SOLUTIONS. x 2 and BV 2 = x

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1 SOLUTIONS /3 SOLUTIONS No problem is ever permanently closed The editor is always pleased to consider new solutions or new insights on past problems Statements of the problems in this section originally appear in 04: 40(3), p Proposed by Michel Bataille Let AUV W be a rectangle with UV In each of the following cases (a) V W 8 and (b) V W 9, is it possible to construct with ruler and compass points B on ray [AU) and C on ray [AW ) such that V and the orthogonal projection of A onto BC are symmetric about the midpoint of BC? We received six correct submissions We present the solution by Titu Zvonaru With D denoting the orthogonal projection of A onto BC, the constructed figure will consist of a right triangle ABC with points V and D on the hypotenuse BC so that BV CD Denoting a AU and x UB, we obtain by similitude ( ABC UBV DAC), Consequently, AC x + a x and CD AC BC BC (x + a) + Now, BV CD then implies that (x + a) x (x + a) (x + ) x and BV x + x (x + a)4 x ( x + a ) + x 4 (x + a) (x + ), or (x + ), x which (because all lengths must be positive) reduces to x 3 a Case a) If a 8, then x and yes, we can construct the point B with ruler and compass so that UB UV (and then C is the point where BV intersects AW ) Case b) If a 9, then we cannot construct with ruler and compass the required points B and C (A segment whose length is the cube root of a rational number cannot be constructed unless it is rational, but 3 9 is not rational) 39 Proposed by Marcel Chiriţă Let M be a point inside a triangle ABC Show that (x + y + z) 9 79xyz a b c,

2 4/ SOLUTIONS where MA x, MB y, MC z and BC a, AC b, AB c We received four correct solutions and one comment We present the solution by AN-anduud Problem Solving Group Observe that Å MA + MB + ã Å MC MA + MB + ã MC 0 x y z x y z Since MA MB xy(cos AMB) (/)(x + y c ), with similar expressions for other dot products, we obtain that 3 + xy (x + y c ) + yz (y + z a ) + zx (z + x b ) 0 Multiplying by xyz leads to (xy + yz + zx)(x + y + z) xa + yb + zc Since x + y + z xy + yz + zx, then (x + y + z) 3(xy + yz + zx), and (x + y + z) 3 (xy + yz + zx)(x + y + z) xa + yb + zc 3 3 a 3 b c xyz Cubing this inequality yields the desired result Editor s Comment Two of the other solvers evaluated a, b, c in terms of x, y, z using the Law of Cosines and reduced the problem to showing that the sum of the cosines of the three angles at M was not less than 3/ Michel Bataille noted that this is Problem J34 in Mathematical Reflections (3) 05, whose solution appears in the same journal (4) Proposed by George Apostolopoulos Prove that in any triangle ABC, sin 3 A sin 3 A + cos3 A + sin 3 B sin 3 B + cos3 B + sin 3 C sin 3 C + cos3 C 3R (r + s), where s, r and R are the semiperimeter, the inradius and the circumradius, respectively, of the triangle ABC Editor s Comments: We received three incorrect solutions (two of them were identical) all proving the given inequality which actually turned out to be false One solver attempted to give a counterexample which is not valid The counterexample below is by Michel Bataille To see that the given inequality is incorrect, consider an isosceles right triangle ABC with A π and side lengths, and Then using the facts that: i) sin π 8 and cos π + 8, Crux Mathematicorum, Vol 4(3), March 05

