SOLUTIONS Proposed by George Apostolopoulos. 32/ SOLUTIONS

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1 3/ SOLUTIONS SOLUTIONS No problem is ever permanently closed. The editor is always pleased to consider for publication new solutions or new insights on past problems. 38. Proposed by George Apostolopoulos. Triangle ABC is isosceles with AB = AC and A =. Let D be the point on AB such that BCD = and let E be the point on BC such that EC = AC. Determine the point K on CD such that triangles KAD and KCE have equal areas. Solved by AN-anduud Problem Solving Group ; Š. Arslanagić ; R. Barbara ; R. Barroso Campos ; M. Bataille ; C. Curtis ; O. Geupel ; N. Hodžić ; D. Jonsson ; V. Konečný ; O. Kouba ; K.W. Lau ; S. Malikić ; M. Modak ; C. Sánchez-Rubio ; Skidmore College Problem Group ; D. Smith ; N. Stanciu and T. Zvonaru ; E. Swylan ; D. Văcaru ; and the proposer. We present two solutions. Solution, by Oliver Geupel. As the point K moves along the segment CD from C to D, the value of [KCE] [KAD] is strictly increasing. Hence, there is a unique location of point K such that [KAD] = [KCE]. Denoting the intersection of the lines AE and CD by K, we prove that K has the required property. AD Using the Law of Sines in triangle ACD, we get sin 3 = AC sin 5. Hence AD = AC sin 5 and BD = AB AD = AB sin 5 sin 5. Also, using the Law of Sines in triangle ABC, we have Thus, BE = BC CE = AB sin 5 AB. BE BD = sin 5 = BC AB. Crux Mathematicorum, Vol. 4(), January 4

SOLUTIONS / 33 Therefore, the lines AC and DE are parallel. Hence the triangles ACK and EDK are homothetic. We conclude AK EK = CK DK and This completes the proof.

2 SOLUTIONS / 33 Therefore, the lines AC and DE are parallel. Hence the triangles ACK and EDK are homothetic. We conclude AK EK = CK DK and This completes the proof. Solution, by Cristóbal Sánchez-Rubio. [KAD] AK DK = [KCE] CK EK =. Let B be the symmetric point of B with respect to the AE-axis. Then, using the given information, we can establish the angle measures as shown in the figure : We show that the desired point K is the meeting point of the lines AE and CD. It is easy to see that the triangle ABB is equilateral, the angles EBB and EB B are both of and BEB = 4. Therefore DEB = 4 o and the line DE is parallel to AC. So the triangles DEK and AKC are similar and we have : KE KA = KD KC KE KC = KA KD Since AKD = EKC, the areas of KAD and KCE are the same, which completes the proof. Editor s Comment. Bataille noticed that this is problem 983, proposed by the same author in The College Mathematics Journal, Vol. 43, No 4, September. 38. Proposed by Marcel Chiriţă. Solve the following system x + + 3y + + 4z + = 5 for x, y, z R. 3 x+ 3y y+ 4z z+ x+ = 3 3 Copyright c Canadian Mathematical Society, 5

3 34/ SOLUTIONS Solved by AN-Anduud Problem Solving Group ; G. Apostolopoulos ; Š. Arslanagić ; M. Bataille ; D. M. Bătineţu-Giurgiu, N. Stanciu and T. Zvonaru ; R. Boukharfane ; M. Coiculescu ; C. Curtis ; J. L. Díaz-Barrero ; C. R. Diminnie ; R. Hess ; N. Hodžić ; S. Malikić ; P. Perfetti ; C. R. Pranesachar ; D. Smith ; D. Văcaru ; and the proposer. We present the solution by Chip Curtis. For simplicity, let u = x +, v = 3y +, w = 4z + and then p = u +v, q = v + w and r = w + u. Then by AM-GM we have : 3 3 = 3 p + 3 q + 3 r p 3 q 3 r = 3 3 p+q+r 3 with equality if and only if p = q = r. On the other hand, by Cauchy-Schwartz p + q + r = (u + v + w) 3 + 3(u + v + w ) = + (u + v + w ) + = 87 with equality if and only if u = v = w. (u + v + w) 3 Hence for any triple (u, v, w) with u + v + w = 5, we have 3 u +v + 3 v +w + 3 w +u 3 3 with equality if and only if u = v = w. Since their sum is 5, this implies that u = v = w = 5 and hence x =, y = 8, z = Proposed by José Luis Díaz-Barrero. Let a, b, and c be positive real numbers. Prove that a + ca + b + ab + c + bc (a + b + c). Solved by A. Alt ; AN-anduud Problem Solving Group ; G, Apostolopoulos ; Š. Arslanagić ; D. Bailey, E. Campbell, and C. Diminnie ; M. Bataille ; D. M. Bătineţu- Giurgiu, N. Stanciu and T. Zvonaru ; R. Boukharfane ; C. Curtis ; J. G. Heuver ; N. Hodžić ; T. Karamfilova ; O. Kouba ; K. Lau ; S. Malikić ( solutions) ; D. E. Manes ; P. McCartney ; C. Mortici ; S. Muralidharan ; P. Perfetti ; C.M. Quang ; Skidmore College Problem Group ; D. Smith ; G. Tsapakadis ; D. Văcaru ( solutions) ; H. Wang and J. Wojdylo ; P. Y. Woo ; and the proposer. We present three solutions. Crux Mathematicorum, Vol. 4(), January 4

