MathDB

1

3 concurrent wanted and given, lines or circles

A, B, C, A', B', C'

be six pairs of different points. Prove that the Circles

BCA'

,

CAB'

and

ABC'

have a common point, then the Circles

B'C'A, C'A'B

and

A'B'C

also share a common point.
Note: For three pairs of different points

X, Y

and

Z

we define the Circle

XYZ

as the circumcircle of the triangle

XYZ

, or - in the case when the points

X, Y

and

Z

lie on a straight line - this straight line.

11

1

concurrency of // lines to sides of triangle, incircle related

The inscribed circle of a triangle

ABC

has the center

O

and touches the triangle sides

BC, CA

and

AB

at points

X, Y

and

Z

, respectively. The parallels to the straight lines

ZX, XY

and

YZ

the straight lines

BC, CA

and

AB

(in this order!) intersect through the point

O

. Points

K, L

and

M

. Then the parallels to the straight lines

CA, AB

and

BC

intersect through the points

K, L

and

M

in one point.

9

1

c (ab) = c (a) + c (b), \phi function

For every natural

n

let

\phi (n)

be the number of coprime numbers

k \in \{1,2,...,n\}

. (Example:

\phi (12) = 4

, because among the numbers

1, 2, ..., 12

there are only the

4

numbers,

1, 5, 7

and

11

coprime to

12.

) If

k

is a natural number, then one defines \phi^k (n)=\underbrace{\strut \phi (\phi ...(\phi (n)) ...)}_{(k \, times \phi)} (Example:

\phi^3 (n)=\phi (\phi (\phi (n)))

) For every whole

n> 2

let

c(n)

be the smallest natural number

k

with

\phi^k (n)= 2

. Prove that

c (ab) = c (a) + c (b)

for odd integers

a

and

b

, both of which are greater than

2

, .

5

1

all p + 1 numbers are equal

Let

p

be a prime number and let further

p + 1

rational numbers

a_0,...,a_p

with the following property given: If one removes any of the

p + 1

numbers, then the remaining may be split in at least two groups , which all have the same mean value (for different distant numbers, however, these mean values may be different). Prove that all

p + 1

numbers are equal.

7

1

ratio of construction steps

Albatross and Frankinfueter both own a circle. Frankinfueter also has stolen from Prof. Trugweg a ruler. Before that, Trugweg had two points with a distance of

1

drawn his (infinitely large) board. For a natural number

n

, let A

(n)

be the number of the construction steps that Albatross needs at least to create two points with a distance of

n

to construct. Similarly, Frankinfueter needs at least

F(n)

steps for this. How big can

\frac{A (n)}{F (n)}

become? There are only the following three construction steps: a) Mark an intersection of two straight lines, two circles or a straight line with one circle. b) Pierce at a marked point

P

and draw a circle around

P

through one marked point . c) Draw a straight line through two marked points (this implies possession of a ruler ahead!).

6

1

no of roads a finite number of cities are connected

An empire has a finite number of cities. Every two cities are connected by a natural number of roads . Every street connects exactly two cities. Show that you have the kingdom can be divided into a maximum of three republics so that within each republic there are just many streets run away. (We say a road runs within a republic if the two cities that it connects, both belonging to this republic. The republics must meet each other be disjoint, and cover all cities of the empire in total.)
[hide=original wording in German]Ein Reich hat endlich viele St¨adte. Je zwei St¨adte sind durch eine natu¨rliche Anzahl von Straßen verbunden. Jede Straße verbindet genau zwei St¨adte. Man zeige, dass man das Reich so in h¨ochstens drei Republiken zerteilen kann, dass innerhalb jeder Republik gerade viele Straßen verlaufen. (Wir sagen, eine Straße verl¨auft innerhalb einer Republik, wenn die zwei St¨adte, die sie verbindet, beide dieser Republik angeh¨oren. Die Republiken mu¨ssen zueinander disjunkt sein, und insgesamt alle St¨adte des Reiches abdecken.)

4

1

\sum {n \choose k} (a + k)^k (b - k)^{n-k} = \sum n! /t! (a + b)^t

Let

a

and

b

be two real numbers and let

n

be a nonnegative integer. Then prove that

\sum_{k=0}^{n} {n \choose k} (a + k)^k (b - k)^{n-k} = \sum_{k=0}^{n} \frac{n!}{t!} (a + b)^t

8

1

sum of a_i = (a_1 + a_2 +... + a_n)/2

Given

n

integers

a_1, a_2, ..., a_n

, which

a_1 = 1

and

a_i \le a_{i + 1} \le 2a_i

for each

i \in \{1,2,...,n-1\}

. Prove that if

a_1 + a_2 +... + a_n

is even, you do select some of the numbers so that their sum equals

\frac{a_1 + a_2 +... + a_n}{2}

.

10

1

2^n dungeons and n Dragon cults in the land of Draconis

Let

n \in N

. The land of Draconis has more than

2^n

dungeons. Between two different Dungeons have exactly one path, but each path is a one-way street. Total

n

Dragon cults share the territory; each path is controlled by exactly one cult. It is said that a dragon cult

K

has established itself in a dungeon

D

if there is both one a path beginning in

D

and one a path ending in

D

, both of which are controlled by

K

. Prove that there is a cult

K

, which has at least one dungeon controlled.

12

1

f (xy + f (x)) = xf (y) + f (x)

Find all functions

f: R\to R

, which satisfy the equation

f (xy + f (x)) = xf (y) + f (x)

.

2

1

color finite no of straight lines in the plane with 3 colors

Let there be a finite number of straight lines in the plane, none of which are three in one point to cut. Show that the intersections of these straight lines can be colored with

3

colors so that that no two points of the same color are adjacent on any of the straight lines. (Two points of intersection are called adjacent if they both lie on one of the finitely many straight lines and there is no other such intersection on their connecting line.)

1

y^5 - x^2 = 4

Solve

y^5 - x^2 = 4

in integers numbers

x,y

.

2009 QEDMO 6th

Subcontests

3 concurrent wanted and given, lines or circles

concurrency of // lines to sides of triangle, incircle related

c (ab) = c (a) + c (b), \phi function

all p + 1 numbers are equal

ratio of construction steps

no of roads a finite number of cities are connected

\sum {n \choose k} (a + k)^k (b - k)^{n-k} = \sum n! /t! (a + b)^t

sum of a_i = (a_1 + a_2 +... + a_n)/2

2^n dungeons and n Dragon cults in the land of Draconis

f (xy + f (x)) = xf (y) + f (x)

color finite no of straight lines in the plane with 3 colors

y^5 - x^2 = 4