Pentominoes Bruce Bguley Cscde Mth Systems, LLC Astrct. Pentominoes nd their reltives the polyominoes, polycues, nd polyhypercues will e used to explore nd pply vrious importnt mthemticl concepts. In this pper, the definition of the pentominoes will e exmined. It will e proved tht there re exctly 12 pentominoes. The pentominoes will then e represented s mtrices with the mnipultions of the pentominoes represented s mtrix ddition nd multipliction. The concept of pentomino s mod numers will e introduced nd, in n ppliction of modulr rithmetic, will e used to prove the existence nd uniqueness of pentominoes nd figures mde with pentominoes s well s similr fcts out polyominoes, polycues, nd polyhypercues. The pentominoes re simple-looking set of ojects through which some powerful mthemticl ides cn e introduced, investigted, nd pplied. The pentominoes cn e expressed in n nlytic geometric setting where they re chnged into vectors nd mtrices with integrl entries nd mnipulted s numers. The questions out pentominoes cn e trnsformed into equivlent questions in different mthemticl setting from which some interesting properties of the pentominoes cn e proved. This is similr to the ides in lgeric topology where questions out spheres nd tori cn e nswered y looking t the lgeric structure of pths on their surfces nd converting the topologicl questions into lgeric ones. With pentominoes nd their reltives, the ojects re trnsformed into mtrices nd vectors to which liner lger nd numer theory re pplied. The pentominoes re puzzle tht hs een used y techers to introduce students to importnt mth concepts such s symmetry, re, nd perimeter. Pentominoes re suggested for use y techers on pge 99 of the NCTM Principles nd Stndrds, in the Geometry Stndrd of the Pre-K-2 section. They pper s well in vrious NCTM rticles, such s The Pentomino Squre Prolem in Mthemtics Teching in the Middle School, Mrch 2003 p. 355 nd the 1
NCTM Illumintions Exploring Cues Activity Sheet. In these rticles, the following questions re investigted: 1. How mny pentominoes re there? 2. Cn certin figure e mde with pentominoes? 3. Cn the pentomino e folded to mke n open ox? This pper will look t the first two questions ut will go into them in much greter depth using more dvnced mth, nmely liner lger nd modulr rithmetic, to prove some surprising results. First there must e definition of pentomino. The ove ctivities give muddled or incorrect definitions. The usul definition, such s the one given in the NCTM s Principles nd Stndrds is incorrect nd yields n infinite numer of pentominoes. On p.99, it is stted, For exmple, second-grde techer might instruct the clss to find ll the different wys to put five squres together so tht one edge of ech squre coincides with n edge of t lest one other squre (see fig. 4.15). By tht definition, the figure t the left is pentomino ecuse one edge of ech squre coincides with n edge of t lest one other squre. Definitions must e mde crefully so tht ll the squres in the figure re connected. A etter definition would e tht polyomino is plne figure mde of squres such tht two different squres cn touch only on sides which coincide nd for every two squres in the polyomino, there is pth tht goes through djcent squres from the first squre chosen to the lst. The figure t the right shows pth from one squre to nother in lrge polyomino. 2
