MathDB

4

Part of 2002 District Olympiad

Problems(6)

if DF + BE = AE then ABCD is square, <DAF=<FAE (2002 Romania District VII P4)

Source:

5/23/2020
Given the rectangle ABCDABCD. The points E,FE ,F lie on the segments (BC),(DC)(BC) , (DC) respectively, such that ∠DAF=∠FAE\angle DAF = \angle FAE. Proce that if DF+BE=AEDF + BE = AE then ABCDABCD is square.
equal anglesgeometrysquarerectangle
perimeter of triangle KLM on surface of a cube of length a > 2a\sqrt3

Source: 2002 Romania District VIII P4

5/24/2020
The cube ABCDA′B′C′D′ABCDA' B' C' D' has of length a. Consider the points K∈[AB],L∈[CC′],M∈[D′A′]K \in [AB], L \in [CC' ], M \in [D'A'].
a) Show that 3KL≥KB+BC+CL\sqrt3 KL \ge KB + BC + CL
b) Show that the perimeter of triangle KLMKLM is strictly greater than 2a32a\sqrt3.
perimetergeometry3D geometrycubegeometric inequality
Exists n numbers whose sum is equal to sum of its squares and arbitrary number

Source: Romanian District Olympiad 2002, Grade IX, Problem 4

10/7/2018
Let n≥2 n\ge 2 be a natural number. Prove the following propositions:
a) a1,a2,…,an∈R∧a1+⋯+an=a12+⋯+an2  ⟹  a1+⋯+an≤an. a_1,a_2,\ldots ,a_n\in\mathbb{R}\wedge a_1+\cdots +a_n=a_1^2+\cdots +a_n^2\implies a_1+\cdots +a_n\le a_n. b) x\in [1,n]\implies\exists b_1,b_2,\ldots ,b_n\in\mathbb{R}_{\ge 0}  x=b_1+\cdots +b_n=b_1^2 +\cdots +b_n^2 .
number theoryalgebra
Sets S s.t. {1,n} in S in {1,2,...,n}

Source: Romanian District Olympiad 2002, Grade X, Problem 4

10/7/2018
For any natural number n≥2, n\ge 2, define m(n) m(n) to be the minimum number of elements of a set S S that simultaneously satisfy: \text{(i)}  \{ 1,n\} \subset S\subset \{ 1,2,\ldots ,n\} \text{(ii)}  any element of S, S, distinct from 1, 1, is equal to the sum of two (not necessarily distinct) elements from S. S.
a) Prove that m(n)\ge 1+\left\lfloor \log_2 n \right\rfloor , \forall n\in\mathbb{N}_{\ge 2} . b) Prove that there are infinitely many natural numbers n≥2 n\ge 2 such that m(n)=m(n+1). m(n)=m(n+1).
⌊⌋ \lfloor\rfloor denotes the usual integer part.
set theorycountingalgebraclosedness
Romania District Olympiad 2002 - Grade XI

Source:

3/18/2011
Consider a function f:R→Rf:\mathbb{R}\rightarrow \mathbb{R} such that:
1. ff has one-side limits in any a∈Ra\in \mathbb{R} and f(a−0)≤f(a)≤f(a+0)f(a-0)\le f(a)\le f(a+0).
2. for any a,b∈R, a<ba,b\in \mathbb{R},\ a<b, we have f(a−0)<f(b−0)f(a-0)<f(b-0).
Prove that ff is strictly increasing.
Mihai Piticari & Sorin Radulescu
functionreal analysisreal analysis unsolved
Shifting endpoints of integral when it&acute;s periodic

Source: Romanian District Olympiad 2002, Grade XII, Problem 4

10/7/2018
Let be a continuous and periodic function f:R⟶[0,∞) f:\mathbb{R}\longrightarrow [0,\infty ) of period 1. 1. Show:
a) a∈R  ⟹  ∫aa+1f(x)dx=∫01f(x)dx. a\in\mathbb{R}\implies\int_a^{a+1} f(x)dx =\int_0^1 f(x) dx .
b) lim⁡n→∞∫01f(x)f(nx)dx=(∫01f(x)dx)2. \lim_{n\to\infty} \int_0^1 f(x)f(nx) dx=\left( \int_0^1 f(x) dx \right)^2 .
C. Mortici
calculusintegrationbreaking integralperiodic functionPeriodicity