Vol. 5 No 10 - Pi Mu Epsilon

Using toothpicks and rearranzing:- Ncte:bined so as to equal the sum of the four like digits.PhobLem Eddoh'b CornmeldIn each of the five cases, the three unlike digits have been com-The multitude of solutions listed belowy in some cases duplicatingrepresentations already listed in trig^'^ solution, display the ingen:ousvariety accomplished by the following solvers: CLAYTON W. DODGE, UnLum.Ltqo# ;.Iaine a,t Oh0n0; JOAN INNES, ch&ighton UnLwui~.Lty, Omaha,Nebhuh ka; RICK JOHNSON, Bmougb cohpo~on, Whington, N. C. ; BRUCELOVETT, Rdgm CoUqe, Nuu B J L U I ~ ~ N. C ~ J.; , JIM METZy SphingdieLd,IL&no.Lh; T. E. MOORE y Wdge~uatm State CoUege, Eitidgui~2.m. !.la~ach u m ; BOB PRIELIPP, The UnLumLty 06 W.Lhconhin, Obhfzo~h; R. ROBIN-SON ROWE, Sawento, C&6oh1&; and the Phcpo~m.The potpoumi 06 ~o.fdLonh bubmitted by the above-mwd AOLLJW .Lh a6oUom :1) 2 ~ 5 + 6 = 2 - 2 ~ 2 * 2 = ~ ~ * ~ ~6/2+5=2+2+2+2256 = [(22)21~ ; 2(.5 6, = 2 + 2 + 2 +. : = 2A + 2:152161 = -1-6:/5] = 2 t 2 + 2 t Z ,where brackets Indicate the greatest integer function.297. [Spring 19731 Ptopo~ed by Rogm E. KueL, Kunha City,ili~bod.A traffic engineer is confronted with the prob:em of connectingtwo non-parallel straight roads by an 5-sha7ed curvs: formed by arcs zftwo equal tangent circles, one tangent to the *:rs? road dt a selectedpoint and the other touching the second road a-FIGURE 5a cave-. ?~int.1) &termhe the radius of the epal chle~ synthet~cally, zr2g-e#dmetrically w analyt3cally.2) If the f e e lends it~elf ta an Ew$idean c~nSructio3, bxMuld Hhe g6 about $t?PhobLem Eddoh'~ NoteBecause of the ehple#tyand dyffh~lty of this problm :he dead-line for consideration of solutions is hereby entende? TO !-io*.*e&r 30,1974. A synthetic solution would be particularly welcore.298. [Spring 19731 Phopobed by Pad Ehdzh, Budc~pe&,Jan hfq&ebL, UbuwmLty 06 CoLohadc, BO~~LYL, Cob.tadu.bove that11) lim ;( 6 t 3 6n-1g) lim ;(nl'lOg+ nn-SoluLLon by Vonn&yBuhe, Ohio.Let E > 0 be given.+ - -. + n&) = 1,J. JO~IIUOI~, hfajc~t, WZ,Using L'Fos~i~~l'sThus, there is a function hf( E) so ?hzznu i m ]L> 1 nl'k - I] < 5 i+k=2H