Problems Author(s): David Gale, Clark Kimberling, Bruce Hanson ...

Problems Author(s): David Gale, Clark Kimberling, Bruce Hanson, Stephen Wayne Coffman, Richard Johnsonbaugh, Allen Schwenk, William P. Wardlaw, Huseyin Demir, Volkhard Schindler, L. Bass, R. Vyborny, V. Thomee, M. S. Klamkin, Nicholas K. Krier, Frank R. Bernhart, Kenneth L. Bernstein, Paul G. Nevai, G. A. Edgar, Howard Morris, J. C. Binz, Andrea Laforgia, Steven R. Weston, Yan-loi Wong, Zachary Franco and Vania D. Mascioni Source: Mathematics Magazine, Vol. 59, No. 2 (Apr., 1986), pp. 111-120 Published by: Mathematical Association of America Stable URL: http://www.jstor.org/stable/2690429 . Accessed: 14/06/2014 12:18 Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at . http://www.jstor.org/page/info/about/policies/terms.jsp

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LOREN C. LARSON, Editor BRUCE

HANSON,

AssociateEditor

St. Olaf College

F. MEYERS, PAST EDirOR The Ohio State University

LEROY

Proposals To be considered forpublication,solutions should be receivedby September1, 1986. 1237. ProposedbyDavid Gale, University ofCalifornia, Berkeley. A two-person manner.Startwithn pointsin theplanesituated gameis playedin thefollowing so thatno threeare collinear.Playerstaketurnsdrawinglinesegments betweenthesepoints.The to intersect the interiorof previously only stipulationis thatline segmentsare not permitted drawnlinesegments The firstpersonwhois unableto draw (theymaysharecommonendpoints). a line segmentin thismanneris theloser of the game. Show thatthe outcomeof thisgame ofpointsand is independent of thestrategies dependsonlyon thegivenconfiguration of thetwo players.Find a "formula"fordetermining thewinner. 1238. ProposedbyClarkKimberling, University ofEvansville. a. Provethattheinterior of a triangleABC containsa pointP forwhichthethreetriangles APB, BPC, CPA havecongruent incircles. *b. Is P uniquelydetermined? Can the radii be determined? What can you say about the of P? properties 1239. ProposedbyBruceHanson,St. OlafCollege. of therationalsin theinterval(0,1). For each j > 1, let Let r1,r2,r3,... be anyenumeration of rj. be a decimal 2 3 *.. representation rj aj 1aj aj a. Provethatthe" maindiagonal".a,1 a22 a3,3 ... is irrational. .

ASSISTANT EDITORS: CLIFTON CORZATT and THEODORE VESSEY,St. Olaf College. We invitereadersto submit and teachersofadvancedundergraduate problemsbelievedto be newand appealingto students mathematics. Proposals should be accompaniedby solutions,if at all possible,and by any otherinformation thatwill assist the editorsand as a Quickieshouldhave an unexpected, succinctsolution.An asterisk(*) next to a referees.A problemsubmitted numberindicatesthatneithertheproposernortheeditorssupplieda solution. problenm Solutionsshould be written in a styleappropriate for MathematicsMagazine. Each solutionshouldbeginon a thesolver'snameandfulladdress. separatesheetcontaining Solutionsand newproposalsshouldbe mailedin duplicateto LorenC. Larson,Department St. Olaf ofMathematics, MN 55057. College,Northfield,

VOL. 59, NO. 2, APRIL 1986


111

b. For arbitrary positiveintegersj and k, provethatthe"diagonal".aj kaj+1, is irrational.

k+ 1aj+2, k+2 ...

MarylandCollege. (student),Western 1240. ProposedbyStephenWayneCoffman Evaluate E n=l

n?'

H,,=l+ f + where +*

and SadahiroSaeki, Kansas State DePaul University, 1241. ProposedbyRichardJohnsonbaugh, University. Does thereexista compactmetricspace M withan isometry thatis into,but not onto,M?

Quickies Answers to the Quickies are on p. 120. Kalamazoo,Michigan. MichiganUniversity, byAllenSchwenk,Western Q708. Submitted Show that E= ( F(sti2t

)

j+?? i?(y) 0 (t+')

2s-i == 2t?s?1.

by WilliamP. Wardlaw,U. S. NavalAcademy,Annapolis. Q709. Submitted Showthatthereis no matrixB over Let A be a real 3 x 3 matrixwithnegativedeterminant. thereal numberssuchthatA is theclassicaladjointof B. Ankara,Turkey. Demir,MiddleEast TechnicalUniversity, byHiiseyin Q710. Submitted If n is any positiveinteger,show that the numberT = (1/8)n(n + 1)(n + 2)(n + 3) is a number. triangular

