MathDB

4

Part of 2002 Junior Balkan Team Selection Tests - Romania

Problems(5)

if triangles ABC and AMN are similar, then parallelogram ABCD is a square

Source: 2002 Romania JBMO TST1 p4

5/16/2020
Let ABCDABCD be a parallelogram of center OO. Points MM and NN are the midpoints of BOBO and CDCD, respectively. Prove that if the triangles ABCABC and AMNAMN are similar, then ABCDABCD is a square.
geometryparallelogramsquaresimilar trianglessimilar
Upper bound k for least area of triangle with segment MN

Source: Romanian TST 2002

2/5/2011
Let ABCDABCD be a unit square. For any interior points M,NM,N such that the line MNMN does not contain a vertex of the square, we denote by s(M,N)s(M,N) the least area of the triangles having their vertices in the set of points {A,B,C,D,M,N}\{ A,B,C,D,M,N\}. Find the least number kk such that s(M,N)≤ks(M,N)\le k, for all points M,NM,N.
Dinu Șerbănescu
geometrysymmetrycombinatorics proposedcombinatorics
just an inequality

Source: Romanian Junior TST 2002, created by Dinu Serbanescu

3/13/2004
0 \sqrt (abc) + \sqrt (1-a)(1-b)(1-c) <1
inequalitiesLaTeXinequalities solved
arithmetic mean of all positive divisors of number n = p^a q^b is integer

Source: 2002 Romania JBMO TST 5.4

5/31/2020
Let p,qp, q be two distinct primes. Prove that there are positive integers a,ba, b such that the arithmetic mean of all positive divisors of the number n=paqbn = p^aq^b is an integer.
number theoryDivisorsInteger
exists a triangle of area greater than 3, (5 points 10 triangles, area &gt;2)

Source: 2002 Romania JBMO TST 4.4

5/31/2020
Five points are given in the plane that each of 1010 triangles they define has area greater than 22. Prove that there exists a triangle of area greater than 33.
geometryareaGeometric Inequalitiescombinatorial geometrycombinatorics