MathDB

4

Part of 1989 China Team Selection Test

Problems(2)

P_1 Q_1 R_1 is congruent to triangle P_2 Q_2 R_2

Source: China TST 1989, problem 4

6/27/2005
Given triangle ABCABC, squares ABEF,BCGH,CAIJABEF, BCGH, CAIJ are constructed externally on side AB,BC,CAAB, BC, CA, respectively. Let AH∩BJ=P1AH \cap BJ = P_1, BJ∩CF=Q1BJ \cap CF = Q_1, CF∩AH=R1CF \cap AH = R_1, AG∩CE=P2AG \cap CE = P_2, BI∩AG=Q2BI \cap AG = Q_2, CE∩BI=R2CE \cap BI = R_2. Prove that triangle P1Q1R1P_1 Q_1 R_1 is congruent to triangle P2Q2R2P_2 Q_2 R_2.
geometrycircumcirclegeometric transformationrotationcongruent trianglescyclic quadrilateral
Sum of partitions

Source: China TST 1989, problem 8

6/27/2005
∀n∈N\forall n \in \mathbb{N}, P(n)P(n) denotes the number of the partition of nn as the sum of positive integers (disregarding the order of the parts), e.g. since 4=1+1+1+1=1+1+2=1+3=2+2=44 = 1+1+1+1 = 1+1+2 = 1+3 = 2+2 = 4, so P(4)=5P(4)=5. "Dispersion" of a partition denotes the number of different parts in that partitation. And denote q(n)q(n) is the sum of all the dispersions, e.g. q(4)=1+2+2+1+1=7q(4)=1+2+2+1+1=7. n≥1n \geq 1. Prove that (1) q(n)=1+∑i=1n−1P(i).q(n) = 1 + \sum^{n-1}_{i=1} P(i). (2) 1+∑i=1n−1P(i)≤2⋅n⋅P(n)1 + \sum^{n-1}_{i=1} P(i) \leq \sqrt{2} \cdot n \cdot P(n).
combinatorics unsolvedcombinatorics