Students could e sked for different tests tht would show this connectedness. Do they hve to find pth etween ny two squres in figure to show tht it is polyomino or cn they find simpler test tht requires fewer pths? The numer of such pths etween ny two squres in polyomino is the hndshke prolem nd requires n(n-1)/2 checks to see if it is true for polyomino with n squres. If one squre is connected y pth to every other squre, only n-1 checks re required. This is sufficient ecuse if there is pth from squre to squre nd pth from squre to squre c, then there is pth from to c going through. Students could e sked to find other tests. For exmple, color squre in polyomino nd color ll the squres in the polyomino tht touch the colored squre. Color ll the squres tht touch colored squre. Continue coloring squres in this mnner in the polyomino s shown elow. If ll the squres in the figure re eventully colored then it is polyomino. This test would require t most n-1 steps in coloring. Students could e sked for proof of this ssertion. The question, Are there exctly 12 pentominoes? cn now e nswered. All the pentominoes cn e found y looking t the tetrminoes, tht is, polyominoes with 4 squres, dding one squre to ech of them, nd throwing out the duplictes. In generl, the n+1-ominoes cn e found y looking t the n-ominoes, dding one squre to ech of them in every possile wy, nd throwing out the duplictes. How is this shown? Tke n n+1-omino. If there is squre in tht n+1-omino tht cn e removed nd the remining n squres re still connected, tht is, is n n-omino, then it hs een shown tht the n+1-omino cn e mde y dding one 3
squre to n n-omino. Tke ll the pths from one squre in the n+1-omino to nother squre in the n+1-omino which do not go through the sme squre twice. Since there re finite numer of squres in polyomino, there is longest such pth. The end squres cn e removed with the remining squres of the originl n+1-omino still eing connected, tht is the remining squres form n n-omino for the following reson. If the remining squres did rek into two disconnected pieces, then the longest pth would e in one of the pieces. But there would e longer pth in the originl n+1-omino y connecting the end point tht we removed to squre in the disconnected piece not contining the pth. This is contrdiction, ecuse the pth originlly choosen ws the longest pth in the n+1-omino nd the constructed pth is longer. Thus, the n+1-omino cn e formed from n n-omino y dding one squre. The pth in the polyomino t the right is the longest possile nd oth its endpoints re removle. There re now specific numer of pentominoes to check y dding one squre to ll of the tetrminoes or 4-ominoes nd throwing out duplictes. This process is itertive, there is only one monomino or 1-omino. Generte the one domino or 2-omino from the 1-omino; the two 3-ominoes from the 2-omino; the five 4-ominoes from the two 3-ominoes; nd finlly the twelve pentominoes from the five 4-ominoes. The prolem comes down to eliminting the duplictes. Here re ll the wys to dd squre to the domino or 2-omino. 4
Students could show ll these nd then decide which re the sme. Hve the students explin why they think tht two of the figures re the sme y hving them show how they would move one figure to the other. This cn led into discussion of rigid motions in the plne, tht is, rottion nd reflection s well s trnsltion. The students cn continue with this process to otin the 12 pentominoes. Some students might wnt to continue nd find the 35 hexominoes. Here re the 12 pentominoes nd their common letter nmes: T U V W X Y Z F L I P N If liner lger nd numer theory re to e used with the pentominoes, then somehow numers must e ttched to ech pentomino. Let the sides of the squres in the pentominoes e length 1. Look t coordinte plne nd tke the grid of ll points with integrl coordintes. If you plce pentomino on the plne so tht the center of ech of its squres is on grid point, then the pentomino piece could e written s mtrix consisting of 5 ordered pirs. Consider the piece in the figure to the right. This is the F pentomino. You cn consider the piece in numer of wys. First, it is the five squres outlined