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Contests
National and Regional Contests
Sweden Contests
Swedish Mathematical Competition
2015 Swedish Mathematical Competition
2015 Swedish Mathematical Competition
Part of
Swedish Mathematical Competition
Subcontests
(6)
6
1
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2 plyer game with a no of positive integers
Axel and Berta play the following games: On a board are a number of positive integers. One move consists of a player exchanging a number
x
x
x
on the board for two positive integers y and
z
z
z
(not necessarily different), such that
y
+
z
=
x
y + z = x
y
+
z
=
x
. The game ends when the numbers on the board are relatively coprime in pairs. The player who made the last move has then lost the game. At the beginning of the game, only the number
2015
2015
2015
is on the board. The two players make do their moves in turn and Berta begins. One of the players has a winning strategy. Who, and why?
5
1
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combo geo with points and reys
Given a finite number of points in the plane as well as many different rays starting at the origin. It is always possible to pair the points with the rays so that they parallell displaced rays starting in respective points do not intersect?
2
1
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x^3 + y^3 + 2015 = 0
Determine all integer solutions to the equation
x
3
+
y
3
+
2015
=
0
x^3 + y^3 + 2015 = 0
x
3
+
y
3
+
2015
=
0
.
4
1
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logarithm 2x3 system
Solve the system of equations
{
x
log
x
+
y
log
y
+
z
log
x
=
0
log
x
x
+
log
y
y
+
log
z
z
=
0
\left\{\begin{array}{l} x \log x+y \log y+z \log x=0\\ \\ \dfrac{\log x}{x}+\dfrac{\log y}{y}+\dfrac{\log z}{z}=0 \end{array} \right.
⎩
⎨
⎧
x
lo
g
x
+
y
lo
g
y
+
z
lo
g
x
=
0
x
lo
g
x
+
y
lo
g
y
+
z
lo
g
z
=
0
3
1
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min of (a^2+2b^2+4c^2)/ b(a+2c) for a,b,c>0
Let
a
a
a
,
b
b
b
,
c
c
c
be positive real numbers. Determine the minimum value of the following expression
a
2
+
2
b
2
+
4
c
2
b
(
a
+
2
c
)
\frac{a^2+2b^2+4c^2}{b(a+2c)}
b
(
a
+
2
c
)
a
2
+
2
b
2
+
4
c
2
1
1
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diameter of circumcircle bisector of BC, divides AC in 1:3 iff tan B = 2 tan C
Given the acute triangle
A
B
C
ABC
A
BC
. A diameter of the circumscribed circle of the triangle intersects the sides
A
C
AC
A
C
and
B
C
BC
BC
, dividing the side
B
C
BC
BC
in half. Show that the same diameter divides the side
A
C
AC
A
C
in a ratio of
1
:
3
1: 3
1
:
3
, calculated from
A
A
A
, if and only if
tan
B
=
2
tan
C
\tan B = 2 \tan C
tan
B
=
2
tan
C
.