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1 Problems Ted Eisenberg, Section Editor ********************************************************* This section of the Journal offers readers an opportunity to echange interesting mathematical problems and solutions. Please send them to Ted Eisenberg, Department of Mathematics, Ben-Gurion University, Beer-Sheva, Israel or fa to: Questions concerning proposals and/or solutions can be sent to Solutions to previously stated problems can be seen at < Solutions to the problems stated in this issue should be posted before June 5, 07 55: Proposed by Kenneth Korbin, New York, NY Find the sides of a triangle with eradii 3,, 5). 56: Proposed by Arsalan Wares, Valdosta State University, Valdosta, GA Polygons ABCD, CEF G, and DGHJ are squares. Moreover, point E is on side DC, X DG EF, and Y BC JH. If GX splits square CEF G in regions whose areas are in the ratio 5:9. What part of square DGHJ is shaded? Shaded region in DGHJ is composed of the areas of triangle Y HG and trapezoid EXGC.)

2 57: Proposed by Iuliana Trască, Scornicesti, Romanai Show that if, y, and z is each a positive real number, then 6 z 3 + y z 6 y 3 y z 3 + y 3 + z y z. 58: Proposed by Yubal Barrios and Ángel Plaza, University of Las Palmas de Gran Canaria, Spain Evaluate: lim n n 0 i,j n i+jn i i ) j j ). 59: Proposed by José Luis Díaz-Barrero, Barcelona Tech, Barcelona, Spain Without the use of a computer, find the real roots of the equation ) : Proposed by Ovidiu Furdui, Technical University of Cluj-Napoca, Cluj-Napoca, Romania Let k be a positive integer. Calculate 0 0 k y k y k ddy, where a denotes the floor the integer part) of a. Solutions 57: Proposed by Kenneth Korbin, New York, NY Rationalize and simplify the fraction + ) ) if Solution by David E. Manes, SUNY at Oneonta, Oneonta, NY Let F + ) / )) and let y Then y and y ). Moreover, 07 + y 07 y, 07 y 07 + y, + 07) 07 y and

3 Therefore, y ) 07 + y) + 07 y) + 07 y 07 y) 07 + y ) 07 y). [ y ) + ) 07 y) 07 + y ] 07 y [ y ) 07 y ) ] 07 y) [ y ] 07 y) [ 0507) + y ] 07) 07 y) Substituting these values into the fraction F and simplifying, we obtain F 07) 07 y ) 07 y) 07 + y)07) ) 0507)+y 07 y) 07)) 3 07 y ) y ) 807) ) ) 807) Solution by Anthony J. Bevelacqua, University of North Dakota, Grand Forks, ND For notational convenience we set d 07 07, y 07 + d, and z 07 d. Thus our is y/z. We have + ) ) y z y z + ) ) y ) y ) ) z z z z y + z) yz06y yz + 06z ) Now y + z 07, 3

4 yz 07 d , and 06y yz + 06z 06y + z ) yz 06y + z) yz) yz 06y + z) 07yz 06 07) ) ). Hence y + z) yz06y yz + 06z ) ) Solution 3 by Jeremiah Bartz, University of North Dakota, Grand Forks, ND Let y 07 and w y y. Observe y + w y w y + y w w y y w y + y y y. Then + ) ) y y w) y w y + w 06 y+w y w ) y+w y w ) + 06 y 06y + w) 3 y w) y + w) y w) + 06y + w)y w) 3 y 05y + y w 07w ) 3 y 3 05y 3 + yw w using y 07 8y 3 05y 3 + yy y) y + y y y) 8y 3 05y 3 + y y 8y 05y + y

5 so that + ) ) 807) 0507) Solution by Arkady Alt, San Jose, CA Let a + a a a a a. Then, + a + a a a a a + a a a a + a a a + a ) a + a a ) a a a + a + a a ) a a a + a + a a + a ) a a a + a + a a a + a + + a a a + a and, therefore, + ) a ) + a )) + ) a ) + ) + a) + + ) a ) + ) + a a ) a a + a a a ) a a + a a 6a a ) a a a a + a)) a a + a) 6a a a a a) a a + a) 8a 3 a a) a 8a a a ). For a 07 we get + ) ) Solution 5 by Kee-Wai Lau, Hong Kong, China We show that Firstly we have + ) ) ) ) ) ) ) ) ) ) ) ) 5