y the hevy lck line. Second, it is the ( 0, 5) ( 1, 5) ( 2, 5) ( 3, 5) ( 4, 5) ( 5, 5) ( 0, 4) ( 1, 4) ( 2, 4) ( 3, 4) ( 4, 4) ( 5, 4) ( 0, 3) ( 1, 3) ( 2, 3) ( 3, 3) ( 4, 3) ( 5, 3) ( 0, 2) ( 1, 2) ( 2, 2) ( 3, 2) ( 4, 2) ( 5, 2) ( 0, 1) ( 1, 1) ( 2, 1) ( 3, 1) ( 4, 1) ( 5, 1) 2 1 1 2 figure represented y the hevy lue lines. Third, it is the mtrix shown on the left. If we ( 0, 0) ( 1, 0) ( 2, 0) ( 3, 0) ( 4, 0) ( 5, 0) move the piece, we would chnge the mtrix representing the piece, ut we would chnge the mtrix in wy tht ws representle s mtrix opertions. 5
For exmple, dding the mtrix on the left to pentomino mtrix would move the pentomino right squres nd up squres. Multiplying pentomino mtrix on the right y these mtrices does the following: 1 0 0 1] Do not move the pentomino. 0-1 1 0] Rotte clockwise 90 0. -1 0 Rotte clockwise 180 0-1] 0 or Reflect through the origin. 0 1 Rotte clockwise 270-1 0] 0 or Rotte counter-clockwise 90 0. -1 0 0 1] Reflect through the y-xis. 1 0 0-1] Reflect through the x-xis. 0 1 1 0] Reflect through the line y = x. 0-1 -1 0] Reflect through the line y = -x. It is the ojective to find some numer or numers tht distinguish ech pentomino. It is not t ll ovious tht the following mtrices ll represent the sme pentomino, the F pentomino: 2 1 1 2-2 -2 1 2-3 -2-3 -4-3 -3 4 3 2 1-4 -3 Even shuffling the rows mkes it difficult to see tht the first nd third mtrices contin the sme points. It is importnt to mke it more ovious. Adding the columns to get single vector will give the sme result even if the rows re shuffled. How will this ffect the other mtrices representing the F pentomino? If the pentomino is trnslted over to the right nd up we hve the following: 2 1 1 2 ( 10, 11 ) + ] = 2 + 1 + 1 + 2 + 2 + 2 + 2 + 3 + 3 + 3 + ( 10+5, 11+5 ) 6
Trnslting the F pentomino over nd up gives vector tht differs from the originl vector y multiple of 5 in oth coordintes. If the vector mod 5 is tken, then the vector of ny trnsltion mod 5 stys the sme. This is clled mod numer for the prticulr pentomino. For the exmple of the F pentmino ove with vector (10,11), the mod numer is (0,1) since (10,11) = (5*2+0, 5*2+1). Any trnslte of this prticulr plcing of the F pentomino would hve mod numer of (0,1). The lck dots on the grid to the right show ll the vector sums of trnsltes tht this prticulr orienttion of the F pentomino could e. The smll red squre in the first qudrnt of the grid shows ll the possile loctions for mod numers of ny pentomino. The red dot is mod numer for this prticulr orienttion of the F pentomino. How do reflections nd rottions ffect the mod numer? Since multiplying vector y mtrix is liner, the mod numer is multiplied y the mtrix. Since we re only deling with interchnging columns nd multiplying y 1 or -1, the mod numer for the new position of the F pentomino will e the mod numer of the originl position multiplied y the mtrix corresponding to the reflection or rottion. Why re the mod numers of ny use in working with the pentominos. The only possile mod numers tht piece cn hve re otined y tking the mod numers for ny orienttion of the piece nd then multiplying those mod numers y the mtrices for the eight reflections nd rottions. Two pentominoes cn e the sme pentomino only if they hve the sme mod 7
