Maximum area of polygon


 Job Perkins
 3 years ago
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1 Mimum re of polygon Suppose I give you n stiks. They might e of ifferent lengths, or the sme length, or some the sme s others, et. Now there re lots of polygons you n form with those stiks. Your jo is to fin the one with mimum (plnr) re. I ws given this prolem y frien t mthemtis meeting. [For mthemtiin new prolem is gift the most wonerful gift frien n estow.] Goo timing, euse t mth meetings there re lwys tlks tht lose me fter while, n it s goo to hve something to lik over in my min. tully this prolem ws entiing enough tht even the goo tlks h serious ompetition. The prolem seems t first to e hr. You re not given muh informtion n there re lots n lots of things you n o. tully there re two ifferent kins of eisions you hve to mke. The first is to put the stiks in the right orer (sequentilly) n the seon is to etermine the ngles etween them. It s hr to know how to egin. One ie is to strt with smll vlue of n, n the smllest nontrivil vlue is n=4. nother is to tke very lrge vlue of n n mke the eges quite short. Tht s wht I first thought of, euse it me it feel lot like the lssi fene prolem. You re given fie length of fening n you wnt to mke n enlosure of mimum re, n most people know tht the nswer is irle. So I figure tht if there were lot of short eges, the nswer might e tht the verties shoul lie in irle. n then I wonere if tht just might e the nswer for ll polygons. Well, mye tht ws it muh to epet. On the other hn, the prolem h een given to me y mthemtiin n mthemtiins like to known y the qulity of the prolems they tlk out, n so this ws likely pretty nie prolem, n irles re pretty nie too. So I kept tht irle ie in the k of my min. I egn y thinking out the orer of the eges. euse tht seeme to mke things quite omplite, n I ws hoping something simple woul pper. oul I imgine sitution in whih I oul inrese the re y rerrnging the orer of the sies? s prtiulr speil se oul I inrese the re y permuting two jent sies? [Suh permuttion is lle trnsposition.] Well the nswer to tht is lerly no. If I tke ny polygon, n reverse the orer of two jent sies, leving the other sies fie in ple, the re remins the sme. Tht insight is rel rekthrough. For emple, if I hve 5sie polygon with sies in orer efg, n I know the mimum re of tht, I on t hve to worry out efg it must hve the sme mimum re. f e g polygon re 1
2 Wht out other permuttions? Wht out gef? Well it s not hr to rgue tht ny permuttion n e otine y sequene of trnspositions of jent terms. For emple, strting with efg, use sequene of jent trnspositions to put first, then sequene to put net, then sequene to put g net, et. It follows from this tht the re nnot e inrese y ny permuttion of the eges. Tht s ig step. It mens we on t hve to worry out the orer of the eges. We n tke ny prtiulr orering of the eges n work with tht. So we re left with the question of wht the ngles shoul e. The se n=4 Perhps this is the moment to fous on the se n=4. Tke 4sie polygon. Wht re the possile shpes we n hve for fie sie lengths,, n? Well we hve one egree of freeom. For emple, if we let e the length of the igonl etween the  sies n the  sies, then the re = () of the qurilterl is etermine y. So the prolem eomes: given,, n, hoose to mimize (). How might we o tht? Well is the sum of the res of the two tringles n we know the sie lengths of eh tringle. Is there formul for the re of tringle in terms of the lengths of the three sies? Yes there is. It s lle Heron s formul. For the upper tringle with sies, n, we let the semiperimeter e s: n then the re is: s = re = s( s )( s )( s ) Do the sme for the lower tringle if the semiperimeter is t: t = then the re of the entire qurilterl is: () = s( s )( s )( s ) t( t )( t )( t ) n we wnt to hoose to mke this mimum. The wy to o this is to set the erivtive of this to e zero. ut ll this is not wht I i. There re ouple of resons for tht. One is tht I i not relly wnt to ifferentite tht firly omple epression with respet to. [Note tht s n t lso epen on, so we hve it of hinrule to nvigte.] n even if I i, oul I solve the result for? n even if I oul o tht, where woul I then e? polygon re
