MathDB

1

Part of 2015 Dutch IMO TST

Problems(2)

moving a pawn on grid points (x,y) to (x\pm a,y \pm a) or (x\pm b,y \pm b)

Source: Dutch IMO TST 2015 day 2 p1

8/30/2019
Let aa and bb be two positive integers satifying gcd(a,b)=1gcd(a, b) = 1. Consider a pawn standing on the grid point (x,y)(x, y). A step of type A consists of moving the pawn to one of the following grid points: (x+a,y+a),(x+a,y−a),(x−a,y+a)(x+a, y+a),(x+a,y-a), (x-a, y + a) or (x−a,y−a)(x - a, y - a). A step of type B consists of moving the pawn to (x+b,y+b),(x+b,y−b),(x−b,y+b)(x + b,y + b),(x + b,y - b), (x - b,y + b) or (x−b,y−b)(x - b,y - b). Now put a pawn on (0,0)(0, 0). You can make a ( nite) number of steps, alternatingly of type A and type B, starting with a step of type A. You can make an even or odd number of steps, i.e., the last step could be of either type A or type B. Determine the set of all grid points (x,y)(x,y) that you can reach with such a series of steps.
gridpointslattice pointscombinatoricscombinatorial geometry
<ADB = < EDC iff MA = MC, ABCD with <A=<C=90^o given

Source: Dutch IMO TST1 2015 P1

8/5/2019
In a quadrilateral ABCDABCD we have ∠A=∠C=90o\angle A = \angle C = 90^o. Let EE be a point in the interior of ABCDABCD. Let MM be the midpoint of BEBE. Prove that ∠ADB=∠EDC\angle ADB = \angle EDC if and only if ∣MA∣=∣MC∣|MA| = |MC|.
right angleequal anglesequal segmentsgeometry