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Problems
Contests
National and Regional Contests
Netherlands Contests
Dutch Mathematical Olympiad
2010 Dutch Mathematical Olympiad
2010 Dutch Mathematical Olympiad
Part of
Dutch Mathematical Olympiad
Subcontests
(5)
3
1
Hide problems
product of integer lengths 6 line segments is a perfect square
Consider a triangle
X
Y
Z
XYZ
X
Y
Z
and a point
O
O
O
in its interior. Three lines through
O
O
O
are drawn, parallel to the respective sides of the triangle. The intersections with the sides of the triangle determine six line segments from
O
O
O
to the sides of the triangle. The lengths of these segments are integer numbers
a
,
b
,
c
,
d
,
e
a, b, c, d, e
a
,
b
,
c
,
d
,
e
and
f
f
f
(see figure). Prove that the product
a
⋅
b
⋅
c
⋅
d
⋅
e
⋅
f
a \cdot b \cdot c\cdot d \cdot e \cdot f
a
⋅
b
⋅
c
⋅
d
⋅
e
⋅
f
is a perfect square.[asy] unitsize(1 cm);pair A, B, C, D, E, F, O, X, Y, Z;X = (1,4); Y = (0,0); Z = (5,1.5); O = (1.8,2.2); A = extension(O, O + Z - X, X, Y); B = extension(O, O + Y - Z, X, Y); C = extension(O, O + X - Y, Y, Z); D = extension(O, O + Z - X, Y, Z); E = extension(O, O + Y - Z, Z, X); F = extension(O, O + X - Y, Z, X);draw(X--Y--Z--cycle); draw(A--D); draw(B--E); draw(C--F);dot("
A
A
A
", A, NW); dot("
B
B
B
", B, NW); dot("
C
C
C
", C, SE); dot("
D
D
D
", D, SE); dot("
E
E
E
", E, NE); dot("
F
F
F
", F, NE); dot("
O
O
O
", O, S); dot("
X
X
X
", X, N); dot("
Y
Y
Y
", Y, SW); dot("
Z
Z
Z
", Z, dir(0)); label("
a
a
a
", (A + O)/2, SW); label("
b
b
b
", (B + O)/2, SE); label("
c
c
c
", (C + O)/2, SE); label("
d
d
d
", (D + O)/2, SW); label("
e
e
e
", (E + O)/2, SE); label("
f
f
f
", (F + O)/2, NW); [/asy]
1
1
Hide problems
area of region enclosed by 3 circles, pairwise tangent, 90-60-30 triangle
Consider a triangle
A
B
C
ABC
A
BC
such that
∠
A
=
9
0
o
,
∠
C
=
6
0
o
\angle A = 90^o, \angle C =60^o
∠
A
=
9
0
o
,
∠
C
=
6
0
o
and
∣
A
C
∣
=
6
|AC|= 6
∣
A
C
∣
=
6
. Three circles with centers
A
,
B
A, B
A
,
B
and
C
C
C
are pairwise tangent in points on the three sides of the triangle. Determine the area of the region enclosed by the three circles (the grey area in the figure).[asy] unitsize(0.2 cm);pair A, B, C; real[] r;A = (6,0); B = (6,6*sqrt(3)); C = (0,0); r[1] = 3*sqrt(3) - 3; r[2] = 3*sqrt(3) + 3; r[3] = 9 - 3*sqrt(3);fill(arc(A,r[1],180,90)--arc(B,r[2],270,240)--arc(C,r[3],60,0)--cycle, gray(0.7)); draw(A--B--C--cycle); draw(Circle(A,r[1])); draw(Circle(B,r[2])); draw(Circle(C,r[3]));dot("
A
A
A
", A, SE); dot("
B
B
B
", B, NE); dot("
C
C
C
", C, SW); [/asy]
4
1
Hide problems
119 pairs of (x,y) such that (x+my) and (mx+y) are integers , 0<x,y<1,
(a) Determine all pairs
(
x
,
y
)
(x, y)
(
x
,
y
)
of (real) numbers with
0
<
x
<
1
0 < x < 1
0
<
x
<
1
and
0
<
y
<
1
0 <y < 1
0
<
y
<
1
for which
x
+
3
y
x + 3y
x
+
3
y
and
3
x
+
y
3x + y
3
x
+
y
are both integer. An example is
(
x
,
y
)
=
(
8
3
,
7
8
)
(x,y) =( \frac{8}{3}, \frac{7}{8})
(
x
,
y
)
=
(
3
8
,
8
7
)
, because
x
+
3
y
=
3
8
+
21
8
=
24
8
=
3
x+3y =\frac38 +\frac{21}{8} =\frac{24}{8} = 3
x
+
3
y
=
8
3
+
8
21
=
8
24
=
3
and
3
x
+
y
=
9
8
+
7
8
=
16
8
=
2
3x+y = \frac98 + \frac78 =\frac{16}{8} = 2
3
x
+
y
=
8
9
+
8
7
=
8
16
=
2
.(b) Determine the integer
m
>
2
m > 2
m
>
2
for which there are exactly
119
119
119
pairs
(
x
,
y
)
(x,y)
(
x
,
y
)
with
0
<
x
<
1
0 < x < 1
0
<
x
<
1
and
0
<
y
<
1
0 < y < 1
0
<
y
<
1
such that
x
+
m
y
x + my
x
+
m
y
and
m
x
+
y
mx + y
m
x
+
y
are integers.Remark: if
u
≠
v
,
u \ne v,
u
=
v
,
the pairs
(
u
,
v
)
(u, v)
(
u
,
v
)
and
(
v
,
u
)
(v, u)
(
v
,
u
)
are different.
2
1
Hide problems
polite numbers, sum of consecutive numbers, power of two
A number is called polite if it can be written as
m
+
(
m
+
1
)
+
.
.
.
+
n
m + (m+1)+...+ n
m
+
(
m
+
1
)
+
...
+
n
, for certain positive integers
m
<
n
m <n
m
<
n
. For example:
18
18
18
is polite, since
18
=
5
+
6
+
7
18 =5 + 6 + 7
18
=
5
+
6
+
7
. A number is called a power of two if it can be written as
2
ℓ
2^{\ell}
2
ℓ
for some integer
ℓ
≥
0
\ell \ge 0
ℓ
≥
0
. (a) Show that no number is both polite and a power of two. (b) Show that every positive integer is polite or a power of two.
5
1
Hide problems
game for two with piles of coins, 2010 totally, winning strategy
Amber and Brian are playing a game using
2010
2010
2010
coins. Throughout the game, the coins are divided into a number of piles of at least 1 coin each. A move consists of choosing one or more piles and dividing each of them into two smaller piles. (So piles consisting of only
1
1
1
coin cannot be chosen.) Initially, there is only one pile containing all
2010
2010
2010
coins. Amber and Brian alternatingly take turns to make a move, starting with Amber. The winner is the one achieving the situation where all piles have only one coin. Show that Amber can win the game, no matter which moves Brian makes.