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Problems
Contests
National and Regional Contests
Bangladesh Contests
Bangladesh Mathematical Olympiad
2016 Bangladesh Mathematical Olympiad
2016 Bangladesh Mathematical Olympiad
Part of
Bangladesh Mathematical Olympiad
Subcontests
(9)
4
1
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Arbitary arrange ment to get the sum
BdMO National 2016 Higher Secondary Problem 4: Consider the set of integers
{
1
,
2
,
.
.
.
.
.
.
.
.
.
,
100
}
\left \{ 1, 2, ......... , 100 \right \}
{
1
,
2
,
.........
,
100
}
. Let
{
x
1
,
x
2
,
.
.
.
.
.
.
.
.
.
,
x
100
}
\left \{ x_1, x_2, ......... , x_{100} \right \}
{
x
1
,
x
2
,
.........
,
x
100
}
be some arbitrary arrangement of the integers
{
1
,
2
,
.
.
.
.
.
.
.
.
.
,
100
}
\left \{ 1, 2, ......... , 100 \right \}
{
1
,
2
,
.........
,
100
}
, where all of the
x
i
x_i
x
i
are different. Find the smallest possible value of the sum,
S
=
∣
x
2
−
x
1
∣
+
∣
x
3
−
x
2
∣
+
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
+
∣
x
100
−
x
99
∣
+
∣
x
1
−
x
100
∣
S = \left | x_2 - x_1 \right | + \left | x_3 - x_2 \right | + ................+ \left |x_{100} - x_{99} \right | + \left |x_1 - x_{100} \right |
S
=
∣
x
2
−
x
1
∣
+
∣
x
3
−
x
2
∣
+
................
+
∣
x
100
−
x
99
∣
+
∣
x
1
−
x
100
∣
.
6
1
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Intersection and finding angle from BdMO 2016
BdMO National 2016 Higher Secondary Problem 6
△
A
B
C
\triangle ABC
△
A
BC
is an isosceles triangle with
A
C
=
B
C
AC = BC
A
C
=
BC
and
∠
A
C
B
<
6
0
∘
\angle ACB < 60^{\circ}
∠
A
CB
<
6
0
∘
.
I
I
I
and
O
O
O
are the incenter and circumcenter of
△
A
B
C
\triangle ABC
△
A
BC
. The circumcircle of
△
B
I
O
\triangle BIO
△
B
I
O
intersects
B
C
BC
BC
at
D
≠
B
D \neq B
D
=
B
.(a) Do the lines
A
C
AC
A
C
and
D
I
DI
D
I
intersect? Give a proof.(b) What is the angle of intersection between the lines
O
D
OD
O
D
and
I
B
IB
I
B
?
5
1
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Martians and Human with Circular table
Bangladesh National Mathematical Olympiad 2016 Higher SecondaryProblem 5: Suppose there are
m
m
m
Martians and
n
n
n
Earthlings at an intergalactic peace conference. To ensure the Martians stay peaceful at the conference, we must make sure that no two Martians sit together, such that between any two Martians there is always at least one Earthling. (a) Suppose all
m
+
n
m + n
m
+
n
Martians and Earthlings are seated in a line. How many ways can the Earthlings and Martians be seated in a line?(b) Suppose now that the
m
+
n
m+n
m
+
n
Martians and Earthlings are seated around a circular round-table. How many ways can the Earthlings and Martians be seated around the round-table?
2
1
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Divisors of 6000 and perfect squares
Bangladesh National Mathematical Olympiad 2016 Higher SecondaryProblem 2: (a) How many positive integer factors does
6000
6000
6000
have?(b) How many positive integer factors of
6000
6000
6000
are not perfect squares?
1
1
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Divisible by 6 & 7 not both
BdMO National
2016
2016
2016
Higher Secondary Problem 1: (a) Show that
n
(
n
+
1
)
(
n
+
2
)
n(n + 1)(n + 2)
n
(
n
+
1
)
(
n
+
2
)
is divisible by
6
6
6
. (b) Show that
1
2015
+
2
2015
+
3
2015
+
4
2015
+
5
2015
+
6
2015
1^{2015} + 2^{2015} + 3^{2015} + 4^{2015} + 5^{2015} + 6^{2015}
1
2015
+
2
2015
+
3
2015
+
4
2015
+
5
2015
+
6
2015
is divisible by
7
7
7
.
3
1
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Angle chasing with 30 and 39 degree
△
A
B
C
\triangle ABC
△
A
BC
is isosceles
A
B
=
A
C
AB = AC
A
B
=
A
C
.
P
P
P
is a point inside
△
A
B
C
\triangle ABC
△
A
BC
such that
∠
B
C
P
=
30
\angle BCP = 30
∠
BCP
=
30
and
∠
A
P
B
=
150
\angle APB = 150
∠
A
PB
=
150
and
∠
C
A
P
=
39
\angle CAP = 39
∠
C
A
P
=
39
. Find
∠
B
A
P
\angle BAP
∠
B
A
P
.
9
1
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Intregration with counting!
