MathDB

4

Part of 2012 India Regional Mathematical Olympiad

Problems(5)

no of unequal subsets of X={1,2,3,...,10} so that A\cap B={2,3,5,7}

Source: CRMO 2012 Region 1 p4

9/30/2018
Let X={1,2,3,...,10}X=\{1,2,3,...,10\}. Find the number of pairs of {A,B}\{A,B\} such that A⊆X,B⊆X,A≠BA\subseteq X, B\subseteq X, A\ne B and A∩B={2,3,5,7}A\cap B=\{2,3,5,7\}.
SubsetscombinatoricsCombinatorics of set
no of unequal subsets of X={1,2,3,...,12} so that A\cap B={2,3,5,7,8}

Source: CRMO 2012 Region 2 p4

9/30/2018
Let X={1,2,3,...,12}X=\{1,2,3,...,12\}. Find the number of pairs of {A,B}\{A,B\} such that A⊆X,B⊆X,A≠BA\subseteq X, B\subseteq X, A\ne B and A∩B={2,3,5,7,8}A\cap B=\{2,3,5,7,8\}.
SubsetscombinatoricsCombinatorics of set
Similar to RMO 1997

Source: RMO 2012

12/2/2012
Let X={1,2,3,...,10}X=\{1,2,3,...,10\}. Find the number of pairs of {A,B}\{A,B\} such that A⊆X,B⊆X,A≠BA\subseteq X, B\subseteq X, A\ne B and A∩B={5,7,8}A\cap B=\{5,7,8\}.
combinatorics unsolvedcombinatorics
area chasing, area of ABE is geometric mean of areas of triangles ABC, ABH

Source: CRMO 2012 region 5 p4 Mumbai

9/30/2018
HH is the orthocentre of an acute–angled triangle ABCABC. A point EE is taken on the line segment CHCH such that ABEABE is a right–angled triangle. Prove that the area of the triangle ABEABE is the geometric mean of the areas of triangles ABCABC and ABHABH.
geometryarea of a triangleareasorthocenter
abc(a+b+c)=3, show that prod(a+b)>=8 [Indian RMO 2012(b) Q4]

Source:

12/2/2012
Let a,b,ca,b,c be positive real numbers such that abc(a+b+c)=3.abc(a+b+c)=3. Prove that we have (a+b)(b+c)(c+a)≥8.(a+b)(b+c)(c+a)\geq 8. Also determine the case of equality.
inequalitiesgeometrycircumcircletrigonometrygeometric inequalityarea of a triangleHeron's formula