Mathematical Association of America is collaborating with JSTOR to digitize, preserve and extend access to The College Mathematics Journal.

Size: px

Start display at page:

Download "Mathematical Association of America is collaborating with JSTOR to digitize, preserve and extend access to The College Mathematics Journal."

Annabel Watson
6 years ago
Views:

PROBLEMS AND SOLUTIONS Source: The College Mathematics Journal, Vol 4, No March, pp 6-69 Published by: Mathematical Association of America Stable URL: http://wwwjstororg/stable/469/74684x4885

1 PROBLEMS AND SOLUTIONS Source: The College Mathematics Journal, Vol 4, No March, pp 6-69 Published by: Mathematical Association of America Stable URL: Accessed: 5/6/4 8:9 Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range of content in a trusted digital archive We use information technology and tools to increase productivity and facilitate new forms of scholarship For more information about JSTOR, please contact support@jstororg Mathematical Association of America is collaborating with JSTOR to digitize, preserve and extend access to The College Mathematics Journal This content downloaded from 498 on Sun, 5 Jun 4 8:9:5 AM

2 PROBLEMS AND SOLUTIONS EDITORS Curtis Cooper Shing S So CMJ Problems CMJ Solutions Department of Mathematics and Department of Mathematics and Computer Science Computer Science University of Central Missouri University of Central Missouri Warrensburg, MO 649 Warrensburg, MO cooper@ucmoedu so@ucmoedu doi:469/74684x4885 This section contains problems intended to challenge students and teachers of college mathematics We urge you to participate actively both by submitting solutions and by proposing problems that are new and interesting To promote variety, the editors welcome problem proposals that span the entire undergraduate curriculum Proposed problems should be sent to Curtis Cooper, either by as a pdf, T E X, or Word attachment preferred or by mail to the address provided above Whenever possible, a proposed problem should be accompanied by a solution, appropriate references, and any other material that would be helpful to the editors Proposers should submit problems only if the proposed problem is not under consideration by another journal Solutions to the problems in this issue should be sent to Shing So, either by as a pdf, T E X, or Word attachment preferred or by mail to the address provided above, no later than June 5, PROBLEMS 9 Proposed by Ovidiu Furdui, Cluj, Romania and Razvan Tudoran, Universitatea de Vest din Timisoara, Timisoara, Romania a Let {x k } k be the sequence defined by x k k + k+ k+ where the fraction k+ appears exactly k + times The first three terms, x, x,andx are: Prove that for k,, k + < x k < b Let {y k } k be the sequence defined by y k k, 7, lnk + where the fraction appears exactly k times What can be said regarding k upper and lower bounds for y k? k k, 6 THE MATHEMATICAL ASSOCIATION OF AMERICA This content downloaded from 498 on Sun, 5 Jun 4 8:9:5 AM

3 9 Proposed by Sadi Abu-Saymeh and Mowaffaq Hajja, Yarmouk University, Irbid, Jordan Let ABC be a triangle in which angles B and C are acute Let the internal angle bisectors of angles B and C meet the altitude AA at B and C, respectively Prove that if BB CC,thenAB AC 9 Proposed by Eugen J Ionascu, Columbus State University, Columbus, GA Let α, be an arbitrary irrational number Construct a sequence {a n } of real numbers with the following two properties: a {a n n N} { n n N} {n n N}, n a b lim +a + +a n n α n 94 Proposed by Joe Howard, Portales, NM and George Apostolopoulos, Messolonghi, Greece Let n, a i > fori,,,n, n i a i, and a n+ a Prove n i ai a i+ n n a i i 95 Proposed by Cezar Lupu student, University of Bucharest, Bucharest, Romania and Tudorel Lupu student, Decebal High School, Constanta, Romania Let f be a twice differentiable function on R with f continuous on [, such that /4 f x dx /4 f x dx Prove that there exists an x, such that f x SOLUTIONS A constrained homogeneous inequality 896 Proposed by Darij Grinberg student, Ludwig Maximilian University, Munchen, Germany and Cezar Lupu student, University of Bucharest, Bucharest, Romania Let a, b, c be positive reals such that a + b + c Show that a + b + c + a + b + c abc Solution by Michael Bataille, Rouen, France Without loss of generality, assume that a b cthen a + b + c + a + b + c abc abc a + b + abc + a + b + c a + b + cabc c VOL 4, NO, MARCH THE COLLEGE MATHEMATICS JOURNAL 6 This content downloaded from 498 on Sun, 5 Jun 4 8:9:5 AM