The integral
Z
(
0
)
=
∫
−
∞
∞
d
x
e
−
x
2
=
π
Z(0)=\int^{\infty}_{-\infty} dx e^{-x^2}= \sqrt{\pi}
Z
(
0
)
=
∫
−
∞
∞
d
x
e
−
x
2
=
π
(a)(3 POINTS:)Show that the integral
Z
(
j
)
=
∫
−
∞
∞
d
x
e
−
x
2
+
j
x
Z(j)=\int^{\infty}_{-\infty} dx e^{-x^{2}+jx}
Z
(
j
)
=
∫
−
∞
∞
d
x
e
−
x
2
+
j
x
Where
j
j
j
is not a function of
x
x
x
,is
Z
(
j
)
=
e
j
2
/
4
a
Z
(
0
)
Z(j)=e^{j^{2}/4a} Z(0)
Z
(
j
)
=
e
j
2
/4
a
Z
(
0
)
(b)(10 POINTS):Show that,
1
Z
(
0
)
=
∫
x
2
n
e
−
x
2
=
(
2
n
−
1
)
!
!
2
n
\dfrac 1 {Z(0)}=\int x^{2n} e^{-x^2}= \dfrac {(2n-1)!!}{2^n}
Z
(
0
)
1
=
∫
x
2
n
e
−
x
2
=
2
n
(
2
n
−
1
)!!
Where
(
2
n
−
1
)
!
!
(2n-1)!!
(
2
n
−
1
)!!
is defined as
(
2
n
−
1
)
(
2
n
−
3
)
×
.
.
.
×
3
×
1
(2n-1)(2n-3)\times...\times3\times 1
(
2
n
−
1
)
(
2
n
−
3
)
×
...
×
3
×
1
(c)(7 POINTS):What is the number of ways to form
n
n
n
pairs from
2
n
2n
2
n
distinct objects?Interept the previous part of the problem in term of this answer.
7
1
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Probability problem
Juli is a mathematician and devised an algorithm to find a husband. The strategy is: • Start interviewing a maximum of
1000
1000
1000
prospective husbands. Assign a ranking
r
r
r
to each person that is a positive integer. No two prospects will have same the rank
r
r
r
. • Reject the first
k
k
k
men and let
H
H
H
be highest rank of these
k
k
k
men. • After rejecting the first
k
k
k
men, select the next prospect with a rank greater than
H
H
H
and then stop the search immediately. If no candidate is selected after
999
999
999
interviews, the
1000
t
h
1000th
1000
t
h
person is selected.Juli wants to find the value of
k
k
k
for which she has the highest probability of choosing the highest ranking prospect among all
1000
1000
1000
candidates without having to interview all
1000
1000
1000
prospects. (a) (6 points:) What is the probability that the highest ranking prospect among all
1000
1000
1000
prospects is the
(
m
+
1
)
t
h
(m + 1)th
(
m
+
1
)
t
h
prospect? (b) (6 points:) Assume the highest ranking prospect is the
(
m
+
1
)
t
h
(m + 1)th
(
m
+
1
)
t
h
person to be interviewed. What is the probability that the highest rank candidate among the first
m
m
m
candidates is one of the first
k
k
k
candidates who were rejected? (c) (6 points:) What is the probability that the prospect with the highest rank is the
(
m
+
1
)
t
h
(m+1)th
(
m
+
1
)
t
h
person and that Juli will choose the
(
m
+
1
)
t
h
(m+1)th
(
m
+
1
)
t
h
man using this algorithm? (d) (16 points:) The total probability that Juli will choose the highest ranking prospect among the
1000
1000
1000
prospects is the sum of the probability for each possible value of
m
+
1
m+1
m
+
1
with
m
+
1
m+1
m
+
1
ranging between
k
+
1
k+1
k
+
1
and
1000
1000
1000
. Find the sum. To simplify your answer use the formula
I
n
N
≈
1
N
−
1
+
1
N
−
2
+
.
.
.
+
1
2
+
1
In N \approx \frac{1}{N-1}+\frac{1}{N-2}+...+\frac{1}{2}+1
I
n
N
≈
N
−
1
1
+
N
−
2
1
+
...
+
2
1
+
1
(e) (6 points:) Find that value of
k
k
k
that maximizes the probability of choosing the highest ranking prospect without interviewing all
1000
1000
1000
candidates. You may need to know that the maximum of the function
x
l
n
A
x
−
1
x ln \frac{A}{x-1}
x
l
n
x
−
1
A
is approximately
A
+
1
e
\frac{A + 1}{e}
e
A
+
1
, where
A
A
A
is a constant and
e
e
e
is Euler’s number,
e
=
2.718....
e = 2.718....
e
=
2.718....
8
1
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Angle Bisector in Circle with Diameter Extensions
Triangle
A
B
C
ABC
A
BC
is inscribed in circle
ω
\omega
ω
with
A
B
=
5
AB = 5
A
B
=
5
,
B
C
=
7
BC = 7
BC
=
7
, and
A
C
=
3
AC = 3
A
C
=
3
. The bisector of angle
A
A
A
meets side
B
C
BC
BC
at
D
D
D
and circle
ω
\omega
ω
at a second point
E
E
E
. Let
γ
\gamma
γ
be the circle with diameter
D
E
DE
D
E
. Circles
ω
\omega
ω
and
γ
\gamma
γ
meet at
E
E
E
and a second point
F
F
F
. Then
A
F
2
=
m
n
AF^2 = \frac mn
A
F
2
=
n
m
, where m and n are relatively prime positive integers. Find
m
+
n
m + n
m
+
n
.