numers. This is not crucil with the pentominoes ecuse visul inspection cn quickly determine if two pentominoes re the sme except for rottions nd reflections. However when deling with something like polyhypercues, sy with 10 cues in 6 dimensions, then the mod numers will e vectors with 6 coordintes mod 10. There might e some overlp of mod numers etween pieces, ut the mod numers considerly reduce the numer of other polyhypercues needed to e checked for dupliction s well s indicting definite procedure to follow for the check, tht is, orient so tht the mod numers re equl, dictionry order, nd check for equlity. A computer could e used to go rpidly through the possiilities. A second use which hs mny more pplictions with the pentominoes would e showing whether shpe is impossile to mke with the pentominoes. Since ny shpe formed with pentominoes consists of multiple of 5 squres, the mod numer for this shpe is defined nd is equl to the sum of the mod numers of the pieces. Tke the 6 y 10 rectngle, which cn e formed with the 12 pentominoes. Let the pentominoes e plced on the grid. If the sum of the mod numers of the pentominoes does not equl (0,0), the mod numer of the rectngle, the pentominoes in those orienttions cn not form the rectngle y trnsltion without rottion or reflection. It is difficult to stop from physiclly rotting or turning over piece when working with the pentominoes, ut the mod numers would e very useful when solving for shpe using computer progrm. On the remining pges, the mod numers for the P pentomino re shown. One position of the P pentomino is shown in the first qudrnt long with the seven other mod numers tht re otined y reflections nd rottions. The mod numer clsses for the different pentominoes re then shown on the following pge. The pper finishes with severl exmples using mod numers to construct shpes with pentominoes nd prove the impossiility of certin constructions with pentominoes nd dominoes. 8
] -1 0 1 0-1 0 = - - -4 3 - - 1 0] 0 1 0 1 = Pentominoes 0 1 1 0 () = (1 3) () = () (-5, 5) (-4, 5) (-3, 5) (-2, 5) (-1, 5) ( 0, 5) ( 1, 5) ( 2, 5) ( 3, 5) ( 4, 5) ( 5, 5) (-5, 4) (-4, 4) (-3, 4) (-2, 4) (-1, 4) ( 0, 4) ( 1, 4) ( 2, 4) ( 3, 4) ( 4, 4) ( 5, 4) (-5, 3) (-4, 3) (-3, 3) (-2, 3) (-1, 3) ( 0, 3) ( 1, 3) ( 2, 3) ( 3, 3) ( 4, 3) ( 5, 3) (-5, 2) (-4, 2) (-3, 2) (-2, 2) (-1, 2) ( 0, 2) ( 1, 2) ( 2, 2) ( 3, 2) ( 4, 2) ( 5, 2) (-5, 1) (-4, 1) (-3, 1) (-2, 1) (-1, 1) ( 0, 1) ( 1, 1) ( 2, 1) ( 3, 1) ( 4, 1) ( 5, 1) (-5, 0) (-4, 0) (-3, 0) (-2, 0) (-1, 0) ( 0, 0) ( 1, 0) ( 2, 0) ( 3, 0) ( 4, 0) ( 5, 0) (-5,-1) (-4,-1) (-3,-1) (-2,-1) (-1,-1) ( 0,-1) ( 1,-1) ( 2,-1) ( 3,-1) ( 4,-1) ( 5,-1) (-5,-2) (-4,-2) (-3,-2) (-2,-2) (-1,-2) ( 0,-2) ( 1,-2) ( 2,-2) ( 3,-2) ( 4,-2) ( 5,-2) (-5,-3) (-4,-3) (-3,-3) (-2,-3) (-1,-3) ( 0,-3) ( 1,-3) ( 2,-3) ( 3,-3) ( 4,-3) ( 5,-3) (-5,-4) (-4,-4) (-3,-4) (-2,-4) (-1,-4) ( 0,-4) ( 1,-4) ( 2,-4) ( 3,-4) ( 4,-4) ( 5,-4) (-5,-5) (-4,-5) (-3,-5) (-2,-5) (-1,-5) ( 0,-5) ( 1,-5) ( 2,-5) ( 3,-5) ( 4,-5) ( 5,-5) -1 0] = -2-2 -3-2 -3-4 -3-3 -2-3 9 0-1] 0-1 1 0-1 0] 0-1] 0-1 1 0 = ] 2-2 2-3 4-3 3-3 3-2 () = (2 1) () = (4 2)
-1 0 0 1 0 1 = - - - - - 0 1] 1 0 0 1] 1 0 = 4 3-1 0 () = (2 4) () = (4 3) (-5, 5) (-4, 5) (-3, 5) (-2, 5) (-1, 5) ( 0, 5) ( 1, 5) ( 2, 5) ( 3, 5) ( 4, 5) ( 5, 5) (-5, 4) (-4, 4) (-3, 4) (-2, 4) (-1, 4) ( 0, 4) ( 1, 4) ( 2, 4) ( 3, 4) ( 4, 4) ( 5, 4) (-5, 3) (-4, 3) (-3, 3) (-2, 3) (-1, 3) ( 0, 3) ( 1, 3) ( 2, 3) ( 3, 3) ( 4, 3) ( 5, 3) (-5, 2) (-4, 2) (-3, 2) (-2, 2) (-1, 2) ( 0, 2) ( 1, 2) ( 2, 2) ( 3, 2) ( 4, 2) ( 5, 2) (-5, 1) (-4, 1) (-3, 1) (-2, 1) (-1, 1) ( 0, 1) ( 1, 1) ( 2, 1) ( 3, 1) ( 4, 1) ( 5, 1) (-5, 0) (-4, 0) (-3, 0) (-2, 0) (-1, 0) ( 0, 0) ( 1, 0) ( 2, 0) ( 3, 0) ( 4, 0) ( 5, 0) (-5,-1) (-4,-1) (-3,-1) (-2,-1) (-1,-1) ( 0,-1) ( 1,-1) ( 2,-1) ( 3,-1) ( 4,-1) ( 5,-1) (-5,-2) (-4,-2) (-3,-2) (-2,-2) (-1,-2) ( 0,-2) ( 1,-2) ( 2,-2) ( 3,-2) ( 4,-2) ( 5,-2) (-5,-3) (-4,-3) (-3,-3) (-2,-3) (-1,-3) ( 0,-3) ( 1,-3) ( 2,-3) ( 3,-3) ( 4,-3) ( 5,-3) (-5,-4) (-4,-4) (-3,-4) (-2,-4) (-1,-4) ( 0,-4) ( 1,-4) ( 2,-4) ( 3,-4) ( 4,-4) ( 5,-4) (-5,-5) (-4,-5) (-3,-5) (-2,-5) (-1,-5) ( 0,-5) ( 1,-5) ( 2,-5) ( 3,-5) ( 4,-5) ( 5,-5) 0-1] -2-2 -2-3 -4-3 -3-3 -3-2 10 = = 1 0] -1 0 0-1 -1 0 0-1 ] 2-2 3-2 3-4 3-3 2-3 0-1 1 0 () = (1 2) () = (3 1)
