MathDB

2

Part of 2015 All-Russian Olympiad

Problems(2)

Parallelogram to find angle

Source: All Russian MO 2015, grade 10, problem 2

8/7/2015
Given is a parallelogram ABCDABCD, with AB<AC<BCAB <AC <BC. Points EE and FF are selected on the circumcircle ω\omega of ABCABC so that the tangenst to ω\omega at these points pass through point DD and the segments ADAD and CECE intersect. It turned out that ∠ABF=∠DCE\angle ABF = \angle DCE. Find the angle ∠ABC\angle{ABC}. A. Yakubov, S. Berlov
geometrycircumcircleparallelogram
Parity of a sum of numerators

Source: All Russian Olympiad 2015 11.2

12/11/2015
Let n>1n > 1 be a natural number. We write out the fractions 1n\frac{1}{n}, 2n\frac{2}{n}, …\dots , n−1n\dfrac{n-1}{n} such that they are all in their simplest form. Let the sum of the numerators be f(n)f(n). For what n>1n>1 is one of f(n)f(n) and f(2015n)f(2015n) odd, but the other is even?
number theory