MathDB

Inequality

Source: Iran 3rd round 2017 first Algebra exam

8/7/2017

Let

a,b,c

and

d

be positive real numbers such that

a^2+b^2+c^2+d^2 \ge 4

. Prove that

(a+b)^3+(c+d)^3+2(a^2+b^2+c^2+d^2) \ge 4(ab+bc+cd+da+ac+bd)

inequalitiesmixing

Number theory: sequence

Source: Iran 3rd round 2017 Number theory first exam-P2

8/9/2017

Consider a sequence

\{a_i\}^\infty_{i\ge1}

of positive integers. For all positvie integers

n

prove that there exists infinitely many positive integers

k

such that there is no pair

(m,t)

of positive integers where

m>n

and

kn+a_n=tm(m+1)+a_m

number theorynumber theory with sequencesSequence

Iran geometry

Source: Iran MO 3rd round 2017 mid-terms - Geometry P2

8/10/2017

Let

ABCD

be a trapezoid (

AB<CD,AB\parallel CD

) and

P\equiv AD\cap BC

. Suppose that

Q

be a point inside

ABCD

such that

\angle QAB=\angle QDC=90-\angle BQC

. Prove that

\angle PQA=2\angle QCD

.

geometrytrapezoidAngle Chasing

easy number theory from iran

Source: Iran third round 2017 number theory , final exam

9/1/2017

For prime number

q

the polynomial

P(x)

with integer coefficients is said to be factorable if there exist non-constant polynomials

f_q,g_q

with integer coefficients such that all of the coefficients of the polynomial

Q(x)=P(x)-f_q(x)g_q(x)

are dividable by

q

; and we write:

P(x)\equiv f_q(x)g_q(x)\pmod{q}

For example the polynomials

2x^3+2,x^2+1,x^3+1

can be factored modulo

2,3,p

in the following way:

\left\{\begin{array}{lll} X^2+1\equiv (x+1)(-x+1)\pmod{2}\\ 2x^3+2\equiv (2x-1)^3\pmod{3}\\ X^3+1\equiv (x+1)(x^2-x+1) \end{array}\right.

Also the polynomial

x^2-2

is not factorable modulo

p=8k\pm 3

.
a) Find all prime numbers

p

such that the polynomial

P(x)

is factorable modulo

p

:

P(x)=x^4-2x^3+3x^2-2x-5

b) Does there exist irreducible polynomial

P(x)

in

\mathbb{Z}[x]

with integer coefficients such that for each prime number

p

, it is factorable modulo

p

?

number theorypolynomialalgebra

P2- first combinatorics exam of 2017 Iran MO 3rd round

Source: 2017 Iran MO 3rd round, first combinatorics exam P2

9/12/2017

An angle is considered as a point and two rays coming out of it. Find the largest value on

n

such that it is possible to place

n

60

degree angles on the plane in a way that any pair of these angles have exactly

4

intersection points.

Irancombinatorics

Iran Geometry

Source: Iran MO 3rd round 2017 finals - Geometry P2

9/3/2017

Assume that

P

be an arbitrary point inside of triangle

ABC

.

BP

and

CP

intersects

AC

and

AB

in

E

and

F

, respectively.

EF

intersects the circumcircle of

ABC

in

B'

and

C'

(Point

E

is between of

F

and

B'

). Suppose that

B'P

and

C'P

intersects

BC

in

C''

and

B''

respectively. Prove that

B'B''

and

C'C''

intersect each other on the circumcircle of

ABC

.

geometrycircumcircle

Complex Polynomial

Source: Iran 3rd round-2017-Algebra final exam-P2

9/2/2017

Let

P(z)=a_d z^d+\dots+ a_1z+a_0

be a polynomial with complex coefficients. The

reverse

of

P

is defined by

P^*(z)=\overline{a_0}z^d+\overline{a_1}z^{d-1}+\dots+\overline{a_d}

(a) Prove that

P^*(z)=z^d \overline{ P\left( \frac{1}{\overline{z}} \right) }

(b) Let

m

be a positive integer and let

q(z)

be a monic nonconstant polynomial with complex coefficients. Suppose that all roots of

q(z)

lie inside or on the unit circle. Prove that all roots of the polynomial

Q(z)=z^m q(z)+ q^*(z)

lie on the unit circle.

algebrapolynomialcomplex numbersIran

P2- second combinatorics exam of 2017 Iran MO 3rd round

Source: 2017 Iran MO 3rd round, second combinatorics exam P2

9/12/2017

Two persons are playing the following game on a

n\times m

table, with drawn lines: Person

\#1

starts the game. Each person in their move, folds the table on one of its lines. The one that could not fold the table on their turn loses the game. Who has a winning strategy?

Irancombinatorics

2

Problems(8)