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

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1 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 Theorem we have and AC BC + AB DC AC AD CC08. In an orthonormal system, the line with equation y 5x crosses the parabola with equation y x in point A. The perpendicular to OA at O intersects the parabola at B. What is the area of triangle AOB? Originally problem 0 from Demi-finale du Concours Maxi de Mathématiques de Belgique 009. We received six correct solutions, and one incorrect solution. solution of Titu Zvonaru. We present the It is easy to deduce that A(5, 5). The slope of OB is /5. Solving the system y 5 x, y x we obtain B( 5, 5 ). It follows that OA » 6, OB 5 + area of the triangle is AOB OA OB Hence the CC09. Let E be the set of reals x for which the two sides of the following equality are defined: cot 8x cot 7x sin kx sin 8x sin 7x. If this equality holds for all the elements of E, what is the value of k? Originally problem from Demi-finale du Concours Maxi de Mathématiques de Belgique 009. We received seven submitted solutions to this problem, one of which was incorrect and five were incomplete. We present the only correct solution by Paolo Perfetti modified by the editor. Note first that E {x R x mπ 8 and x mπ 7 for any m Z. For x E, the given equality is equivalent to sin 8x sin 7x(cot 8x cot 7x) sin kx. () We shall prove that the only value of k for which () holds for all x E is k 9. Crux Mathematicorum, Vol. 4(), February 05

2 THE CONTEST CORNER /53 Since sin 8x sin 7x(cot 8x cot 7x) sin 7x cos 8x cos 7x sin 8x k 9 satisfies (). sin (7x 8x) sin 9x, Next, suppose () holds for all x E and some k Z with k 9. If k 9, then from () we have sin 9x 0 for all x E, which is false (for example, if x π 38, then x E, but sin 9x sin π 0). Hence k 9. From (), we also have ã 9 k sin x cos 9 + k ã x 0. () Since sin ( 9 k x ) 0 if and only if 9 k x mπ or x mπ 9 k and cos ( 9+k x ) 0 if and only if 9+k x (m + (m+)π )π or x 9+k for some m Z, there must be some x E that does not satisfy (). (To be more precise, the set of all x such that x mπ (m+)π 9 k or x 9+k for some m Z is countable while E is clearly uncountable.) This is a contradiction and our proof is complete. CC0. What is the number of real solutions to the equation: + x x x x x. Originally problem 6 from Demi-finale du Concours Maxi de Mathématiques de Belgique 009. We have received four correct solutions and one incorrect submission. We present the solution by Henry Ricardo. We compute the left-hand side (LHS) and the right-hand side (RHS) on three intervals that cover the real number line. Case. Suppose that 0 x. Then When x [, ], and when x (, ], RHS x ( x) x + x. + x x ( x) + x ( x) 3x + x x + x (x ) x x so that LHS ß 3x if 0 x, x if < x. Copyright c Canadian Mathematical Society, 05

3 SOLUTIONS /83 Solution, abridged version of the solution by the Missouri State University Problem Solving Group. Let A (0, 0), B (, 0), C (0, ), M (a, 0), and N (0, b), where a and b are rational and 0 < a < b <. The equations of the lines BN and CM have rational coefficients, so the coordinates of O are rational. The area of a triangle with vertices (x, y ), (x, y ), and (x 3, y 3 ) is Ñ x y det x y x 3 y 3 Therefore, the areas of MBO, BCO, CNO, and AMON are all rational. By stretching the triangle ABC, the corresponding areas can be made to be integers. Since stretching does not alter the ratios AM/M B and AN/N C, the configurations are not affinely equivalent for distinct choices of a and b. Solution 3, by Titu Zvonaru. Let a, b, m, n be positive integers and let ABC be a triangle with BC a and h A b(m + )(n + )(m + n + ). Choose the points M and N on AB and AC such that BM BA m +, é. CN CA n +. Denote by [XY... Z] the area of the polygon XY... Z. Then [BMC] [ABC] m +, [ABC] [CNB] n +. Suppose that AO intersects BC at A. By Van Aubel s Theorem for Cevian triangles we obtain AO OA AM MB + AN NC m + n and therefore OA AA /(m + n + ). It follows that [BOC] [ABC] m + n +. Thus the areas [ABC], [BMC], [CNB], and [BOC] are all integers and by taking differences of these areas so are [MBO], [CNO], and [AMON] Proposed by Nathan Soedjak. Let a, b, c be positive real numbers. Prove that ã ab ã bc ( ca ) ã ab + bc + ca c a b a + b + c There were 3 correct solutions, with two solutions from one solver, as well as a Maple verification. We present a sampling of the different approaches. Copyright c Canadian Mathematical Society, 05

