MathDB

2

Part of 2018 Bulgaria JBMO TST

Problems(2)

(a^2+b^2)/(2a^5b^5) + (81a^2b^2)/4 + 9ab \geq 81

Source: Bulgaria JBMO TST 2018, Day 1, Problem 2

6/25/2018
For all positive reals aa and bb, show that a2+b22a5b5+81a2b24+9ab>18.\frac{a^2+b^2}{2a^5b^5} + \frac{81a^2b^2}{4} + 9ab > 18.
inequalities
CD=BE [Bulgaria JBMO TST 2018]

Source: Bulgaria JBMO TST 2018, Day 2, Problem 2

6/25/2018
Let ABCABC be a triangle and AA1AA_1 be the angle bisector of AA (A1∈BCA_1 \in BC). The point PP is on the segment AA1AA_1 and MM is the midpoint of the side BCBC. The point QQ is on the line connecting PP and MM such that MM is the midpoint of PQPQ. Define DD and EE as the intersections of BQBQ, ACAC, and CQCQ, ABAB. Prove that CD=BECD=BE.
geometryangle bisector