MathDB

1

Part of 2006 Romania Team Selection Test

Problems(4)

Two isosceles triangles become equilateral

Source: Romanian TST 1 2006, Problem 1

4/19/2006
Let ABCABC and AMNAMN be two similar triangles with the same orientation, such that AB=ACAB=AC, AM=ANAM=AN and having disjoint interiors. Let OO be the circumcenter of the triangle MABMAB. Prove that the points OO, CC, NN, AA lie on the same circle if and only if the triangle ABCABC is equilateral. Valentin Vornicu
geometrycircumcirclegeometric transformationrotationsimilar trianglescomplex numbersgeometry proposed
Rather known sequence

Source: Romanian IMO TST 2006, day 2, problem 1

4/22/2006
Let {an}n≥1\{a_n\}_{n\geq 1} be a sequence with a1=1a_1=1, a2=4a_2=4 and for all n>1n>1, an=an−1an+1+1. a_{n} = \sqrt{ a_{n-1}a_{n+1} + 1 } . a) Prove that all the terms of the sequence are positive integers. b) Prove that 2anan+1+12a_na_{n+1}+1 is a perfect square for all positive integers nn. Valentin Vornicu
inductionquadraticsalgebraalgebra proposed
A circle inscribed in a quadrilateral

Source: Romanian IMO TST 2006, day 3, problem 1

5/16/2006
The circle of center II is inscribed in the convex quadrilateral ABCDABCD. Let MM and NN be points on the segments AIAI and CICI, respectively, such that ∠MBN=12∠ABC\angle MBN = \frac 12 \angle ABC. Prove that ∠MDN=12∠ADC\angle MDN = \frac 12 \angle ADC.
geometryincentertrigonometryconicshyperbolaromania
Functional equation on rational numbers

Source: Romanian IMO TST 2006, day 4, problem 1

5/19/2006
Let rr and ss be two rational numbers. Find all functions f:Q→Qf: \mathbb Q \to \mathbb Q such that for all x,y∈Qx,y\in\mathbb Q we have f(x+f(y))=f(x+r)+y+s. f(x+f(y)) = f(x+r)+y+s.
functionLaTeXalgebra proposedalgebra