4 84/ SOLUTIONS Solution, by Mohammed Aassila. Note that x + y + z xy + yz + zx and (x + y + z) 3(xy + yz + zx) for real x, y, z. The left side of the inequality is not less than b + c + a. However (a + b + c )(a + b + c) (ab + bc + ca)[3(ab + bc + ca)] 3(ab + bc + ca), and the desired result follows. Solution, by Michel Bataille. By homogeneity, we may suppose that a + b + c. equivalent to The inequality is then (ab) 4 + (bc) 4 + (ca) 4 3(a b c )(ab + bc + ca). Observe that x 4 + y 4 + z 4 4 [(x4 + x 4 + y 4 + z 4 ) + (x 4 + y 4 + y 4 + z 4 ) + (x 4 + y 4 + z 4 + z 4 )] x yz + xy z + xyz xyz(x + y + z), and (x + y + z) 3(xy + yz + zx). Applying these inequalities leads to as desired. (ab) 4 + (bc) 4 + (ca) 4 [(ab) 4 + (bc) 4 + (ca) 4 ][(a + b + c) ] [a b c (ab + bc + ca)][3(ab + bc + ca)] 3(a b c )(ab + bc + ca), Solution 3, by Dionne Bailey, Elsie Campbell, and Charles Dimminnie; Angel Plaza; Cao Minh Quang; and Edmund Swylan, independently. Since x + y + z xy + yz + zx, ab c + bc a + ca b (ab) + (bc) + (ca) abc(a + b + c) a + b + c. abc abc Using either the convexity of the function x or the inequality of the root-meansquare and arithmetic mean, we find that ã ab + c ã bc ( ca ) ab + a b 3 c + bc a + ca ã b 3 (a + b + c) (a + b + c) 4 3 (a + b + c) [3(ab + bc + ca)] 3(a + b + c) 3 ab + bc + ca a + b + c ã. Crux Mathematicorum, Vol. 4(), February 05

5 SOLUTIONS /85 Solution 4, by Paolo Perfetti. Since x y 4 z + x y z + y z x + z x ã y xy by the arithmetic-geometric means inequality, we can follow the strategy of Solution to obtain ã ab ã bc ( ca ) 3(ab + bc + ca) 3(ab + bc + ca) + + ab + bc + ca c a b 3(ab + bc + ca) (a + b + c) as desired. Solution 5 by Henry Ricardo. We have ã ab + c ã bc ( ca ) b + ïa ã a b c + c a b + b ã c + c b a + c ãò a + a b a + b + c (a + b + c) 3 ã ab + bc + ca 3. a + b + c 397. Proposed by Peter Y. Woo. Given a circle Z, its center O, and a point A on Z, with only a long unmarked ruler, and no compass, can you draw: i) points B,C and D on Z so that ABCD is a square? ii) the square AOBA? iii) the points B, W, W and W on Z such that angles AOB, AOW, AOW and AOW are 90, 60, 45 and 30? There were five correct solutions to this problem. We feature the one by the Missouri State University Problem Solving Group. We need the following basic construction: Given three collinear points A, B, C such that AB BC and a point P not on AC, we want to construct a line through P parallel to AC. To do this, we choose a point Q on the ray AP such that P is between A and Q. Denote the intersection of BQ and CP by R and denote the intersection of AR and QC by S. We claim that P S is the line we seek. By Ceva s theorem, AB BC CS SQ QP P A, Copyright c Canadian Mathematical Society, 05

