SOLUTIONS. Let P be an arbitrary interior point of an equilateral Š triangle ABC. Prove that P BC P CB arcsin 2 sin P AB

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1 98/ SOLUTIONS SOLUTIONS No problem is ever permaetly closed. The editor is always pleased to cosider for publicatio ew solutios or ew isights o past problems. Due to a filig error, a umber of readers solutios got misplaced ad were ever ackowledged. The followig solutios were received by the editor-i-chief: ARKADY ALT, Sa Jose, CA, USA(364); ŠEFKET ARSLANAGIĆ, Uiversity of Sarajevo, Sarajevo, Bosia ad Herzegovia (366, 3634, 3635, 3636); MICHEL BATAILLE, Roue, Frace (364); VÁCLAV KONEČNÝ, Big Rapids, MI, USA (354); PAOLO PERFETTI, Dipartimeto di Matematica, Uiversità degli studi di Tor Vergata Roma, Rome, Italy (3639); EDMUND SWYLAN, Riga, Latvia (36, 3634, 3635, 3638); ad PETER Y. WOO, Biola Uiversity, La Mirada, CA, USA (366, 367, 368, 369, 363, 3634, 3635, 3638). The editor apologizes sicerely for the oversight [ : 499; : 559] Proposed by Michael Lambrou, Uiversity of Crete, Crete, Greece. Let P be a arbitrary iterior poit of a equilateral Š triagle ABC. Prove P AB P AC that P BC P CB arcsi si P AB P AC P AB P AC. Show that the left iequality caot be improved i the sese that there is a positio Q of P o the ray AP givig a equality. (Thus the iequality i 55 [997: 3; 998: ; 999: 3-4] is improved.) Solutio by Tomasz Cieśla, studet, Uiversity of Warsaw, Polad. Whe BAP = P AC the give relatios hold because the three quatities beig compared are all zero; therefore, without loss of geerality we shall assume that BAP > P AC (ad, cosequetly, P CB P BC). Defie l to be that portio of the lie AP i the iterior of ABC. We first will prove that the positio of P o l that maximizes the quatity o the left is where BP C = π 3. O l choose poit Q such that BQC = π 3. Deote by P ad Q the reflectios of P ad Q i the bisector of agle BAC. The we have P CB P BC = P BP ad QCB QBC = QBQ. Our claim is that QBQ P BP for all positios of P o l. Cosider homothety cetered at A which seds P ito Q. The P is set to Q, ad B is set to some poit B lyig o lie AB. We have P BP = QB Q. Sice circle (BQ QC) is taget to lie AB at B (because BQC = π 3 ), we see that poit B lies outside the circle o the same side of lie QQ as B. This implies that QBQ QB Q = P BP as claimed. Crux Mathematicorum, Vol. 38(5), May

2 SOLUTIONS / 99 A A Q P P Q B Q P Q P B C B C B Next we will see that QBQ QAB QAC = arcsi si P AB P AC QAB QAC. () Š P AB P AC is costat for Because the differece arcsi si all poits P o l, this will prove that the left iequality holds ad, moreover, it caot be improved. A Q S O M Q B C J R Deote the circumcetre of ABC by O, ad the reflectios of Q ad O i BC by R ad J. Because triagle ABC is equilateral, poits R, J lie o the circumcircle of ABC ad J is the circumceter of trapezoid BCQQ. Note that O is midpoit of arc QQ of circle (BQQ C). Agle chasig gives us Q OQ = π QBQ = π QJQ = π QJO = π JOR = ROA. I additio, OQ = OQ ad OR = OA. Thus there exists a rotatio about O which maps Q to Q ad R to A; deote by S the image of Q uder this rotatio. Copyright c Caadia Mathematical Society, 3

