2017 Dutch Mathematical Olympiad

Part of Dutch Mathematical Olympiad

Subcontests

(4)

1

equal segments in a parallelogram given, angle wanted

A parallelogram

ABCD

|AD| =|BD|

has been given. A point

E

lies on line segment

|BD|

in such a way that

|AE| = |DE|

. The (extended) line

AE

intersects line segment

BC

F

DF

is the angle bisector of angle

CDE

. Determine the size of angle

ABD

. [asy] unitsize (3 cm);
pair A, B, C, D, E, F;
D = (0,0); A = dir(250); B = dir(290); C = B + D - A; E = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), B, D); F = extension(A, E, B, C);
draw(A--B--C--D--cycle); draw(A--F--D--B);
dot("

A

", A, SW); dot("

B

", B, SE); dot("

C

", C, NE); dot("

D

", D, NW); dot("

E

", E, S); dot("

F

", F, SE); [/asy]

1

smallest number of circles through 8 points, (9x9 square - center)

The eight points below are the vertices and the midpoints of the sides of a square. We would like to draw a number of circles through the points, in such a way that each pair of points lie on (at least) one of the circles. Determine the smallest number of circles needed to do this. [asy] unitsize(1 cm);
dot((0,0)); dot((1,0)); dot((2,0)); dot((0,1)); dot((2,1)); dot((0,2)); dot((1,2)); dot((2,2)); [/asy]

1

sum of remainders of n divided by 99,132, 229 is n (n>229)

If we divide the number

13

by the three numbers

5, 7

9

, then these divisions leave remainders: when dividing by

5

the remainder is

3

, when dividing by

7

the remainder is

6

, and when dividing by

9

the remainder is 4. If we add these remainders, we obtain

3 + 6 + 4 = 13

, the original number. (a) Let

n

be a positive integer and let

a

b

be two positive integers smaller than

n

. Prove: if you divide

n

a

b

, then the sum of the two remainders never equals

n

. (b) Determine all integers

n > 229

having the property that if you divide

n

99, 132

229

, the sum of the three remainders is

n

1

even-steven and oddball integers

We consider positive integers written down in the (usual) decimal system. Within such an integer, we number the positions of the digits from left to right, so the leftmost digit (which is never a

0

) is at position

1

. An integer is called even-steven if each digit at an even position (if there is one) is greater than or equal to its neighbouring digits (if these exist). An integer is called oddball if each digit at an odd position is greater than or equal to its neighbouring digits (if these exist). For example,

3122

is oddball but not even-steven,

7

is both even-steven and oddball, and

123

is neither even-steven nor oddball. (a) Prove: every oddball integer greater than

9

can be obtained by adding two oddball integers. (b) Prove: there exists an oddball integer greater than

9

that cannot be obtained by adding two even-steven integers.