SOLUTIONS Proposed by Paolo Perfetti.

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1 60/ 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 Statements of the problems in this section originally appear in 07: 4(6, p Proposed by Paolo Perfetti Let 8x y and z 0 be the equations of a parabola in the (x, y plane of R Let C be the cone of vertex in P (0, 0, 4 and generate the segments from P to the points of the parabola Let S be the sphere of equation x + y + z 4z 0 a Calculate the area of that portion of C inside S b Calculate the area of that portion of S inside C We received correct solutions We present the solution by the proposer Note that the proposer solves the slightly more general problem in which the equation of the parabola is ax y a The parametric equations of the cone are (x, y, z (ϕ, ϕ, ϕ (t, v, where ï v (ϕ, ϕ, ϕ (t, v (0, 0, a + t a, v, 0 ò (0, 0, a tv a, tv, a ta, where 0 t, < v < + The point (x, y, z of the cone lies inside S iff Now, t v 4 4a + v t + a ( t a ( t t a ( v a + a < ϕ t, ϕ t, ϕ ϕ t v, ϕ v, ϕ ate tve v t v a e, where e k is the kth standard unit vector in R Thus (ϕ, ϕ, ϕ t (ϕ, ϕ, ϕ v a t + t v + v4 t 4a The area of C inside S is given by the integral v t a + ta dv a 0 ( v a +a v a + a t dt 8a7 dv (v + a π a 6 Crux Mathematicorum, Vol 44(6, June 08

2 SOLUTIONS /6 b We adopt stereographic parametric equations for the sphere x ϕ (u, v a u u + v + a, y a v ϕ (u, v u + v + a, z ϕ (u, v a(u + v u + v + a The point (x, y, z of the sphere lies inside the cone C if and only if v au with u (, ϕ ϕ so that u, ϕ u, ϕ u ï a 6 u (u + v + a The area of S inside C is ò e + v, ϕ v, ϕ v ï a 6 v (u + v + a (ϕ, ϕ, ϕ u (ϕ, ϕ, ϕ v a 4 dv v a ò e + du (u + v + a, ï a 4 (a u v ò (a + v + u e, a 4 (u + v + a which we claim is πa 4 To prove it we introduce polar coordinates in the plane (u, v We have u r cos ϑ, v r sin ϑ, π/ ϑ π/, and v au becomes r a cos ϑ/(sin ϑ Thus the integral is a 4 π/ that is, π/ a cos ϑ sin dϑ ϑ 0 Evaluating this integral yields r π/ (r + a dr a4 π/ a π π/ a π/ 4 cos ϑ sin 4 ϑ + dϑ a π ( π a π a π 4 4 Ç å a dϑ; 4a cos ϑ + a sin 4 ϑ 45 Proposed by Leonard Giugiuc and Marian Cucoanes Let ABCD be a tetrahedron with BAC CAD DAB 60 Denote by R a, R b, R c the circumradii of the triangles BAC, CAD and DAB, respectively Prove that R a + R b + R c AB + AC + BC We received correct solutions We present the solution by Oliver Geupel Editor s comment The statement above contain a typo that was corrected in the proposers solution The intended result (proven below is as following: Copyright c Canadian Mathematical Society, 08

3 6/ SOLUTIONS Let ABCD be a tetrahedron with BAC CAD DAB 60 Denote by R b, R c, and R d the circumradii of the triangles CAD, DAB, and BAC, respectively Then, R b + R c + R d AB + AC + AD Let b AB, c AC, and d AD By the law of cosines we have CD R b sin CAD c + d cd cos CAD c cd + d sin CAD with similar identities for the other circumradii Applying the Cauchy-Schwarz inequality, we deduce and two similar inequalities Finally, ( R b R c d c + 4 c ( d c d b + 4 bc d + bc cd db, d b + 4 b (R b + R c + R d (R b R c + R c R d + R d R b d + bc cd db cyc Hence the result 45 Proposed by Titu Zvonaru b + c + d Let ABC be a triangle with A 90 and 45 < C < 60 Let M be the midpoint of BC The perpendicular from C to AM intersects the leg AB at D On the side AC we take a point E and let K be the intersection of the lines CD and BE If BK AE, then prove that the triangle CEK is isosceles We received seven submissions, all correct Chow We present the solution by Steven Let (0, 0 A, (b, 0 B, (0, c C and (0, e E such that b, c, e > 0 Then CD is y b cx + c and BE is y e b (x b Since K is CD BE, by solving the previous system of equations we get the coordinates of the point K: bc (c e (b + c (b c e b, ce b ce Crux Mathematicorum, Vol 44(6, June 08