3 / SOLUTIONS The QSA = Q QR = π. Sice O is the circumceter of isosceles triagle Q QS, SQQ = OQQ = OBQ = QBQ. () Let M be midpoit of QQ. Poits Q, M, S, A lie o the circle with diameter QA, because QMA = π = QSA. Thus SQQ = SQM = SAM. (3) Observe that si SAQ = SQ AQ = Q Q AQ = MQ AQ = si MAQ = si OAQ. From that we get SAQ = arcsi( si OAQ). (4) From OAQ = QAB QAC ad equatios () through (4), QBQ = SQQ = SAM = SAQ OAQ, which is equatio (), as claimed. For the iequality o the right, simply ote that we have proved that the middle differece is the maximum of P CB P BC over all poits P l, while Problem 55 established that this differece is at most P AB P AC. This observatio cocludes the proof. Also solved by the proposer; o solutio was published before ow. For a alterative proof of the right iequality, let x = P AB P AC, x < π 3. The iequality to prove reduces to arcsi si x 3x, for x < π, which is a elemetary 3 exercise. It is iterestig to ote that accordig to the solutio of Problem 55, the iequality there, amely P AB P AC P CB P BC, holds for all isosceles triagles ABC for which A π 3 (ad B = C π ), while the iequality fails for some positios of P i 3 isosceles triagles with A < π 3. Note that arcsi si x is o loger real for x > π, so that 3 there are positios of P for which the right iequality of the preset problem fails for isosceles triagles with A > π [ : 34, 37] Proposed by José Luis Díaz-Barrero, Uiversitat Politècica de Cataluya, Barceloa, Spai. Let x, x,..., x < π/ be real umbers. Prove that sec(x k )! Ž / si(x k )!. I. Composite of similar solutios by Arkady Alt, Sa Jose, CA, USA; ad Paolo Perfetti, Dipartimeto di Matematica, Uiversità degli studi di Tor Vergata Roma, Rome, Italy. Let f(x) = sec x, g(x) = si x ad set x = x k. Sice f (x) = + si x > ad g (x) = si x < for < x < π, f is covex ad g is cos 3 x cocave o the iterval (, ). Crux Mathematicorum, Vol. 38(5), May

4 SOLUTIONS / Hece Jese s Iequality esures that Therefore we have sec(x k ) sec x k sec( x)! ad si x k si( x). Ž / si(x k )! sec( x)( si ( x)) / = sec( x) cos( x) =. II. Composite of virtually idetical solutios by Šefket Arslaagić, Uiversity of Sarajevo, Sarajevo, Bosia ad Herzegovia; ad Salem Malikić, studet, Simo Fraser Uiversity, Buraby, BC. so Hece, By Cauchy-Schwarz Iequality we have si (x k ) sec(x k )!! = = = Y!! si(x k ) si(x k ) sec(x k ) sec(x k )!!! cos(x k ) cos(x k ) si (x k )! Ž /! si (x k ). si (x k ) ( si (x k )) cos (x k )! / si(x k )! /! /!! / Y / Ž / cos (x k )! = by the AM-GM Iequality. Clearly, equality holds if ad oly if x = x = = x. Also solved by OLIVER GEUPEL, Brühl, NRW, Germay; ALBERT STADLER, Herrliberg, Switzerlad; ad the proposer. Copyright c Caadia Mathematical Society, 3

5 / SOLUTIONS 364. [ : 35, 37] Proposed by Michel Bataille, Roue, Frace. Evaluate R lim (x 5x ) dx R (x 4x ) dx. Solutio by Paul Bracke, Uiversity of Texas, Ediburg, TX, USA; modified by the editor. R Write the limit as lim I R ()/I (), where I () = ( + 5x x ) dx ad I () = ( + 4x x ) dx, ad let φ(x) = l( + 5x x ). Itegratio by parts gives I () = Z = e φ(t) e φ(t) dt = Z = d dt Z Z d φ (t) dt [eφ(t) ] dt e φ(t) dt φ (t) d dt φ (t) e φ(t) dt. The fuctio d/dt[/φ (t)] = / + (33/)(5 4t) icreases o [, ], takig the value 9/5 at t = ad the value 7 at t =. It follows that Z 9 Z e φ(t) d Z dt 5 dt φ e φ(t) dt 7 e φ(t) dt, (t) ad hece there is a costat C with 9/5 C 7 ad such that Z d Z dt φ e φ(t) dt = C e φ(t) dt. (t) Thus, ad solvig for I () gives I () = C I () I () = C By similar calculatios there is a costat C with 9/8 C 3 such that I () = C 4. 4 Fially, I () lim I () = lim C Š C Š Š = Š Crux Mathematicorum, Vol. 38(5), May