6 ) Net, we have and + + ) ) ) ) ) + Since + ) ) 06 + ), so ) follows. Also solved by Bruno Salgueiro Fanego, Viveiro, Spain; Ed Gray, Highland Beach, FL; Telman Rashidov, Azerbaijan Medical University, Baku Azerbaijan; Boris Rays, Brooklyn, NY; Albert Stadler, Herrliberg, Switzerland; Toshihiro Shimizu, Kawasaki, Japan; David Stone and John Hawkins, Georgia Southern University, Statesboro, GA, and the proposer. 58: Proposed by Nicusor Zlota, Traian Vuia Technical College, Focsani, Romania If > 0, then [] [] + [] + {}) + {} {} + [] + {}), where [.] and {.} respectively denote the integer part and the fractional part of. Solution by Soumava Chakraborty, Kolkata, India Case : 0 < < [] 0. Therefore, LHS {} 7{} > 7. Case : [] and {} 0. Therefore, LHS [] [] >. Case 3: [] and 0 < {} <. Therefore, ) {} < [] {} < [] {} + [] + [] < 8[] LHS > [] [] + [] + {}) >, and 8 8 >. {} > 0, and therefore {} + [] + {}) 6

7 Combining the 3 cases, the LHS is always > 8 which is > Solution by Bruno Salgueiro Fanego, Viveiro, Spain Since [] + {} and [] < [] +, we have that [] + {} + {} and {} [] <, so [] + {} + {} + and, thus, since > 0, + {}) < ) ; hence, [] + + {}) < + 6 and {} + + {}) < + 6. It follows that 0 < [] + + {}) < 7 and 0 < {} + + {}) < 7 so 0 < [] + + {}) and 0 < 7 [] {]} + + {}) 0 < {} {]} + + {}) [] 7 with equality iff [] 0 and {} + + {}) and hence, 7 {} 7 with equality iff {} 0, so [] [] + [] + {}) + {} [{} + [] + {}) [] [] + + {}) + {} [{} + + {}) [] 7 + {} [] + {} with equality iff [] 0 and {} 0, that is, iff 0 < < and N, with is impossible. Hence, we have proved the more general and strict inequality [] [] + [] + {}) + {} {} + [] + {}) > 7 which implies, because >, the initial result.) Also solved by Moti Levy, Rehovot, Israel; Nirapada Pal-India, and the proposer. 59: Proposed by Titu Zvonaru, Comănesti, Romania and Neculai Stanciu, George Emil Palade School, Buzău, Romania Prove that there are infinitely many positive integers a, b such that 8a b 6a b 0. Solution by Ulrich Abel, Technische Hochschule Mittelhessen, Germany Define and g a, b) 8a 6a b b f a, b) 577a + 36b 8, 8a + 577b 0). 7

8 By direct computation we see that g f a, b)) g a, b). If g a 0, b 0 ) 0 with a 0, b 0 N then the iterates a n, b n ) f a n, b n ) are in N N and satisfy g a n, b n ) 0, for all n N. Since g, 3) 0, starting with a 0, b 0 ), 3) we obtain the infinite sequence of solutions, 3), 957, 059), 085, 68659), 783, ), , ),... Since g 5, 0) 0, starting with a 0, b 0 ) 5, 0) we obtain another infinite sequence of solutions: 5, 0), 5577, 3660), 63566, 73096), , ), , ),... Solution by Trey Smith, Angelo State University, San Angelo, TX Solution by Trey Smith, Angelo State University, San Angelo, TX We start by observing that 8a b 6a b 0 b + ) 6a ) which is suspiciously close to being Pell s Equation. Our particular equation is of the form y. Notice that 7, 5) 7 and y 5) is a solution to y. We will now create a sequence of solutions starting with c 0, d 0 ) 7, 5) in the following recursive manner. For n 0, let c n+ c 3 n + 6c n d n, d n+ 3c nd n + d 3 n. We prove the following facts regarding this sequence. Fact. For all n, c n, d n ) is a solution to y. Proof: We use induction to prove this. In the ground case, it is clear that c 0, d 0 ) 7, 5) is a solution to y. Assume that c n, d n ) is a solution. c n+ d n+ c 3 n + 6c n d n) 3c nd n + d 3 n) c 6 n + c nd n + 36c nd n 9c nd n + c nd n + d 6 n) c 6 n + c nd n + 36c nd n 8c nd n c nd n 8d 6 n c 6 n 6c nd n + c nd n 8d 6 n c n d n) 3. 8

