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Problems
Contests
National and Regional Contests
Netherlands Contests
Dutch Mathematical Olympiad
2018 Dutch Mathematical Olympiad
2018 Dutch Mathematical Olympiad
Part of
Dutch Mathematical Olympiad
Subcontests
(5)
1
1
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10-digit shuffle numbers, divisible by 11, 12 (at any order of the digits)
We call a positive integer a shuffle number if the following hold: (1) All digits are nonzero. (2) The number is divisible by
11
11
11
. (3) The number is divisible by
12
12
12
. If you put the digits in any other order, you again have a number that is divisible by
12
12
12
. How many
10
10
10
-digit shuffle numbers are there?
2
1
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coloring red or blue the numbers 1 to 15
The numbers
1
1
1
to
15
15
15
are each coloured blue or red. Determine all possible colourings that satisfy the following rules: • The number
15
15
15
is red. • If numbers
x
x
x
and
y
y
y
have different colours and
x
+
y
≤
15
x + y \le 15
x
+
y
≤
15
, then
x
+
y
x + y
x
+
y
is blue. • If numbers
x
x
x
and
y
y
y
have different colours and
x
⋅
y
≤
15
x \cdot y \le 15
x
⋅
y
≤
15
, then
x
⋅
y
x \cdot y
x
⋅
y
is red.
3
1
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3x3 system, x^2 + y^2 = -x + 3y + z, x^2 + y^2 = -x + 3y + z, x^2 + z^2 = 2x + 2
Determine all triples
(
x
,
y
,
z
)
(x, y,z)
(
x
,
y
,
z
)
consisting of three distinct real numbers, that satisfy the following system of equations:
{
x
2
+
y
2
=
−
x
+
3
y
+
z
y
2
+
z
2
=
x
+
3
y
−
z
x
2
+
z
2
=
2
x
+
2
y
−
z
\begin {cases}x^2 + y^2 = -x + 3y + z \\ y^2 + z^2 = x + 3y - z \\ x^2 + z^2 = 2x + 2y - z \end {cases}
⎩
⎨
⎧
x
2
+
y
2
=
−
x
+
3
y
+
z
y
2
+
z
2
=
x
+
3
y
−
z
x
2
+
z
2
=
2
x
+
2
y
−
z
4
1
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equal segments wanted, old school geometry style
In triangle
A
B
C
,
∠
A
ABC, \angle A
A
BC
,
∠
A
is smaller than
∠
C
\angle C
∠
C
. Point
D
D
D
lies on the (extended) line
B
C
BC
BC
(with
B
B
B
between
C
C
C
and
D
D
D
) such that
∣
B
D
∣
=
∣
A
B
∣
|BD| = |AB|
∣
B
D
∣
=
∣
A
B
∣
. Point
E
E
E
lies on the bisector of
∠
A
B
C
\angle ABC
∠
A
BC
such that
∠
B
A
E
=
∠
A
C
B
\angle BAE = \angle ACB
∠
B
A
E
=
∠
A
CB
. Line segment
B
E
BE
BE
intersects line segment
A
C
AC
A
C
in point
F
F
F
. Point
G
G
G
lies on line segment
A
D
AD
A
D
such that
E
G
EG
EG
and
B
C
BC
BC
are parallel. Prove that
∣
A
G
∣
=
∣
B
F
∣
|AG| =|BF|
∣
A
G
∣
=
∣
BF
∣
.[asy] unitsize (1.5 cm);real angleindegrees(pair A, pair B, pair C) { real a, b, c; a = abs(B - C); b = abs(C - A); c = abs(A - B); return(aCos((a^2 + c^2 - b^2)/(2*a*c))); };pair A, B, C, D, E, F, G;B = (0,0); A = 2*dir(190); D = 2*dir(310); C = 1.5*dir(310 - 180); E = extension(B, incenter(A,B,C), A, rotate(angleindegrees(A,C,B),A)*(B)); F = extension(B,E,A,C); G = extension(E, E + D - B, A, D);filldraw(anglemark(A,C,B,8),gray(0.8)); filldraw(anglemark(B,A,E,8),gray(0.8)); draw(C--A--B); draw(E--A--D); draw(interp(C,D,-0.1)--interp(C,D,1.1)); draw(interp(E,B,-0.2)--interp(E,B,1.2)); draw(E--G);dot("
A
A
A
", A, SW); dot("
B
B
B
", B, NE); dot("
C
C
C
", C, NE); dot("
D
D
D
", D, NE); dot("
E
E
E
", E, N); dot("
F
F
F
", F, N); dot("
G
G
G
", G, SW); [/asy]
5
1
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quiz show, 3 doors, only in 1 prize, lies at most 10 times, min no of questions
At a quiz show there are three doors. Behind exactly one of the doors, a prize is hidden. You may ask the quizmaster whether the prize is behind the left-hand door. You may also ask whether the prize is behind the right-hand door. You may ask each of these two questions multiple times, in any order that you like. Each time, the quizmaster will answer ‘yes’ or ‘no’. The quizmaster is allowed to lie at most
10
10
10
times. You have to announce in advance how many questions you will be asking (but which questions you will ask may depend on the answers of the quizmaster). What is the smallest number you can announce, such that you can still determine with absolute certainty the door behind which the prize is hidden?