6 SOLUTIONS / 3 Also solved by Albert Stadler, Herrliberg, Switzerlad; ad the proposer. Oe icorrect solutio ad oe icomplete solutio were received [ : 35, 38] Proposed by Pham Va Thua, Haoi Uiversity of Sciece, Haoi, Vietam. Let u ad v be positive real umbers. Prove that 8 7 uv u + v É u É v r(u 3 v + 3 u + v) u v + For each iequality, determie whe equality holds. Editor s ote: Perfetti poited out that the very same problem (by the same proposer) has appeared i Mathematics Magazie (Vol. 8, No. 3, 9) ad a solutio was published i Vol. 83, No. 3, pp However, we decide to publish a differet solutio which is completely elemetary. Solutio by Titu Zvoaru, Comáeşti, Romaia. È Let x = 3 u v. The x > ad x3 = u v. The left iequality is equivalet, i successio, to 8 7 x3 x 6 + x + x 7x 6 x x 6 8x x 8x 8 7x 7 + 8x 6 + x 4 + 8x 7x + 8 (x ) (8x 6 x 5 x 4 3x 3 x x + 8) (x ) ((8x 6 x 5 x 4 x 3 x x + 8) + 7x 3 ) (x ) ((x ) (8x 4 + 5x 3 + x + 5x + 8) + 7x 3 ) which is clearly true. ad To establish the right iequality ote that (u + v) u + vš = x 3 + x 3 + x + x rx 3 + x 3 + x + x + x3 + x 3 + x 6 x 5 x + (x )(x 5 ) (x ) (x 4 + x 3 + x + x + ) which clearly holds. Note that equality holds i either iequality if ad oly if x = ; that is, if ad oly if u = v. Also solved by ARKADY ALT, Sa Jose, CA, USA; ŠEFKET ARSLANAGIĆ, Uiversity of Sarajevo, Sarajevo, Bosia ad Herzegovia; PRITHWIJIT DE, Homi Bhabha Cetre for Sciece Educatio, Mumbai, Idia; OLIVER GEUPEL, Brühl, NRW, Germay; KEE-WAI LAU, Hog Kog, Chia; SALEM MALIKIĆ, studet, Simo Fraser Uiversity, Buraby, BC; PAOLO PERFETTI, Dipartimeto di Matematica, Uiversità degli studi di Tor Vergata Roma, Rome, Italy; ALBERT STADLER, Herrliberg, Switzerlad; ad the proposer. Copyright c Caadia Mathematical Society, 3

7 4/ SOLUTIONS [ : 35, 38] Proposed by George Apostolopoulos, Messologhi, Greece. We trisect the sides AB ad AC of triagle ABC with the poits D, E ad F, G respectively such that AE = ED = DB ad AF = F G = GC. The lie BF itersects CD, CE i the poits K, L respectively, while BG itersects CD, CE i N, M respectively. Prove that: (a) KM is parallel to BC; (b) Area(KLM) = 5 7 Area(KLMN). Composite of complemetary solutios by Edmud Swyla, Riga, Latvia; ad Titu Zvoaru, Comáeşti, Romaia. Let K, L, M, ad N be the poits where the side BC meets the lies joiig A to K, L, M, ad N, respectively. (Oe ca easily show that, i fact, L = N is the midpoit of BC, but this is ot relevat to our work here.) A E F L M G D K B N L = N K M C By applyig Va Aubel s theorem (see, for example, F. G.-M., Exercices de géométrie compreat l exposé des méthodes géométriques et questios résolues, quatrième éditio, J. De Gigord, Paris (97), paragraph 4j, page 54) four times, we have AK KK = AD DB + AF F C = +, AL LL = AE EB + AF F C = + AM MM = AE EB + AG GC = +, AN NN = AD DB + AG GC = +, KK or AK = 7 ;, or LL AL = ; MM or AM = 7 ; NN or AN = 5. AK KK = AM From the first ad third of these equatios we get MM, whece KM K M, ad part (a) follows immediately. For part (b) we assume without loss Crux Mathematicorum, Vol. 38(5), May

