MathDB

3

Part of 2008 Mathcenter Contest

Problems(2)

remainder from division sum_{i=1}^{2006} x_i^2 with 2551

Source: Mathcenter Contest / Oly - Thai Forum 2008 R1 p3 https://artofproblemsolving.com/community/c3196914_mathcenter_contest

11/10/2022
Set M={1,2,⋯ ,2550} M= \{1,2,\cdots,2550\} and min⁡A, max⁡A\min A ,\ \max A represents the minimum and maximum values of the elements in the set AA. For k∈{1,2,⋯2006} k \in \{1,2,\cdots 2006\} define xk=12008(∑A⊂M:n(A)=k( minA+max⁡A) ) x_k = \frac{1}{2008} \bigg (\sum_{A \subset M : n(A)= k} (\ min A + \max A) \, \bigg) . Find remainder from division ∑i=12006xi2\sum_{i=1}^{2006} x_i^2 with 25512551.
(passer-by)
number theory
(absin2C+bcsin2A+casin2B){ab+bc+ca)<=\sqrt3/2

Source: Mathcenter Contest / Oly - Thai Forum 2008 R3 p3 https://artofproblemsolving.com/community/c3196914_mathcenter_contest

11/9/2022
Let ABCABC be a triangle whose side lengths are opposite the angle A,B,CA,B,C are a,b,ca,b,c respectively. Prove that absin⁡2C+bcsin⁡2A+casin⁡2Bab+bc+ca≤32\frac{ab\sin{2C}+bc\sin{ 2A}+ca\sin{2B}}{ab+bc+ca}\leq\frac{\sqrt{3}}{2}.
(nooonuii)
geometrytrigonometryTrigonometric inequalityinequalitiesgeometric inequality