MathDB

279

Part of 1954 Moscow Mathematical Olympiad

Problems(1)

MMO 279 Moscow MO 1954 OB <= OA_1/ 2, starting with 4 concurrent lines

Source:

8/13/2019
Given four straight lines, m1,m2,m3,m4m_1, m_2, m_3, m_4, intersecting at OO and numbered clockwise with OO as the center of the clock, we draw a line through an arbitrary point A1A_1 on m1m_1 parallel to m4m_4 until the line meets m2m_2 at A2A_2. We draw a line through A2A_2 parallel to m1m_1 until it meets m3m_3 at A3A_3. We also draw a line through A3A_3 parallel to m2m_2 until it meets m4m_4 at A4A_4. Now, we draw a line throughA4 A_4 parallel to m3m_3 until it meets m1m_1 at BB. Prove that
a) OB<OA12OB< \frac{OA_1}{2} .
b) OB≤OA14OB \le \frac{OA_1}{4} . https://cdn.artofproblemsolving.com/attachments/5/f/5ea08453605e02e7e1253fd7c74065a9ffbd8e.png
geometric inequalityconcurrentgeometry