MathDB

3

Part of 2018 Saudi Arabia IMO TST

Problems(4)

DP // angle bisector of <BEC , ABCD cyclic, DB = DA + DC, AP = BC, BE _|_AD

Source: 2018 Saudi Arabia IMO TST I p3

7/27/2020
Let ABCDABCD be a convex quadrilateral inscibed in circle (O)(O) such that DB=DA+DCDB = DA + DC. The point PP lies on the ray ACAC such that AP=BCAP = BC. The point EE is on (O)(O) such that BE⊥ADBE \perp AD. Prove that DPDP is parallel to the angle bisector of ∠BEC\angle BEC.
geometryangle bisectorcyclic quadrilateralparallelequal segmentsperpendicular
ab+1 is perfect square

Source: 2018 Saudi Arabia IMO TST II p3

7/28/2020
Two sets of positive integers A,BA, B are called connected if they are not empty and for all a∈A,b∈Ba \in A, b \in B, number ab+1ab + 1 is a perfect square. i) Given A={1,2,3,4}A =\{1, 2,3, 4\}. Prove that there does not exist any set BB such that A,BA, B are connected. ii) Suppose that A,BA, B are connected with ∣A∣,∣B∣≥2|A|,|B| \ge 2. For any a1>a2∈Aa_1 > a_2 \in A and b1>b2∈Bb_1 > b_2 \in B, prove that a1b1>13a2b2a_1b_1 > 13a_2b_2.
Perfect Squarenumber theory
x_0 > 2^{2018} where xo_ is max of f (x) = (x - F_1)(x - F_2) ...(x -F_{3030})

Source: 2018 Saudi Arabia IMO TST III p3

7/28/2020
Consider the function f(x)=(x−F1)(x−F2)...(x−F3030)f (x) = (x - F_1)(x - F_2) ...(x -F_{3030}) with (Fn)(F_n) is the Fibonacci sequence, which defined as F1=1,F2=2F_1 = 1, F_2 = 2, Fn+2=Fn+1+FnF_{n+2 }=F_{n+1} + F_n, n≥1n \ge 1. Suppose that on the range (F1,F3030)(F_1, F_{3030}), the function ∣f(x)∣|f (x)| takes on the maximum value at x=x0x = x_0. Prove that x0>22018x_0 > 2^{2018}.
fibonacci numberalgebrainequalitiesmax
exists permutation a_i of i=1,1000 such that |a_i - i| = k

Source: 2018 Saudi Arabia IMO TST IV p3

7/28/2020
Find all positive integers kk such that there exists some permutation of (1,2,...,1000)(1, 2,...,1000) namely (a1,a2,...,a1000)(a_1, a_2,..., a_{1000}) and satisfy ∣ai−i∣=k|a_i - i| = k for all i=1,1000i = 1,1000.
permutationcombinatorics