4 SOLUTIONS / 35 Solution, by AN-anduud Problem Solving Group. By the AM-GM inequality, we have : a + ca =»a(a + c) Clearly, equality holds if and only if a = b = c. Solution, by Kee-Wai Lau. a + (a + c) = (3a + c) = (a + b + c). We have a + ca = (3a + c ( a a + c) ) 3a + c. Similarly, b + ab 3b + a and c + bc 3c + b. The result follows by adding up the three inequalities. Solution 3, by D. M. Bătineţu-Giurgiu, Neculai Stanciu and Titu Zvonaru. By the Cauchy-Schwarz inequality, we have : ( a a + c + b b + a + c c + b) (a + b + c)(a + c + b + a + b + c) or a + ca + b + ab + c + bc (a + b + c). Editor s Comment. D. M. Bătineţu-Giurgiu, N. Stanciu and T. Zvonaru proved the following generalization : let m, n N with m n, b j [, ), j =,,..., m, x k (, ), k =,,..., n. Let B m = m j= b j, X n = n k= x k and X k,m = m i= b ix k x k+i where the indices are taken modulo n and k =,,..., n. Then n Xk,m B m X n. k= The inequality in the problem is the special case when m = n = 3, b =, b = b 3 = and (x, x, x 3 ) = (a, b, c) Proposed by Václav Kone cný. Let ABCD be a convex quadrilateral. Construct, using only compass and straightedge, the line parallel to one side of the quadrilateral which bisects its area. Solved by O. Geupel, C. Sánchez-Rubio, E. Swylan, P. Woo, and the proposer. We present the partial solution by Peter Woo, modified and completed by the editor. Copyright c Canadian Mathematical Society, 5

5 36/ SOLUTIONS We shall use square brackets to denote area. Label the given quadrilateral ABCD so that D is further from AB than C is. Let E be the point of the side AD for which CE AB, and F be the point on the line AD for which BF CA. Note that the points lie in the order D, E, A, F and, because [ABC] = [AF C], the area of triangle F CD equals the area of the given quadrilateral ABCD. Our goal is to construct the bisecting line XY parallel to AB with X on AD and Y falling on either BC or CD. Construct M to be the midpoint of F D. Then [MCD] = [F CD] = [ABCD]. If M coincides with E, then MC is the bisecting line parallel to AB (so that X = M and Y = C). Should E lie between D and M, then XY will necessarily lie between EC and AB (with Y on the segment BC) ; otherwise, EC will lie between XY and AB (and Y will lie on CD). Consider first the case where E lies between M and D. Should the lines BC and AD meet in a point N, then we construct X so that NX is the geometric mean of NM and NE ; that is, NX = NM NE. For the usual Euclidean construction when E lies between N and M as in the figure, let P be either point where the perpendicular to AD through E meets the circle whose diameter is MN ; then X is the point between M and E where the circle with centre N and radius NP intersects AD. (When N is on the other side so that M lies between E and N, use the same construction reversing the roles of E and M.) Let the line through X parallel to AB meet CB at Y. Because we have [XY N] = [MCN], whence [XY N] [ECN] = NX NE = NM NE = [MCN] [ECN], [XY CD] = [MCD] = [ABCD], as desired. This argument breaks down when the sides BC and AD are parallel (so that N is at infinity). A straightforward continuity argument places X at the midpoint of the segment ME ; alternatively, one can argue directly (using the parallelograms AF BC and EABC) to show that when X is the midpoint of ME, the base DM of triangle CDM is twice the length of the base XA of parallelogram XABY while these two polygons have the same altitude. It remains to construct the point X in the case where M lies between D and E. Here we want [XY D] = [MCD], so that now DX is the geometric mean of DE Crux Mathematicorum, Vol. 4(), January 4

