MathDB

2

Part of 2018 Saudi Arabia IMO TST

Problems(4)

no of 0's on the table is odd then max odd number on table is perfect square

Source: 2018 Saudi Arabia IMO TST I p2

7/28/2020
Let nn be an even positive integer. We fill in a number on each cell of a rectangle table of nn columns and multiple rows as following: i. Each row is assigned to some positive integer aa and its cells are filled by 00 or aa (in any order); ii. The sum of all numbers in each row is nn. Note that we cannot add any more row to the table such that the conditions (i) and (ii) still hold. Prove that if the number of 00’s on the table is odd then the maximum odd number on the table is a perfect square.
number theorycombinatoricsPerfect Square
arithmetic and geometric sequence inequality problems

Source: 2018 Saudi Arabia IMO TST II p2

7/28/2020
a) For integer n≥3n \ge 3, suppose that 0<a1<a2<...<an0 < a_1 < a_2 < ...< a_n is a arithmetic sequence and 0<b1<b2<...<bn0 < b_1 < b_2 < ... < b_n is a geometric sequence with a1=b1,an=bna_1 = b_1, a_n = b_n. Prove that a_k > b_k for all k=2,3,...,n−1k = 2,3,..., n -1. b) Prove that for every positive integer n≥3n \ge 3, there exist an integer arithmetic sequence (an)(a_n) and an integer geometric sequence (bn)(b_n) such that 0<b1<a1<b2<a2<...<bn<an0 < b_1 < a_1 < b_2 < a_2 < ... < b_n < a_n.
geometric sequenceinequalitiesarithmetic sequencealgebra
number of arabic subsets of {1,2, ..., n} has the same parity as n

Source: 2018 Saudi Arabia IMO TST III p2

7/28/2020
A non-empty subset of {1,2,...,n}\{1,2, ..., n\} is called arabic if arithmetic mean of its elements is an integer. Show that the number of arabic subsets of {1,2,...,n}\{1,2, ..., n\} has the same parity as nn.
number theory
right angle wanted, circumcircle and another circle related, angle bisector

Source: 2018 Saudi Arabia IMO TST IV p2

7/27/2020
Let ABCABC be an acute-angled triangle inscribed in circle (O)(O). Let GG be a point on the small arc ACAC of (O)(O) and (K)(K) be a circle passing through AA and GG. Bisector of ∠BAC\angle BAC cuts (K)(K) again at PP. The point EE is chosen on (K)(K) such that AEAE is parallel to BCBC. The line PKPK meets the perpendicular bisector of BCBC at FF. Prove that ∠EGF=90o\angle EGF = 90^o.
geometrycircumcircleright angleangle bisector