4 SOLUTIONS /6 Therefore, (e (AE b (b + c (b c (b + c (b c e BK b + ce b, ce and expanding, simplifying, and factoring reduces the equation to ( ce b + c ( ce ( b + c e + b ce + b 4 b c 0 If ce b + c 0, and AE BK, from here we have CK EK Let f (x cx ( b + c x + b cx + b 4 b c for all x Since C > 45, b > c > 0 so 0 < b 4 b c f (0, and as x, f (x, so f (x has a root less than 0 Since f (c ( b c > 0 and f (b b (b c < 0, and as x, f (x, f (x cannot have a positive root less than c 454 Proposed by George Apostolopoulos Let ABC be a triangle Prove that sin A sin B sin C (sin A + sin B cyc sin C 4 We received 0 submissions, all correct, and we present the solution by Michel Bataille Let S denote the left side of the given inequality From sin A+sin B sin A+B we deduce that cos A B cos C cos A B and sin C sin C cos C, so (sin A + sin B sin C cot C cos A B (sin A + sin B sin C tan A cyclic cyclic cot C, cos A cos B cos C sin A sin B sin C since it is well known that tan A cyclic tan B Since it is also well known that cos A cos B cos C 8 [Ed: cf eg, item 8 on p6 of the book Geometric Inequalities by O Bottema et al], it follows that Ç å S 8 4, completing the proof Copyright c Canadian Mathematical Society, 08

5 64/ SOLUTIONS Editor s comments By using the concavity of sin x and Jensen s Inequality, Bailey, Campbell, and Diminnie (jointly gave a slight improvement to the inequality by proving that S (sin A + sin B + sin C, which implies the given inequality since it is well known [see item on p 8 of the book mentioned in the solution presented above] that sin A + sin B + sin C 455 Proposed by Michel Bataille Let ABC be an isosceles triangle with AB AC and P a point of its circumcircle (with P A The reflection about AP of the circle with diameter AB intersects the circle with diameter AP at A and Q Prove that AQ and QC are perpendicular Five complete and four incomplete solutions were obtained One other submission was incorrect One of the solvers used barycentric coordinates We present two solutions Solution, by Ivko Dimitrić Let A, B, C be the angles of triangle ABC at the eponymous vertices Let Γ be the circle with AB as diameter, Γ its reflection about AP and Γ the circle with diameter AP Let R be the point of intersection of Γ and Γ other than A Then ARP 90 ARB, so that B, P, R are collinear Since Γ and Γ are reflections of each other about AP and Γ is its own reflection, Q and R are reflected images so that AP Q AP R We note that P should be distinct not only from A but from B as well When P B, the three circles Γ, Γ and Γ coincide and Q is indeterminate If P C, then QC QP is perpendicular to AQ If P belongs to one of the open arcs AC or BC that does not contain the third vertex of the triangle, then B and R are on the same side of AP (and Q is on the other side Thus, AP R AP B C and AP Q AP R C B, so that in the cyclic kite ARP Q, QAR 80 QP R 80 (B + C A If P belongs to the open arc AB, then B and Q are on one side of AP and R is on the other, so that P is between B and R Then AP R is an exterior angle of triangle ABP so that AP R ABP + P AB ACP + P CB ACB C Then AP Q C B and hence QAR 80 (B + C A, again In all cases, the counterclockwise rotation about A through angle A takes the points A, R, B to the points A, Q, C Thus AQC ARB 90, so that AQ QC We note that, since AQC AQP 90, the points Q, P and C are collinear Crux Mathematicorum, Vol 44(6, June 08

