MathDB

1

Part of 2010 IMO Shortlist

Problems(3)

In a concert, 20 singers will perform

Source: IMO Shortlist 2010, Combinatorics 1

7/17/2011
In a concert, 20 singers will perform. For each singer, there is a (possibly empty) set of other singers such that he wishes to perform later than all the singers from that set. Can it happen that there are exactly 2010 orders of the singers such that all their wishes are satisfied?
Proposed by Gerhard Wöginger, Austria
combinatoricsposetLinear extensionscountingIMO Shortlistfree
IMO Shortlist 2010 - Problem G1

Source:

7/17/2011
Let ABCABC be an acute triangle with D,E,FD, E, F the feet of the altitudes lying on BC,CA,ABBC, CA, AB respectively. One of the intersection points of the line EFEF and the circumcircle is P.P. The lines BPBP and DFDF meet at point Q.Q. Prove that AP=AQ.AP = AQ.
Proposed by Christopher Bradley, United Kingdom
geometrycircumcirclegeometric transformationreflectionIMO Shortlist
IMO Shortlist 2010 - Problem N1

Source:

7/17/2011
Find the least positive integer nn for which there exists a set {s1,s2,…,sn}\{s_1, s_2, \ldots , s_n\} consisting of nn distinct positive integers such that (1−1s1)(1−1s2)⋯(1−1sn)=512010. \left( 1 - \frac{1}{s_1} \right) \left( 1 - \frac{1}{s_2} \right) \cdots \left( 1 - \frac{1}{s_n} \right) = \frac{51}{2010}.
Proposed by Daniel Brown, Canada
number theoryequationIMO Shortlistnumber theory solvedmultiplication