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Problems
Contests
National and Regional Contests
France Contests
France Team Selection Test
2004 France Team Selection Test
2004 France Team Selection Test
Part of
France Team Selection Test
Subcontests
(3)
2
1
Hide problems
Inequality for the incircle
Let
P
P
P
,
Q
Q
Q
, and
R
R
R
be the points where the incircle of a triangle
A
B
C
ABC
A
BC
touches the sides
A
B
AB
A
B
,
B
C
BC
BC
, and
C
A
CA
C
A
, respectively. Prove the inequality
B
C
P
Q
+
C
A
Q
R
+
A
B
R
P
≥
6
\frac{BC} {PQ} + \frac{CA} {QR} + \frac{AB} {RP} \geq 6
PQ
BC
+
QR
C
A
+
RP
A
B
≥
6
.
3
2
Hide problems
Same as German TST
Each point of the plane with two integer coordinates is the center of a disk with radius
1
1000
\frac {1} {1000}
1000
1
. Prove that there exists an equilateral triangle whose vertices belong to distinct disks. Prove that such a triangle has side-length greater than 96.
Best wishes from Romania
Let
P
P
P
be the set of prime numbers. Consider a subset
M
M
M
of
P
P
P
with at least three elements. We assume that, for each non empty and finite subset
A
A
A
of
M
M
M
, with
A
≠
M
A \neq M
A
=
M
, the prime divisors of the integer
(
∏
p
∈
A
)
−
1
( \prod_{p \in A} ) - 1
(
∏
p
∈
A
)
−
1
belong to
M
M
M
. Prove that
M
=
P
M = P
M
=
P
.
1
2
Hide problems
Partition into two sets with equal product
If
n
n
n
is a positive integer, let
A
=
{
n
,
n
+
1
,
.
.
.
,
n
+
17
}
A = \{n,n+1,...,n+17 \}
A
=
{
n
,
n
+
1
,
...
,
n
+
17
}
. Does there exist some values of
n
n
n
for which we can divide
A
A
A
into two disjoints subsets
B
B
B
and
C
C
C
such that the product of the elements of
B
B
B
is equal to the product of the elements of
C
C
C
?
A trivial inequality
Let
n
n
n
be a positive integer, and
a
1
,
.
.
.
,
a
n
,
b
1
,
.
.
.
,
b
n
a_1,...,a_n, b_1,..., b_n
a
1
,
...
,
a
n
,
b
1
,
...
,
b
n
be
2
n
2n
2
n
positive real numbers such that
a
1
+
.
.
.
+
a
n
=
b
1
+
.
.
.
+
b
n
=
1
a_1 + ... + a_n = b_1 + ... + b_n = 1
a
1
+
...
+
a
n
=
b
1
+
...
+
b
n
=
1
. Find the minimal value of
a
1
2
a
1
+
b
1
+
a
2
2
a
2
+
b
2
+
.
.
.
+
a
n
2
a
n
+
b
n
\frac {a_1^2} {a_1 + b_1} + \frac {a_2^2} {a_2 + b_2} + ...+ \frac {a_n^2} {a_n + b_n}
a
1
+
b
1
a
1
2
+
a
2
+
b
2
a
2
2
+
...
+
a
n
+
b
n
a
n
2
.