3 SOLUTIONS /5 ii) R a sin A, iii) s +, iv) r K s +, where K denotes the area of the triangle, we compute and find that the left side of the proposed inequality is 0636 while the right side is (both rounded to 4 decimal places) Arslanagić stated, without proof, that the given inequality would hold if the right side is replaced by 03R r + s 394 Proposed by Michel Bataille Let {F k } be the Fibonacci sequence defined by F 0 0, F and F k+ F k + F k for every positive integer k If m and n are positive integers with m odd and n not a multiple of 3, prove that 5F m 3 divides 5F mn + 3( ) n We received three correct solutions and one incorrect solution Solution, by Oliver Geupel Let α + 5, β 5 Let {L n } denote the Lucas sequence defined by L 0, L, and L n+ L n +L n+ for every nonnegative integer n Binet s formulas state that F n αn β n α β and L n α n + β n for every n 0 Let m and k be nonnegative integers with m odd By Binet s formulas and αβ, we obtain and furthermore 5F k + ( ) k 5 Ç å α k + β k + ( ) k α k + β k L k, () α β L m L m(k+) L mk + L m(k+) () Let P (n) denote the assertion that for every odd natural number m, L mn ß ( ) n+ (mod L m ) if n 0 (mod 3), ( ) n (mod L m ) if n 0 (mod 3) We will show by induction on n that P (n) is valid for every n 0 For the start cases, L 0 (mod L m ) and L m (mod L m ), which shows P (0) and P () Given any n, we show P (n) under the hypotheses P (n ) and P (n ) in three cases

4 6/ SOLUTIONS First consider n 0 (mod 3) By () and the induction hypothesis, we obtain L mn L m L m(n ) L m(n ) ( ) n ( ) n (3) Next let n (mod 3) Then ( ) n (mod L m ) (4) L mn ( ) n ( ) n ( ) n+ (mod L m ) Finally consider n (mod 3) Then L mn ( ) n ( ) n ( ) n+ (mod L m ) Hence P (n) follows and the induction is complete Suppose that m and n are positive integers with m odd and n not a multiple of 3 By (), we obtain 5F m 3 L m and 5F mn + 3( ) n L mn ( ) n+ By P (n), we conclude that 5F m 3 divides 5F mn + 3( ) n Solution, by CR Pranesachar, slightly expanded by the editor For all positive integers m, n with m odd and n 0 (mod 3), let Q mn 5F mn + 3( ) n 5Fm 3 Let x + 5 and y 5, such that F t xt y t 5 For any odd integer m, Furthermore 5F m 3 (x m y m ) 3 x m + y m x m y m 3 x m + y m ( ) m 3 x m + y m x m + y m x m y m 5F mn + 3( ) n (x mn y mn ) + 3( ) n x mn + y mn x mn y mn + 3(xy) mn x mn + x mn y mn + y mn We know that for any positive integer n, the polynomial x n + x n + is divisible by x +x+ if and only if n 0 (mod 3) This is obtained by a simple application of the Remainder Theorem: if ω is a nonreal cube root of unity, then (x ω) and (x ω ) are factors of x n + x n + Similarly, the polynomial x n + x n y n + y n is divisible by x +xy +y for n 0 (mod 3) as it has factors (x ωy) and (x ω y) as a polynomial in x Consequently, Q mn is a polynomial in x and y with integer coefficients Since both its numerator and denominator are symmetric polynomials in x and y, so is Q mn Hence Q mn can be expressed as a polynomial with integer Crux Mathematicorum, Vol 4(3), March 05

5 SOLUTIONS /7 coefficients in terms of x + y and xy But x + y and xy Consequently Q mn is an integer, as desired As a remark, in the case of n 0 (mod 3), the term 5F m 3 divides F mn 395 Proposed by Ilker Can Çiçek Let ABC be a scalene triangle Let D be the foot of the altitude from the vertex A Let P be the point on the segment AD (P A, P D), such that for the points E and F defined by BP AC E and CP AB F, the equality BF CD BD CE holds Let G be the intersection point of the circumcircle of the triangle DEF and the segment BC with G lying between D and C: Prove that AB + AC BC + AE if and only if BF + CG CE + BD We received three correct submissions We present the solution by Peter Y Woo By Ceva s theorem we have BF F A AE EC CD, () DB so that the given equality BF CD BD CE is equivalent to AE AF We shall now see that there is exactly one possibility for the location of P on AD: There is a unique position of E on AC and F on AB satisfying both AE AF and equation () when, as in our problem, the sides b AC and c AB have different lengths (and, consequently, CD DB) Letting x denote the common length of AE AF (and assuming, as the proposer intended, that the order of the points on BC is BDGC), we see that () reduces to x b DB c CD DB CD, which is a fixed quantity Note also that the center of any circle through E and F necessarily lies on the bisector of BAC But we can say more: The angle bisector AG is the diameter of the circumcircle of DEF (see Figure ) To see this let Γ be the circle whose diameter is AG Then Γ contains D (the foot of the altitude from A) It remains to prove that Γ intersects the sides AC