4 a + b + cbc + ac + ab abc a + b + c a + b + cabc since a + b + c a b + ab + bc + c a + ca + b c a + b + c a + b + cabc By the Arithmetic-Geometric Mean inequality, a b + ab + bc + c a + ca + b c a b + ab + bc c a + ca + b c Since a b c and a b + ab + bc c a + ca + b c a + b + c a + b + cabc a b c + b c ab c ab c b c a ba c, the desired inequality follows from Also solved by ARKADY ALT, San Jose, CA; GEORGE APOSTOLOPOULOS, Messolonghi, Greece; BRIAN BRADIE, Christopher Newport U; ROBIN CHAPMAN, U of Exeter, Exeter, UK; HONGWEI CHEN solutions, Christopher Newport U; MARGARET CIBES, Hartford, CT; CHIP CURTIS, Missouri Southern State U; BILL DUNN, Lone Star C, Montgomery; REBECCA EVERDING student, Southeast Missouri State U; FISHER PROBLEM SOLVING GROUP, St John Fisher C; EUGEN IONASCU, Columbus State U; KEE-WAI LAU, Hong Kong, China; NANCY MUELLER student, Southeast Missouri State U; MICHAEL NEUBAUER, California State U Northridge; PAOLO PERFETTI, Dipartimento di Matematica, Università degli studi di Tor Vergata Roma, Rome, Italy; WILLIAM SEAMAN, Albright C; HAOHAO WANG and JERZY WOJDYLO jointly, Southeast Missouri State U; ALBERT WHITCOMB, Castle Shannon, PA; and the proposers One incorrect solution was received A computer generated solution was submitted by STAN WAGON, Macalester C Editor s Note Robert Clay of Dalton State College noted the following geometric interpretations of this problem: Consider a rectangular prism with linear dimensions a, b, c satisfying a + b + c, volume V, length of a diagonal d, and total surface area S; then the proposed inequality can be interpreted as S V V max + d d min V If a, a,,a n are linear dimensions of an n-prism with n faces of dimension n satisfying a + a + +a n n, volume V, length of a diagonal d, and total surface area S; then a generalization of the above geometric interpretation is S V V max + n d d min V Nagel point, Euler line, and Fuhrmann circle 897 Proposed by R S Tiberio, Wellesley High School, Wellesley, ME Let G be the centroid, O the circumcenter, and N the Nagel point of a scalene triangle ABC Show that OG ON, if triangle ABC contains an angle of 6 Solution by John Sumner and Aida Kadic-Galeb, University of Tampa, Tampa, FL More generally if ABC has circumradius R, side lengths, a, b, c for sides BC, AC, AB, and angles α, β, γ at A, B, C, respectively, then we shall prove that i the dot product OG ON R cosα cosβ cosγ ; ii the dot product NO NG a+b ca+c bb+c a a+b+c ; 6 THE MATHEMATICAL ASSOCIATION OF AMERICA This content downloaded from 498 on Sun, 5 Jun 4 8:9:5 AM

