A collection of cardboard circles, each with a diameter of at most $1$, lie on a $5\times 8$ table without overlapping or overhanging the edge of the table. A cardboard circle of diameter $2$ is added to the collection. Prove that this new collection of cardboard circles can be placed on a $7\times 7$ table without overlapping or overhanging the edge.
Problem
Source: XIII Rioplatense Mathematical Olympiad (2004), Level 3
Tags: geometry, rectangle, combinatorics unsolved, combinatorics
mavropnevma
26.07.2011 20:06
Let the rectangular $5\times 8$ table be $ABCD$, with $AB=CD=5$ and $BC = DA = 8$. Consider points $C'$ on $BC$ and $D'$ on $DA$ so that $CC'= DD'= 2$. Consider point $E$ such that $B$ on $AE$ and $AE=7$, point $F$ on $AD$ such that $AF=7$, and points $G$ such that $AEGF$ is a $7\times 7$ square, and $H$ on $BC$ such that $BH = 7$. Rectangle $CC'D'D$ fits inside rectangle $EGHB$, leaving room for a $2\times 2$ square to contain the circle of diameter $2$. All old discs meeting $CHFD$ were contained in $CC'D'D$, and are moved to $EGHB$.
horizon
26.11.2011 17:43
mavropnevma wrote: Let the rectangular $5\times 8$ table be $ABCD$, with $AB=CD=5$ and $BC = DA = 8$. Consider points $C'$ on $BC$ and $D'$ on $DA$ so that $CC'= DD'= 2$. Consider point $E$ such that $B$ on $AE$ and $AE=7$, point $F$ on $AD$ such that $AF=7$, and points $G$ such that $AEGF$ is a $7\times 7$ square, and $H$ on $BC$ such that $BH = 7$. Rectangle $CC'D'D$ fits inside rectangle $EGHB$, leaving room for a $2\times 2$ square to contain the circle of diameter $2$. All old discs meeting $CHFD$ were contained in $CC'D'D$, and are moved to $EGHB$. your solution probably has some bugs.when rectangle $FHCD$ translates to rectangle $BEGH$, some circles with diameter at most $1$ which cover the $5*8$ initially may be divided into half!
mavropnevma
26.11.2011 18:43
NO; maybe I did not make myself perfectly clear. All discs contained in $AFHB$ remain there (this is part of the $7\times 7$ new table $AFGE$). Only the discs touching $FH$, thus contained in $CC'D'D$, are moved into $EGHB$, leaving room for a $2\times 2$ square to contain the diameter $2$ disc. No disc is ever cut, since I never said rectangle $CHFD$ is fully moved; the only things that move are the discs contained in $CC'D'D$.
horizon
28.11.2011 18:19
mavropnevma wrote: NO; maybe I did not make myself perfectly clear. All discs contained in $AFHB$ remain there (this is part of the $7\times 7$ new table $AFGE$). Only the discs touching $FH$, thus contained in $CC'D'D$, are moved into $EGHB$, leaving room for a $2\times 2$ square to contain the diameter $2$ disc. No disc is ever cut, since I never said rectangle $CHFD$ is fully moved; the only things that move are the discs contained in $CC'D'D$. Thanks mavropnevma.Initially I don't understand it correctly.here you mean moving the rectangle $D'C'CD$ to $BEGH$,the circles which intersect segment $FH$ are all in $D'C'CD$