6 SOLUTIONS / 37 and DM ; that is, DX = DE DM. The proof of the claim proceeds as before with D in the role of N : whence [XY D] = [MCD], as desired Proposed by Mehmet Şahin. [XY D] [ECD] = DX DE = DM DE = [MCD] [ECD], Let ABC be a triangle with incentre I. Let A be on ray IA beyond A such that A A = BC. Let B and C be similarly defined, such that B B = CA and C C = AB. Prove that [A B C ] ( + 3), [ABC] where [ ] denotes the area. Solved by M. Bataille ; N. Hodžic ; O. Kouba ; S. Malikić ; M. Modak ; C. R. Pranesacher ; and the proposer. We present the solution of Madhav Modak, modified by the editor. We note that [A B C ] = [A IB ] + [B IC ] + [C IA ]. Since AIB = 8 (A + B) = 9 + C, we have with usual notation, [A IB ] = A I B I sin(9 + C) = (a + AI)(b + BI) cos(c/) = (ab + b AI + a BI + AI BI) cos(c/). () The Law of Sines for AIB gives AI/ sin(b/) = c/ cos(c/), so that b AI cos(c/) = sin(b/) sin(b/) bc sin A = [ABC] sin A sin A. Similarly, sin(a/) a BI cos(c/) = [ABC] sin B. Hence () can be written as : [A IB ] = [ABC] cos(c/) sin C sin(b/) + [ABC] sin A + [ABC] sin(a/) sin B + [AIB]. We have similar expressions for [B IC ] and [C IA ]. Adding gives [A B C ] =[A IB ] + [B IC ] + [C IA ] =[ABC] (E + E + ), Copyright c Canadian Mathematical Society, 5

7 38/ SOLUTIONS where E = cos(a/) sin A + cos(b/) sin B + cos(c/) sin C, sin(b/) + sin(c/) sin(c/) + sin(a/) E = + sin A sin B Hence [A B C ] [ABC] + sin(a/) + sin(b/). sin C = E + E +. () We now prove that E 3 and E 3, which will prove the claim. First, for E we have E = sin(a/) + sin(b/) + sin(c/) = The convexity of f(x) = csc(x/) on (, π) implies that csc(a/). cyclic E = [f(a) + f(b) + f(c)] 3 f[(a + B + C)/3] = 3 csc(π/6) = 3. Next, by the AM-GM inequality, ñ sin (A/) sin (B/) sin (C/) E 6 sin A sin B sin C ï = 6 64 cos (A/) cos (B/) cos (C/) ï ò /3 = 3. cos(a/) cos(b/) cos(c/) cyclic ô /6 ò /6 Another application of the AM-GM inequality gives cos(a/) cos(b/) cos(c/) 3 cos(a/), 7 and the concavity of the function g(x) = cos(x/) on (, π) gives cyclic Hence, cos( A ) = g(a)+g(b)+g(c) 3 g[(a+b+c)/3)] = 3 cos(π/6) = 3 3. implying that E 3. cos(a/) cos(b/) cos(c/) 3 3 8, Crux Mathematicorum, Vol. 4(), January 4

8 SOLUTIONS / Proposed by Michel Bataille. Let triangle ABC with angles α, β, γ 9 be inscribed in a circle with centre O and radius R, and let U, V, W be the centres of the hyperbolas with parameter R, focus O and associated directrices BC, CA, AB, respectively. Prove that where [ ] denotes the area. [UV W ] [ABC] = R 4 (cos α + cos β + cos γ), Solved by C.R. Pranesachar ; D. Văcaru, and the proposer. We present the solution by C.R. Pranesachar. Because the letters a, b, c are reserved for the sides of the given triangle ABC, we shall let x a + y b = be the equation of the hyperbola whose centre is U, eccentricity is e, and directrix (corresponding to its focus O) is BC. If OU intersects the directrix BC at D, then O = (a e, ), D = Ä a e, ä, and OD = OU DU = a ( e ) = R cos α. () e The parameter of the hyperbola (which equals half the length of its latus rectum) equals R = b a, or (because b = a ( e ) ), a ( e ) = R. () From () and () we have e = sec α and a = R cot α, so that Similarly, OV = R cos β sin β OU = a e = R cos α sin α. and OW = R cos γ sin γ. But UOV = α + β and sin(α + β) = sin γ, so that [OUV ] = OU OV sin γ, etc. Consequently, [UV W ] = [OUV ] + [OV W ] + [OW U] = cyclic R cos α cos β sin α sin sin γ. β Letting F = [ABC] and using the formulae cos α = b +c a bc has [UV W ] = c (b + c a )(c + a b ). R 64F 3 cyclic, sin α = F bc, etc., one Copyright c Canadian Mathematical Society, 5

