Angle Bisectors Bisect Arcs

Transcription

1 Volume, Number prl 006 My 006 Olympd orer elow ws the Fd Roud of the 6th ustr Mth Olympd 005. gle sectors sect rcs K Y. L rt (My 0, 005) roblem. Show tht fte umber of multples of 005 exst, whch ech of the 0 dgts 0,,,,9 occurs the sme umber of tmes, ot coutg ledg zeros. roblem. For how my teger vlues of wth 005 does the system of equtos x = y +, y = x + hve teger solutos? roblem. We re gve rel umbers, b d c d defe s s the sum s = + b + c of ther -th powers for o-egtve tegers. t s kow tht s =, s = 6 d s = 4 hold. Show tht s s s+ holds for ll tegers >. = 8 roblem 4. We re gve two equlterl trgles d QR wth prllel sdes, oe potg up d oe potg dow. The commo re of the trgles teror s hexgo. Show tht the les jog opposte corers of ths hexgo re cocurret. (cotued o pge 4) Edtors: (HEUNG k-hog), Musg ollege, HK (KO Tsz-Me) (LEUNG Tt-Wg) (L K-Y), ept. of Mth., HKUST (NG Keg-o Roger), T, HKU rtst: (YEUNG Su-Yg mlle), MF, U ckowledgmet: Thks to El hu, Mth. ept., HKUST for geerl ssstce. O-le: The edtors welcome cotrbutos from ll techers d studets. Wth your submsso, plese clude your me, ddress, school, eml, telephoe d fx umbers (f vlble). Electroc submssos, especlly MS Word, re ecourged. The dedle for recevg mterl for the ext ssue s ugust 6, 006. For dvdul subscrpto for the ext fve ssues for the cdemc yer, sed us fve stmped self-ddressed evelopes. Sed ll correspodece to: r. K-Y L eprtmet of Mthemtcs The Hog Kog Uversty of Scece d Techology ler Wter y, Kowloo, Hog Kog Fx: (85) Eml: mkyl@ust.hk geerl, gle bsectors of trgle do ot bsect the sdes opposte the gles. However, gle bsectors lwys bsect the rcs opposte the gles o the crcumcrcle of the trgle! mth compettos, ths fct s very useful for problems cocerg gle bsectors or ceters of trgle volvg the crcumcrcle. Recll tht the ceter of trgle s the pot where the three gle bsectors cocur. Theorem. Suppose the gle bsector of tersect the crcumcrcle of t X. Let be pot o the le segmet X. The s the ceter of f d oly f X = X = X. roof. Note X = X = X. So X = X. The s the ceter of = X X = X X X = X X = X = X. Exmple. (98 ustrl Mth Olympd) Let be trgle, d let the terl bsector of the gle meet the crcumcrcle g t. efe Q d R smlrly. rove tht + Q + R > + +. R Q X Soluto. Let be the ceter of. y the theorem, we hve R = R + R > d smlrly >, Q >. lso + >, + > d + >. ddg ll these equltes together, we get ( + Q + R) > ( + + ). Exmple. (978 MO), =. crcle s tget terlly to the crcumcrcle of d lso to the sdes, t, Q, respectvely. rove tht the mdpot of segmet Q s the ceter of the crcle of. Soluto. Let be the mdpot of le segmet Q d X be the tersecto of the gle bsector of wth the rc ot cotg. y symmetry, X s dmeter of the crcumcrcle of d X s the mdpot of the rc XQ o the sde crcle, whch mples X bsects Q. Now X = 90 = X so tht X,,, re cocyclc. The X Q X = X = X = X. So X = X. y the theorem, s the ceter of. Exmple. (00 MO) Let be dmeter of the crcle Γ wth ceter O. Let be pot o Γ such tht 0 < O < 0. Let be the mdpot of the rc ot cotg. The le through O prllel to meets the le t J. The perpedculr bsector of O meets Γ t E d t F. rove tht J s the ceter of the trgle EF.