6 SOLUTIONS /65 Solution, by Steven Chow There are essentially three configurations according as P lies on one of the short arcs AB, BC and CA of the circumcircle of triangle ABC To unify their treatment, we use directed angles with addition modulo 80 Thus UV W is equal to W V U or 80 W V U, depending on the situation Thus, when U, V, X, Y are concyclic, we can always write UXV UY V even if X and Y are on opposite sides of U V (and the two angles are measured in opposite senses The reflection Q of Q about AP lies on the circles with diameters AB and AP, so that BQ P BQ A + AQ P so that B, P and Q are collinear Thus Therefore AP Q Q P A BP A BCA CP Q CP B + BP A + AP Q CAB + BCA + BCA CAB + BCA + ABC 0, so that C, P and Q are collinear Since AQ QP, then AQ QC Editor s comments The reflection in AP of the circle with diameter AB coincides with the reflection about AQ of circle with diameter AC Is there a nice way of seeing this? 456 Proposed by Daniel Sitaru Let a, b, c R such that a + b + c Prove that e b e a b a + ec e b c b + ea e c a c > 4 We received solutions, of which were correct and complete We present solutions Solution, by Dionne Bailey, Elsie Campbell, and Charles Diminnie We will prove the slight improvement that e b e a b a + ec e b c b + ea e c a c > e > 4 for distinct a, b, c R, which satisfy the condition a + b + c Note first that the last inequality follows from the fact that < e ( 7 Copyright c Canadian Mathematical Society, 08

7 66/ SOLUTIONS For the remainder of our solution, we will utilize Hadamard s Inequality which states that if f (x is continuous and convex on [p, q], then q p q p ( p + q f (x dx f ( A proof of this result can be found in R P Boas, Jr, A Primer of Real Functions (rd ed, Carus Mathematical Monograph No, The Mathematical Association of America, 98, pg 74 Since a and b must be distinct and e b e a b a ea e b a b, we may assume without loss of generality that a < b Then, since f (x e x is continuous and convex on R, ( implies that Similar arguments show that e b e a b a b a b a e x dx e a+b ( e c e b c b e b+c and e a e c a c e a+c (4 Further, because f (x e x is strictly convex on R, Jensen s Theorem and the distinct values of a, b, and c imply that e a+b + e b+c + e a+c > e ( a+b + b+c + a+c e a+b+c e (5 Finally, it follows from (, (, (4, and (5 that e b e a b a + ec e b c b + ea e c a c e a+b + e b+c + e a+c > e 4 > 4 Solution, by M Bello, M Benito, Ó Ciaurri, E Fernández, and L Roncal We prove a more general result Let a, a,, a n R such that a + a + + a n, then k e a k+ e a k a k+ a k ne /n, ( with a n+ a Moreover, the equality holds if and only if a i n, for i,, n (in this case the left hand side has to be understood as a limit Crux Mathematicorum, Vol 44(6, June 08

8 SOLUTIONS /67 The proposed inequality follows taking n, a a, a b, and a c and using that e / > 4 Let us prove ( From the inequality sinh x x, for x R, with equality for x 0 only, taking x (a k+ a k /, we deduce that e a k+ e a k a k+ a k e (a k++a k /, with equality when a k+ a k In this way, applying the AM-GM inequality, we have e a k+ e a k e (a k++a k / ne (a+ +an/n ne /n a k+ a k k k and the equality holds when a i /n, for i,, n, only Solution, by Paul Bracken By Taylor s theorem, we have the expansion with remainder e b e a + e a (b a + ea (b a + eτ 6 (b a, where τ in the remainder is between a and b This implies that e b e a b a ea + ea (b a + eτ 6 (b a e a + ea (b a, since e τ > 0 and (b a 0 always holds In exactly the same way, we obtain the inequalities e c e b c b e a e c a c eb + eb (c b + eτ 6 (c b e b + eb (c b, ec + ec (a c + eτ 6 (a c e c + ec (a c Adding these three results, the following lower bound for the function in ( is obtained, h(a, b, c eb e a e b e c b a +ec c b +ea a c ea +e b +e c + (ea (b a+e b (c b+e c (a c ( This result holds for all a, b, c and is independent of the constraint which has not been used Let us minimize the function on the right of (, f(a, b, c e a + e b + e c + (eb (c b + e b (c b + e c (a c, by introducing a Lagrange multiplier λ L f(a, b, c λ(a + b + c Copyright c Canadian Mathematical Society, 08