8 SOLUTIONS / 5 of geerality that the altitude from A to BC has legth. The above equatios the imply that the lie segmets perpedicular to BC from K, L, M, N equal 7,, 7, 5, respectively. Thus Area(KLM) Area(KMN) = = 5, ad therefore Area(KLM) Area(KLMN) = 5 7. Also solved by ŠEFKET ARSLANAGIĆ, Uiversity of Sarajevo, Sarajevo, Bosia ad Herzegovia; OLIVER GEUPEL, Brühl, NRW, Germay; JOHN G. HEUVER, Grade Prairie, AB; VÁCLAV KONEČNÝ, Big Rapids, MI, USA( solutios); ad MICHAEL PARMENTER, Memorial Uiversity of Newfoudlad, St. Joh s, NL [ : 35, 38] Proposed by José Luis Díaz-Barrero ad Jua José Egozcue, Uiversitat Politècica de Cataluya, Barceloa, Spai. that Let a, b, ad c be positive umbers such that a + b + c + abc =. Prove X r a b b c c >. cyclic I. Solutio by Arkady Alt, Sa Jose, CA, USA. a Sice È Observe that < a, b, c < so that abc. The iequality is equivalet to ( b )( c ) + b È ( c )( a ) + c È ( a )( b ) > abc. () ( b )( c ) = b c + b c = a + abc + (bc) = (a + bc), ad, similarly, ( c )( a ) = (b + ca) ad ( a )( b ) = (c + ab), the left side is equal to a(a + bc) + b(b + ca) + c(c + ab) = a + b + c + 3abc = + abc > abc by the Arithmetic-Geometric Meas Iequality. II. Solutio usig ideas from Šefket Arslaagić, Uiversity of Sarajevo, Sarajevo, Bosia ad Herzegovia; Oliver Geupel, Brühl, NRW, Germay; Salem Malikić, studet, Simo Fraser Uiversity, Buraby, BC; ad Albert Stadler, Herrliberg, Switzerlad. Copyright c Caadia Mathematical Society, 3

9 6/ SOLUTIONS We ca select acute agles A, B, C for which a = cos A, b = cos B, c = cos C. The cos(a + B + C) = cos A cos B cos C cos A si B si C si A cos B si C so that A + B + C = π. Therefore a È = abc a(a + bc) b(b + ca) c(c + ab) =, ( b )( c ) = cos A[cos(B C) cos(b + C)] = cos(b + C) cos(b C) + cos A = (cos B + cos C) + cos A = cos B cos C + + cos A = + a b c, si A si B cos C with similar equatios for the other two terms of the left side of () i the first solutio. Therefore, the left side of () is equal to Also solved by the proposer. (3 a b c ) = + abc > abc [ : 35, 38] Proposed by Ovidiu Furdui, Campia Turzii, Cluj, Romaia. Let α ad let β be a positive umber. Fid the limit L(α, β) = lim + kα β k!. Solutio by Aastasios Kotroois, Athes, Greece; modified slightly by the editor. We will prove that L(α, β) = 8 >< >: by cosiderig three cases separately. if β α < if β α > β if β α = Case (i) β α <. By Beroulli s Iequality, we have k + kα β + kα+ β Crux Mathematicorum, Vol. 38(5), May

10 SOLUTIONS / 7 so Note that + kα β k lim k + kα+ = α β+ α+ = Z β = k α+! k α+ β. () x α+ dx = α +. () Sice α β + > ad α+ >, we coclude from () ad () that L(α, β) =. Case (ii) β α >. Note first that for all k =,,..., ad i =,, we have I particular, < kα+i β β α i < i. < kα β < ad < kα+ β <. (3) It is well kow that as x + we have ad Usig (3), (4) ad (5), we have, as, that so l( + x) = x + O(x ) (4) e x = + x + O(x ). (5) k + kα β = exp k l + kα k α = exp k β + O (α β)š k α+ = exp + O (α β)+š + kα β k = = β = + kα+ β k α+ β β α β + O (α β)+š + O (α β)+3š k α+! + O (α β)+3š. (6) Sice β α > ad (α β) + 3 <, it follows from () ad (6) that L(α, β) =. Copyright c Caadia Mathematical Society, 3

11 8/ SOLUTIONS Case (iii) β α =. We proceed as i case (ii). Sice β α =, it follows from () ad (6) agai that L(α, β) = α+ = β This completes our proof. Also solved by OLIVER GEUPEL, Brühl, NRW, Germay; RICHARD I. HESS, Racho Palos Verdes, CA, USA; ad the proposer [ : 36, 38] Proposed by Paagiote Ligouras, Leoardo da Vici High School, Noci, Italy. Show that i triagle ABC with exradii r a, r b ad r c, X cyclic where AB = c, BC = a, ad CA = b. (r a + r b )(r b + r c ) ac 9, Similar solutios by Prithwijit De, Homi Bhabha Cetre for Sciece Educatio, Mumbai, Idia; ad Kee-Wai Lau, Hog Kog, Chia. We kow that r a = s a, r b = s b, ad r c = a+b+c s c, where s = is the È semiperimeter of ABC ad = s(s a)(s b)(s c) is its area. It follows that r a + r b = s a + s b = c (s a)(s b) ad whece Similarly, r b + r c = (r a + r b )(r b + r c ) ac s b + = s c = a (s b)(s c), (s a)(s b) (s c) = s s b. (r b + r c )(r c + r a ) ba = s s c These last three equatios give us X cyclic (r a + r b )(r b + r c ) ac But by the AM-HM iequality, 3 s a + s b + s c which, whe multiplied by 3s, yields s s a + ad = s s a + s s b + Crux Mathematicorum, Vol. 38(5), May (r c + r a )(r a + r b ) cb s s b + = s s a. 3 s a + s b + s c = 3 s, s s c. () s 9. () s c