( 0, 4) ( 1, 4) ( 2, 4) ( 3, 4) ( 4, 4) The pentominoes seprte into 6 clsses of pieces. ( 0, 3) ( 1, 3) ( 2, 3) ( 3, 3) ( 4, 3) ( 0, 2) ( 1, 2) ( 2, 2) ( 3, 2) ( 4, 2) The first clss is the clss contining I, X, nd Z. This clss ( 0, 1) ( 1, 1) ( 2, 1) ( 3, 1) ( 4, 1) ( 0, 0) ( 1, 0) ( 2, 0) ( 3, 0) ( 4, 0) hs the mod numers (0 0). ( 0, 4) ( 1, 4) ( 2, 4) ( 3, 4) ( 4, 4) ( 0, 3) ( 1, 3) ( 2, 3) ( 3, 3) ( 4, 3) The second is the clss contining F. This clss hs the mod numers, (0 1), (0 4), (1 0), nd (4 0). ( 0, 4) ( 1, 4) ( 2, 4) ( 3, 4) ( 4, 4) ( 0, 2) ( 1, 2) ( 2, 2) ( 3, 2) ( 4, 2) ( 0, 1) ( 1, 1) ( 2, 1) ( 3, 1) ( 4, 1) ( 0, 0) ( 1, 0) ( 2, 0) ( 3, 0) ( 4, 0) ( 0, 3) ( 1, 3) ( 2, 3) ( 3, 3) ( 4, 3) ( 0, 2) ( 1, 2) ( 2, 2) ( 3, 2) ( 4, 2) The third is the clss contining T nd U. This clss hs the mod numers, (0 2), (0 3), (2 0), nd (3 0). ( 0, 1) ( 1, 1) ( 2, 1) ( 3, 1) ( 4, 1) ( 0, 0) ( 1, 0) ( 2, 0) ( 3, 0) ( 4, 0) The fourth is the clss contining W nd L. This clss hs the mod numers, (1 1), (1 4), (4 1), nd (1 4). ( 0, 4) ( 1, 4) ( 2, 4) ( 3, 4) ( 4, 4) ( 0, 3) ( 1, 3) ( 2, 3) ( 3, 3) ( 4, 3) ( 0, 2) ( 1, 2) ( 2, 2) ( 3, 2) ( 4, 2) ( 0, 1) ( 1, 1) ( 2, 1) ( 3, 1) ( 4, 1) ( 0, 4) ( 1, 4) ( 2, 4) ( 3, 4) ( 4, 4) ( 0, 0) ( 1, 0) ( 2, 0) ( 3, 0) ( 4, 0) ( 0, 3) ( 1, 3) ( 2, 3) ( 3, 3) ( 4, 3) ( 0, 2) ( 1, 2) ( 2, 2) ( 3, 2) ( 4, 2) ( 0, 1) ( 1, 1) ( 2, 1) ( 3, 1) ( 4, 1) ( 0, 0) ( 1, 0) ( 2, 0) ( 3, 0) ( 4, 0) The fifth is the clss contining Y nd P. This clss hs the mod numers, (1 2), (1 3), (2 1), (2 4), (3 1), (), (4 2), nd (4 3). ( 0, 4) ( 1, 4) ( 2, 4) ( 3, 4) ( 4, 4) ( 0, 3) ( 1, 3) ( 2, 3) ( 3, 3) ( 4, 3) The sixth is the clss contining V nd N. This clss hs the mod numers, (), (), (), nd (). ( 0, 2) ( 1, 2) ( 2, 2) ( 3, 2) ( 4, 2) ( 0, 1) ( 1, 1) ( 2, 1) ( 3, 1) ( 4, 1) ( 0, 0) ( 1, 0) ( 2, 0) ( 3, 0) ( 4, 0) 11
The mod numers cn e used to see whether certin constuctions with the pentominoes re impossile. It ll depends on the fct tht the sum of the mod numers is the mod numer of the sum, tht is the figure formed y the two pentominoes. Hve students discuss why this is so. Here re some exmples. (0,1) (1 3)+(4 3)=(0 1) Y + P (1 1)+(4 0)=(0 1) L + F (2 4)+()=(0 1) P + N The mod numers cn e used to show tht it is impossile to solve puzzle. Consider this exmple using mod numers to show tht it is impossile to solve certin pentomino prolems. The figure t the right shows tht the figure, 5 x 6 rectngle is constructed of 6 I s. Cn the figure formed y moving one of the squres s shown elow e constructed if ny numer of I s, X s, nd Z s s well s t most one other piece other thn n F is used?. (0 0) Oddly enough, there is no solution to this puzzle. I, X, nd Z ll hve mod numers (0 0). The sum of the mod numers for the finl shpe must e the mod numers of the other piece. The mod numer of the one other piece must e (1 0). Since F is the only piece with this mod numer, the (1 0) shpe on the left cnnot e constructed with one piece tht is not n F nd ll the other five pieces eing I s, X s, or F s. 12
As n exercise, tke this solution of the 6 x 10 using the twelve pentominoes. Flip the P pentomino s shown nd prove the twelve pentominoes cnnot e ressemled into the 6 x 10 through trnsltions lone. There is well-known prolem involving checkerord nd dominoes. If ech domino covers exctly two squres on checkerord nd two digonl corners re cut out, cn it e covered using 31 dominoes? The nswer is no ecuse ech domino covers red nd lck squre on the checkerord. The figure on the right hs 30 red squres nd 32 lck squres, so 30 dominoes will lwys leve two lck squres to e covered y domino 31. This will not work. Hence there is no covering. There is similr prolem using mod numers. Cover checkerord with oriented dominoes such s shown on the right. There re two possile orienttions, horizontl nd verticl. Tke one of the horizontl dominoes nd replce it with verticl one s shown elow. Cn the dominoes e put ck without chnging their orienttions nd still cover the checkerord? The nswer is no. Apply the sme rguments (0 0) 13