9 68/ SOLUTIONS Differentiating L with respect to a, b, c and λ, the following nonlinear system results, e a + e c + e a (b a λ 0, e b + e a + e b (c b λ 0, ( e c + e b + e c (a c λ 0, a + b + c 0 This set of equations maps into itself under a cyclic permutation of the variables The first three equations of ( can be put in the form, + b a + e c a λe a, + c b + e a b λe b, + a c + e c b λe c For example, adding these three equations, an expression for λ results, λ ea b + e c a + e c b + (e a + e b + e c In fact, the solution to the system ( is given by a b c, λ e/ The minimum value of f is found to be f,, e / > 4 (4 This will correspond to a minimum since a maximum is not expected Take for example a N, b N + and c 0, then e N as N, so h can be made as large as we please Combining ( and (4, these imply ( Letting c b and then b a in h and the constraint, or using Taylor s formula, it can be seen that h reduces to e / which matches the minimum (4 Thus the absolute minimum of h under the constraint is e / 457 Proposed by Leonard Giugiuc and Dan Stefan Marinescu Calculate the following limit lim n Ç n(n+ å!! n! n We received 9 submissions, all correct, and we present the solution by the Missouri State University Problem Solving Group, modified slightly by the editor Let A n n(n+!! n! n Crux Mathematicorum, Vol 44(6, June 08

10 SOLUTIONS /69 Then where and log A n n(n + n(n + n(n + n(n + n(n + log k! log n k (n k + log k log n k ( n n(n + (n k + log k log n k ( n ( n (n k + log k k log n k ( n (n k + log k k ( n (n k + log k n(n + n k ( n n k + log k n + n n k k log k n + n n k k n + n k k (n k + log n k log k n + n log k n n + B n + n + C n ( B n k C n n Note that for all n, we have C n 0 and so Hence, C n > n k log k n n log k n k log n log log n, n k log n n + n + C n 0 lim n n + C n 0 ( Copyright c Canadian Mathematical Society, 08

11 70/ SOLUTIONS On the other hand, lim n n + B n lim n From ( to ( we then obtain lim A n exp n e 4 n n + n B n lim n n B n lim k n n 0 lim n k log k n n ( x log xdx 4 ( n + B n + lim n n + C n 458 Proposed by Mihaela Berindeanu Let ABC be an acute triangle with circumcircle O, orthocentre H, D BC, AD BC, E AC, BE AC Define points F and G to be the fourth vertices of parallelograms CADF and CBEG If X is the midpoint of F G, and Y is the point where XC intersects the circumcircle again, prove that AHBY is a parallelogram All ten submissions were complete and correct They came in two varieties, so we feature an example of each Crux Mathematicorum, Vol 44(6, June 08