2 Mthemtcl Exclbur, Vol., No., pr My 06 ge E J O F Soluto. The codto O < 0 esures s sde EF (whe O creses to 0, wll cocde wth ). Now rdus O d chord EF re perpedculr d bsect ech other. So EOF s rhombus. Hece s the mdpot of rc EF. The bsects EF. Sce O = O, O = / O = O. The O s prllel to J. Hece OJ s prllelogrm. The J = O = EO = E. y the theorem, J s the ceter of EF. Exmple 4. (996 MO) Let be pot sde trgle such tht =. Let, E be the ceters of trgles, respectvely. Show tht, d E meet t pot. H J K F Soluto. Let les,, tersect the crcumcrcle of g t F, G, H respectvely. Now E G = FG G = FG. Smlrly, = FH. So F bsects HG. Let K be the ceter of HG. The K s o F d les HK, GK pss through the mdpots, J of mor rcs G, H respectvely. Note les, E lso pss through, J s they bsect, respectvely. pplyg scl s theorem (see vol.0, o. of Mth Exclbur) to, G, J,, H, o the crcumcrcle, we see tht =G H, K=GJ H d J= E re coller. Hece, E s o le K, whch s the sme s le. Exmple 5. (006 MO) Let, be two dstct pots o gve crcle O d let be the mdpot of le segmet. Let O be the crcle tget to the le t d tget to the crcle O. Let l be the tget le, dfferet from the le, to O pssg through. Let be the tersecto pot, dfferet from, of l d O. Let Q be the mdpot of the le segmet d O be the crcle tget to the le t Q d tget to the le segmet. rove tht the crcle O s tget to the crcle O. K L J M Soluto. Let the perpedculr to through tersect crcle O t N d M wth N d o the sme sde of le. y symmetry, segmet N s dmeter of the crcle of O d ts mdpot L s the ceter of O. Let le L tersect crcle O g t Z. Let le ZQ tersect le M t J d crcle O g t K. Sce d re tget to crcle O, L bsects so tht Z s the mdpot of rc. Sce Q s the mdpot of segmet, ZQ = 90 = L d JQ = 90 = M. Next ZQ = Z = Z = L. So ZQ, L re smlr. Sce M s the mdpot of rc M, JQ = M = M = M. N Q So JQ, M re smlr. y the tersectg chord theorem, = N M = L M. Usg the smlr trgles bove, we hve Z L M ZQ JQ = =. Q Q y the tersectg chord theorem, KQ ZQ = Q Q so tht KQ = (Q Q)/ZQ = JQ. Ths mples J s the mdpot of KQ. Hece the crcle wth ceter J d dmeter KQ s tget to crcle O t K d tget to t Q. Sce J s o the bsector of, ths crcle s lso tget to. So ths crcle s O. Exmple 6. (989 MO) cute-gled trgle the terl bsector of gle meets the crcumcrcle of the trgle g t. ots d re defed smlrly. Let 0 be the pot of tersecto of the le wth the exterl bsectors of gles d. ots 0 d 0 re defed smlrly. rove tht: () the re of the trgle s twce the re of the hexgo, () the re of the trgle s t lest four tmes the re of the trgle Soluto. () Let be the ceter of. Sce terl gle bsector d exterl gle bsector re perpedculr, we hve 0 0 = 90. y the theorem, =. So must be the mdpot of the hypoteuse 0 of rght trgle 0. So the re of 0 s twce the re of. uttg the hexgo to sx trgles wth commo vertex d pplyg smlr re fct lke the lst sttemet to ech of the sx trgles, we get the cocluso of (). () Usg (), we oly eed to show the re of hexgo s t lest twce the re of. H (cotued o pge 4)

3 Mthemtcl Exclbur, Vol., No., pr My 06 ge roblem orer We welcome reders to submt ther solutos to the problems posed below for publcto cosderto. The solutos should be preceded by the solver s me, home (or eml) ddress d school fflto. lese sed submssos to r. K Y. L, eprtmet of Mthemtcs, The Hog Kog Uversty of Scece & Techology, ler Wter y, Kowloo, Hog Kog. The dedle for submttg solutos s ugust 6, 006. roblem 5. eterme wth proof the lrgest umber x such tht cubcl gft of sde x c be wrpped completely by foldg ut squre of wrppg pper (wthout cuttg). roblem 5. Fd ll polyomls f(x) wth teger coeffcets such tht for every postve teger, s dvsble