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Problems
Contests
National and Regional Contests
India Contests
Regional Mathematical Olympiad
2012 India Regional Mathematical Olympiad
2012 India Regional Mathematical Olympiad
Part of
Regional Mathematical Olympiad
Subcontests
(8)
8
1
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1/(x-2)(y-2)(z-2)+8/(x+2)(y+2)(z+2)\le 1/ 32 if 2(xy + yz + zx) = xyz
Let
x
,
y
,
z
x, y, z
x
,
y
,
z
be positive real numbers such that
2
(
x
y
+
y
z
+
z
x
)
=
x
y
z
2(xy + yz + zx) = xyz
2
(
x
y
+
yz
+
z
x
)
=
x
yz
. Prove that
1
(
x
−
2
)
(
y
−
2
)
(
z
−
2
)
+
8
(
x
+
2
)
(
y
+
2
)
(
z
+
2
)
≤
1
32
\frac{1}{(x-2)(y-2)(z-2)} + \frac{8}{(x+2)(y+2)(z+2)} \le \frac{1}{32}
(
x
−
2
)
(
y
−
2
)
(
z
−
2
)
1
+
(
x
+
2
)
(
y
+
2
)
(
z
+
2
)
8
≤
32
1
7
1
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angle chasing in Mumbai
On the extension of chord
A
B
AB
A
B
of a circle centroid at
O
O
O
a point
X
X
X
is taken and tangents
X
C
XC
XC
and
X
D
XD
X
D
to the circle are drawn from it with
C
C
C
and
D
D
D
lying on the circle, let
E
E
E
be the midpoint of the line segment
C
D
CD
C
D
. If
∠
O
E
B
=
14
0
o
\angle OEB = 140^o
∠
OEB
=
14
0
o
then determine with proof the magnitude of
∠
A
O
B
\angle AOB
∠
A
OB
.
3
3
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Given a+b=1; show expression<=1(Well-known)
Let
a
a
a
and
b
b
b
be positive real numbers such that
a
+
b
=
1
a+b=1
a
+
b
=
1
. Prove that
a
a
b
b
+
a
b
b
a
≤
1
a^ab^b+a^bb^a\le 1
a
a
b
b
+
a
b
b
a
≤
1
.
2^{2x} \cdot 2^{3\{x\}} = 11 \cdot 2^{5\{x\}} + 5 \cdot 2^{2[x]}
Solve for real
x
x
x
:
2
2
x
⋅
2
3
{
x
}
=
11
⋅
2
5
{
x
}
+
5
⋅
2
2
[
x
]
2^{2x} \cdot 2^{3\{x\}} = 11 \cdot 2^{5\{x\}} + 5 \cdot 2^{2[x]}
2
2
x
⋅
2
3
{
x
}
=
11
⋅
2
5
{
x
}
+
5
⋅
2
2
[
x
]
(For a real number
x
,
[
x
]
x, [x]
x
,
[
x
]
denotes the greatest integer less than or equal to x. For instance,
[
2.5
]
=
2
[2.5] = 2
[
2.5
]
=
2
,
[
−
3.1
]
=
−
4
[-3.1] = -4
[
−
3.1
]
=
−
4
,
[
π
]
=
3
[\pi ] = 3
[
π
]
=
3
. For a real number
x
,
{
x
}
x, \{x\}
x
,
{
x
}
is defined as
x
−
[
x
]
x - [x]
x
−
[
x
]
.)
(2^x-1)(2^y-1)=2^{2^z}+1[Indian RMO 2012(b) Q3]
Find all natural numbers
x
,
y
,
z
x,y,z
x
,
y
,
z
such that
(
2
x
−
1
)
(
2
y
−
1
)
=
2
2
z
+
1.
(2^x-1)(2^y-1)=2^{2^z}+1.
(
2
x
−
1
)
(
2
y
−
1
)
=
2
2
z
+
1.
4
5
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5
6
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1
6
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2
6
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6
6
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