MathDB

Why is this a triangle geometry problem?

Source: Problem 3 of the German pre-TST 2004, written in December 03

2/28/2004

Given six real numbers

a

,

b

,

c

,

x

,

y

,

z

such that

0 < b-c < a < b+c

and

ax + by + cz = 0

.
What is the sign of the sum

ayz + bzx + cxy

?

geometryanalytic geometrycircumcircleincentertrigonometryratioconics

IMO ShortList 2003, geometry problem 5

Source: German pre-TST 2004, problem 6; Singapore TST 2004; Swiss TST 2004

5/10/2004

Let

ABC

be an isosceles triangle with

AC=BC

, whose incentre is

I

. Let

P

be a point on the circumcircle of the triangle

AIB

lying inside the triangle

ABC

. The lines through

P

parallel to

CA

and

CB

meet

AB

at

D

and

E

, respectively. The line through

P

parallel to

AB

meets

CA

and

CB

at

F

and

G

, respectively. Prove that the lines

DF

and

EG

intersect on the circumcircle of the triangle

ABC

.
Proposed by Hojoo Lee, Korea

geometryincentercircumcircleIMO ShortlistTriangle

You'll hate this one...

Source: German TST 2004, exam III

5/18/2004

Let

n \geq 2

be a natural number, and let

\left( a_{1};\;a_{2};\;...;\;a_{n}\right)

be a permutation of

\left(1;\;2;\;...;\;n\right)

. For any integer

k

with

1 \leq k \leq n

, we place

a_k

raisins on the position

k

of the real number axis. [The real number axis is the

x

-axis of a Cartesian coordinate system.] Now, we place three children A, B, C on the positions

x_A

,

x_B

,

x_C

, each of the numbers

x_A

,

x_B

,

x_C

being an element of

\left\{1;\;2;\;...;\;n\right\}

. [It is not forbidden to place different children on the same place!] For any

k

, the

a_k

raisins placed on the position

k

are equally handed out to those children whose positions are next to

k

. [So, if there is only one child lying next to

k

, then he gets the raisin. If there are two children lying next to

k

(either both on the same position or symmetric with respect to

k

), then each of them gets one half of the raisin. Etc..] After all raisins are distributed, a child is unhappy if he could have received more raisins than he actually has received if he had moved to another place (while the other children would rest on their places). For which

n

does there exist a configuration

\left( a_{1};\;a_{2};\;...;\;a_{n}\right)

and numbers

x_A

,

x_B

,

x_C

, such that all three children are happy?

analytic geometrycombinatorics proposedcombinatorics

Easy but nice regular hexagon procedure

Source: German TST 2004, Exam V, Problem 3

5/1/2004

We attach to the vertices of a regular hexagon the numbers

1

,

0

,

0

,

0

,

0

,

0

. Now, we are allowed to transform the numbers by the following rules: (a) We can add an arbitrary integer to the numbers at two opposite vertices. (b) We can add an arbitrary integer to the numbers at three vertices forming an equilateral triangle. (c) We can subtract an integer

t

from one of the six numbers and simultaneously add

t

to the two neighbouring numbers. Can we, just by acting several times according to these rules, get a cyclic permutation of the initial numbers? (I. e., we started with

1

,

0

,

0

,

0

,

0

,

0

; can we now get

0

,

1

,

0

,

0

,

0

,

0

, or

0

,

0

,

1

,

0

,

0

,

0

, or

0

,

0

,

0

,

1

,

0

,

0

, or

0

,

0

,

0

,

0

,

1

,

0

, or

0

,

0

,

0

,

0

,

0

,

1

?)

vectorinvariantgeometrycomplex numberscombinatorics solvedcombinatorics

Ramsey won't help: switching a black and white graph

Source: German TST 2004, exam VII, problem 3, by Eric MÃ¼ller

6/1/2004

We consider graphs with vertices colored black or white. "Switching" a vertex means: coloring it black if it was formerly white, and coloring it white if it was formerly black. Consider a finite graph with all vertices colored white. Now, we can do the following operation: Switch a vertex and simultaneously switch all of its neighbours (i. e. all vertices connected to this vertex by an edge). Can we, just by performing this operation several times, obtain a graph with all vertices colored black? [It is assumed that our graph has no loops (a loop means an edge connecting one vertex with itself) and no multiple edges (a multiple edge means a pair of vertices connected by more than one edge).]

linear algebramatrixvectorinductionalgorithmpigeonhole principlecombinatorics solved

3

Problems(5)