by f(). roblem 5. Suppose the bsector of tersect the rc opposte the gle o the crcumcrcle of t. Let d be defed smlrly. rove tht the re of s t lest the re of. roblem 54. rove tht f, b, c > 0, the bc ( + b + c) + ( + b + c) 4 bc ( + b + c). roblem 55. Twelve drm groups re to do seres of performces (wth some groups possbly mkg repeted performces) seve dys. Ech group s to see every other group s performce t lest oce oe of ts dy-offs. Fd wth proof the mmum totl umber of performces by these groups. ***************** Solutos **************** roblem 46. spy ple s flyg t the speed of 000 klometers per hour log crcle wth ceter d rdus 0 klometers. rocket s fred from t the sme speed s the spy ple such tht t s lwys o the rdus from to the spy ple. rove such pth for the rocket exsts d fd how log t tkes for the rocket to ht the spy ple. (Source: 965 Sovet Uo Mth Olympd) Soluto. Jeff HEN (Vrg, US), Koyrts G. HRYSSOSTOMOS (Lrss, Greece, techer), G.R.. 0 Mth roblem Group (Rom, tly) d lex O K-ht (STF heg Yu Tug Secodry School). O L Let the spy ple be t Q whe the rocket ws fred. Let L be the pot o the crcle obted by rottg Q by 90 the forwrd drecto of moto wth respect to the ceter. osder the semcrcle wth dmeter L o the sme sde of le L s Q. We wll show the pth from to L log the semcrcle stsfes the codtos. For y pot o the rc QL, let the rdus tersect the semcrcle t R. Let O be the mdpot of L. Sce R Q Q = RL = / RO d L = O, the legth of rc R s the sme s the legth of rc Q. So the codtos re stsfed. Flly, the rocket wll ht the spy ple t L fter 5π/000 hour t ws fred. ommets: Oe solver guessed the pth should be curve d decded to try crculr rc to strt the problem. The other solvers derved the equto of the pth by dfferetl equto s follows: usg polr coordtes, sce the spy ple hs costt gulr velocty of 000/0 = 00 rd/sec, so t tme t, the spy ple s t (0, 00t) d the rocket s t (r(t), θ(t)). Sce the rocket d the spy ple re o the sme rdus, so θ(t) = 00t. Now they hve the sme speed, so The 6 ( r'( t)) + ( r( t) θ '( t)) = 0. r'( t) 00 r( t) = 00. tegrtg both sdes from 0 to t, we get the equto r = 0 s(00t) = 0 s θ, whch descrbes the pth bove. roblem 47. () Fd ll possble postve tegers k such tht there re k postve tegers, every two of them re ot reltvely prme, but every three of them re reltvely prme. (b) eterme wth proof f there exsts fte sequece of postve tegers stsfyg the codtos () bove. (Source: 00 elruss Mth Olympd) Soluto. G.R.. 0 Mth roblem Group (Rom, tly) d YUNG F. () We shll prove by ducto tht the codtos re true for every postve teger k. For k =, the umbers 6, 0, 5 stsfy the codtos. ssume t s true for some k wth the umbers beg,,, k. Let p, p,, p k be dstct prme umbers such tht ech p s greter th k. For = to k, let b = p d let b k+ = p p p k. The gcd(b, b j )=gcd(, j ) > for < j k, gcd(b, b k+ ) = p > for k, gcd(b h, b, b j ) = gcd( h,, j ) = for h < j k d gcd(b, b j, b k+ ) = for < j k, completg the ducto. (b) ssume there re ftely my postve tegers,,, stsfyg the codtos (). Let hve exctly m prme dvsors. For = to m +, sce ech of the m + umbers gcd(, ) s dvsble by oe of these m prmes, by the pgeohole prcple, there re, j wth < j m + such tht gcd(, ) d gcd(, j ) re dvsble by the sme prme. The gcd(,, j ) >, cotrdcto. ommeded solvers: HN Ng Y (rmel ve Grce Foudto Secodry School, Form 6) d HN Yt Sg (rmel ve Grce Foudto Secodry School, Form 6). roblem 48. Let be covex qudrlterl such tht le s tget to the crcle wth sde s dmeter. rove tht le s tget to the crcle wth sde s dmeter f d oly f les d re prllel. Soluto. Jeff HEN (Vrg, US) d Koyrts G. HRYSSOSTOMOS (Lrss, Greece, techer).