3 ut there s muh more signifint reson I i not wnt to go the Heron s formul route. I still h the fene prolem in min n I ws hoping tht the mimum re woul e foun when the verties of the qurilterl ly in irle. Now qurilterl with tht property is lle yli, n stnr theorem sys tht qurilterl is yli when opposite ngles to 180. In the igrm t the right, tht mens tht φ = 180. So tht ws the onition I ws gunning for, n therefore I wnte ngles to pper in my nlysis. n Heron s formul oesn t use ngles. stnr geometry result is tht qurilterl is yli if opposite ngles to 180. Oky. So wht I wnt is formul for the re of tringle tht involves one of the ngles. Sine n re the given sie lengths, we ll look for formul tht involves n n the ontine ngle. [It s ler enough tht the tringle n therefore its re re etermine y these three quntities.] Tht s not so hr to o. Tking to e the se, the ltitue h of the tringle will stisfy: n sine sin(180 ) = sin: h/ = sin(180 ). h = sin. 180 h Finlly, the re is onehlf se times height: re = sin. s n immeite hek on this formul, if =0 we shoul get zero re (n we o) n if =90, will e n ltitue n the re shoul e / (n it is). Now the prolem eomes: hoose n φ to mimize the totl re: (, φ) = 1 = sin sin φ The troule with this formultion is tht it looks s if n φ re two inepenent vriles, ut they re not. If we hoose one of them, the other is etermine, so there s relly only one vrile here. We oul try to fin the reltionship etween them (hene epress one of them in terms of the other) ut in ft it s geometrilly (n intuitively) nier to work (s efore) with the single vrile. Tht is (for fie,, n ), we will regr n φ s funtions of : 1 = () φ = φ() These in turn etermine the res of the two tringles n if we those together we get. 1 { φ polygon re 3
4 Fining to mimize The formul is: () = sin ( ) sin φ( ). s efore, to fin the vlue of whih mimizes, we set the erivtive / equl to zero. Using the hin rule: os = osφ φ Now we nee / n φ/. Let s o the first. We n write s funtion of using the osine lw: = os. Now ifferentite with respet to (treting s funtion of ): = sin (/) Solve for /: Similrly: = sin φ =. sin φ If we put these into the eqution for /, we get: os = sin osφ sin φ tn tn φ tn = tnφ. Now when to two ngles (in qurnts 1 n ) hve tngents whih re negtives of one nother? when they to 180. Thus: φ = 180. Tht s the result we were fter. It tells us tht the qurilterl is yli. Now it s time to move on to more sies. One ie is to use n inutive pproh. If we ut the polygon with hor, we hve two polygons eh with fewer sies thn the originl. If we know the result for these (tht the verties hve to lie on irle) mye we n put these together to get the result for the originl. It s goo ie n it works. There re lots of wys to o it, though, n it turns out to e it of hllenge (t lest it ws for me) to fin one tht feels right. tully the niest one I know of is not tritionl inution t ll, ut goes iretly from n=4 to the generl result. Tke some time to ply with this it. polygon re 4
5 It might e esier to tke one step t time. Try to prove the n=5 result. We ve one four sies. n you eten tht result to o five? The se n=5. Suppose we hve pentgon DE n the verties hve een positione to give miml re. I show tht the verties must lie on irle. Now the three verties etermine unique irle, so it s question of showing the other two must lie on this irle. E Tke D. Drw the hor D. Now onsier the polygon D with fie sie lengths. I lim tht it must e of mimum re. Inee, if this were not so, I oul inrese the re y justing the position of n leving n D in position (sine the istne D won t hnge) n this woul inrese the re of the originl polygon DE. Thus, if D is of miml re, our n=4 result tells us tht D is yli n therefore D lies on the irle etermine y. D E similr rgument works for E, using the hor E. D The generl se. s I si there re mny wys to generlize the ove rgument to ny vlue of n. Here is the most elegnt I hve enountere. Suppose we hve n ngon with fie sie lengths whose verties hve een positione to give miml re. Tke three jent verties. These etermine unique irle (it s esy to rgue tht they nnot lie in stright line), n I show tht ll other verties must lie on this irle. Tke ny other verte X n onsier the polygon X with fie sie lengths. I lim tht it must e of mimum re. Inee, if this were not so, I oul inrese the re y justing the position of the verties. Sine X n X won t hnge in length in this justment, the other verties n e tthe to X n X giving us new ngon with the originl sie lengths ut of greter re. Tht s ontrition, so X must e of miml re. y the n=4 result, X is yli n X must lie on the irle etermine y. n we re finishe. some verties X some verties polygon re 5
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