IMO Shortlist 1997 problem 8


Kvaliteta:
  Avg: 3,0
Težina:
  Avg: 6,0
Dodao/la: arhiva
2. travnja 2012.
LaTeX PDF
It is known that \angle BAC is the smallest angle in the triangle ABC. The points B and C divide the circumcircle of the triangle into two arcs. Let U be an interior point of the arc between B and C which does not contain A. The perpendicular bisectors of AB and AC meet the line AU at V and W, respectively. The lines BV and CW meet at T.

Show that AU = TB + TC.


Alternative formulation:

Four different points A,B,C,D are chosen on a circle \Gamma such that the triangle BCD is not right-angled. Prove that:

(a) The perpendicular bisectors of AB and AC meet the line AD at certain points W and V, respectively, and that the lines CV and BW meet at a certain point T.

(b) The length of one of the line segments AD, BT, and CT is the sum of the lengths of the other two.
Izvor: Međunarodna matematička olimpijada, shortlist 1997