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PrintWinter Mathematical Competition
Bulgaria geometry
Problem
Let be a right triangle with , and . Given a point such that we construct a sequence of points , in the following way. Let be the intersection point of and the line through and parallel to , and be the foot of the perpendicular from to . Then is the intersection point of and the line through and parallel to . Using we construct in the same way, etc. Find: a) ; b) .
Solution
a) Set . Then it follows from that . Since , we find that . Hence and therefore .
b) Since , and we get . But a) implies that is a geometric progression with ratio . Hence and therefore .
b) Since , and we get . But a) implies that is a geometric progression with ratio . Hence and therefore .
Final answer
a) -1/5; b) 5/36
Techniques
TrianglesConstructions and lociRecurrence relations