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Undergraduate contests
IMC
2011 IMC
2011 IMC
Part of
IMC
Subcontests
(5)
4
2
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IMC Day 1 Problem 4
Let
A
1
,
A
2
,
…
,
A
n
A_1,A_2,\dots, A_n
A
1
,
A
2
,
…
,
A
n
be finite, nonempty sets. Define the function
f
(
t
)
=
∑
k
=
1
n
∑
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
(
−
1
)
k
−
1
t
∣
A
i
1
∪
A
i
2
∪
⋯
∪
A
i
k
∣
.
f(t)=\sum_{k=1}^n \sum_{1\leq i_1<i_2<\dots<i_k\leq n} (-1)^{k-1}t^{|A_{i_1}\cup A_{i_2}\cup \dots\cup A_{i_k}|}.
f
(
t
)
=
k
=
1
∑
n
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
∑
(
−
1
)
k
−
1
t
∣
A
i
1
∪
A
i
2
∪
⋯
∪
A
i
k
∣
.
Prove that
f
f
f
is nondecreasing on
[
0
,
1
]
.
[0,1].
[
0
,
1
]
.
(
∣
A
∣
|A|
∣
A
∣
denotes the number of elements in
A
.
A.
A
.
)
IMC 2011 Day 2, Problem 4
Let
f
f
f
be a polynomial with real coefficients of degree
n
n
n
. Suppose that
f
(
x
)
−
f
(
y
)
x
−
y
\displaystyle \frac{f(x)-f(y)}{x-y}
x
−
y
f
(
x
)
−
f
(
y
)
is an integer for all
0
≤
x
<
y
≤
n
0 \leq x<y \leq n
0
≤
x
<
y
≤
n
. Prove that
a
−
b
∣
f
(
a
)
−
f
(
b
)
a-b | f(a)-f(b)
a
−
b
∣
f
(
a
)
−
f
(
b
)
for all distinct integers
a
,
b
a,b
a
,
b
.
3
2
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IMC 2011 Day 1 Problem 3
Let
p
p
p
be a prime number. Call a positive integer
n
n
n
interesting if
x
n
−
1
=
(
x
p
−
x
+
1
)
f
(
x
)
+
p
g
(
x
)
x^n-1=(x^p-x+1)f(x)+pg(x)
x
n
−
1
=
(
x
p
−
x
+
1
)
f
(
x
)
+
p
g
(
x
)
for some polynomials
f
f
f
and
g
g
g
with integer coefficients. a) Prove that the number
p
p
−
1
p^p-1
p
p
−
1
is interesting. b) For which
p
p
p
is
p
p
−
1
p^p-1
p
p
−
1
the minimal interesting number?
IMC 2011 Day 2, Problem 3
Calculate
∑
n
=
1
∞
ln
(
1
+
1
n
)
ln
(
1
+
1
2
n
)
ln
(
1
+
1
2
n
+
1
)
\displaystyle \sum_{n=1}^\infty \ln \left(1+\frac{1}{n}\right) \ln\left( 1+\frac{1}{2n}\right)\ln\left( 1+\frac{1}{2n+1}\right)
n
=
1
∑
∞
ln
(
1
+
n
1
)
ln
(
1
+
2
n
1
)
ln
(
1
+
2
n
+
1
1
)
.
5
2
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IMC 2011 Day 1 Problem 5
Let
n
n
n
be a positive integer and let
V
V
V
be a
(
2
n
−
1
)
(2n-1)
(
2
n
−
1
)
-dimensional vector space over the two-element field. Prove that for arbitrary vectors
v
1
,
…
,
v
4
n
−
1
∈
V
,
v_1,\dots,v_{4n-1} \in V,
v
1
,
…
,
v
4
n
−
1
∈
V
,
there exists a sequence
1
≤
i
1
<
⋯
<
i
2
n
≤
4
n
−
1
1\leq i_1<\dots<i_{2n}\leq 4n-1
1
≤
i
1
<
⋯
<
i
2
n
≤
4
n
−
1
of indices such that
v
i
1
+
⋯
+
v
i
2
n
=
0.
v_{i_1}+\dots+v_{i_{2n}}=0.
v
i
1
+
⋯
+
v
i
2
n
=
0.
IMC 2011 Day 2, Problem 5
Let
F
=
A
0
A
1
.
.
.
A
n
F=A_0A_1...A_n
F
=
A
0
A
1
...
