MathDB
Problems
Contests
National and Regional Contests
Sweden Contests
Swedish Mathematical Competition
1979 Swedish Mathematical Competition
1979 Swedish Mathematical Competition
Part of
Swedish Mathematical Competition
Subcontests
(6)
6
1
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aAB < AG < b AB and c AB < BG < d AB if G centroid , G and GA > GB > GC
Find the sharpest inequalities of the form
a
⋅
A
B
<
A
G
<
b
⋅
A
B
a\cdot AB < AG < b\cdot AB
a
⋅
A
B
<
A
G
<
b
⋅
A
B
and
c
⋅
A
B
<
B
G
<
d
⋅
A
B
c\cdot AB < BG < d\cdot AB
c
⋅
A
B
<
BG
<
d
⋅
A
B
for all triangles
A
B
C
ABC
A
BC
with centroid
G
G
G
such that
G
A
>
G
B
>
G
C
GA > GB > GC
G
A
>
GB
>
GC
.
5
1
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min a such that ax^2 - bx + c has two distinct zeros in the interval (0,1)
Find the smallest positive integer
a
a
a
such that for some integers
b
b
b
,
c
c
c
the polynomial
a
x
2
−
b
x
+
c
ax^2 - bx + c
a
x
2
−
b
x
+
c
has two distinct zeros in the interval
(
0
,
1
)
(0,1)
(
0
,
1
)
.
4
1
Hide problems
\int\limits_0^\pi f(x)dx=0, \qquad \int\limits_0^\pi f(x)\cos x dx=0
f
(
x
)
f(x)
f
(
x
)
is continuous on the interval
[
0
,
π
]
[0, \pi]
[
0
,
π
]
and satisfies
∫
0
π
f
(
x
)
d
x
=
0
,
∫
0
π
f
(
x
)
cos
x
d
x
=
0
\int\limits_0^\pi f(x)dx=0, \qquad \int\limits_0^\pi f(x)\cos x dx=0
0
∫
π
f
(
x
)
d
x
=
0
,
0
∫
π
f
(
x
)
cos
x
d
x
=
0
Show that
f
(
x
)
f(x)
f
(
x
)
has at least two zeros in the interval
(
0
,
π
)
(0, \pi)
(
0
,
π
)
.
3
1
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x^{13} + 1/x^{13} as a polynomial in y = x + 1/x
Express
x
13
+
1
x
13
x^{13} + \frac{1}{x^{13}}
x
13
+
x
13
1
as a polynomial in
y
=
x
+
1
x
y = x + \frac{1}{x}
y
=
x
+
x
1
.
2
1
Hide problems
rational x in (3,4) such that \sqrt{x-3} and \sqrt{x+1} are rational
Find rational
x
x
x
in
(
3
,
4
)
(3,4)
(
3
,
4
)
such that
x
−
3
\sqrt{x-3}
x
−
3
and
x
+
1
\sqrt{x+1}
x
+
1
are rational.
1
1
Hide problems
nxn system x_1 + 2 x_2 + 3 x_3 + \cdots + (n-1) x_{n-1} + n x_n = n
Solve the equations:
{
x
1
+
2
x
2
+
3
x
3
+
⋯
+
(
n
−
1
)
x
n
−
1
+
n
x
n
=
n
2
x
1
+
3
x
2
+
4
x
3
+
⋯
+
n
x
n
−
1
+
x
n
=
n
−
1
3
x
1
+
4
x
2
+
5
x
3
+
⋯
+
x
n
−
1
+
2
x
n
=
n
−
2
⋯
⋯
⋯
⋯
⋅
⋯
⋯
⋯
⋯
⋅
⋯
⋯
⋯
⋯
⋅
(
n
−
1
)
x
1
+
n
x
2
+
x
3
+
⋯
+
(
n
−
3
)
x
n
−
1
+
(
n
−
2
)
x
n
=
2
n
x
1
+
x
2
+
2
x
3
+
⋯
+
(
n
−
2
)
x
n
−
1
+
(
n
−
1
)
x
n
=
1
\left\{ \begin{array}{l} x_1 + 2 x_2 + 3 x_3 + \cdots + (n-1) x_{n-1} + n x_n = n \\ 2 x_1 + 3 x_2 + 4 x_3 + \cdots + n x_{n-1} + x_n = n-1 \\ 3 x_1 + 4 x_2 + 5 x_3 + \cdots + x_{n-1} + 2 x_n = n-2 \\ \cdots \cdots \cdots \cdots\cdot\cdots \cdots \cdots \cdots\cdot\cdots \cdots \cdots \cdots\cdot \\ (n-1) x_1 + n x_2 + x_3 + \cdots + (n-3) x_{n-1} + (n-2) x_n = 2 \\ n x_1 + x_2 + 2 x_3 + \cdots + (n-2) x_{n-1} + (n-1) x_n = 1 \end{array} \right.
⎩
⎨
⎧
x
1
+
2
x
2
+
3
x
3
+
⋯
+
(
n
−
1
)
x
n
−
1
+
n
x
n
=
n
2
x
1
+
3
x
2
+
4
x
3
+
⋯
+
n
x
n
−
1
+
x
n
=
n
−
1
3
x
1
+
4
x
2
+
5
x
3
+
⋯
+
x
n
−
1
+
2
x
n
=
n
−
2
⋯⋯⋯⋯
⋅
⋯⋯⋯⋯
⋅
⋯⋯⋯⋯
⋅
(
n
−
1
)
x
1
+
n
x
2
+
x
3
+
⋯
+
(
n
−
3
)
x
n
−
1
+
(
n
−
2
)
x
n
=
2
n
x
1
+
x
2
+
2
x
3
+
⋯
+
(
n
−
2
)
x
n
−
1
+
(
n
−
1
)
x
n
=
1