MathDB

1

Part of 2004 Germany Team Selection Test

Problems(5)

Easy functional equation

Source: 0

5/17/2004
A function ff satisfies the equation f(x)+f(1−1x)=1+xf\left(x\right)+f\left(1-\frac{1}{x}\right)=1+x for every real number xx except for x=0x = 0 and x=1x = 1. Find a closed formula for ff.
functionFunctional EquationsClosed FormulaGermanyTeam Selection TestTST2004
I wouldn't exactly call it trivial...

Source: German TST 2004, exam III

5/18/2004
Let ABCABC be an acute triangle, and let MM and NN be two points on the line ACAC such that the vectors MNMN and ACAC are identical. Let XX be the orthogonal projection of MM on BCBC, and let YY be the orthogonal projection of NN on ABAB. Finally, let HH be the orthocenter of triangle ABCABC. Show that the points BB, XX, HH, YY lie on one circle.
vectorgeometrygeometric transformationsimilar trianglesgeometry proposed
Asymmetric polynomials

Source: German TST 2004, exam IV, invented by Eric M�ller, I think

5/18/2004
Let n be a positive integer. Find all complex numbers x1x_{1}, x2x_{2}, ..., xnx_{n} satisfying the following system of equations: x1+2x2+...+nxn=0x_{1}+2x_{2}+...+nx_{n}=0, x12+2x22+...+nxn2=0x_{1}^{2}+2x_{2}^{2}+...+nx_{n}^{2}=0, ... x1n+2x2n+...+nxnn=0x_{1}^{n}+2x_{2}^{n}+...+nx_{n}^{n}=0.
algebrapolynomiallinear algebramatrixcomplex numberssystem of equationsalgebra proposed
Excircles & angle bisectors

Source: German TST 2004, Exam V, Problem 1

5/1/2004
The AA-excircle of a triangle ABCABC touches the side BCBC at the point KK and the extended side ABAB at the point LL. The BB-excircle touches the lines BABA and BCBC at the points MM and NN, respectively. The lines KLKL and MNMN meet at the point XX. Show that the line CXCX bisects the angle ACNACN.
geometryparallelogramtrigonometryangle bisectorgeometry proposed
A mathematically experienced flea

Source: German TST 2004, exam VII, problem 1, by Arthur Engel

6/1/2004
Consider the real number axis (i. e. the xx-axis of a Cartesian coordinate system). We mark the points 11, 22, ..., 2n2n on this axis. A flea starts at the point 11. Now it jumps along the real number axis; it can jump only from a marked point to another marked point, and it doesn't visit any point twice. After the (2n−12n-1)-th jump, it arrives at a point from where it cannot jump any more after this rule, since all other points are already visited. Hence, with its 2n2n-th jump, the flea breaks this rule and gets back to the point 11. Assume that the sum of the (non-directed) lengths of the first 2n−12n-1 jumps of the flea was n(2n−1)n\left(2n-1\right). Show that the length of the last (2n2n-th) jump is nn.
analytic geometrynumber theory proposednumber theory