T Unvrsty o Syny MATH 2009 APH THEOY Tutorl 7 Solutons 2004 1. Lt t sonnt plnr rp sown. Drw ts ul, n t ul o t ul ( ). Sow tt s sonnt plnr rp, tn s onnt. Du tt ( ) s not somorp to. ( ) A onnt rp s on n w tr xsts pt twn ny pr o vrts. Lt sonnt plnr rp, n onsr ts ul. In, tr s lrly pt twn ny two vrts w orrspon to s wtn on o t omponnts o. Lt v t vrtx n orrsponn to t nnt o. Tn v s nt to t lst on vrtx (orrsponn to ) n omponnt o. So u n w r two vrts o w orrspon to s n rnt omponnts o, tn tr s pt rom u to w, v v. Hn s onnt. Sn s onnt, ts ul ( ) s lso onnt. ut s sonnt, n so ( ) s not somorp to. 2. A rtn polyron s s w r trnls n pntons, wt trnl surroun y pntons n pnton surroun y trnls. I vry vrtx s t sm r, p sy, sow tt 1 = 1 p 7. Du tt p = 4, n tt tr r 20 trnls n 12 pntons. Cn you onstrut su polyron? Stt t ul rsult. Lt T t numr o trnls, n P t numr o pntons. Tn t numr o s, = T + P. Sn trnl s oun y 3 s, n pnton y 5, w v 3T + 5P = 2. ut ons xtly on trnl n xtly on pnton,
2 so 3T = 5P. Hn, T = 2 6 = 3, P = 2 10 = 5, = P + T = 3 + 5 = 8 15. Also, ssumn vry vrtx s r p, w v pv = 2, or v = 2/p. Now susttut nto Eulr s ormul: Dv y 2 n smply: 2 p + 8 15 = 2. 1 = 1 p 7 Clrly, 1 must postv numr, so w must v 1 p 7 0, or p /7. ut p must n ntr 3. So t only possl vlus or p r 3 or 4. I p = 3, = 10, ut tn v = 2/p = 20/3, w s not n ntr. (Atully, t s sy nou to s tt t s not possl to put totr trnls n pntons n t wy sp, su tt t r o vrtx s tr. Try rwn t n s!) I p = 4, = 60 n T = 20 n P = 12, s rqur. Su polyron my onstrut n two lvs: W tn stt totr lon t s w orrspon n t lln. It s rtnly possl to rw t ull rp, wt lttl mor ort: Not tt t outs ron s on o t pntons. T ul rsult s otn y rpln vrts y s n s y vrts: A rtn polyron s vrts w v rs 3 n 5, wt vrtx o r 3 nt to 3 vrts o r 5, n vrtx o r
3 5 nt to 5 vrts o r 3. I vry s oun y t sm numr o s, tn tt numr s 4, n tr r 20 vrts o r 3 n 12 vrts o r 5. (E vrtx n t ul rp orrspons to ron n t ornl rp, n n nnt to two nt rons n t ornl s n n t ul onn t orrsponn vrts o t ul.) 3. Dtrmn t romt numr o o t ollown rps: k k l () (v) A rp s k-olourl ts vrts n olour usn k olours n su wy tt no two nt vrts v t sm olour. T romt numr o smpl rp, wrttn χ(), s n to t smllst ntr k or w s k-olourl. Usul ts: I ontns ( surp somorp to) K n, tn χ() n. I ontns n o rut, tn χ() 3. I s t lst two vrts n no o ruts, tn χ() = 2. Ts rp ontns svrl trnls, so χ() 3. On t otr n w n sly n 3-olourn (.,,, r;,, wt; lu), so χ() = 3. T rp s no o yls, so χ() = 2. (A 2-olourn s sly oun (.,,,,, wt,,,,,, k lk). () Sn tr r trnls, χ() 3. W n n 3-olourn (.,,,, k r;,,, l wt;,,, lu), so χ() = 3. (v) Sn tr r trnls, (., {,, }), χ() 3, ut s 3-olourl? I so, wtout loss o nrlty lt,, r, wt, lu rsptvly. Tn, nt to ot n, must wt. Also, nt to ot n, must lu. Also, nt to ot n, must r. Tn, owvr, ourt olour s n or, w s nt to, n. Hn χ() = 4. 4. For o t ollown rps, wt os rooks Torm tll you out t romt numr o t rp? Fn t romt numr o rp. T omplt rp K 20. T prtt rp K 10,20. () A yl wt 20 s. (v) A yl wt 29 s.
(v) T u rp Q 3. (v) T ul o Q 3. 4 Lt () t mxmum o t rs o t vrts o rp. rooks Torm stts tt, or ll rps otr tn omplt rps or o yls, t romt numr o rp, χ() (). For omplt rps, n or o yls, χ() = 1 + (). (K 20 ) = 19, χ(k 20 ) = 1 + 19 = 20. (Clrly, o t 20 vrts must olour rntly or propr olourn.) (K 10,20 ) = 20, so rooks Torm sys tt χ(k 10,20 ) 20. O ours, t romt numr o ny prtt rp s 2, so χ(k 10,20 ) = 2. () In yl, ll vrts v r 2. y rooks Torm, t romt numr o yl wt n vn numr o s s t most 2, n lrly 2 olours r rqur, so t romt numr s qul to 2. (v) y rooks Torm, t romt numr o yl wt n o numr o s s 3. (v) Q 3 s rulr o r 3, n so y rooks Torm χ(q 3 ) 3. In t, Q 3 s prtt rp wt romt numr 2. (v) T ul o Q 3 s t otron, n w vry vrtx s r 4, so y rooks Torm ts romt numr s t most 4. Not tt t rp ontns trnls, so ts romt numr s t lst 3. It s sy to n 3-olourn, n so χ(t otron) = 3. 5. Dtrmn t mnmum numr o olours rqur to olour t s o Q 3 n su wy tt onn s v rnt olour. pt prt or t ul o Q 3. T rp Q 3 s t polyrl rp orrsponn to u. It s lr tt t lst 3 olours r n to olour t s o u so tt no two onn s v t sm olour. T rm low nts on wy n w to olour t s o t u usn tr olours (wr =r, =lu, =rn). T otron s t ul o t u. For n otron only two olours r nssry. T rm sows 2-olourn o t s o n otron. Not tt t mnmum numr o olours rqur to olour t s o t u s qul to t romt numr o t otron, n v vrs.
5 6. Sow tt smpl onnt plnr rp wt 17 s n 10 vrts nnot proprly olour wt two olours. (Hnt: Sow tt su rp must ontn trnl.) Suppos su rp no trnls. Tn, sn t s not tr, must oun y t lst 4 s, n so 4 2, or 2. Howvr, y Eulr s ormul, = 2 v + = 2 10 + 17 = 9, so 2 = 2 9 > 17 = ontrton. Hn t rp s t lst on trnl n so s not 2-olourl. 7. Lt T tr wt t lst 2 vrts. Prov tt χ(t )=2. Not tt w n t lst two olours to proprly olour T, so w prov tt T s 2-olourl tn χ(t )=2. Us nuton on t numr o vrts. I tr r 2 vrts, tn t tr s lrly 2-olourl. Suppos tt tr wt k vrts s 2-olourl. Lt T tr wt k + 1 vrts, n rmov rom T on o ts vrts wt r 1. (Evry tr s t lst two vrts wt r 1.) Ts lvs tr wt k vrts, w s 2-olourl y t nuton ypotss. Colour t smllr tr wt 2 olours, rstor t rmov vrtx n olour t wt t olour not us on t (on) vrtx to w t s nt. Hn T s 2-olourl, n t rsult ollows.