5 iii the dot product GO GN a +b +c a+b+ca +b +c ; 9a+b+c iv in order for OGN to be a right triangle it is necessary and sufficient that ABC contains a 6 angle or the length of one side of ABC is the average of the lengths of the other two sides; and v the points O, G,andN are distinct and collinear if and only if ABC is isosceles and not equilateral Since every triangle in Euclidean geometry must contain an acute angle, we may assume that β is acute Position ABC in the first quadrant with B at the origin and A on the x-axis Let a and c represent vectors from B to C and B to A, respectively Then a a, c c, andb c a SinceG is the intersection of the medians of ABC and lies of the way along each median, we have BG a + c a a + c Let r and t be the unique real numbers such that BO r a + t c It is well-known that O is the intersection of the perpendicular bisectors of ABC This implies a BO a and c BO c Therefore, [ a r [ a + t c c r a + t c Since a a a, a c ac cos β, and c c c, it follows that and Solving and yields r ra + tccos β a, racos β + tc c a c cos β a sin β a c cos β BO a sin β c a cos β and t, and hence c sin β a + c a cos β c sin β c Let s a+b+c be the semiperimeter of ABC It is well-known that the Nagel point of ABC is the intersection of the lines extending from each vertex to the point opposite the vertex that lies s units around ABC In particular, the vector from C to the opposite point is a + s a c Similarly, the vector from A to the opposite point is c + s c a or c a Thus, there exist unique real numbers r and t such that BN a + r [ a + s a c c + t c s c a [ r t c a [ t r [ c + s a c s c a a, VOL 4, NO, MARCH THE COLLEGE MATHEMATICS JOURNAL 6 This content downloaded from 498 on Sun, 5 Jun 4 8:9:5 AM

6 Since a and c satisfy and are linearly independent, each coefficient is zero Hence, r and t ar + ts c a, rs a + ct c Solving and 4 yields r ac cs c c ac as a and t ac s as c a+b+c ac s as c Thus, BN a + r [ s a a + a + b c c a + b + c a + c c a + b + c a + b + c Using OG BG BO, NG BG BN, ON BN BO, and the above results, we have a + c cot β csc β a csc β OG 6a NG a c + a cot β csc β c csc β + c, 6c a + b c a b c a + a + b + c a + b + c c, ca + b + c cos β a[c + a + b c cos β csc β ON aa + b + c a aa + b + c cos β c[a + a + b + c cos β csc β + ca + b + c It is well-known that R abc 4A,where A 4 a + b + ca b + ca + b c a + b + c a a+b+c c is the area of ABC Using a a a, a c ac cos β, and c c c, cos β a +c b, and the above results, we have ac i the dot product b bc + c a a ac + c b a ab + b c OG ON a + b + ca + b ca b + ca b c ii the dot product R [cosα cosβ cosγ ; NO NG [ a + a b + ab b + a c abc + b c + ac + bc c a + b + c a + b ca + c bb + c a ; and a + b + c 64 THE MATHEMATICAL ASSOCIATION OF AMERICA This content downloaded from 498 on Sun, 5 Jun 4 8:9:5 AM

