MathDB

5

Part of 2014 Sharygin Geometry Olympiad

Problems(4)

Ratio of Sides of Triangle given Congruent Segments

Source: Sharygin Geometry Olympiad 2014 - Problem 5

11/15/2014
In an acute-angled triangle ABCABC, AMAM is a median, ALAL is a bisector and AHAH is an altitude (HH lies between LL and BB). It is known that ML=LH=HBML=LH=HB. Find the ratios of the sidelengths of ABCABC.
ratiogeometry
triangle 30-70-80, cut with a straight line so that 1 angle bisector // 1 median

Source: 2014 Sharygin Geometry Olympiad Final Round 8.5

8/3/2018
A triangle with angles of 30,7030, 70 and 8080 degrees is given. Cut it by a straight line into two triangles in such a way that an angle bisector in one of these triangles and a median in the other one drawn from two endpoints of the cutting segment are parallel to each other. (It suffices to find one such cutting.)
(A. Shapovalov )
geometryangle bisector
construct a triange given intersections of A altitude, B bisector, C median

Source: 2014 Sharygin Geometry Olympiad Final Round 10.5

8/3/2018
The altitude from one vertex of a triangle, the bisector from the another one and the median from the remaining vertex were drawn, the common points of these three lines were marked, and after this everything was erased except three marked points. Restore the triangle. (For every two erased segments, it is known which of the three points was their intersection point.)
(A. Zaslavsky)
constructiongeometry
common chord of circumcircles perpendicular to a triangle side

Source: 2014 Sharygin Geometry Olympiad Final Round 9.5

8/3/2018
In triangle ABCABC ∠B=60o,O\angle B = 60^o, O is the circumcenter, and LL is the foot of an angle bisector of angle BB. The circumcirle of triangle BOLBOL meets the circumcircle of ABCABC at point D≠BD \ne B. Prove that BD⊥ACBD \perp AC.
(D. Shvetsov)
geometrycirclesperpendicular