12 SOLUTIONS / 9 The desired iequality follows from () ad (). Equality holds if ad oly if the triagle is equilateral. Also solved by ARKADY ALT, Sa Jose, CA, USA; ŠEFKET ARSLANAGIĆ, Uiversity of Sarajevo, Sarajevo, Bosia ad Herzegovia; OLIVER GEUPEL, Brühl, NRW, Germay; JOHN G. HEUVER, Grade Prairie, AB; PAOLO PERFETTI, Dipartimeto di Matematica, Uiversità degli studi di Tor Vergata Roma, Rome, Italy; JUAN-BOSCO ROMERO MÁRQUEZ, Uiversidad de Valladolid, Valladolid, Spai; EDMUND SWYLAN, Riga, Latvia; TITU ZVONARU, Comáeşti, Romaia; ad the proposer [ : 36, 39] Proposed by Michel Bataille, Roue, Frace. Fid all real umbers x, y, z such that xyz = ad x 3 + y 3 + z 3 = S(S 4) 4 where S = x y + y x + y z + z y + z x + x z. I. Solutio by Paolo Perfetti, Dipartimeto di Matematica, Uiversità degli studi di Tor Vergata Roma, Rome, Italy. We assume the equatio x 3 + y 3 + z 3 = S(S 4) (xyz), 4 without the restrictio o xyz. This is equivalet to X (x 4 y + x 3 y 3 + x y z ) = X (x 4 yz + x 3 y z), where both sums, take over the six permutatios of the variables, are symmetric. By the Arithmetic-Geometric Meas Iequality, X x 4 y = x 4 (y + z ) + y 4 (z + x ) + z 4 (x + y ) x 4 yz + y 4 zx + z 4 xy = X x 4 yz, with equality if ad oly if x = y = z. Recall Schur s Iequality that, for a, b, c, (a 3 + b 3 + c 3 ) + 3abc (a b + a c + b a + b c + c a + c b) = a(a b)(a c) + b(b a)(b c) + c(c a)(c b). Settig (a, b, c) = (yz, zx, xy), we obtai that X (x 3 y 3 + x y z ) X x 3 y z, where agai each sum is symmetric with six terms. Therefore X(x 4 y + x 3 y 3 + x y z ) X (x 4 yz + x 3 y z). Sice we are assumig that equality holds ad that xyz =, the give equatio is satisfied if ad oly if x = y = z =. Copyright c Caadia Mathematical Society, 3

13 / SOLUTIONS II. Solutio by the proposer. The give coditios imply that S = xy + yx + yz + zy + zx + xz. Without loss of geerality, let x be the maximum of x, y, z. Defie T = 4(xy + yx + yz )(zx + xz + zy ). The ad also whece T = S (y z) (x + xy + xz yz) T = 4xyz(x 3 + y 3 + z 3 + S) + 4y z (x y)(x z), S = 4(x 3 + y 3 + z 3 + S) + 4y z (x y)(x z) + (y z) (x + xy + xz yz). Sice, by hypothesis, 4(x 3 + y 3 + z 3 + S) = S, we deduce that 4y z (x y)(x z) + (y z) (x + xy + xz yz) =. Sice x y, z, both terms o the left are oegative ad therefore must vaish. If, say, x = y, the x + xy + xz yz = x, so that y = z. Sice xyz =, we must have that x = y = z =. No other solutios were received [ : 36, 39] Proposed by Pham Va Thua, Haoi Uiversity of Sciece, Haoi, Vietam. Prove that Let a, b, ad c be three positive real umbers ad let k = (a + b + c) (a 3 + b 3 + c 3 ) a 3 + b 3 + c 3 a + b + c k3 5k + 63k 45, 4 k 5 ± k k + 9 ad equality holds if ad oly if (a, b, c) = 4 its permutatios. Solutio by Oliver Geupel, Brühl, NRW, Germay..,, or ay of P P P P a All sums shall be cyclic. Let x = b, y = b a, m = a bc, ad = bc We have x + y = k 3, hece 4xy (k 3). Usig the relatios X a 3 b 3 = x3 3(m + ) 6, X b 3 a 3 = y3 3(m + ) 6, a. Crux Mathematicorum, Vol. 38(5), May

