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Contests
National and Regional Contests
Ireland Contests
Ireland National Math Olympiad
1991 Irish Math Olympiad
1991 Irish Math Olympiad
Part of
Ireland National Math Olympiad
Subcontests
(5)
4
2
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Eight Politicians Attending Parliament
Eight politicians stranded on a desert island on January 1st, 1991, decided to establish a parliament. They decided on the following rules of attendance:(a) There should always be at least one person present on each day.(b) On no two days should the same subset attend.(c) The members present on day
N
N
N
should include for each
K
<
N
K<N
K
<
N
,
(
K
≥
1
)
(K\ge 1)
(
K
≥
1
)
at least one member who was present on day
K
K
K
.For how many days can the parliament sit before one of the rules is broken?
Function Equation on Positive Rationals
Let
P
\mathbb{P}
P
be the set of positive rational numbers and let
f
:
P
→
P
f:\mathbb{P}\to\mathbb{P}
f
:
P
→
P
be such that
f
(
x
)
+
f
(
1
x
)
=
1
f(x)+f\left(\frac{1}{x}\right)=1
f
(
x
)
+
f
(
x
1
)
=
1
and
f
(
2
x
)
=
2
f
(
f
(
x
)
)
f(2x)=2f(f(x))
f
(
2
x
)
=
2
f
(
f
(
x
))
for all
x
∈
P
x\in\mathbb{P}
x
∈
P
. Find, with proof, an explicit expression for
f
(
x
)
f(x)
f
(
x
)
for all
x
∈
P
x\in \mathbb{P}
x
∈
P
.
3
1
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Constructing all Integers with 3 Functions
Three operations
f
,
g
f,g
f
,
g
and
h
h
h
are defined on subsets of the natural numbers
N
\mathbb{N}
N
as follows:
f
(
n
)
=
10
n
f(n)=10n
f
(
n
)
=
10
n
, if
n
n
n
is a positive integer;
g
(
n
)
=
10
n
+
4
g(n)=10n+4
g
(
n
)
=
10
n
+
4
, if
n
n
n
is a positive integer;
h
(
n
)
=
n
2
h(n)=\frac{n}{2}
h
(
n
)
=
2
n
, if
n
n
n
is an even positive integer. Prove that, starting from
4
4
4
, every natural number can be constructed by performing a finite number of operations
f
f
f
,
g
g
g
and
h
h
h
in some order.
[
[
[
For example:
35
=
h
(
f
(
h
(
g
(
h
(
h
(
4
)
)
)
)
)
)
.
]
35=h(f(h(g(h(h(4)))))).]
35
=
h
(
f
(
h
(
g
(
h
(
h
(
4
))))))
.
]
5
2
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find all polynomials 2 (IrMO 1991)
Find all polynomials
f
(
x
)
=
x
n
+
a
1
x
n
−
1
+
⋯
+
a
n
f(x) = x^{n} + a_{1}x^{n-1} + \cdots + a_{n}
f
(
x
)
=
x
n
+
a
1
x
n
−
1
+
⋯
+
a
n
with the following properties(a) all the coefficients
a
1
,
a
2
,
.
.
.
,
a
n
a_{1}, a_{2}, ..., a_{n}
a
1
,
a
2
,
...
,
a
n
belong to the set
{
−
1
,
1
}
\{ -1, 1 \}
{
−
1
,
1
}
; and (b) all the roots of the equation
f
(
x
)
=
0
f(x)=0
f
(
x
)
=
0
are real.
Subset of the Rationals
Let
Q
\mathbb{Q}
Q
denote the set of rational numbers. A nonempty subset
S
S
S
of
Q
\mathbb{Q}
Q
has the following properties:(a)
0
0
0
is not in
S
S
S
;(b) for each
s
1
,
s
2
s_1,s_2
s
1
,
s
2
in
S
S
S
, the rational number
s
1
/
s
2
s_1/s_2
s
1
/
s
2
is in
S
S
S
;(c) there exists a nonzero number
q
∈
Q
\
S
q\in \mathbb{Q} \backslash S
q
∈
Q
\
S
that has the property that every nonzero number in
Q
\
S
\mathbb{Q} \backslash S
Q
\
S
is of the form
q
s
qs
q
s
for some
s
s
s
in
S
S
S
.Prove that if
x
x
x
belongs to
S
S
S
, then there exists elements
y
,
z
y,z
y
,
z
in
S
S
S
such that
x
=
y
+
z
x=y+z
x
=
y
+
z
.
1
2
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Construct Triangle Given 3 Points
Three points
X
,
Y
X,Y
X
,
Y
and
Z
Z
Z
are given that are, respectively, the circumcenter of a triangle
A
B
C
ABC
A
BC
, the mid-point of
B
C
BC
BC
, and the foot of the altitude from
B
B
B
on
A
C
AC
A
C
. Show how to reconstruct the triangle
A
B
C
ABC
A
BC
.
Sum of consecutive squares
Problem. The sum of two consecutive squares can be a square; for instance
3
2
+
4
2
=
5
2
3^2 + 4^2 = 5^2
3
2
+
4
2
=
5
2
.(a) Prove that the sum of
m
m
m
consecutive squares cannot be a square for
m
∈
{
3
,
4
,
5
,
6
}
m \in \{3, 4, 5, 6\}
m
∈
{
3
,
4
,
5
,
6
}
. (b) Find an example of eleven consecutive squares whose sum is a square.Can anyone help me with this? Thanks.
2
2
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Functional equations problem
Problem: Find all polynomials satisfying the equation
f
(
x
2
)
=
(
f
(
x
)
)
2
f(x^2) = (f(x))^2
f
(
x
2
)
=
(
f
(
x
)
)
2
for all real numbers x. I'm not exactly sure where to start though it doesn't look too difficult. Thanks!
Sequence and Inequality
Let a_n=\frac{n^2+1}{\sqrt{n^4+4}}, n=1,2,3,\dots and let
b
n
b_n
b
n
be the product of
a
1
,
a
2
,
a
3
,
…
,
a
n
a_1,a_2,a_3,\dots ,a_n
a
1
,
a
2
,
a
3
,
…
,
a
n
. Prove that
b
n
2
=
n
2
+
1
n
2
+
2
n
+
2
,
\frac{b_n}{\sqrt{2}}=\frac{\sqrt{n^2+1}}{\sqrt{n^2+2n+2}},
2
b
n
=
n
2
+
2
n
+
2
n
2
+
1
,
and deduce that
1
n
3
+
1
<
b
n
2
−
n
n
+
1
<
1
n
3
\frac{1}{n^3+1}<\frac{b_n}{\sqrt{2}}-\frac{n}{n+1}<\frac{1}{n^3}
n
3
+
1
1
<
2
b
n
−
n
+
1
n
<
n
3
1
for all positive integers
n
n
n
.