A
n
be a convex polygon in the plane. Define for all
1
≤
k
≤
n
−
1
1 \leq k \leq n-1
1
≤
k
≤
n
−
1
the operation
f
k
f_k
f
k
which replaces
F
F
F
with a new polygon
f
k
(
F
)
=
A
0
A
1
.
.
A
k
−
1
A
k
′
A
k
+
1
.
.
.
A
n
f_k(F)=A_0A_1..A_{k-1}A_k^\prime A_{k+1}...A_n
f
k
(
F
)
=
A
0
A
1
..
A
k
−
1
A
k
′
A
k
+
1
...
A
n
where
A
k
′
A_k^\prime
A
k
′
is the symmetric of
A
k
A_k
A
k
with respect to the perpendicular bisector of
A
k
−
1
A
k
+
1
A_{k-1}A_{k+1}
A
k
−
1
A
k
+
1
. Prove that
(
f
1
∘
f
2
∘
f
3
∘
.
.
.
∘
f
n
−
1
)
n
(
F
)
=
F
(f_1\circ f_2 \circ f_3 \circ...\circ f_{n-1})^n(F)=F
(
f
1
∘
f
2
∘
f
3
∘
...
∘
f
n
−
1
)
n
(
F
)
=
F
.
2
2
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IMC 2011 Day 1 Problem 2
Does there exist a real
3
×
3
3\times 3
3
×
3
matrix
A
A
A
such that
tr
(
A
)
=
0
\text{tr}(A)=0
tr
(
A
)
=
0
and
A
2
+
A
t
=
I
?
A^2+A^t=I?
A
2
+
A
t
=
I
?
(
tr
(
A
)
\text{tr}(A)
tr
(
A
)
denotes the trace of
A
,
A
t
A,\ A^t
A
,
A
t
the transpose of
A
,
A,
A
,
and
I
I
I
is the identity matrix.)Proposed by Moubinool Omarjee, Paris
IMC 2011 Day 2, Problem 2
An alien race has three genders: male, female and emale. A married triple consists of three persons, one from each gender who all like each other. Any person is allowed to belong to at most one married triple. The feelings are always mutual ( if
x
x
x
likes
y
y
y
then
y
y
y
likes
x
x
x
). The race wants to colonize a planet and sends
n
n
n
males,
n
n
n
females and
n
n
n
emales. Every expedition member likes at least
k
k
k
persons of each of the two other genders. The problem is to create as many married triples so that the colony could grow.a) Prove that if
n
n
n
is even and
k
≥
1
/
2
k\geq 1/2
k
≥
1/2
then there might be no married triple. b) Prove that if
k
≥
3
n
/
4
k \geq 3n/4
k
≥
3
n
/4
then there can be formed
n
n
n
married triple ( i.e. everybody is in a triple).
1
2
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IMC 2011 Day 1 Problem 1
Let
f
:
R
→
R
f:\mathbb{R} \to \mathbb{R}
f
:
R
→
R
be a continuous function. A point
x
x
x
is called a shadow point if there exists a point
y
∈
R
y\in \mathbb{R}
y
∈
R
with
y
>
x
y>x
y
>
x
such that
f
(
y
)
>
f
(
x
)
.
f(y)>f(x).
f
(
y
)
>
f
(
x
)
.
Let
a
<
b
a<b
a
<
b
be real numbers and suppose that
∙
\bullet
∙
all the points of the open interval
I
=
(
a
,
b
)
I=(a,b)
I
=
(
a
,
b
)
are shadow points;
∙
\bullet
∙
a
a
a
and
b
b
b
are not shadow points. Prove that a)
f
(
x
)
≤
f
(
b
)
f(x)\leq f(b)
f
(
x
)
≤
f
(
b
)
for all
a
<
x
<
b
;
a<x<b;
a
<
x
<
b
;
b)
f
(
a
)
=
f
(
b
)
.
f(a)=f(b).
f
(
a
)
=
f
(
b
)
.
Proposed by José Luis Díaz-Barrero, Barcelona
IMC 2011 Day 2, Problem 1
Let
(
a
n
)
⊂
(
1
2
,
1
)
(a_n)\subset (\frac{1}{2},1)
(
a
n
)
⊂
(
2
1
,
1
)
. Define the sequence
x
0
=
0
,
x
n
+
1
=
a
n
+
1
+
x
n
1
+
a
n
+
1
x
n
x_0=0,\displaystyle x_{n+1}=\frac{a_{n+1}+x_n}{1+a_{n+1}x_n}
x
0
=
0
,
x
n
+
1
=
1
+
a
n
+
1
x
n
a
n
+
1
+
x
n
. Is this sequence convergent? If yes find the limit.