6 8/ SOLUTIONS Figure : The angle bisector AG is the diameter of the circumcircle of DEF and AB at the unique points E and F that satisfy () together with AE AF Then, as in the figure, AF G and AEG are congruent right triangles, and we let α BAC and δ DF G DAG DEG Then BF D 90 δ and BDF F AG α; hence (by the sine law applied to BDF ), BF sin BDF BD sin BF D sin α cos δ Similarly, DEC 90 + δ and CDE GDE GAE α, and CE sin CDE CD sin DEC sin α cos δ Consequently () holds for these positions of E and F, and we proved the claim that these are the points on BP and CP described in the statement of the problem As in the figure, we have F GD F AD α δ and CGE DAE α + δ, so that AB AD cos(α δ), AC AD cos(α + δ), BC BD + DC AD AD tan(α δ) + tan(α + δ), AE AD AE AG AG AD cos α cos δ The first of the given equations thereby becomes 0 AB + AC BC AE Å AD cos(α δ) + cos(α + δ) ã cos α (tan(α δ) + tan(α + δ)) cos δ Crux Mathematicorum, Vol 4(3), March 05

7 SOLUTIONS /9 After multiplying and dividing by convenient nonzero terms this becomes, in succession, 0 cos(α + δ) + cos(α δ) sin α cos α cos δ (cos α + cos δ) cos α cos δ sin α cos α (cos α + cos δ), cos δ 0 4 cos α cos δ sin α cos δ cos α cos α cos α cos δ, 0 4 cos δ 4 sin α cos δ cos α cos δ, 0 4 cos δ 4 sin α cos δ + sin α cos δ +, 0 cos δ sin α cos δ + sin α (cos δ sin α) By a similar calculation using the ratios BF GF tan(α δ), CE GF CE GE tan(α + δ), CG GF CE GE cos(α + δ), BD and GF BD AD AD AG AG GF tan(α δ) cos δ sin α, the second of the given equations becomes whence 0 BF CE + CG BD Å GF tan(α δ) tan(α + δ) + cos(α + δ) cos δ ã tan(α δ), sin α 0 sin δ + (sin α cos(α δ) cos δ cos(α + δ) sin(α δ)), sin α 0 sin α sin δ + sin α(cos α cos δ + sin α sin δ) cos δ (sin α sin δ) sin α sin δ + sin α cos α cos δ + sin α sin δ cos δ sin α cos α + 0 sin α + sin α cos δ + cos δ, 0 sin α cos δ + sin α + cos δ (cos δ sin α) cos δ sin δ, We conclude, finally, that AB + AC BC + AE and BF + CG CE + BD are both equivalent to cos δ sin α Editor s Comments It seems that after all this work, we have managed to discover yet another fascinating property of the empty set: The required condition (namely, cos δ sin α) can never occur (unless, of course, this editor has made an error, which can occur frequently) Note that in ADG we have cos δ AD AG, while in AGE we have sin α GE AG But equality would imply that AD GE Letting D be the point of the circle Γ diametrically opposed to D, we have AD GD > GE (because D is outside the circle with centre G that meets Γ in E and F ) In other

8 30/ SOLUTIONS words, there exists no triangle that satisfies all the requirements of the problem Specifically, our assumptions (namely AC > AB, BF CD BD CE, and the points along BC lie in the order BDGC) imply that AB + AC > BC + AE and BF + CG > CE + BD 396 Proposed by George Apostolopoulos Let a, b and c be positive real numbers such that a + b + c Find the minimum value of the expression a + b + a + b + b + c + b + c + a + c + a + c We received twelve correct submissions We present 3 different solutions Solution, by Michel Bataille For x, y > 0, we have (x + y ) (x + y), hence x + y + x + Å y (x + y ) + ã Å (x + y) x y + ã (xy) It follows that the given expression E satisfies E E where E Ç a + b + a b + b + c + b c + c + a + å c a Consider the function f(t) + t, t > 0 Then straightforward calculations show that f t (t) and f (t) > 0, + t ( + t ) 3/ so f is convex on (0, ) Since Jensen s inequality yields Since a + b + b + c + c + a, E Å a + b + ab + b + c bc + c + a ã Å + ca a + b + ã c a + b + c 9 a + b + c 9 by the AM-HM inequality, we finally obtain E E 8 4 It is easily checked that equality holds if and only if a b c 3 Crux Mathematicorum, Vol 4(3), March 05