9 4/ SOLUTIONS Since F = [ABC] = abc 4R, we have after simplification, [UV W ] [ABC] = But R4 4a b c (a6 +b 6 +c 6 a 4 (b +c ) b 4 (c +a ) c 4 (a +b )+6a b c ). cos α + cos β + cos γ = = (b + c a a ) a (b + c a a ) 4b c = 4a b c Thus, as desired. cyclic = a6 + b 6 + c 6 a 4 (b + c ) b 4 (c + a ) c 4 (a + b ) + 6a b c 4a b c. [UV W ] [ABC] = R 4 (cos α + cos β + cos γ), 387. Proposed by George Apostolopoulos. Let ABC be a triangle with incentre I through which an arbitrary line passes meeting sides AB and AC at the points D and E respectively. Show that where r denotes the inradius of ABC. r AD + AE Solved by A. Alt ; M. Amengual Covas ; AN-anduud Problem Solving Group ; Š. Arslanagić ; R. Barroso Campos ; M. Bataille ; R. Barbara ; D. M. Bătineţu-Giurgiu, N. Stanciu, and T. Zvonaru ; C. Curtis ; P. De ; O. Geupel ; D. Jonsson ; O. Kouba ; S. Malikić ; M. Modak ; C.R. Pranesachar ; C. M. Quang ; C. Sánchez-Rubio ; E. Swylan ; G. Tsapakidis ; D. Văcaru ; P.Y. Woo ; and the proposer. We present the solution by Oliver Geupel. Denoting area by [ ], we have AD AE AD AE sin A = [ADE] = [ADI] + [AEI] = AD r + AE r. Crux Mathematicorum, Vol. 4(), January 4

SOLUTIONS / 4 Hence AD AE r AD + AE and the result follows immediately. The equality holds if and only if A is a right angle. 388. Proposed by Mehmet Şahin.

10 SOLUTIONS / 4 Hence AD AE r AD + AE and the result follows immediately. The equality holds if and only if A is a right angle Proposed by Mehmet Şahin. Let ABC be a triangle with area ; circumradius R ; exradii r a, r b, r c ; and excenters I a, I b, I c. The excircle with centre I a touches the sides of ABC at K, L, and M. Let represent the area of triangle KLM and let and 3 be similarly defined. Prove that = r a + r b + r c. R Solved by A. Alt ; M. Amengual Covas ; Š. Arslanagić ; M. Bataille ; P. De ; O. Geupel ; J. Heuver ; O. Kouba ; S. Malikić ; C.R. Pranesachar ; C. Sánchez-Rubio ; G. Tsapakidis ; D. Văcaru ; P. Y. Woo ; T. Zvonaru ; and the proposer. We present a composite of similar solutions by Arkady Alt, Miguel Amengual Covas, and Oliver Geupel. We use the common notation a = BC, b = CA, c = AB, s = a + b + c. Since quadrilaterals MI a KB, I a LCK, and MI a LA are cyclic, we have It follows that MI a K = B, KI a L = C, and MI a L = 8 A. = [I a KM] + [I a LK] [I a LM] = r a (sin B + sin C sin(8 A)) = r a (sin B + sin C sin A) = r a b + c a = r a R R r a(s a) = r a R. Copyright c Canadian Mathematical Society, 5

11 4/ SOLUTIONS Analogously, hence the result. = r b R, 3 = r c R, 389. Proposed by Michel Bataille. For positive real numbers x, y, let Prove that G(x, y) = xy, A(x, y) = x + y, Q(x, y) = G(x x, y y ) (Q(x, y)) A(x,y). x + y. Solved by AN-anduud Problem Solving Group ; R. Boukharfane ; C. Curtis ; P. Deiermann and H. Wang ; O. Kouba ; K. W. Lau ; P. Perfetti ; D. Smith ; and the proposer. One incorrect solution was received. We present the solution by Paolo Perfetti. The given inequality is equivalent to Ç å x+y x x y x + y x x x y x + y x+y y x+y, which upon being divided by x becomes y y x+y x y x+y Å ( y ã +. () x) Without loss of generality, we assume that x y. Let t = y x. Then t, y x+y = t +t and () becomes t t +t + t To prove (), let f(t) = t find : +t t ln t ln + t Å + t ã. () +t ln t ln, t. Then by routine calculations, we f (t) = Å t + ( + t ) ln t ( + t) ( + t ) ã. We claim that t + ( + t ) ln t for all t. (3) Let h(t) = ln t t + t = ln t + + t. Then h (t) = t 4t ( + t ) = ( t ) t( + t ), Crux Mathematicorum, Vol. 4(), January 4