14 SOLUTIONS / ad m + = xy 3, we deduce that (a 3 + b 3 + c 3 ) This proves the iequality. a 3 + b 3 + c 3 = x 3 + y xy = (x + y) 3 3(x + y)xy + 9 6xy = (k 3) (k )xy (k 3) (k ) = k3 5k + 63k (k 3) 4 Equality holds if ad oly if = x y = (a b)(b c)(c a)/abc, that is, if two of a, b, c coicide. Without loss of geerality, suppose that b = c. The equality is the equivalet to (k 3)/ = x = + a/b + b/a. Š However, this holds if ad oly if p = a/b is a root of the quadratic p k 5 p +. The coditio for equality (up to permutatio) therefore eeds to be corrected to where λ >. (a, b, c) = λ k 5 ± k k + 9 4, λ, λ, Also solved by the proposers. Our featured solver said his solutio was similar to ad ispired by the solutio to problem 75 i Secrets i Iequalities (Vol. I) by Pham Kim Hug, GIL Publishig House, Zalău, 7, pp [ : 38, 3] Replacemet. Proposed by Michel Bataille, Roue, Frace. Let ABC be a triagle ad R, O, G ad K its circumradius, circumcetre, cetroid ad Lemoie poit, respectively. Prove that BC KA GA = CA KB GB = AB KC GC = È 3(R OK ). Recall that a symmedia of a triagle is the reflectio of the media from a vertex i the agle bisector of the same vertex. The Lemoie poit of a triagle is the poit of itersectio of the three symmedias. Solutio by Edmud Swyla, Riga, Latvia. Let the legth of BC be deoted by a ad its midpoit by A. Let the symmedia to BC meet the circumcircle agai at A ad, fially, let G c be the foot of the perpedicular from G to AB ad K b be the foot of the perpedicular from K to AC. Copyright c Caadia Mathematical Society, 3

15 / SOLUTIONS A G C K B G K O B A C A Usig, i tur, the similar right triagles AG c G ad AK b K, the fact that GG c equals a third of the altitude from C, ad the sie law, we deduce that with aalogous formulas for KB GB We take KK b b KA GA = KK b = KK b GG c 3 a si B = KK b R 3, ab KC ad GC. = abc R(a + b + c ) to be a kow property of the Lemoie poit (see, for example, Roger A. Johso, Advaced Euclidea Geometry, Dover reprit (96), page 4, paragraph 34) ad obtai O the other had, a KA GA = bkb GB = ckc GC = 3abc a + b + c. () R OK = (R OK)(R + OK) = KA KA = KA(AA KA). () Because ABA AA C, From () we have KA = Next, Stewart s theorem says that AA = bc A A. (3) bc 3GA a + b + c = bc A A a + b + c. (4) A A = 4 a + b + c Š. (5) Crux Mathematicorum, Vol. 38(5), May

16 SOLUTIONS / 3 Puttig together equatios () through (5), we deduce that as desired. R OK = = = b c a + b + c 4b c AA (a + b + c ) b c a + b + c 4b c (a + b + c ) 4 ( a + b + c ) 3a b c (a + b + c ), Also solved by TITU ZVONARU, Comáeşti, Romaia; ad the proposer. Crux Mathematicorum Foudig Editors / Rédacteurs-fodateurs: Léopold Sauvé & Frederick G.B. Maskell Former Editors / Acies Rédacteurs: G.W. Sads, R.E. Woodrow, Bruce L.R. Shawyer Crux Mathematicorum with Mathematical Mayhem Former Editors / Acies Rédacteurs: Bruce L.R. Shawyer, James E. Totte, Václav Liek, Shaw Godi Copyright c Caadia Mathematical Society, 3

SOLUTIONS. The edges of the tetrahedron are denoted a, b, c, d, e, f. Prove or disprove that 4 6 GD 1 9. a + 1 b + 1 c + 1 d + 1 e + 1 f.

SOLUTIONS. The edges of the tetrahedron are denoted a, b, c, d, e, f. Prove or disprove that 4 6 GD 1 9. a + 1 b + 1 c + 1 d + 1 e + 1 f. 8/ SOLUTIONS SOLUTIONS No problem is ever permaetly closed. The editor is always pleased to cosider for publicatio ew solutios or ew isights o past problems. 49. [995 : 58; 996 : 83-84] Proposed by Ja