MYSTERIES OF THE EQUILATERAL TRIANGLE - HIKARI Ltd

Mathematical Competitions 145
Figure 5.25: Mathematical Olympiad of Moldova 1999a
Problem 57 (Mathematical Olympiad of Moldova 1999a). On the sides
BC and AB of the equilateral triangle ABC, the points D and E, respectively,
are taken such that CD : DB = BE : EA = ( √ 5+1)/2. The straight lines AD
and CE intersect in the point O. The points M and N are interior points of the
segments OD and OC, respectively, such that MN � BC and AN = 2OM.
The parallel to the straight line AC, drawn throught the point O, intersects
the segment MC in the point P (Figure 5.25). Prove that the half-line AP
is the bisectrix of the angle MAN. (Note: This problem is ill-posed in that
AN = 2OM cannot be true if the other conditions are true!) [306, p. 44]
Figure 5.26: Mathematical Olympiad of Moldova 1999b
Problem 58 (Mathematical Olympiad of Moldova 1999b). On the sides
BC, AC and AB of the equilateral triangle ABC, consider the points M, N
and P, respectively, such that AP : PB = BM : MC = CN : NA = λ (Figure
5.26). Show that the circle with diameter AC covers the triangle bounded by
the straight lines AM, BN and CP if and only if 1 ≤ λ ≤ 2. (In the case of
2
concurrent straight lines, the triangle degenerates into a point.) [306, p. 44]