12 SOLUTIONS /7 Solution, by Oliver Geupel If the points C, O, and X are collinear then the line segment CY is a diameter of the circumcircle (O of triangle ABC, and by Thales Theorem the angles Y AC and CBY are right angles This implies that AY EB and BY DA; that is, the quadrilateral AHBY is a parallelogram So all we have to do is to prove that the points C, O, and X are collinear Consider the problem in the plane of complex numbers where (O is the unit circle, and capital letters serve as the complex numbers that represent their corresponding points It is an easily verifiable fact that the foot of the perpendicular from an arbitrary point Z to the chord through points P and Q of the unit circle is the point (P + Q + Z P Q Z/ Hence, D A + B + C BC A and Similarly, Thus, F C + D A G X (F + G 4 A + B + C BC A A B + C AC B 6 B A A C B The complex number B A + A B is a real number because it is equal to its complex conjugate Hence the complex number X is a real multiple of the complex number C, which proves that the points C, O, and X are collinear Solution, by Titu Zvonaru As usual, we let a BC, b CA, c AB, h a AD, h b BE, and we use square brackets to represent areas Suppose that OC intersects F G at X Since ACG 90 and AOC B, we have OCA 90 B and X CG B Similarly, F CX A It follows that [GCX ] [CF X ] CG CX sin X CG CF CX sin F CX h b sin B h a sin A bh b ah a We deduce that the point X coincides with X, the midpoint of F G, whence CY is a diameter of the circumcircle Consequently, Y A AC, implying that Y A BH; and Y B BC, implying that Y B AH; hence the quadrilateral AHBY is a parallelogram Editor s comments Muralidharan observed that there is no need to place restrictions on the angles of the given triangle: no matter what the angles of ABC Copyright c Canadian Mathematical Society, 08

13 7/ SOLUTIONS might be, AHBY will be a parallelogram, although if there is a right angle at A or B then the resulting parallelogram will degenerate into the line segment AB Either of the featured solutions apply to an arbitrary triangle (although in Solution, angles should be interpreted as directed angles 459 Proposed by Mihály Bencze Prove that ( n k p p k k p p n + n We received correct solutions We present the solution by AN-anduud Problem Solving Group The result is clear for n Let n > We have A k k p p k p p + k p p k p p > + + k k p p p p On the other hand, we get k p p > k + k + k + + k p p k p p A n > n + n, n A n > (n, A > 4, A >, A, which implies that n k A k > n + n Crux Mathematicorum, Vol 44(6, June 08

14 SOLUTIONS /7 460 Proposed by Leonard Giugiuc and Diana Trailescu Let a i, i,, 6 be positive numbers such that a + a + a a 4 a 5 a 6 and a + a + a + a 4 + a 5 + a 6 9 Prove that a a a a 4 a 5 a 6 We received 8 solutions Solving Group We present the solution by the AN-anduud Problem Using the Cauchy-Schwarz inequality, we have» + a 4 + a 5 + a 6 a + a + a ( + + (a + a + a» (a + a + a» (9 (a 4 + a 5 + a 6» 7 ( + + (a 4 + a 5 + a 6» 7 (a 4 + a 5 + a 6 Hence ( + (a 4 + a 5 + a 6 7 (a 4 + a 5 + a 6 (a 4 + a 5 + a 6 + (a 4 + a 5 + a a 4 + a 5 + a a 4 + a 5 + a Applying the AM-GM inequality, we have ( From (, we get a a 4 a 5 a 4 + a 5 + a ( a + a + a + (a 4 + a 5 + a which implies that a + a + a Using the AM-GM inequality, we get , a a a a + a + a + 5 ( Copyright c Canadian Mathematical Society, 08

74/ SOLUTIONS From ( and (, we have or a a a a 4 a 5 a 6 + 5 a a a a 4 a 5 a 6 + 5

15 74/ SOLUTIONS From ( and (, we have or a a a a 4 a 5 a a a a a 4 a 5 a Equality holds only when a a a + 5, a 4 a 5 a Crux Mathematicorum, Vol 44(6, June 08

Originally problem 21 from Demi-finale du Concours Maxi de Mathématiques de

Originally problem 21 from Demi-finale du Concours Maxi de Mathématiques de 5/ THE CONTEST CORNER Originally problem 6 from Demi-finale du Concours Maxi de Mathématiques de Belgique 00. There were eight solution submitted for this question, all essentially the same. By the Pythagorean