112

MATHEMATICS MAGAZINE


Solutions March 1985

Isopticof an Ellipse

Ankara,Turkey. Demir,MiddleEast TechnicalUniversity, 1211. ProposedbyHiiseyin Find thelocus ofpointsunderwhichan ellipseis seenundera constantangle. Berlin,East Germany. Schindler, Solutionby Volkhard (x, y) coordinatesystem.It is well We considertheellipsex2/a2+ y2/b2= 1 in a rectangular knownthatthe tangentto theellipseat thepoint(xl, y1) has equationx1x/a2 + y1y/b2= 1. a 2/x1 and y-intercept froma Sincethetangenthas.x-intercept b2/yl,theslope m ofthetangent point(x, y) outsidetheellipseis givenby m

-

a

y-b2/y

so that x1 a

ma mx-y

b and Y1 b y-mx(

Since (xl, yl) lies on the ellipse,we have [ma/(mx -y)]2 + [b/(y - Mx)]2 becomes simplification (X2 _ a2)m2

-

2xym+ (y2 - b2) = 0.

=1,

whichafter

(2)

If a is theconstantanglesubtended bytheellipse,thenwe can numbertherootsmI, m2of(2) so thattana = (iml - m2)/(1 + mIm2). Hence 2

(iml (1

-

M2)2

+ mIm2)2

(iml

+ M2)2 (1

-

4mlm2

+ mIm2)2

whichremainsvalidif a is replacedby 1800 - a. Since ml + m2= 2xy/(x2 (y2 - b2)/(x2 - a2), we obtain tan2a= 4 b2X2+a2y2-a2b2 (X2+y2

a2

b2)2

-

a2) and m1M2=

(

In particular, if a = 1800,then(3) reducesto theequationof theoriginalellipse,as it should.If a = 900, then(3) reducesto X2 + y2 = a2 + b2,whichis theequationofa circleofradius /a2+ b2. we introduce (x = rcos 0, Since equation(3) is not convenient forplotting, polarcoordinates y = r sin0). Then(3) becomesr4 - 2Ar2+ B = 0, where A = a2 + b2+ 2(b2cos2O+ a2sin2O)cot2a, B=

(a2 + b2)2 + 4a2b2cot2a,

fromwhichwe obtain r2=A+ lA2-B.

(4)

Since forfixed 0, r2 decreasesas a increases,we see thatthe plus signin (4) is used when

VOL. 59, NO. 2, APRIL 1986


113

theloci are < a < 900, and theminussignwhen900 < a < 1800. As seen fromthe figures, of Boothwhen when900 < a < 1800,and are nearlyovalsof Cassinior lemniscates near-ellipses 00 < a < 900. Of course,if a = b, all loci are circles. 00

a/b = 4.

a/b = 2.

In each figurethe values of a forthe fivecurves,starting are 300, 60?, 90?, 120?, and 180?. fromthe outermost,

Also solvedby Michael V. Finn, J. T. Groenman(The Netherlands),L. Kuipers(Switzerland),Vania Mascioni (studenit,Switzerland), WilliamA. Newcomb,RichardParris, StephanieSloyan, and RobertL. Young. Solved by Zachary Franco (student)and theproposer. pacrtiallv J. W. M. S. Klamkin (Canada) foundthe resultin R. C. Yates, A Handbookon CurvesatndtheirProperties, NCTM, 1974), pp. 138-140, wherethe terms Edwards,Ann Arbor,1947 (reprintedas Curvesand theirProperties, are defined.None of thesolversconsideredtheexceptionalcases arisingwhen,forexample,x, isopticand orthoptic XIl,.', yj or ni is zero or m is infinitein (1).

March 1985

An InequalityfortheLogarithm

The University of Queensland,Australia;and V. 1212. Proposedby L. Bass and R. Vyborny', Sweden. ofTechnology, Thome'e,ChalmersInstitute Provethatif x > 1 and 0 < u < 1 < v, then v(x-1)(xVl

1) < log x < u(x - )(

-

xul1)

SolutionbyM. S. Klamkin,University ofAlberta,Canada. as The inequalitycan be rewritten v

F(tv -1)l) 2i, thenthere in theinterval is servedby a lowernumbered are at least i receivers [1,2i - 1]. But each receiver On theotherhand,xi_ xi. Thus,pairingxi withyi foreach i givesone schedulerealizingX as a servingset. thatno interval[1,j] has morereceiversthanservers. The conditionxi_ 1 - 2 t/q > 0 for O< t < gIr2.Hence sec2t< (1 - 2t/'i))-2, and so f0sec2tdt