12 SOLUTIONS / 43 so h(t) is an increasing function. Since h() =, we have h(t), from which (3) follows. Hence, f (t), which implies that f(t) is an increasing function. Since f() =, we conclude that f(t) for all t, which establishes () and completes the proof. 38. Proposed by Ovidiu Furdui. Let k > be a positive real number. Find the value of x k dxdy, y where {a} = a a denotes the fractional part of a. Solved by Š. Arslanagić ; R. Boukharfane ; C. Curtis ; O. Geupel ; R. I. Hess ; O. Kouba ; J. Ling ; D. Stone and J. Hawkins ; and the proposer. One incorrect solution was received, although the error was of a purely algebraic variety. We present two solutions. Solution, by Oliver Geupel. ß x For x, let f(x) = dy. Let y = x y t, then dy = x dt and we obtain : t {t} f(x) = x x t dt = x = x ( ( x dt t + lim n log x + lim n l= n l= = x log x + x lim n = x log x + x( γ). Hence, f(x k ) = x k ( γ) kx k log x. n l+ l t l t dt ) Å log(l + ) log l + Å log(n + ) Let I be the integral to be evaluated. Then we have : I = f(x k ) dx = γ k + k = γ k + k lim u + = γ k + k lim u + = γ k + + k (k + ), x k log x dx u x k log x dx ñ k + xk+ log x l ã ) l + Å n + ô xk+ (k + ) u ãã Copyright c Canadian Mathematical Society, 5

13 44/ SOLUTIONS because lim u + uk+ log u = by l Hôpital s rule. Solution, by Chip Curtis slightly modified by the editor. We consider this as an integral over a two-dimensional region, and perform a change of variables. Let u = xk y, v = y; let α = k, so that we obtain x = (uv)α. The Jacobian is given by : (x, y) (u, v) = αuα v α αu α v α = αuα v α. The image S of the region R = [, ] [, ] is ß S = (u, v) : v and u v ß = {(u, v) : u and v } (u, v) : u and v. u The integral then becomes : I(k) = α {u}u α v α dvdu = α = α S {u}u α v α dvdu + α u α v α dvdu + α u {u}u α v α dvdu Ç å {u}u α u v α dv du Å ã α = (α + ) + α {u}u α (α + )u α+ du α = (α + ) + α {u}u du α + α = (α + ) + α ( γ) α + k = (k + ) + γ k +, where the integral at the end of the computation is part of the one performed in the previous solution, presented above. Editor s comments. Two main methods of proof were utilized. One method involved using a substitution ; some did a D swap just to compute the inside integral, and others used a full D change of variables, complete with Jacobian factor. The other method featured a computation of the branches of { xk y } in the unit Crux Mathematicorum, Vol. 4(), January 4

14 SOLUTIONS / 45 square, then summing the integrals over each region. The reader may wish to prove that the limit defining the Euler-Mascheroni number γ is convergent. Two comments regarding this problem were received, both stating that the problem and the proposer s solution appear (on pages 4 and 9, respectively) in the proposer s 3 book, Limits, Series, and Fractional Part Integrals, published by Springer. A Taste Of Mathematics Aime-T-On les Mathématiques ATOM ATOM Volume I : Mathematical Olympiads Correspondence Program (995-96) by Edward J. Barbeau This volume contains the problems and solutions from the Mathematical Olympiads Correspondence Program. This program has several purposes. It provides students with practice at solving and writing up solutions to Olympiad-level problems, it helps to prepare student for the Canadian Mathematical Olympiad and it is a partial criterion for the selection of the Canadian IMO team. There are currently 3 booklets in the series. For information on tiles in this series and how to order, visit the ATOM page on the CMS website : Copyright c Canadian Mathematical Society, 5

SOLUTIONS Proposed by Marcel Chiriţă. SOLUTIONS / 217

SOLUTIONS Proposed by Marcel Chiriţă. SOLUTIONS / 217 SOLUTIONS / 7 SOLUTIONS No problem is ever permanently closed. The editor is always pleased to consider for publication new solutions or new insights on past problems. The editor would like to acknowledge