2015 Silk Road

Part of Silk Road

Subcontests

(5)

1

set of all sequences of length n consisting of zeros and ones, sum related

B_n

be the set of all sequences of length

n

, consisting of zeros and ones. For every two sequences

a,b \in B_n

(not necessarily different) we define strings

\varepsilon_0\varepsilon_1\varepsilon_2 \dots \varepsilon_n

\delta_0\delta_1\delta_2 \dots \delta_n

\varepsilon_0=\delta_0=0

and \varepsilon_{i+1}=(\delta_i-a_{i+1})(\delta_i-b_{i+1}), \delta_{i+1}=\delta_i+(-1)^{\delta_i}\varepsilon_{i+1} (0 \leq i \leq n-1). . Let

w(a,b)=\varepsilon_0+\varepsilon_1+\varepsilon_2+\dots +\varepsilon_n

f(n)=\sum\limits_{a,b \in {B_n}} {w(a,b)}

1

(a_n^2+b_n^2)(a_{n+1}^2+b_{n+1}^2), (a_n^2+a_{n+1}^2)(b_n^2+b_{n+1}^2)

\left\{ {{a}_{n}} \right\}_{n \geq 1}

\left\{ {{b}_{n}} \right\}_{n \geq 1}

be two infinite arithmetic progressions, each of which the first term and the difference are mutually prime natural numbers. It is known that for any natural

n

, at least one of the numbers

\left( a_n^2+a_{n+1}^2 \right)\left( b_n^2+b_{n+1}^2 \right)

\left( a_n^2+b_n^2 \right) \left( a_{n+1}^2+b_{n+1}^2 \right)

is an perfect square. Prove that

{{a}_{n}}={{b}_{n}}

, for any natural

n

1

maybe easy but good problem

Prove that there is no positive real numbers

a,b,c,d

such that both of the following equations hold.

\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=6 , \frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d}=32

1

Radius is twice the other radius

Let O be a circumcenter of an acute-angled triangle ABC. Consider two circles ω and Ω inscribed in the angle BAC in such way that ω is tangent from the outside to the arc BOC of a circle circumscribed about the triangle BOC; and the circle Ω is tangent internally to a circumcircle of triangle ABC. Prove that the radius of Ω is twice the radius ω.

1

SRMC 2015 inequality P1

Given positive real numbers

a,b,c,d

such that
\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=6 \text{and} \frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d}=36.
Prove the inequality

{{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}}>ab+ac+ad+bc+bd+cd.