4 Mthemtcl Exclbur, Vol., No., pr My 06 ge 4 E Let E be the mdpots of. Sce s tget to the crcle, the dstce from E to le s h = /. Let F be the mdpot of d let h be the dstce from F to le. Observe tht the res of EF d EF = /8. Now le s tget to the crcle wth sde s dmeter h =/ res of EF, EF, EF d EF re equl to /8 EF, EF. roblem 49. For postve teger, f,,, b,, b re [,] d tht F + L + = b + L + b, the 7 + L + ( + L+ ). b b 0 prove Soluto. Jeff HEN (Vrg, US). For x, y [,], we hve / x/y y/ x y (y/ x)(y x) 0 x + y 5xy/. Let x = d y = b, the + b 5 b /. Summg d mpultg, we get b 5 ( 4 + b ) = 5. Let x = ( /b ) / d y = ( b ) /. The x/y = /b [,]. So /b + b 5 /. Summg, we get b + 5 b ddg the two dsplyed equltes, we get 7 + L + ( + L+ ). b b 0. roblem 50. rove tht every rego wth covex polygo boudry cot be dssected to ftely my regos wth ocovex qudrlterl boudres. Soluto. YUNG F. ssume the cotrry tht there s dssecto of the rego to ocovex qudrlterl R, R,, R. For ocovex qudrlterl R, there s vertex where the gle s θ > 80, whch we refer to s the lrge vertex of the qudrlterl. The three other vertces, where the gles re less th 80 wll be referred to s smll vertces. Sce the boudry of the rego s covex polygo, ll the lrge vertces re the teror of the rego. t lrge vertex, oe gle s θ > 80, whle the remg gles re gles of smll vertces of some of the qudrlterls d dd up to 60 θ. Now o ( 60 θ ) ccouts for ll the gles ssocted wth ll the smll vertces. Ths s cotrdcto sce ths wll leve o more gles from the qudrlterls to form the gles of the rego. Olympd orer rt, y (Jue 8, 005) (cotued from pge ) roblem. eterme ll trples of postve tegers (,b,c), such tht + b +c s the lest commo multple of, b d c. roblem. Let, b, c, d be postve rel umbers. rove + b + c + d bcd b c d roblem. cute-gled trgle, crcle k wth dmeter d k wth dmeter re drw. Let E be the foot of o d F be the foot of o. Furthermore, let L d N be the pots whch the le E tersects wth k (wth L lyg o the segmet E) d K d M be the pots whch the le F tersects wth k (wth K o the segmet F). rove tht KLMN s cyclc qudrlterl. rt, y (Jue 9, 005) roblem 4. The fucto f s defed for ll tegers {0,,,, 005}, ssumg o-egtve teger vlues ech cse. Furthermore, the followg codtos re fulflled for ll vlues of x for whch the fucto s defed: f(x + ) = f(x), f(x + ) = f(x) d f(5x + ) = f(5x). How my dfferet vlues c the fucto ssume t most? roblem 5. eterme ll sextuples (,b,c,d,e,f) of rel umbers, such tht the followg system of equtos s fulflled: 4=(b+c+d+e) 4, 4b=(c+d+e+f) 4, 4c=(d+e+f+) 4, 4d=(e+f++b) 4, 4e=(f++b+c) 4, 4f=(+b+c+d) 4. roblem 6. Let Q be pot the teror of cube. rove tht fte umber of les pssg through Q exsts, such tht Q s the md-pot of the le-segmet jog the two pots d R whch the le d the cube tersect. gle sectors sect rcs (cotued from pge ) Let H be the orthoceter of. Let le H tersect t d the crcumcrcle of g t. Note = = = 90 = H. Smlrly, we hve = H. The H. Sce s the mdpot of rc, t s t lest s fr from chord s. So the re of s t lest the re of. The the re of qudrlterl H s t lest twce the re of H. uttg hexgo to three qudrlterls wth commo vertex H d comprg wth cuttg to three trgles wth commo vertex H terms of res, we get the cocluso of (). Remrks. the soluto of (), we sw the orthoceter H of hs the property tht H (hece, lso H = ). These re useful fcts for problems relted to the orthoceters volvg the crcumcrcles.

