MathDB

4

Part of 1999 Romania National Olympiad

Problems(3)

EF/BC+EG / AD=1 1999 Romania NMO VII p4

Source:

8/14/2024
In the triangle ABCABC, let D∈(BC)D \in (BC), E∈(AB)E \in (AB), EF∥BCEF \parallel BC, F∈(AC)F \in (AC), EG∥ADEG\parallel AD, G∈(BC)G\in (BC) and M,NM,N be the midpoints of (AD)(AD) and (BC)(BC), respectively. Prove that:
a) EFBC+EGAD=1\frac{EF}{BC}+\frac{EG}{AD}=1
b) the midpoint of [FG][FG] lies on the line MN MN.
geometryratio
_|_ planes wanted, regular pyramid 1999 Romania NMO VIII p4

Source:

8/14/2024
Let SABCSABC be a regular pyramid, OO the center of basis ABCABC, and MM the midpoint of [BC][BC]. If N∈[SA]N \in [SA] such that SA=25⋅NSSA = 25 \cdot NS and SO∩MN={P}SO \cap MN=\{P\}, AM=2⋅SOAM=2\cdot SO, prove that the planes (ABP)(ABP) and (SBC)(SBC) are perpendicular.
geometry3D geometrypyramid
f(m x+(1- m)y) < m f{x)+(1- m)f(y), parallelogram on graph

Source: 1999 Romania NMO IX p4

8/15/2024
a) Let a,b∈Ra,b\in R, a<ba <b. Prove that x∈(a,b)x \in (a,b) if and only if there exists λ∈(0,1)\lambda \in (0,1) such that x=λa+(1−λ)bx=\lambda a +(1-\lambda)b.
b) If the function f:R→Rf: R \to R has the property: f(λx+(1−λ)y)<λf(x)+(1−λ)f(y),∀x,y∈R,x≠y,∀λ∈(0,1),f (\lambda x+(1-\lambda) y) < \lambda f(x) + (1-\lambda)f(y), \forall x,y \in R, x\ne y, \forall \lambda \in (0,1), prove that one cannot find four points on the function’s graph that are the vertices of a parallelogram
geometryparallelogramalgebrainequalities