olymmath

SILK-ROAD 2002

Silk Road Mathematical Competition 2002.

Other problems

Problem 1

Let △ABC\triangle ABC be a triangle with incircle ω(I,r)\omega(I,r) and circumcircle ζ(O,R)\zeta(O,R) .Let lal_{a} be the angle bisector of ∠BAC\angle BAC .Denote P=la∩ζP=l_{a}\cap\zeta .Let DD be the point of tangency ω\omega with [BC][BC] .Denote Q=PD∩ζQ=PD\cap\zeta .Show that PI=QIPI=QI if PD=rPD=r .

Problem 2

I tried to search SRMC problems,but i didn't find them(I found only SRMC 2006).Maybe someone know where on this site i could find SRMC problems?I have all SRMC problems,if someone want i could post them, :wink: Here is one of them,this is one nice inequality from first SRMC: Let nn be an integer with n>2n>2 and a1,a2,…,an∈R+a_{1},a_{2},\dots,a_{n}\in R^{+} .Given any positive integers t,k,pt,k,p with 1<t<n1<t<n ,set m=k+pm=k+p ,prove the following inequalities: a) a1pa2k+a3k+⋯+atk+a2pa3k+a4k+⋯+at+1k+⋯+an−1pank+a1k+⋯+at−2k+anpa1k+a2k+⋯+at−1k≥(a1p+a2p⋯+anp)2(t−1)(a1m+a2m+⋯+anm)\frac{a_{1}^{p}}{a_{2}^{k}+a_{3}^{k}+\dots+a_{t}^{k}}+\frac{a_{2}^{p}}{a_{3}^{k}+a_{4}^{k}+\dots+a_{t+1}^{k}}+\dots+\frac{a_{n-1}^{p}}{a_{n}^{k}+a_{1}^{k}+\dots+a_{t-2}^{k}}+\frac{a_{n}^{p}}{a_{1}^{k}+a_{2}^{k}+\dots+a_{t-1}^{k}}\geq\frac{(a_{1}^{p}+a_{2}^{p}\dots+a_{n}^{p})^{2}}{(t-1) ( a_{1}^{m}+a_{2}^{m}+\dots+a_{n}^{m})} b) a2k+a3k⋯+atka1p+a3k+a4k⋯+at+1ka2p+⋯+a1k+a2k⋯+at−1kanp≥(t−1)(a1k+a2k⋯+ank)2(a1m+a2m+⋯+anm)\frac{a_{2}^{k}+a_{3}^{k}\dots+a_{t}^{k}}{a_{1}^{p}}+\frac{a_{3}^{k}+a_{4}^{k}\dots+a_{t+1}^{k}}{a_{2}^{p}}+\dots+\frac{a_{1}^{k}+a_{2}^{k}\dots+a_{t-1}^{k}}{a_{n}^{p}}\geq\frac{(t-1)(a_{1}^{k}+a_{2}^{k}\dots+a_{n}^{k})^{2}}{( a_{1}^{m}+a_{2}^{m}+\dots+a_{n}^{m})} :wink:

Problem 3

In each unit cell of a finite set of cells of an infinite checkered board, an integer is written so that the sum of the numbers in each row, as well as in each column, is divided by 20022002 . Prove that every number α\alpha can be replaced by a certain number α′\alpha' , divisible by 20022002 so that ∣α−α′∣<2002|\alpha-\alpha'| <2002 and the sum of the numbers in all rows, and in all columns will not change.

Problem 4

Observe that the fraction 17=0,142857\frac{1}{7}=0,142857 is a pure periodical decimal with period 6=7−16=7-1 ,and in one period one has 142+857=999142+857=999 .For n=1,2,…n=1,2,\dots find a sufficient and necessary condition that the fraction 12n+1\frac{1}{2n+1} has the same properties as above and find two such fractions other than 17\frac{1}{7} .