5 Soluto:, _ _ We frst ote tht 005 = Lettg M := ud Nk := M - (lo.(k- ) + Jo +) ) ) we see tht ech Nk eds 5 d s therefore dvsble by 5, d lso tht ech Nk s composed of the sme umber of ech of the te dgts 0,...,9. Sce we hve fte umber of such umbers Nk but oly fte umber of cogruece clsses modulo 40, some cogruece clss modulo 40 must cot fte umber of umbers Nb. Let N,,, be the smllest mog these umbers. Tkg ll vlues of Nb from ths cogruece clss, we obt u te umber.of umbers of the form Nk - V,,., 5: NL_,,,.!O *oh, ll of whch must be dvsble by 40.. Sce 0 d 40 re reltvely prme, ll resultg umbers Nk_,,, must be dvsble by 40, d sce ech of them eds the dgt 5, they must ll be dvsble by. 005, yeldg &rte umber pf umbers wth, the requred TheeerootsreOdl~f... Wthout loss of geerhty, settg = + $, b = - f d c = 0, we hve $ + $r = 6 d &= - d therefore.' "..., ",+ (f-l-+b~-~)(~"+'+~~+l),=~+b~+4"--~(.'+~~j %-lht whch yelds the requred result., Soluto: =, - '&?b" -(cd)."-'s =s-.(b) - 6.(b) = 8: -8.(-),. _ -. Soluto: f (z, y) s u teger soluto of the gve system of equtos, subtrctg the secod equto from the frst shows us tht. x-yz=.y-z w (z- y)(z+y)+(z--y)=o w (z-y)(z+y+l)=o must hold. We therefore cosder two cres. se:z-y=o ths c&r we hve z = y = m. Substtutg o=me- oe of the gve equtos yelds m=m(m-l), ud we see tht s the product of two cosecutve tegers. s such, must be o-egtve, d ll such umbers ot greter th 005 yeld solutos of ths type. Sce = 980 < 005 d = 070 > 005 there re 45 umbers we c substtute for 4, sce m c sssume y vlue 5 m 5 45, yeldg dfferet vlue of ech cse. (Note tht egtve vlues of m yeld the sme vlues for.) se:z+y+l=o _ ths cee we hve z = m d y = -(m + l), d substtutg ~=mgr+=m(mf)+. yelds ths cse, s oe greter th the product of two cosecutve tegers, d we see tht, every vlue of from se yelds secod possble vlue of 4 by ddg. Ths mes tht there re lso 45 vlues of ths type tht yeld teger solutos of,the system of equtos, mkg the totl umber of such vlues to be 90. ~-- Soluto: osderg symmetrc fuctos yelds the vlues for, 6 d c. Sce +b+c = s,=, b+&+c = +(s;- sg)=-l ud, 6 d c re the roots of the cubc equto ubc = -f * (8; -5 -s,(6+k+co))=0, v R : Numberg the vertces of the hexgo s show tlmfgure, we see tht sdes d 45 of the hexgo le o prke ler d RQ. Sce the lttudes d of the trkrglm ud 845 ech result by subtr&tg the dstce betwee the prllel les d RJ from the equl lttudes d V of tr&+ QR d respectvely, they re of equl legth. t therefore follows tht the sde of the equlterl trgles d 45 re lso of equl legth, d d 45 sre therefore ot oly prllel, but lso of equl legth. The qudrlterl 45 s therefore prllelogrm, d ts dgols tersect ther commo md-pot. t therefore follows tht the dgol 4 of the hexgo pssses through themd-pot of the dgol 5.. logously, the qudrlterl 56 must lso be prllelogrm, d the dgol 6 must therefore lso p& through the md-pot of 5, whch s therefore the commo pot of ll. three dgols of the hexgo. Soluto: We frst ote tht ll three umbers cot be equl, sce ths would mply lcm(, b, c) =, whch cotrdcts cm(, b, c) = + b + c. We ssume o 5 6 -< c wthout loss of geerlty. t follows tht + b < c must hold, ud we therefore hve c<+b+c<c, d therefore + b + c = c (sce kr(, 6, c) must be multple of c) d + b = c. Sce b dvdes cm(, b, c) = 4 + 6, we see tht b dvdes 4. From 5 b we coclude tht ether b = or 6 =. must hold. f b =, we hve c = + 6 = 4, ud therefore fcm(, 6, c) = cm(,, ) = 4, whch s certlyotequlto+b+c=++=4. s - - = 0. f;;f Y