s efore with the mod numers ut now use mod 2 insted of mod 5. The verticl domino hs mod numer (0 1); the horizontl domino hs mod numer (1 0). The overll shpe hs mod numer of (0 0). If we tke out horizontl domino the remining 31 dominos will form shpe tht hs mod numer (1 0), when we dd the verticl domino we get shpe with mod numer (1 0) (0 1) of (1 1). The dominoes with exctly one with chnged orienttion cnnot e put ck onto the checkerord in their new orienttions ecuse the shpe they mke hs mod numers (1 1) nd the mod numers of the checkerord re (0 0). In this rticle, pentominoes nd their close reltives hve een explored. Pentominoes hve een defined more crefully nd the ide of pths through the figures hs een discussed. Using the ide of longest pth, it hs een shown tht n+1-omines re derived from n-ominoes. This ws used to show how to prove tht there re exctly 12 pentominoes. It ws shown how to ssign coordintes to polyomino nd develop mod numers for ech pentomino. Properties of mod numers were developed nd used to prove severl constructions would e impossile. An exmple ws used to show how to develop nd use mod numers for dominoes nd it ws indicted how mod numers could e extended to more dimensions nd numers of squres, cues, hypercues... These investigtions could lso e used to introduce students to different importnt techniques nd rnches of mthemtics. Proof y contrdiction, itertion, nd mth induction were some of the techniques employed. Coordintes from nlyticl geometry, liner lger, 14
modulr rithmetic from numer theory, nd group theory figure prominently in the rguments. For exmple, the mod numers could e used to explore symmetry from the point of view of liner lger. The pentominoes lso open n venue into group theory to explore the opertion of group, the eight mtrices for rottion nd reflection, on set, the mod numers, to look t the vrious orits of the mod numers nd the reking down of the mod numers into equivlence clsses. Even prolems in computer science could e investigted. If the pentominoes were to e ssemled into vrious constructions, the mod numers could e used to eliminte mny possile cses to e tried nd improve the efficiency of lgorithms looking for solutions. The use of the mtrices could extend the prolems investigted with more efficient computer progrms from the pentominoes to prolems with polyominoes s well s higher dimension cues nd hypercues. There is more to pentominoes thn re, perimeter, nd simple symmetry nd the techniques used to study them cn extend well eyond wht is tught in middle school. Bio - Bruce Bguley works for Cscde Mth Systems, LLC. He received BA in Mthemtics from Tulne University, n MS in Mthemtics from MIT, nd his techer trining from Heritge College in Toppenish, WA. While teching elementry nd middle school students, he ecme interested in showing mth concepts using mnipultives rther thn relying on memorizing formuls. He hs given numerous workshops t mth conferences over the pst few yers, showing people how to use mnipultives to represent mth concepts from counting, through whole numer, rtionl, nd integer opertions, to solving nd grphing liner equtions s well s proving numer theory prolems. References NCTM Principles nd Stndrds, NCTM, Reston, VA, 2000 The Pentomino Squre Prolem, Mthemtics Teching in the Middle School, NCTM, Mrch 2003 Exploring Cues Activity Sheet, http://illumintions.nctm.org/lessons/6-8/geometry/geomiddle- AS-Cues.pdf 15