9 SOLUTIONS /3 Solution, by John G Heuver By Minkowski s inequality and the AM-GM inequality, we have: a + b + a + b + b + c + b + c + a + c + a + c Å (a + b + c) + (b + c + a) + a + b + c Å + a + b + ã Å ã 3 + c 3 abc Å ã a + b + c ã Solution 3, by Titu Zvonaru Since (x + y ) (x + y), we have x + y (x + y) By the AM-HM inequality, we also have a + b + c 9 a + b + c Hence by using the triangle inequality for complex numbers, we obtain: a + b + a + b + b + c + b + c + a + c + a + c a + i a + b + i b + b + i b + c + i c + a + i a Å a + i a + b + i b + c + i ã c Å (a + b + c) + i Å + 9 a + b + c a + b + c ã c + i c ã Å (a + b + c) + a + b + ã c 397 Proposed by Marcel Chiriţă Let ABCO be a tetrahedron with the face angles at O all right angles If we denote the altitude from O by h, the inradius by r, and the angles that the lines OA, OB, OC make with the face ABC by x, y, z, show that r h(cos x + cos y + cos z) We received three correct submissions We present the solution by Titu Zvonaru

10 3/ SOLUTIONS The proposed inequality is false; we shall see that instead r( + 3) h(cos x + cos y + cos z) We use square brackets to denote areas and denote by V the volume of ABCO; letting a OA, b OB, and c OC, we have V [OAB] OC 3 [OBC] OA 3 [OCA] OB 3 Because AB a + b, BC b + c, CA c + a, we have abc 6 6[ABC] (a + b )(b + c ) + (b + c )(c + a ) + (c + a )(a + b ) (a + b ) (b + c ) (c + a ) 4(a b + b c + c a ) In terms of h and r the volume is consequently, h V [ABC] h 3 r([oab] + [OBC] + [OCA] + [ABC]) ; 3 abc abc and r a b + b c + c a ab + bc + ca + a b + b c + c a Let H be the foot of the altitude from O to the face ABC Then sin x OH OA h a, and sin x h a b c a b + b c + c a, so that cos x sin x b c a b + b c + c a Add this last equation to the analogous expressions for cos y and cos z to get We can now prove our initial claim: cos x + cos y + cos z r( + 3) h(cos x + cos y + cos z) + 3 ab + bc + ca + a b + b c + c a a b + b c + c a» 3(a b + b c + c a ) ab + bc + ca 3(a b + b c + c a ) (ab + bc + ca) (ab bc) + (bc ca) + (ca ab) 0 Equality holds if and only if a b c Crux Mathematicorum, Vol 4(3), March 05

11 SOLUTIONS /33 Editor s Comments The editors misinterpreted Chiriţă s original proposal: he intended r to be the radius of the incircle of the face ABC (not the inradius of the tetrahedron) Using an argument similar to that of the featured solution with his inradius, he correctly obtained the inequality of his proposal, namely r h(cos x + cos y + cos z), with equality if and only if a b c 398 Proposed by Michel Bataille Let A M n (C) with rank(a) and complex numbers x, x,, x n such that n x k If k Ü ê 0 x x n B x x n and I n+ is the unit matrix of size n +, prove that A Ö x è det(i n+ + B) ( x x n ) A x n We received three correct solutions We present the solution submitted by Oliver Geupel, slightly modified by the editor Since rank(a), there are complex numbers a,, a n and λ,, λ n such that Ö è Ö è λ λ a λ a n ( ) A a a n λ n λ n a λ n a n Denote by I n the unit matrix of size n We construct two intermediate matrices C and D, and relate the determinants of C, D and B to each other by showing how each matrix can be built up from the previous via elementary row operations Let á ë a a n C λ I n + A λ n ( C 0 C n ) (that is, C 0,, C n denote the columns of C); we obtain det C det ( ) C 0 C a C 0 C n a n C 0 á ë 0 0 det λ I n λ n