6 .for postve tegers. Q(,b,c)=Zcm(;,0)~~~=++= +b+c We see tht.possble trples wth. the requred property re those of the form.(,+,4 wth >$. Soluto:,We rst ote tht the left bud sde c be wrtte s 0 f&+c+d 0 6.c d. =-&+~+-& d +--&=~+~+-J-p& - -bed f we ow cosder the sequece ( f, f, ;, t) tbree tmes, the rerrgemet &@mty mmedtely yelds _.. : 88 requred. Soluto: 0 k..--- Sce LE = 90 d s the dmeter of ke., E s como pot of d L. Smlrly, F s commo pot of d k. so, sce ZE F LF = 900, both E d F lb o the crcle wth dmeter, s show the fgure. t therefore follows tht E - = F. Sce L s rght trgle wth lttude Lh,.t follows tht E.k K we hve F - = cfl. We therefore hve = L!, d smlrly Furthermore, sce LN s perpedculr to the dmeter of kl, t lso follows, tht L = N must hold, dthe logous rgumet for KM d re, yelds K 5 M., _. e&q, we see tht 8 segmets K, L, M d N re of equl legth. The por& K, L, J4 d Nl therefore le o commo crcle wth md-pot, d KLMN s deed cyclc qudrter s requred. Soluto: For y E {6,,,... ; 005} d rv+ble. by 4, or 5 we hve f(v) = f($ + lj. The oly vhres for y such tht j (y) d f(y - ) c be dfferet re those retv&y. prme to 90. She ~(0) = 8, the fucto f c ssume t most 8 d&ret vhres for y {O;,,. : 9}, d smlrly for uy of the~folowug sets of thrtmmtw. tegers. ; _. Sce t e 0 * , f c ssume d dermf vlues ech of the 66 msecttve e& of 0 comtve tegers. For the lst 5 ;t = 6 elemets of the.& 0,,,...,005], t & sme moth& 8 vlues, sce the lrgest tegk e& th 9 d retvey prm& to 0 s, whch s tself less th 5. t folows tht the totl mxmum umber of vlues the fucto f cu ssume s 67-8 = 56. _. 0 6 Soluto: Frst of h, we ote tht ll vrbles re equ8 to 4. tmes fourth power, d c _ tberefore.ot be egtve. : We ote tht the system s cyclc. f y two successve vrbks re ot equl, wthout loss of geerty sy o : b, comprg the hst two equtos yekls _:L.- _.._ b+c+d+e<c+d6e+f.b< j ud <b<,f.. Sce <,~co&rg the frst d l&t equtos y&ds _ b+ +d&ek+b+c+d =+ :e< d e <.<b<f. *_ Sce e.< f, comprg the lst two,equtos &ds f+fbfc.<+b+c+d =+ f<d d e<<b<f<$. :._ Sce e <.d, 6orprg the fourth d ffth &rtos yelds _- - f++b+c<eff++b. & ccc d c<e<c.b<f<d. Flly,%~e c < d, comprg the thrd d four@ equtos yehls :.. d.+e+f+<e+j++b d<b md: c<e-z<b<f<d<,... whch s cotrdcto for b. t -follows tht sx vrbles must be equl, d sce we therefore hve. h.= (40) d o.0, t folks tht ether 4u = ld therefore = f or 0 = 0 must bold. The oly solutos.of the ~~,ofequtosretb~f?~.(!,,t,t,,) d(o,o,o,o,o,). Soluto: Let be thexf ll &t o the surfce of the cube d the set of ll ports obted by reflectg the pots of c Q. For the sk of,sm&ty,~ we & spek of the cube d the reflected cube c. - *... Sce Q s teror pot of ; t must &. be &or hve commo teror rego. pot of. The two cubes therefore v e cosder the set. Ths set must cot fte umber of pots. f ths were ot the cse, d would oly hve fte umber of pots commo (possbly oe t ). Let us ssume tht ths s the cse. The cubes d cmot tersect commo le sevety or hve ple sectos commo, but rthr oe of the cubes (sy c) must the be j the teso~ of the other (sy ), wth t most some orl of the wttce of beg pots of!. y resos of symmetry; 0 must the & be complete& the teror of, whch s oly possble X = (.e. f Q s the md-pout of ). the cs d hve fte umber of pots commo, cotrdctg the ssumpto tht oly cots. fte tber of pots. We see tht f must cot fte mm&r of pots. Tkg uy of these pots s pt o the surfce of the cube,. we see tht s leo pot o:, d the symmetrc pot to whh espect to Q.must therefore r&o.le o the cube, d c thus be chose s pot R Sde Q s certly the md-pot of y such le segmet R, t see: tht te umber of les wth the requred propertes exst. : _._ c