MathDB
Problems
Contests
National and Regional Contests
India Contests
India National Olympiad
2012 India National Olympiad
2012 India National Olympiad
Part of
India National Olympiad
Subcontests
(6)
3
1
Hide problems
Polys with int coefficients
Define a sequence
<
f
0
(
x
)
,
f
1
(
x
)
,
f
2
(
x
)
,
⋯
>
<f_0 (x), f_1 (x), f_2 (x), \dots>
<
f
0
(
x
)
,
f
1
(
x
)
,
f
2
(
x
)
,
⋯
>
of functions by
f
0
(
x
)
=
1
f_0 (x) = 1
f
0
(
x
)
=
1
f
1
(
x
)
=
x
f_1(x)=x
f
1
(
x
)
=
x
(
f
n
(
x
)
)
2
−
1
=
f
n
+
1
(
x
)
f
n
−
1
(
x
)
(f_n(x))^2 - 1 = f_{n+1}(x) f_{n-1}(x)
(
f
n
(
x
)
)
2
−
1
=
f
n
+
1
(
x
)
f
n
−
1
(
x
)
for
n
≥
1
n \ge 1
n
≥
1
. Prove that each
f
n
(
x
)
f_n (x)
f
n
(
x
)
is a polynomial with integer coefficients.
4
1
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P 4 number of good points 27th Indian NMO 2012
Let
A
B
C
ABC
A
BC
be a triangle. An interior point
P
P
P
of
A
B
C
ABC
A
BC
is said to be good if we can find exactly
27
27
27
rays emanating from
P
P
P
intersecting the sides of the triangle
A
B
C
ABC
A
BC
such that the triangle is divided by these rays into
27
27
27
smaller triangles of equal area. Determine the number of good points for a given triangle
A
B
C
ABC
A
BC
.
5
1
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Indian National Mathematical Olympiad Problem 5
Let
A
B
C
ABC
A
BC
be an acute angled triangle. Let
D
,
E
,
F
D,E,F
D
,
E
,
F
be points on
B
C
,
C
A
,
A
B
BC, CA, AB
BC
,
C
A
,
A
B
such that
A
D
AD
A
D
is the median,
B
E
BE
BE
is the internal bisector and
C
F
CF
CF
is the altitude. Suppose that
∠
F
D
E
=
∠
C
,
∠
D
E
F
=
∠
A
\angle FDE=\angle C, \angle DEF=\angle A
∠
F
D
E
=
∠
C
,
∠
D
EF
=
∠
A
and
∠
E
F
D
=
∠
B
.
\angle EFD=\angle B.
∠
EF
D
=
∠
B
.
Show that
A
B
C
ABC
A
BC
is equilateral.
1
1
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Find max area of quadrilateral
Let
A
B
C
D
ABCD
A
BC
D
be a quadrilateral inscribed in a circle. Suppose
A
B
=
2
+
2
AB=\sqrt{2+\sqrt{2}}
A
B
=
2
+
2
and
A
B
AB
A
B
subtends
135
135
135
degrees at center of circle . Find the maximum possible area of
A
B
C
D
ABCD
A
BC
D
.
2
1
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P 2 Prove 30 \ p_1 - q_1 27th Indian NMO 2012
Let
p
1
<
p
2
<
p
3
<
p
4
p_1<p_2<p_3<p_4
p
1
<
p
2
<
p
3
<
p
4
and
q
1
<
q
2
<
q
3
<
q
4
q_1<q_2<q_3<q_4
q
1
<
q
2
<
q
3
<
q
4
be two sets of prime numbers, such that
p
4
−
p
1
=
8
p_4 - p_1 = 8
p
4
−
p
1
=
8
and
q
4
−
q
1
=
8
q_4 - q_1= 8
q
4
−
q
1
=
8
. Suppose
p
1
>
5
p_1 > 5
p
1
>
5
and
q
1
>
5
q_1>5
q
1
>
5
. Prove that
30
30
30
divides
p
1
−
q
1
p_1 - q_1
p
1
−
q
1
.
6
1
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P 6 Functional equation - 27th Indian NMO 2012
Let
f
:
Z
→
Z
f : \mathbb{Z} \to \mathbb{Z}
f
:
Z
→
Z
be a function satisfying
f
(
0
)
≠
0
f(0) \ne 0
f
(
0
)
=
0
,
f
(
1
)
=
0
f(1) = 0
f
(
1
)
=
0
and
(
i
)
f
(
x
y
)
+
f
(
x
)
f
(
y
)
=
f
(
x
)
+
f
(
y
)
(i) f(xy) + f(x)f(y) = f(x) + f(y)
(
i
)
f
(
x
y
)
+
f
(
x
)
f
(
y
)
=
f
(
x
)
+
f
(
y
)
(
i
i
)
(
f
(
x
−
y
)
−
f
(
0
)
)
f
(
x
)
f
(
y
)
=
0
(ii)\left(f(x-y) - f(0)\right ) f(x)f(y) = 0
(
ii
)
(
f
(
x
−
y
)
−
f
(
0
)
)
f
(
x
)
f
(
y
)
=
0
for all
x
,
y
∈
Z
x,y \in \mathbb{Z}
x
,
y
∈
Z
, simultaneously.
(
a
)
(a)
(
a
)
Find the set of all possible values of the function
f
f
f
.
(
b
)
(b)
(
b
)
If
f
(
10
)
≠
0
f(10) \ne 0
f
(
10
)
=
0
and
f
(
2
)
=
0
f(2) = 0
f
(
2
)
=
0
, find the set of all integers
n
n
n
such that
f
(
n
)
≠
0
f(n) \ne 0
f
(
n
)
=
0
.