MathDB

3

Part of 1997 Romania National Olympiad

Problems(3)

altitudes wanted in triangle - easy roman geo - 1997 Romania NMO VII p3

Source:

8/13/2024
The triangle ABCABC has ∠ACB=30o\angle ACB = 30^o, BC=4BC = 4 cm and AB=3AB = 3 cm . Compute the altitudes of the triangle.
geometryaltitude
parallelepiped with min CF + FM - 1997 Romania NMO VIII p3

Source:

8/13/2024
ABCDA′B′CD′ABCDA'B'CD' is a rectangular parallelepiped with AA′=2AB=8aAA'= 2AB = 8a , EE is the midpoint of (AB)(AB) and MM is the point of (DD′)(DD') for which DM=a(1+ADAC)DM = a \left( 1 + \frac{AD}{AC}\right).
a) Find the position of the point. FF on the segment (AA′)(AA') for which the sum CF+FMCF + FM has the minimum possible value.
b) Taking FF as above, compute the measure of the angle of the planes (D,E,F)(D, E, F) and (D,B′,C′)(D, B', C').
c) Knowing that the straight lines AC′AC' and FDFD are perpendicular, compute the volume of the parallelepiped ABCDA′B′C′D′ABCDA'B'C'D'.
geometry3D geometryparallelepiped
f(u)+f(v)=7

Source: Romania 1997

8/23/2005
Suppose that a,b,c,d∈Ra,b,c,d\in\mathbb{R} and f(x)=ax3+bx2+cx+df(x)=ax^3+bx^2+cx+d such that f(2)+f(5)<7<f(3)+f(4)f(2)+f(5)<7<f(3)+f(4). Prove that there exists u,v∈Ru,v\in\mathbb{R} such that u+v=7,f(u)+f(v)=7u+v=7 , f(u)+f(v)=7
algebrapolynomialalgebra unsolvedfunction