12 34/ SOLUTIONS Let á 0 a a n ë D x I n + A x n ( D 0 D n ) (that is, D 0,, D n denote the columns of D) Note that D i C i for i On the other hand, a simple calculation shows that D 0 x á D x n D n (a x + + a n x n ) ë λ (a x + + a n x n ) λ n (a x + + a n x n ) (a x + + a n x n )C 0 Using the properties of determinants, det D det ( ) D 0 x D x n D n D D n det ( ) (a x + + a n x n )C 0 C C n (a x + + a n x n ) det C (a x + + a n x n ) Let Ö R0 è I n+ + B R n so R 0,, R n denote the rows of I n+ + B Note that, except for the first row, the rows of I n+ + B are the same as the rows of D Moreover, R 0 x R x n R n ( 0 a (λ x + + λ n x n ) a n (λ x + + λ n x n ) ), which is (λ x + + λ n x n ) times the first row of D Note that for the above n calculation we needed to use the hypothesis that to get that the first x k k Crux Mathematicorum, Vol 4(3), March 05

13 SOLUTIONS /35 entry is zero By the properties of determinants, it follows that á R0 x R x n R n ë det(i n+ + B) det R R n (λ x + + λ n x n ) det D (λ x + + λ n x n )(a x + + a n x n ) è ( x x n ) Ö λ λ n Ö x ( a a n ) è Ö x x n è ( x x n ) A x n This completes the proof 399 Proposed by Péter Ivády Show that for all 0 < x < π/, the following inequality holds: Å + ã Å + ã ( π ) 4 ò 5 ï + x 4 sin x cos x x There were eight submissions for this problem, all of which were correct present the solution by CR Pranesachar We We shall prove that if f(x) Å + ã Å + ã ( π ) 4, and g(x) 5 + x 5 sin(x) cos(x) 4 x π, 0 < x <, then min f(x) > 58 > max g(x) 0<x< π 0<x< π We shall give a calculus-free proof Since f(x) is symmetric about the point x π 4

14 36/ SOLUTIONS in (0, π ), we may use the substitution x π 4 t, where π 4 < t < π 4 Then Ç å Ç å f(x) + sin( π 4 t) + cos( π 4 Ä t) (cos(t) sin(t)) + ä Ä (cos(t) + sin(t)) + ä sin( π 4 t) cos( π 4 t) ( + cos(t) sin(t))( + cos(t) + sin(t)) sin( π 4 t) cos( π 4 t) ( + cos(t)) sin (t) sin( π t) + cos(t) + cos(t) cos(t) + ( cos(t) + ) cos (t) + cos(t) For f(x) to be at a minimum, cos(t) is at a maximum, and so cos(t) This happens for t 0, that is, x π 4 Thus min f(x) > 3 + (4) 58 Now, the maximum of x( π π x) is 6, which is attained at x π 4, as ( π ) ( x x π π ) 6 4 x So Å π ã 4 max g(x) π Since π < 0 (this follows from the fact that π < 35), we see that: ã Å max g(x) < 5 Å ã Å + (03 ) ã 6 00 < 5 Å + ( 0) ã 6 00 ã 5 Å + 4 5( + 06) Hence the inequality follows 3930 Proposed by José Luis Díaz Barrero In a triangle ABC, let a, b and c denote the lengths of the sides BC, CA and AB Show that a sin / B 4a + b + c + b sin/ C a + 4b + c + c sin/ A a + b + 4c Crux Mathematicorum, Vol 4(3), March 05

15 SOLUTIONS /37 We received nine correct submissions Quang Let F We present the solution by Cao Minh a sin / B 4a + b + c Then by Cauchy-Schwarz inequality, we have F a 4a + b + c sin / A () Since (x + y) 4xy, we have x+y Ä 4 x + ä y for all x > 0, y > 0 Hence, a 4a + b + c a 3a + (a + b + c) Å ã a 4 3a + Å ã a + b + c a a + b + c () Since it is well known that sin A 3 3 we have, by using Cauchy-Schwarz inequality again, that Ñ é / Ñ sin / A sin A From () (3), we obtain that é / (3) therefore, F F ,