MathDB

Circles in trapezoid

Source: China TST 2006 (1)

3/24/2006

ABCD

is a trapezoid with

AB || CD

. There are two circles

\omega_1

and

\omega_2

is the trapezoid such that

\omega_1

is tangent to

DA

,

AB

,

BC

and

\omega_2

is tangent to

BC

,

CD

,

DA

. Let

l_1

be a line passing through

A

and tangent to

\omega_2

(other than

AD

), Let

l_2

be a line passing through

C

and tangent to

\omega_1

(other than

CB

).
Prove that

l_1 || l_2

.

geometrytrapezoidincentergeometric transformationhomothetyratiogeometry unsolved

Find general term

Source: China TST Quiz 2006

6/18/2006

Two positive valued sequences

\{ a_{n}\}

and

\{ b_{n}\}

satisfy: (a):

a_{0}=1 \geq a_{1}

,

a_{n}(b_{n+1}+b_{n-1})=a_{n-1}b_{n-1}+a_{n+1}b_{n+1}

,

n \geq 1

. (b):

\sum_{i=1}^{n}b_{i}\leq n^{\frac{3}{2}}

,

n \geq 1

. Find the general term of

\{ a_{n}\}

.

inductionratioinequalitieslimitalgebra unsolvedalgebra

Concurrency

Source: China TST 2006

6/18/2006

The centre of the circumcircle of quadrilateral

ABCD

is

O

and

O

is not on any of the sides of

ABCD

.

P=AC \cap BD

. The circumecentres of

\triangle{OAB}

,

\triangle{OBC}

,

\triangle{OCD}

and

\triangle{ODA}

are

O_1

,

O_2

,

O_3

and

O_4

respectively. Prove that

O_1O_3

,

O_2O_4

and

OP

are concurrent.

geometrycircumcirclegeometric transformationreflectionprojective geometrypower of a pointradical axis

Divalent radical

Source: China TST 2006

6/18/2006

Let

A

be a non-empty subset of the set of all positive integers

N^*

. If any sufficient big positive integer can be expressed as the sum of

2

elements in

A

(The two integers do not have to be different), then we call that

A

is a divalent radical. For

x \geq 1

, let

A(x)

be the set of all elements in

A

that do not exceed

x

, prove that there exist a divalent radical

A

and a constant number

C

so that for every

x \geq 1

, there is always

\left| A(x) \right| \leq C \sqrt{x}

.

inequalitiescombinatorics unsolvedcombinatorics

Cyclic points [variations on a Fuhrmann generalization]

Source: China TST 2006

6/18/2006

H

is the orthocentre of

\triangle{ABC}

.

D

,

E

,

F

are on the circumcircle of

\triangle{ABC}

such that

AD \parallel BE \parallel CF

.

S

,

T

,

U

are the semetrical points of

D

,

E

,

F

with respect to

BC

,

CA

,

AB

. Show that

S, T, U, H

lie on the same circle.

geometrycircumcirclegeometry unsolved

Two intersecting circles

Source: China TST 2006

6/18/2006

Let the intersections of

\odot O_1

and

\odot O_2

be

A

and

B

. Point

R

is on arc

AB

of

\odot O_1

and

T

is on arc

AB

on

\odot O_2

.

AR

and

BR

meet

\odot O_2

at

C

and

D

;

AT

and

BT

meet

\odot O_1

at

Q

and

P

. If

PR

and

TD

meet at

E

and

QR

and

TC

meet at

F

, then prove:

AE \cdot BT \cdot BR = BF \cdot AT \cdot AR

.

geometryparallelogramgeometry unsolved

Existence of a polynomial

Source: China TST 2006

6/18/2006

Let

k

be an odd number that is greater than or equal to

3

. Prove that there exists a

k^{th}

-degree integer-valued polynomial with non-integer-coefficients that has the following properties: (1)

f(0)=0

and

f(1)=1

; and. (2) There exist infinitely many positive integers

n

so that if the following equation:

n= f(x_1)+\cdots+f(x_s),

has integer solutions

x_1, x_2, \dots, x_s

, then

s \geq 2^k-1

.

algebrapolynomialinductionmodular arithmeticpigeonhole principlealgebra unsolved

Collinearity of orthocentres

Source: China TST 2006

6/18/2006

Let

K

and

M

be points on the side

AB

of a triangle

\triangle{ABC}

, and let

L

and

N

be points on the side

AC

. The point

K

is between

M

and

B

, and the point

L

is between

N

and

C

. If

\frac{BK}{KM}=\frac{CL}{LN}

, then prove that the orthocentres of the triangles

\triangle{ABC}

,

\triangle{AKL}

and

\triangle{AMN}

lie on one line.

geometrycircumcircleparallelogram3D geometrytetrahedrongeometric transformationhomothety

1

Problems(8)