50
50
apples,
50
50
50
oranges,
50
50
50
bananas into two boxes, such that the products of the number of apples, oranges, and bananas in each box are nonzero and equal? p6. Sujay and Rishabh are taking turns marking lattice points within a square board in the Cartesian plane with opposite vertices
(
1
,
1
)
(1, 1)
(
1
,
1
)
,
(
n
,
n
)
(n, n)
(
n
,
n
)
for some constant
n
n
n
. Sujay loses when the two-point pattern
P
P
P
below shows up:https://cdn.artofproblemsolving.com/attachments/1/9/d1fe285294d4146afc0c7a2180b15586b04643.png That is, Sujay loses when there exists a pair of points
(
x
,
y
)
(x, y)
(
x
,
y
)
and
(
x
+
2
,
y
+
1
)
(x + 2, y + 1)
(
x
+
2
,
y
+
1
)
. He and Rishabh stop marking points when the pattern
P
P
P
appears on the board. If Rishabh goes first, let
S
S
S
be the set of all integers
3
≤
n
≤
100
3 \le n \le 100
3
≤
n
≤
100
such that Rishabh has a strategy to always trick Sujay into being the one who creates
P
P
P
. Find the sum of all elements of
S
S
S
. p7. Let
a
a
a
be the shortest distance between the origin
(
0
,
0
)
(0, 0)
(
0
,
0
)
and the graph of
y
3
=
x
(
6
y
−
x
2
)
−
8
y^3 = x(6y -x^2)-8
y
3
=
x
(
6
y
−
x
2
)
−
8
. Find
⌊
a
2
⌋
\lfloor a^2 \rfloor
⌊
a
2
⌋
. (
⌊
x
⌋
\lfloor x\rfloor
⌊
x
⌋
is the largest integer not exceeding
x
x
x
) p8. Find all real solutions to the following equation:
2
2
x
2
+
x
−
1
−
x
2
−
2
=
0.
2\sqrt2x^2 + x -\sqrt{1 - x^2 } -\sqrt2 = 0.
2
2
x
2
+
x
−
1
−
x
2
−
2
=
0.
p9. Given the expression
S
=
(
x
4
−
x
)
(
x
2
−
x
3
)
S = (x^4 - x)(x^2 - x^3)
S
=
(
x
4
−
x
)
(
x
2
−
x
3
)
for
x
=
cos
2
π
5
+
i
sin
2
π
5
x = \cos \frac{2\pi}{5 }+ i\sin \frac{2\pi}{5 }
x
=
cos
5
2
π
+
i
sin
5
2
π
, find the value of
S
2
S^2
S
2
.p10. In a
32
32
32
team single-elimination rock-paper-scissors tournament, the teams are numbered from
1
1
1
to
32
32
32
. Each team is guaranteed (through incredible rock-paper-scissors skill) to win any match against a team with a higher number than it, and therefore will lose to any team with a lower number. Each round, teams who have not lost yet are randomly paired with other teams, and the losers of each match are eliminated. After the
5
5
5
rounds of the tournament, the team that won all
5
5
5
rounds is ranked
1
1
1
st, the team that lost the 5th round is ranked
2
2
2
nd, and the two teams that lost the
4
4
4
th round play each other for
3
3
3
rd and
4
4
4
th place. What is the probability that the teams numbered
1
,
2
,
3
1, 2, 3
1
,
2
,
3
, and
4
4
4
are ranked
1
1
1
st, 2nd, 3rd, and 4th respectively? If the probability is
m
n
\frac{m}{n}
n
m
for relatively prime integers
m
m
m
and
n
n
n
, find
m
m
m
. PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here.
2022 DMM Individual Round - Duke Math Meet
p1. Sujay sees a shooting star go across the night sky, and took a picture of it. The shooting star consists of a star body, which is bounded by four quarter-circle arcs, and a triangular tail. Suppose
A
B
=
2
AB = 2
A
B
=
2
,
A
C
=
4
AC = 4
A
C
=
4
. Let the area of the shooting star be
X
X
X
. If
6
X
=
a
−
b
π
6X = a-b\pi
6
X
=
a
−
bπ
for positive integers
a
,
b
a, b
a
,
b
, find
a
+
b
a + b
a
+
b
. https://cdn.artofproblemsolving.com/attachments/0/f/f9c9ff23416565760df225c133330e795b9076.pngp2. Assuming that each distinct arrangement of the letters in
D
I
S
C
U
S
S
I
O
N
S
DISCUSSIONS
D
I
SC
U
SS
I
ONS
is equally likely to occur, what is the probability that a random arrangement of the letters in
D
I
S
C
U
S
S
I
O
N
S
DISCUSSIONS
D
I
SC
U
SS
I
ONS
has all the
S
S
S
’s together? p3. Evaluate
(
1
+
2022
)
(
1
+
202
2
2
)
(
1
+
202
2
4
)
.