7 iii the dot product a ab a b + b a c b c ac bc + c GO GN 9a + b + c a + b + c a + b + ca + b + c 9a + b + c iv Since ABC is scalene, O, G,andN are distinct points Since by Chebyshev s inequality a + b + c a + b + ca + b + c, with equality if and only if a b c, it follows that GO GN for any distinct a, b, c Thus, iv follows from i and ii In particular, the proposed result follows from i v Note that O, G, andn are collinear if and only if OG NG Since a a c c and c a a c,wehave [ a + c cot β csc β a csc β OG NG 6a c + a cot β csc β c csc β 6c a ba cb c a + b ca + c bb + c a a b c a + b + c a + b c a + b + c a c and hence, O, G,andN are collinear, if and only if a ba cb c, a + b ca + c bb + c a a c if and only if ABC is isosceles Since O, G, andn are distinct, if and only if ABC is not equilateral; the result follows Solution by Jerry Minkus, San Francisco, CA; Ruthven Murgatroyd, Albany, OR; and Albert Whitcomb, Castle Shannon, PA; independently It is well-known that the circumcenter O, the orthocenter H, and the centroid G of any triangle ABC lie on the same line called the Euler line LetƔ denote the circumcircle of ABCLetM a denote the midpoint of the arc BC and F a the reflection of M a with respect to BC; M b the midpoint of the arc AC and F b the reflection of M b with respect to AC; andm c the midpoint of the arc AB and F c the reflection of M c with respect to AB Then the circumcircle Ɣ F of F a F b F c is called the Fuhrmann circle of ABC It is also well-known that the Nagel point N and the orthocenter H of ABC both lie on the Fuhrmann circle Ɣ F and HN is a diameter of Ɣ F [ The proof will be complete when one shows that the circumcenter O of ABC lies on Ɣ F Referring to the figure below, let X M a be the point of intersection of the angle bisector of BAC and the circumcircle of ABC ThenBX CX Without loss of generality assume BAC 6 Since BX C and BAC are opposite angles of the cyclic quadrilateral ABX C, BX C and BAC are supplementary, and hence BX C LetA be the midpoint of BCThenAX BCLetX F a be the reflection of X with respect to BC and so X lies on the Fuhrmann circle of ABC Since BXC BX C, BXC NowO also lies on A X and BOC BAC Since BOC BXC, andx and O both lie on the perpendicular VOL 4, NO, MARCH THE COLLEGE MATHEMATICS JOURNAL 65 This content downloaded from 498 on Sun, 5 Jun 4 8:9:5 AM

8 bisector of BC and on the same side of line BC, X and O coincide, and hence O lies on the Fuhrmann circle if ABC contains an angle of 6 Reference Ross Honsberger, Episodes in Nineteenth and Twentieth Century Euclidean Geometry, New Mathematical Library, MAA, Washington, DC, 995 Also solved by MICHEL BATAILLE, Rouen, France; ROBIN CHAPMAN, U of Exeter, Exeter, UK; CHIP CURTIS, Missouri Southern State U; MICHAEL GOLDENBERGand MARK KAPLAN jointly, Baltimore Poly Inst; KEE- WAI LAU, Hong Kong, China; JOHN ZACHARIAS, Melbourne, FL; and the proposer An inequality with definite integrals of a convex function 898 Proposed by Mihály Bencze, Sacele, Brasov, Romania Suppose f is a function defined on an open interval I such that f x forall x I,and[a, b I Prove that f a + b aydy a + b f 4 + b a y dy Solution by John Sumner and Aida Kadic-Galeb, University of Tampa, Tampa, FL More generally, for λ, we prove that f λa + λb + λ b ay dy f a + b ay dy The inequality is trivial for λ For λ, the substitution u y shows that f b b ay dy f a + b ay dy, and hence holds for λ Suppose <λ< Since f x >forallx I, f is convex on I and λa + λb + λ b ay λa + b ay + λb b ay 66 THE MATHEMATICAL ASSOCIATION OF AMERICA This content downloaded from 498 on Sun, 5 Jun 4 8:9:5 AM

