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Problems
Contests
National and Regional Contests
Sweden Contests
Swedish Mathematical Competition
1994 Swedish Mathematical Competition
1994 Swedish Mathematical Competition
Part of
Swedish Mathematical Competition
Subcontests
(6)
1
1
Hide problems
decimal expansions fo roots of x\sqrt8 + 1/(x\sqrt8) = \sqrt8 in place 1994
x
8
+
1
x
8
=
8
x\sqrt8 + \frac{1}{x\sqrt8} = \sqrt8
x
8
+
x
8
1
=
8
has two real solutions
x
1
,
x
2
x_1, x_2
x
1
,
x
2
. The decimal expansion of
x
1
x_1
x
1
has the digit
6
6
6
in place
1994
1994
1994
. What digit does
x
2
x_2
x
2
have in place
1994
1994
1994
?
3
1
Hide problems
sin^2c = sin^2a + sin^2b
The vertex
B
B
B
of the triangle
A
B
C
ABC
A
BC
lies in the plane
P
P
P
. The plane of the triangle meets the plane in a line
L
L
L
. The angle between
L
L
L
and
A
B
AB
A
B
is a, and the angle between
L
L
L
and
B
C
BC
BC
is
b
b
b
. The angle between the two planes is
c
c
c
. Angle
A
B
C
ABC
A
BC
is
9
0
o
90^o
9
0
o
. Show that
sin
2
c
=
sin
2
a
+
sin
2
b
\sin^2c = \sin^2a + \sin^2b
sin
2
c
=
sin
2
a
+
sin
2
b
. https://cdn.artofproblemsolving.com/attachments/9/e/c0608e5408fd27a5f907a3488cce7dc2af6953.png
6
1
Hide problems
f(a+b) = f(f(a)+b), f(a+b) = f(a)+f(b) for a+b < 10, f(10) = 1
Let
N
N
N
be the set of non-negative integers. The function
f
:
N
→
N
f:N\to N
f
:
N
→
N
satisfies
f
(
a
+
b
)
=
f
(
f
(
a
)
+
b
)
f(a+b) = f(f(a)+b)
f
(
a
+
b
)
=
f
(
f
(
a
)
+
b
)
for all
a
,
b
a, b
a
,
b
and
f
(
a
+
b
)
=
f
(
a
)
+
f
(
b
)
f(a+b) = f(a)+f(b)
f
(
a
+
b
)
=
f
(
a
)
+
f
(
b
)
for
a
+
b
<
10
a+b < 10
a
+
b
<
10
. Also
f
(
10
)
=
1
f(10) = 1
f
(
10
)
=
1
. How many three digit numbers
n
n
n
satisfy
f
(
n
)
=
f
(
N
)
f(n) = f(N)
f
(
n
)
=
f
(
N
)
, where
N
N
N
is the "tower"
2
,
3
,
4
,
5
2, 3, 4, 5
2
,
3
,
4
,
5
, in other words, it is
2
a
2^a
2
a
, where
a
=
3
b
a = 3^b
a
=
3
b
, where
b
=
4
5
b = 4^5
b
=
4
5
?
5
1
Hide problems
a_1^2 > 2ka_2/(k-1) if a polynomial had k distinct real roots
The polynomial
x
k
+
a
1
x
k
−
1
+
a
2
x
k
−
2
+
.
.
.
+
a
k
x^k + a_1x^{k-1} + a_2x^{k-2} +... + a_k
x
k
+
a
1
x
k
−
1
+
a
2
x
k
−
2
+
...
+
a
k
has
k
k
k
distinct real roots. Show that
a
1
2
>
2
k
a
2
k
−
1
a_1^2 > \frac{2ka_2}{k-1}
a
1
2
>
k
−
1
2
k
a
2
.
4
1
Hide problems
2n^3 - m^3 = mn^2 + 11
Find all integers
m
,
n
m, n
m
,
n
such that
2
n
3
−
m
3
=
m
n
2
+
11
2n^3 - m^3 = mn^2 + 11
2
n
3
−
m
3
=
m
n
2
+
11
.
2
1
Hide problems
cot B + cot C >= 2/3 when medians from B,C are perpendicular
In the triangle
A
B
C
ABC
A
BC
, the medians from
B
B
B
and
C
C
C
are perpendicular. Show that
cot
B
+
cot
C
≥
2
3
\cot B + \cot C \ge \frac23
cot
B
+
cot
C
≥
3
2
.