.
.
(
1
+
202
2
2
2022
)
1
+
2022
+
202
2
2
+
.
.
.
+
202
2
2
2023
−
1
.
\frac{(1 + 2022)(1 + 2022^2)(1 + 2022^4) ... (1 + 2022^{2^{2022}})}{1 + 2022 + 2022^2 + ... + 2022^{2^{2023}-1}} .
1
+
2022
+
202
2
2
+
...
+
202
2
2
2023
−
1
(
1
+
2022
)
(
1
+
202
2
2
)
(
1
+
202
2
4
)
...
(
1
+
202
2
2
2022
)
.
p4. Dr. Kraines has
27
27
27
unit cubes, each of which has one side painted red while the other five are white. If he assembles his cubes into one
3
×
3
×
3
3 \times 3 \times 3
3
×
3
×
3
cube by placing each unit cube in a random orientation, what is the probability that the entire surface of the cube will be white, with no red faces visible? If the answer is
2
a
3
b
5
c
2^a3^b5^c
2
a
3
b
5
c
for integers
a
a
a
,
b
b
b
,
c
c
c
, find
∣
a
+
b
+
c
∣
|a + b + c|
∣
a
+
b
+
c
∣
. p5. Let S be a subset of
{
1
,
2
,
3
,
.
.
.
,
1000
,
1001
}
\{1, 2, 3, ... , 1000, 1001\}
{
1
,
2
,
3
,
...
,
1000
,
1001
}
such that no two elements of
S
S
S
have a difference of
4
4
4
or
7
7
7
. What is the largest number of elements
S
S
S
can have? p6. George writes the number
1
1
1
. At each iteration, he removes the number
x
x
x
written and instead writes either
4
x
+
1
4x+1
4
x
+
1
or
8
x
+
1
8x+1
8
x
+
1
. He does this until
x
>
1000
x > 1000
x
>
1000
, after which the game ends. What is the minimum possible value of the last number George writes? p7. List all positive integer ordered pairs
(
a
,
b
)
(a, b)
(
a
,
b
)
satisfying
a
4
+
4
b
4
=
281
⋅
61
a^4 + 4b^4 = 281 \cdot 61
a
4
+
4
b
4
=
281
⋅
61
. p8. Karthik the farmer is trying to protect his crops from a wildfire. Karthik’s land is a
5
×
6
5 \times 6
5
×
6
rectangle divided into
30
30
30
smaller square plots. The
5
5
5
plots on the left edge contain fire, the
5
5
5
plots on the right edge contain blueberry trees, and the other
5
×
4
5 \times 4
5
×
4
plots of land contain banana bushes. Fire will repeatedly spread to all squares with bushes or trees that share a side with a square with fire. How many ways can Karthik replace
5
5
5
of his
20
20
20
plots of banana bushes with firebreaks so that fire will not consume any of his prized blueberry trees? p9. Find
a
0
∈
R
a_0 \in R
a
0
∈
R
such that the sequence
{
a
n
}
n
=
0
∞
\{a_n\}^{\infty}_{n=0}
{
a
n
}
n
=
0
∞
defined by
a
n
+
1
=
−
3
a
n
+
2
n
a_{n+1} = -3a_n + 2^n
a
n
+
1
=
−
3
a
n
+
2
n
is strictly increasing. p10. Jonathan is playing with his life savings. He lines up a penny, nickel, dime, quarter, and half-dollar from left to right. At each step, Jonathan takes the leftmost coin at position
1
1
1
and uniformly chooses a position
2
≤
k
≤
5
2 \le k \le 5
2
≤
k
≤
5
. He then moves the coin to position
k
k
k
, shifting all coins at positions
2
2
2
through
k
k
k
leftward. What is the expected number of steps it takes for the half-dollar to leave and subsequently return to position
5
5
5
? PS. You had better use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here.