MathDB

2

Part of 2019 Taiwan TST Round 1

Problems(4)

Radical center of Five circles

Source: 2019 Taiwan TST Round 1, Quiz 3, Problem 2

3/31/2019
Given a convex pentagon ABCDE. ABCDE. Let A1 A_1 be the intersection of BD BD with CE CE and define B1,C1,D1,E1 B_1, C_1, D_1, E_1 similarly, A2 A_2 be the second intersection of ⊙(ABD1),⊙(AEC1) \odot (ABD_1),\odot (AEC_1) and define B2,C2,D2,E2 B_2, C_2, D_2, E_2 similarly. Prove that AA2,BB2,CC2,DD2,EE2 AA_2, BB_2, CC_2, DD_2, EE_2 are concurrent.
Proposed by Telv Cohl
geometrygeometry proposed
Denominator of a Product of Factorials

Source: 2019 Taiwan TST Round 1

3/31/2020
Given a positive integer n n , let A,B A, B be two co-prime positive integers such that BA=(n(n+1)2)!⋅∏k=1nk!(2k)! \frac{B}{A} = \left(\frac{n\left(n+1\right)}{2}\right)!\cdot\prod\limits_{k=1}^{n}{\frac{k!}{\left(2k\right)!}} Prove that A A is a power of 2 2 .
factorialnumber theory
Pudding Game

Source: 2019 Taiwan TST Round 1

3/31/2020
Alice and Bob play a game on a Cartesian Coordinate Plane. At the beginning, Alice chooses a lattice point (x0,y0) \left(x_{0}, y_{0}\right) and places a pudding. Then they plays by turns (B goes first) according to the rules
a. If A A places a pudding on (x,y) \left(x,y\right) in the last round, then B B can only place a pudding on one of (x+2,y+1),(x+2,y−1),(x−2,y+1),(x−2,y−1) \left(x+2, y+1\right), \left(x+2, y-1\right), \left(x-2, y+1\right), \left(x-2, y-1\right)
b. If B B places a pudding on (x,y) \left(x,y\right) in the last round, then A A can only place a pudding on one of (x+1,y+2),(x+1,y−2),(x−1,y+2),(x−1,y−2) \left(x+1, y+2\right), \left(x+1, y-2\right), \left(x-1, y+2\right), \left(x-1, y-2\right)
Furthermore, if there is already a pudding on (a,b) \left(a,b\right) , then no one can place a pudding on (c,d) \left(c,d\right) where c≡a(modn),d≡b(modn) c \equiv a \pmod{n}, d \equiv b \pmod{n} .
1. Who has a winning strategy when n=2018 n = 2018
1. Who has a winning strategy when n=2019 n = 2019
analytic geometrycombinatorics
Sum of Consecutive Squares is Square

Source: 2019 Taiwan TST Round 1

3/31/2020
Find all positive integers n n such that there exists an integer m m satisfying 1n∑k=mm+n−1k2 \frac{1}{n}\sum\limits_{k=m}^{m+n-1}{k^2} is a perfect square.
number theory