MathDB

Easy one on angle bisectors

Source: 0

5/17/2004

In a triangle

ABC

, let

D

be the midpoint of the side

BC

, and let

E

be a point on the side

AC

. The lines

BE

and

AD

meet at a point

F

.
Prove: If

\frac{BF}{FE}=\frac{BC}{AB}+1

, then the line

BE

bisects the angle

ABC

.

geometryangle bisectorMenelausGermanyTSTTeam Selection Test2004

Old functional equation

Source: German TST 2004, Exam V, Problem 2

5/1/2004

Find all functions

f: \Bbb{R}_{0}^{+}\rightarrow \Bbb{R}_{0}^{+}

with the following properties:
(a) We have

f\left( xf\left( y\right) \right) \cdot f\left( y\right) =f\left( x+y\right)

for all

x

and

y

. (b) We have

f\left(2\right) = 0

. (c) For every

x

with

0 < x < 2

, the value

f\left(x\right)

doesn't equal

0

.
NOTE. We denote by

\Bbb{R}_{0}^{+}

the set of all non-negative real numbers.

functionalgebra proposedalgebraFunctional EquationsGermanyTSTTeam Selection Test

Diophantine equation.

Source:

1/30/2004

Find all pairs of positive integers

\left(n;\;k\right)

such that

n!=\left( n+1\right)^{k}-1

.

modular arithmeticinductionAMCUSAMOnumber theory

For all fans of case-distinction [< MPA = < BPN]

Source: German TST 2004, exam VI, problem 2

5/30/2004

Let

d

be a diameter of a circle

k

, and let

A

be an arbitrary point on this diameter

d

in the interior of

k

. Further, let

P

be a point in the exterior of

k

. The circle with diameter

PA

meets the circle

k

at the points

M

and

N

.
Find all points

B

on the diameter

d

in the interior of

k

such that \measuredangle MPA = \measuredangle BPN \text{and} PA \leq PB. (i. e. give an explicit description of these points without using the points

M

and

N

).

geometry proposedgeometryLocus problemsGeometric InequalitiesGermanyTSTTeam Selection Test

OMKN is a parallelogram (easy geometry)

Source: German TST 2004, exam VII, problem 2, by Arthur Engel; part of AMM problem #10874

6/1/2004

Let two chords

AC

and

BD

of a circle

k

meet at the point

K

, and let

O

be the center of

k

. Let

M

and

N

be the circumcenters of triangles

AKB

and

CKD

. Show that the quadrilateral

OMKN

is a parallelogram.

geometryparallelogramcircumcirclegeometric transformationreflectionperpendicular bisectorgeometry proposed

2

Problems(5)