9 Thus, f λa + λb + λ b ay dy λ λ f a + b ay dy + λ f a + b ay dy + λ f a + b ay dy by the convexity of f In particular, if λ, then becomes 4 f a + b aydy which is the proposed inequality a + b f 4 f b b ay dy f b b ay dy + b a y dy, Also solved by ARKADY ALT, San Jose, CA; MICHAEL ANDREOLI,MiamiDadeC;GEORGE APOSTOLOPOU- LOS, Messolonghi, Greece; MICHEL BATAILLE, Rouen, France; BRIAN BRADIE, Christopher Newport U; ROBIN CHAPMAN, U of Exeter, Exeter, UK; HONGWEI CHEN, Christopher Newport U; MARGARET CIBES, Hartford, CT; CHIP CURTIS, Missouri Southern State U; CHARLES DIMINNIE, Angelo State U; JAMES DUEM- MEL, Bellingham, WA; FISHER PROBLEM SOLVING GROUP solutions, St John Fisher C; OVIDIU FURDUI, Campia Turzii, Cluj, Romania; DAVID GETLING, Berlin, Germany; GRA PROBLEM SOLVING GROUP, Rome, Italy; CODY GUINAN, Southeast Missouri State U; EUGENE HERMAN, Grinnell C; EUGEN IONASCU Columbus State U; PARVIZ KHALILI, Christopher Newport U; YOUNG HO KIM, Inst of Sci Ed for Gifted and Talented, Yonsei U, Seoul, Korea; PANAGIOTIS KRASOPOULOS, Athens, Greece; ELIAS LAMPAKIS, Kiparissia, Greece; KEE-WAI LAU, Hong Kong, China; VINCE MATSKO, Illinois Math and Sci Academy; MATTHEW MCMULLEN, Otterbein C; BRUCE O NEILL, Milwaukee School of Engineering; JENNIFER PAJDA student, Southeast Missouri State U; PAOLO PERFETTI, Dipartimento di Matematica, Università degli studi di Tor Vergata Roma, Rome, Italy; ÁNGEL PLAZA and SERGIO FALCÓN jointly, U of Las Palmas de Gran Canaria, Spain; JOEL SCHLOSBERG, Bayside, NY; WILLIAM SEAMAN, Albright C; SERGE VARJABEDIAN, Douai, France; HAOHAO WANG and JERZY WOJDYLO jointly, Southeast Missouri State U; ALBERT WHITCOMB, Castle Shannon, PA; JOHN ZACHARIAS, Melbourne, FL; and the proposer A partial solution was received Editor s Note Vince Matsko of Illinois Mathematics and Science Academy and James Duemmel of Bellingham, WA independently proved the proposed inequality using the following ideas: Define gλ b a f a + λ y + b λy dy for λ [, and show that g is decreasing Then the proposed inequality is proved by showing g g b a 4 Editor s Note The GRA Problem Solving Group of Rome, Italy and Serge Varjabedian of Douai, France independently proved the proposed inequality using the following ideas: Define Hλ f λa y + λb + yλb + λa dy for λ [, and show that H is nondecreasing Then the proposed inequality is proved by showing H H 4 A definite integral involving a geometric series 899 Proposed by Ovidiu Furdui, University of Toledo, Toledo, OH Let k be a natural number Prove that x n + x k + x k + +x dx k nk lim n n m + km VOL 4, NO, MARCH THE COLLEGE MATHEMATICS JOURNAL 67 This content downloaded from 498 on Sun, 5 Jun 4 8:9:5 AM

10 Solution by Bruce Davis, St Louis Community College at Florissant Valley, St Louis, MO Let I ɛ x n +x k +x k + x nk dx,where<ɛ< Since, for x <, + x k + x k + +x nk x kn+ x k x n x k + x k + x nk x k x k x n x, kn+ ɛ x k x n I dx x kn+ Noting that x kn+ m x kmn+ for x [, ɛ, and substituting this uniformly convergent geometric series into, yields [ ɛ I x k x n x kmn+ dx m Since m x k x n x kmn+ m x kmn+, we may integrate term-by-term: [ ɛ I [ x k x n x kmn+ dx m [ ɛ m [x kmn++n x kmn++n+k dx [ kmn + + n + k + [kmn + + n + ɛ ɛ +kmn+ k n + + km[n + + km + k m By Abel s Limit Theorem, x n [ + x k + x k + +x dx k nk Hence, the proposed limit is lim n n lim n k x n m + x k + x k + +x [ m n + + km[n + + km + k nk dx n n + + km[n + + km + k Also solved by ARKADY ALT, San Jose, CA; GEORGE APOSTOLOPOULOS, Messolonghi, Greece; MICHEL BATAILLE, Rouen, France; PAUL BRACKEN, U of Texas, Edinburg; BRIAN BRADIE, Christopher Newport U; ROBIN CHAPMAN, U of Exeter, Exeter, UK; HONGWEI CHEN, Christopher Newport U; GORDON CRANDALL, LaGuardia CC; RICHARDDAQUILA, Muskingum C; JAMESDUEMMEL, Bellingham, WA; DAVID FELDMAN, C of the Atlantic; PARVIZ KHALILI, Christopher Newport U; HARRIS KWONG, SUNY Fredonia; KEE-WAI LAU, Hong Kong, China; KIM MCINTURFF, Santa Barbara, CA; LUIS MEDINA, Rutgers U; PAOLO PERFETTI, Dipartimento di Matematica, Università degli studi di Tor Vergata Roma, Rome, Italy; JOEL SCHLOSBERG, Bayside, NY; WILLIAM SEAMAN, Albright C; PETER SIMONE, U of Nebraska Medical Center; JOHN SUMNER and AIDA KADIC-GALEB jointly, U of Tampa; NORA THORNBER, Raritan Valley CC; BOB TOMPER, U of North Dakota; SERGE VARJABEDIAN, Douai, France; NGUYEN VAN VINH, Belarusian State U, Belarus; ALBERT WHITCOMB, Castle Shannon, PA; JOHN ZACHARIAS, Melbourne, FL; and the proposer 68 THE MATHEMATICAL ASSOCIATION OF AMERICA This content downloaded from 498 on Sun, 5 Jun 4 8:9:5 AM