6 88/ SOLUTIONS 390. Proposed by Alina Sîntǎmǎrian. Evaluate 6n + 0n + 7. (4n + )! There were 5 submitted solutions for this problem, 4 of which were correct. We present three solutions, representative of the two main solution methods utilized together with one variant. Solution, by the AN-anduud Problem Solving Group. Consider the following two power series, sin x Hence, we have and sin ( ) n x n+ (n + )!, and ex ( ) n (n + )! e Using the above considerations, we get 6n + 0n + 7 (4n + )! n n!. x n n!, x R. ã (4n + )!, (4n + 3)! (4n + )(4n + ) + (4n + ) + (4n + )! ã (4n)! + (4n + )! + (4n + )! n! + ã (4n + )! (4n + 3)! e + sin. Solution, by the group of Dionne Bailey, Elsie Campbell, and Charles Diminnie. To begin, we note that for n 0, 6n + 0n + 7 (4n + )! (4n + ) (4n + ) + (4n + ) + (4n + )! (4n)! + (4n + )! + (4n + )!, Crux Mathematicorum, Vol. 4(), February 05

7 SOLUTIONS /89 and hence, 6n + 0n + 7 (4n + )! (4n)! + (4n + )! + (4n + )! (since the Ratio Test easily confirms that each of the three series on the right converges). The remainder of this solution depends on the following known series: ( ) k sin (k + )!, cos ( ) k (k)!, Since we have ( ) k + we obtain: sinh k0 k0 if k is even 0 if k is odd k0 (k + )!, cosh (k)!. k0 and ( ) k+ + 0 if k is even if k is odd, sin + sinh cos + cosh cos + cosh k0 k0 k0 Therefore, we obtain, 6n + 0n + 7 (4n + )! ( ) k + (k + )! ( ) k + (k)! ( ) k+ + (k)! cos + cosh [(n) + ]! (4n + )!, [(n)]! (4n)!, sin + sinh + cosh sin + e e sin + e. [(n + )]! (4n + )!. + (sin + sinh ) + + e + e cos + cosh Solution 3, by Paolo Perfetti. First, we have: 6n + 0n + 7 (4n + )! (4n)! + (4n + )! + (4n + )!. Copyright c Canadian Mathematical Society, 05

8 90/ SOLUTIONS Let so that we obtain: f (x) f iv (x) n n f(x) x 4n (4n )!, f (x) n x 4n (4n)!, x 4n 4 (4n 4)! x 4n (4n)! f(x). x 4n (4n )!, f (x) n x 4n (4n )! Thus f(x) satisfies f iv (x) f(x), f(0), f (0) 0, f (0) 0, f (0) 0, whose unique solution is f(x) cosh x + cos x. Evaluating, we get Moreover, if we define f() cosh + cos g(x) x 4n+ (4n + )!, (4n)!. we get g() (4n+)! and g (x) f(x), g(0) 0. This implies Finally, defining g(x) sinh x + sin x, g() sinh + sin. h(x) x 4n+ (4n + )!, we get h() (4n+)! and h (x) g(x), h(0) 0. This implies h(x) cosh x cos x, h() cosh cos. Summing up the terms, we obtain f() + g() + h() e + sin. Editor s Comment. The presented solutions illustrate three techniques: rearrange the summations wisely to get a simple expression, rearrange the summations and then recall other atypical power series that make things work, and solve a couple of DEs to avoid having to work too much with power series. Wagon commented that the sum can be explicitly computed when the numerator is an arbitrary quadratic in n. Crux Mathematicorum, Vol. 4(), February 05