IMO Shortlist 1997 problem 8
Kvaliteta:
Avg: 3,0Težina:
Avg: 6,0 It is known that is the smallest angle in the triangle . The points and divide the circumcircle of the triangle into two arcs. Let be an interior point of the arc between and which does not contain . The perpendicular bisectors of and meet the line at and , respectively. The lines and meet at .
Show that .
Alternative formulation:
Four different points are chosen on a circle such that the triangle is not right-angled. Prove that:
(a) The perpendicular bisectors of and meet the line at certain points and respectively, and that the lines and meet at a certain point
(b) The length of one of the line segments and is the sum of the lengths of the other two.
Show that .
Alternative formulation:
Four different points are chosen on a circle such that the triangle is not right-angled. Prove that:
(a) The perpendicular bisectors of and meet the line at certain points and respectively, and that the lines and meet at a certain point
(b) The length of one of the line segments and is the sum of the lengths of the other two.
Izvor: Međunarodna matematička olimpijada, shortlist 1997