MathDB

1

Part of 1997 Romania Team Selection Test

Problems(4)

Triangle of the three centres is obtuse

Source: Romanian TST 1997

9/17/2011
We are given in the plane a line ℓ\ell and three circles with centres A,B,CA,B,C such that they are all tangent to ℓ\ell and pairwise externally tangent to each other. Prove that the triangle ABCABC has an obtuse angle and find all possible values of this this angle.
Mircea Becheanu
trigonometryinequalitiesgeometry proposedgeometry
Plane through the edges of pyramid must be parallel to base

Source: Romanian TST 1997

9/17/2011
Let VA1A2…AnVA_1A_2\ldots A_n be a pyramid, where n≥4n\ge 4. A plane Π\Pi intersects the edges VA1,VA2,…,VAnVA_1,VA_2,\ldots, VA_n at the points B1,B2,…,BnB_1,B_2,\ldots,B_n respectively such that the polygons A1A2…AnA_1A_2\ldots A_n and B1B2…BnB_1B_2\ldots B_n are similar. Prove that the plane Π\Pi is parallel to the plane containing the base A1A2…AnA_1A_2\ldots A_n.
Laurentiu Panaitopol
geometry3D geometrypyramidgeometry proposed
Romania 1997

Source:

4/18/2011
Let ABCDEFABCDEF be a convex hexagon, and let P=AB∩CDP= AB \cap CD, Q=CD∩EFQ = CD \cap EF, R=EF∩ABR = EF \cap AB, S=BC∩DES = BC \cap DE, T=DE∩FAT = DE \cap FA, U=FA∩BCU = FA \cap BC. Prove that
PQCD=QREF=RPAB\frac{PQ}{CD} = \frac{QR}{EF} = \frac{RP}{AB} if and only if STDE=TUFA=USBC\frac{ST}{DE} = \frac{TU}{FA} = \frac{US}{BC}
vectorratiogeometry proposedgeometry
P(X)^2-Q(X)^2 has a rational root

Source: Romanian TST 1997

9/17/2011
Let P(X),Q(X)P(X),Q(X) be monic irreducible polynomials with rational coefficients. suppose that P(X)P(X) and Q(X)Q(X) have roots α\alpha and β\beta respectively, such that α+β\alpha + \beta is rational. Prove that P(X)2−Q(X)2P(X)^2-Q(X)^2 has a rational root.
Bogdan Enescu
algebrapolynomialinequalitiesalgebra proposed