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Duke Math Meet (DMM)
1999 / 2000 Duke Math Meet
1999 / 2000 Duke Math Meet
Part of
Duke Math Meet (DMM)
Subcontests
(1)
3
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1999 / 2000 DMM Individual Round - Duke Math Meet
p1. The least prime factor of
a
a
a
is
3
3
3
, the least prime factor of
b
b
b
is
7
7
7
. Find the least prime factor of
a
+
b
a + b
a
+
b
. p2. In a Cartesian coordinate system, the two tangent lines from
P
=
(
39
,
52
)
P = (39, 52)
P
=
(
39
,
52
)
meet the circle defined by
x
2
+
y
2
=
625
x^2 + y^2 = 625
x
2
+
y
2
=
625
at points
Q
Q
Q
and
R
R
R
. Find the length
Q
R
QR
QR
. p3. For a positive integer
n
n
n
, there is a sequence
(
a
0
,
a
1
,
a
2
,
.
.
.
,
a
n
)
(a_0, a_1, a_2,..., a_n)
(
a
0
,
a
1
,
a
2
,
...
,
a
n
)
of real values such that
a
0
=
11
a_0 = 11
a
0
=
11
and
(
a
k
+
a
k
+
1
)
(
a
k
−
a
k
+
1
)
=
5
(a_k + a_{k+1}) (a_k - a_{k+1}) = 5
(
a
k
+
a
k
+
1
)
(
a
k
−
a
k
+
1
)
=
5
for every
k
k
k
with
0
≤
k
≤
n
−
1
0 \le k \le n-1
0
≤
k
≤
n
−
1
. Find the maximum possible value of
n
n
n
. (Be careful that your answer isn’t off by one!) p4. Persons
A
A
A
and
B
B
B
stand at point
P
P
P
on line
ℓ
\ell
ℓ
. Point
Q
Q
Q
lies at a distance of
10
10
10
from point
P
P
P
in the direction perpendicular to
ℓ
\ell
ℓ
. Both persons intially face towards
Q
Q
Q
. Person
A
A
A
walks forward and to the left at an angle of
2
5
o
25^o
2
5
o
with
ℓ
\ell
ℓ
, when he is again at a distance of
10
10
10
from point
Q
Q
Q
, he stops, turns
9
0
o
90^o
9
0
o
to the right, and continues walking. Person
B
B
B
walks forward and to the right at an angle of
5
5
o
55^o
5
5
o
with line
ℓ
\ell
ℓ
, when he is again at a distance of
10
10
10
from point
Q
Q
Q
, he stops, turns
9
0
o
90^o
9
0
o
to the left, and continues walking. Their paths cross at point
R
R
R
. Find the distance
P
R
PR
PR
. p5. Compute
l
c
m
(
1
,
2
,
3
,
.
.
.
,
200
)
l
c
m
(
102
,
103
,
104
,
.
.
.
,
200
)
.
\frac{lcm (1,2, 3,..., 200)}{lcm (102, 103, 104, ..., 200)}.
l
c
m
(
102
,
103
,
104
,
...
,
200
)
l
c
m
(
1
,
2
,
3
,
...
,
200
)
.
p6. There is a unique real value
A
A
A
such that for all
x
x
x
with
1
<
x
<
3
1 < x < 3
1
<
x
<
3
and
x
≠
2
x \ne 2
x
=
2
,
∣
A
x
2
−
x
−
2
+
1
x
2
−
6
x
+
8
∣
<
1999.
\left| \frac{A}{x^2-x - 2} +\frac{1}{x^2 - 6x + 8} \right|< 1999.
x
2
−
x
−
2
A
+
x
2
−
6
x
+
8
1
<
1999.
Compute
A
A
A
. p7. Nine poles of height
1
,
2
,
.
.
.
,
9
1, 2,..., 9
1
,
2
,
...
,
9
are placed in a line in random order. A pole is called dominant if it is taller than the pole immediately to the left of it, or if it is the pole farthest to the left. Count the number of possible orderings in which there are exactly
2
2
2
dominant poles. p8.
tan
(
11
x
)
=
tan
(
3
4
o
)
\tan (11x) = \tan (34^o)
tan
(
11
x
)
=
tan
(
3
4
o
)
and
tan
(
19
x
)
=
tan
(
2
1
o
)
\tan (19x) = \tan (21^o)
tan
(
19
x
)
=
tan
(
2
1
o
)
. Compute
tan
(
5
x
)
\tan (5x)
tan
(
5
x
)
. PS. You had better use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here.