11 An identity with the harmonic series 9 Proposed by Michel Bataille, Rouen, France Show that for p >, n H n n + p+ n + p+ where H n n k k Solution by Rob Pratt, Raleigh, NC Using the summation by parts formula: n n H n n n, p n + p a n+ b n+ b n a b a n+ a n b n, with a n Hn and b n yields n p H n n H p n + p H p n H n n p + n n n n n n n n p H n + H n H n H n n p+ H n + H n H n n + H n p+ n p+ n H n n + H n p+ n + n p+ H n n + p+ n + p+ Also solved by ARKADY ALT, San Jose, CA; BRIAN BRADIE, Christopher Newport U; ROBIN CHAPMAN, U of Exeter, Exeter, UK; HONGWEI CHEN, Christopher Newport U; DONAL CONNON, Kent, UK; CHIP CURTIS, Missouri Southern State U; CHARLES DIMINNIE, Angelo State U; BRAD EMMONS, Utica C; DAVID FELDMAN, C of the Atlantic; OVIDIU FURDUI, Campia Turzii, Cluj, Romania; DAVID GETLING, Berlin, Germany; GRA PROBLEM SOLVING GROUP, Rome, Italy; EUGENE HERMAN, Grinnell C; PARVIZ KHALILI, Christopher Newport U; KEE-WAI LAU, Hong Kong, China; NORTHWESTERN UNIVERSITY MATH PROBLEM SOLVING GROUP, NorthwesternU; PAOLO PERFETTI, Dipartimento di Matematica, Universitàdegli studi di Tor Vergata Roma, Rome, Italy; JOEL SCHLOSBERG, Bayside, NY; WILLIAM SEAMAN, Albright C; PETER SIMONE, U of Nebraska Medical Center; JOHN SUMNER and AIDA KADIC-GALEB jointly, U of Tampa; BOB TOMPER, U of North Dakota; ALEXEY VOROBYOV, Irvine, CA; JOHN ZACHARIAS, Melbourne, FL; and the proposer A computer generated solution was submitted by STAN WAGON, Macalester C VOL 4, NO, MARCH THE COLLEGE MATHEMATICS JOURNAL 69 This content downloaded from 498 on Sun, 5 Jun 4 8:9:5 AM

Let ABCD be a tetrahedron and let M be a point in its interior. Prove or disprove that [BCD] AM 2 = [ABD] CM 2 = [ABC] DM 2 = 2

Let ABCD be a tetrahedron and let M be a point in its interior. Prove or disprove that [BCD] AM 2 = [ABD] CM 2 = [ABC] DM 2 = 2 SOLUTIONS / 343 In other words we have shown that if a face of our tetrahedron has a nonacute angle at one vertex the face of the tetrahedron opposite that vertex must be an acute triangle. It follows