Problems(2)
Tiling a chessboard with 1x3 triominos and T-tetrominos
Source: XII Rioplatense Mathematical Olympiad (2003), Level 3
8/8/2011
An chessboard is to be tiled (i.e., completely covered without overlapping) with pieces of the following shapes:
[asy]
unitsize(.6cm);
draw(unitsquare,linewidth(1));
draw(shift(1,0)*unitsquare,linewidth(1));
draw(shift(2,0)*unitsquare,linewidth(1));
label("\footnotesize rectangle",(1.5,0),S);
draw(shift(8,1)*unitsquare,linewidth(1));
draw(shift(9,1)*unitsquare,linewidth(1));
draw(shift(10,1)*unitsquare,linewidth(1));
draw(shift(9,0)*unitsquare,linewidth(1));
label("\footnotesize T-shaped tetromino",(9.5,0),S);
[/asy] The rectangle covers exactly three squares of the chessboard, and the T-shaped tetromino covers exactly four squares of the chessboard. (a) What is the maximum number of pieces that can be used?
(b) How many ways are there to tile the chessboard using this maximum number of pieces?
geometryrectanglesymmetrygeometric transformationreflectionrotationcombinatorics unsolved
Packing hexagons in an isosceles right triangle
Source: XII Rioplatense Mathematical Olympaid (2003), Level 3
8/9/2011
Without overlapping, hexagonal tiles are placed inside an isosceles right triangle of area whose hypotenuse is horizontal. The tiles are similar to the figure below, but are not necessarily all the same size.[asy]
unitsize(.85cm);
draw((0,0)--(1,0)--(1,1)--(2,2)--(-1,2)--(0,1)--(0,0),linewidth(1));
draw((0,2)--(0,1)--(1,1)--(1,2),dashed);
label("\footnotesize ",(0.5,0),S);
label("\footnotesize ",(0,0.5),W);
label("\footnotesize ",(1,0.5),E);
label("\footnotesize ",(0,1.5),E);
label("\footnotesize ",(1,1.5),W);
label("\footnotesize ",(-0.5,2),N);
label("\footnotesize ",(0.5,2),N);
label("\footnotesize ",(1.5,2),N);
[/asy] The longest side of each tile is parallel to the hypotenuse of the triangle, and the horizontal side of length of each tile lies between this longest side of the tile and the hypotenuse of the triangle. Furthermore, if the longest side of a tile is farther from the hypotenuse than the longest side of another tile, then the size of the first tile is larger or equal to the size of the second tile. Find the smallest value of such that every such configuration of tiles has a total area less than .
geometrycombinatorics unsolvedcombinatorics