1999 / 2000 DMM Team Round - Duke Math Meet
p1. Function
f
f
f
is defined by
f
(
x
)
=
a
x
+
b
f (x) = ax+b
f
(
x
)
=
a
x
+
b
for some real values
a
,
b
>
0
a, b > 0
a
,
b
>
0
. If
f
(
f
(
x
)
)
=
9
x
+
5
f (f (x)) = 9x + 5
f
(
f
(
x
))
=
9
x
+
5
for all
x
x
x
, find
b
b
b
. p2. At some point during a game, Will Avery has made
1
/
3
1/3
1/3
of his shots. When he shoots once and makes a basket, his average increases to
2
/
5
2/5
2/5
. Find his average (expressed as a fraction) after a second additional basket. p3. A dealer has a deck of
1999
1999
1999
cards. He takes the top card off and “ducks” it, that is, places it on the bottom of the deck. He deals the second card onto the table. He ducks the third card, deals the fourth card, ducks the fifth card, deals the sixth card, and so forth, continuing until he has only one card left; he then ducks the last card with itself and deals it. Some of the cards (like the second and fourth cards) are not ducked at all before being dealt, while others are ducked multiple times. The question is: what is the average number of ducks per card? p4. Point
P
P
P
lies outside circle
O
O
O
. Perpendicular lines
ℓ
\ell
ℓ
and m intersect at
P
P
P
. Line
ℓ
\ell
ℓ
is tangent to circle
O
O
O
at a point
6
6
6
units from
P
P
P
. Line
m
m
m
crosses circle
O
O
O
at a point
4
4
4
units from
P
P
P
. Find the radius of circle
O
O
O
. p5. Define
f
(
n
)
f(n)
f
(
n
)
by
f
(
n
)
=
{
n
/
2
if
n
i
s
e
v
e
n
(
n
+
1023
)
/
2
if
n
i
s
o
d
d
f(n) = \begin{cases} n/2 \,\,\,\text{if} \,\,\, n\,\,\,is\,\,\, even \\ (n + 1023)/2\,\,\, \text{if} \,\,\, n\,\,\,is\,\,\, odd \end{cases}
f
(
n
)
=
{
n
/2
if
n
i
s
e
v
e
n
(
n
+
1023
)
/2
if
n
i
s
o
dd
Find the least positive integer
n
n
n
such that
f
(
f
(
f
(
f
(
f
(
n
)
)
)
)
)
=
n
.
f(f(f(f(f(n))))) = n.
f
(
f
(
f
(
f
(
f
(
n
)))))
=
n
.
p6. Write
10001
\sqrt{10001}
10001
to the sixth decimal place, rounding down. p7. Define
(
a
n
)
(a_n)
(
a
n
)
recursively by
a
1
=
1
a_1 = 1
a
1
=
1
,
a
n
=
20
cos
(
a
n
−
1
o
)
a_n = 20 \cos (a_{n-1}^o)
a
n
=
20
cos
(
a
n
−
1
o
)
. As
n
n
n
tends to infinity,
(
a
n
)
(a_n)
(
a
n
)
tends to
18.9195...
18.9195...
18.9195...
. Define
(
b
n
)
(b_n)
(
b
n
)
recursively by
b
1
=
1
b_1 = 1
b
1
=
1
,
b
n
=
800
+
800
cos
(
b
n
−
1
o
)
b_n =\sqrt{800 + 800 \cos (b_{n-1}^o)}
b
n
=
800
+
800
cos
(
b
n
−
1
o
)
. As
n
n
n
tends to infinity,
(
b
n
)
(b_n)
(
b
n
)
tends to
x
x
x
. Calculate
x
x
x
to three decimal places. p8. Let
m
o
d
d
(
k
)
mod_d (k)
m
o
d
d
(
k
)
be the remainder of
k
k
k
when divided by
d
d
d
. Find the number of positive integers
n
n
n
satisfying
m
o
d
n
(
1999
)
=
n
2
−
89
n
+
1999
mod_n(1999) = n^2 - 89n + 1999
m
o
d
n
(
1999
)
=
n
2
−
89
n
+
1999
p9. Let
f
(
x
)
=
x
3
+
x
f(x) = x^3 + x
f
(
x
)
=
x
3
+
x
. Compute
∑
k
=
1
10
1
1
+
f
−
1
(
k
−
1
)
2
+
f
−
1
(
k
−
1
)
f
−
1
(
k
)
+
f
−
1
(
k
)
2
\sum^{10}_{k=1} \frac{1}{1 + f^{-1}(k - 1)^2 + f^{-1}(k - 1)f^{-1}(k) + f^{-1}(k)^2}
k
=
1
∑
10
1
+
f
−
1
(
k
−
1
)
2
+
f
−
1
(
k
−
1
)
f
−
1
(
k
)
+
f
−
1
(
k
)
2
1
(
f
−
1
f^{-1}
f
−
1
is the inverse of
f
f
f
:
f
(
f
−
1
1
(
x
)
)
=
f
−
1
1
(
f
(
x
)
)
=
x
f (f^{-1}1 (x)) = f^{-1}1 (f (x)) = x
f
(
f
−
1
1
(
x
))
=
f
−
1
1
(
f
(
x
))
=
x
for all
x
x
x
.) PS. You had better use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here.
1999 DMM Tiebreaker Round - Duke Math Meet
p1A. Compute
1
+
1
2
3
+
1
3
3
+
1
4
3
+
1
5
3
+
.
.
.
1 + \frac{1}{2^3} + \frac{1}{3^3} + \frac{1}{4^3} + \frac{1}{5^3} + ...
1
+
2
3
1
+
3
3
1
+
4
3
1
+
5
3
1
+
...
1
−
1
2
3
+
1
3
3
−
1
4
3
+
1
5
3
−
.
.
.
1 - \frac{1}{2^3} + \frac{1}{3^3} - \frac{1}{4^3} + \frac{1}{5^3} - ...
1
−
2
3
1
+
3
3
1
−
4
3
1
+
5
3
1
−
...
p1B. Real values
a
a
a
and
b
b
b
satisfy
a
b
=
1
ab = 1
ab
=
1
, and both numbers have decimal expansions which repeat every five digits:
a
=
0.
(
a
1
)
(
a
2
)
(
a
3
)
(
a
4
)
(
a
5
)
(
a
1
)
(
a
2
)
(
a
3
)
(
a
4
)
(
a
5
)
.
.
.
a = 0.(a_1)(a_2)(a_3)(a_4)(a_5)(a_1)(a_2)(a_3)(a_4)(a_5)...
a
=
0.
(
a
1
)
(
a
2
)
(
a
3
)
(
a
4
)
(
a
5
)
(
a
1
)
(
a
2
)
(
a
3
)
(
a
4
)
(
a
5
)
...
and
b
=
1.
(
b
1
)
(
b
2
)
(
b
3
)
(
b
4
)
(
b
5
)
(
b
1
)
(
b
2
)
(
b
3
)
(
b
4
)
(
b
5
)
.
.
.
b = 1.(b_1)(b_2)(b_3)(b_4)(b_5)(b_1)(b_2)(b_3)(b_4)(b_5)...
b
=
1.
(
b
1
)
(
b
2
)
(
b
3
)
(
b
4
)
(
b
5
)
(
b
1
)
(
b
2
)
(
b
3
)
(
b
4
)
(
b
5
)
...
If
a
5
=
1
a_5 = 1
a
5
=
1
, find
b
5
b_5
b
5
.p2.
P
(
x
)
=
x
4
−
3
x
3
+
4
x
2
−
9
x
+
5
P(x) = x^4 - 3x^3 + 4x^2 - 9x + 5
P
(
x
)
=
x
4
−
3
x
3
+
4
x
2
−
9
x
+
5
.
Q
(
x
)
Q(x)
Q
(
x
)
is a
3
3
3
rd-degree polynomial whose graph intersects the graph of
P
(
x
)
P(x)
P
(
x
)
at
x
=
1
x = 1
x
=
1
,
2
2
2
,
5
5
5
, and
10
10
10
. Compute
Q
(
0
)
Q(0)
Q
(
0
)
. p3. Distinct real values
x
1
x_1
x
1
,
x
2
x_2
x
2
,
x
3
x_3
x
3
,
x
4
x_4
x
4
all satisfy
∣
∣
x
−
3
∣
−
5
∣
=
1.34953
||x - 3| - 5| = 1.34953
∣∣
x
−
3∣
−
5∣
=
1.34953
. Find
x
1
+
x
2
+
x
3
+
x
4
x_1 + x_2 + x_3 + x_4
x
1
+
x
2
+
x
3
+
x
4
. p4. Triangle
A
B
C
ABC
A
BC
has sides
A
B
=
8
AB = 8
A
B
=
8
,
B
C
=
10
BC = 10
BC
=
10
, and
C
A
=
11
CA = 11
C
A
=
11
. Let
L
L
L
be the locus of points in the interior of triangle
A
B
C
ABC
A
BC
which are within one unit of either
A
A
A
,
B
B
B
, or
C
C
C
. Find the area of
L
L
L
. p5. Triangles
A
B
C
ABC
A
BC
and
A
D
E
ADE
A
D
E
are equilateral, and
A
D
AD
A
D
is an altitude of
A
B
C
ABC
A
BC
. The area of the intersection of these triangles is
3
3
3
. Find the area of the larger triangle
A
B
C
ABC
A
BC
. PS. You had better use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here.