9 For the net two facts, we use the notation q m t to represent the statement q t mod m). Fact. For all n, c n 3 and c n. Proof: We proceed by induction noting, first, that c 0 3 and c 0. Then assuming that c n 3 we have that Also, assuming that c n, we have c n+ c 3 n + 6c n d n c n+ c 3 n + 6c n d n Fact 3. For all n, d n. Proof: Clearly d 0. Assuming that d n, we have d n+ 3c nd n + d 3 n Fact. For all n, d n 3. Proof: Certainly d 0 3. Assume that for n, d n 3. Then so that d n+ 3c nd n + d 3 n , d n+) d n+ 3c n+d n+ + d 3 n Using the facts above, we show that there are infinitely many pairs a, b) that satisfy b + ) 6a ). Fi an even number m. Then c m, d m ) satisfies y. Since c m we have that c m is even and greater than 0) so that b c m is an integer. Also, d m 3 which tells us that d m + is divisible by 3, and since d m, d m + is divisible by. Hence d m + is divisible by 6. Then a d m + 6 is an integer. Thus, the pair a, b) is a solution to 8a b 6a b 0. Solution 3 by Jeremiah Bartz, University of North Dakota, Grand Forks, ND 9

10 Observe two such solutions a, b) are given by, 3) and 5, 0). We claim that if a i, b i ) is a solution in positive integers, then so is a i+, b i+ ) where a i+ 577a i + 36b i 8 b i+ 8a i + 577b i 0. To see this, note that a i+, b i+ ) are clearly positive integers and 8a i+ b i+ 6a i+ b i+ 8577a i + 36b i 8) 8a i + 577b i 0) 6577a i + 36b i 8) 8a i + 577b i 0) 8a i b i 6a i b i 0. The solutions, 3) and 5, 0) are seeds which produce two infinite families of solutions. The first four solutions in each family is given below. i a i b i a i b i Solution by Ángel Plaza, University of Las Palmas de Gran Canaria, Spain The proposed equation may be written as follows: 8 a 6 8a b 6a b 0 ) b + ) a b + 6) ) 7 a b + 6) ) b + ) 6a ). The last equation is a Pell-type equation y, by doing b + and y 6a. The smallest solution of y is, ) and therefore all its solutions ) i+ are given by i + y i +. Note that i and y i are allways odd so b is an integer. Also 6a + ) i +. Since the epression + ) i + is even and k + k + k 0 k 0 multiple of 3 for i of the form i 6m, for m integer, the proposed equation has infinitely many positive integral solutions. Solution 5 by David E. Manes, SUNY at Oneonta, NY Solution. Writing the equation as a quadratic in b, one obtains b + b 6a3a ) 0 and, since we want positive integer solutions, b + + 7a a. 0

11 Note that the above fraction is a positive integer provided that 7a a + c for some integer c. This last equation is equivalent to a negative Pell equation c d, where d 6a. This equation is solvable and the positive integer solutions are given by the odd powers of +. More precisely, if n is a positive integer and c n, d n ) is a solution of c d, then c n + d n + ) n. The problem is that not all the solutions for d n yield solutions for a n. Observe: ) if n 0 mod ), then c n 5 mod 6) and d n mod 6), ) if n mod ), then c n d n mod 6), 3) if n mod ), then c n mod 6) and d n 5 mod 6), ) if n 3 mod ), then c n d n 5 mod 6). The above observations provide straightforward inductive arguments for the following consequences. If n 0 or mod ), then there are no solutions since d n mod 6) implies no integer solution for a n. On the other hand, if n or 3 mod ), then a n d n + is a positive integer and b n + 7a 6 n a n + )/. Since there are infinitely many positive integers congruent to or 3 modulo, the result follows. Some of the infinitely many solutions are: if n, then c 7, d 5 and a, b ), 3); if n 3, then c 3, d 3 9 and a 3, b 3 ) 5, 0); if n 6, then c 6 89, d 6 57 and a 6, b 6 ) 957, 059); if n 7, then c 7 73, d and a 7, b 7 ) 5577, 3660); if n 0, then c , d and a 0, b 0 ) 085, 68659); if n, then c , d and a, b ) 63566, 73096). Also solved by Arkady Alt, San Jose, CA; Hatef I. Arshagi, Guilford Technical Community College, Jamestown, NC; Dionne Bailey, Elsie Campbell, and Charles Diminnie, Angelo State University, San Angelo, TX; Anthony J. Bevelacqua, University of North Dakota, ND; Ed Gray, Highland Beach, FL; Moti Levy, Rehovot, Israel; Kenneth Korbin, NY, NY; Kee-Wai Lau, Hong Kong, China; Albert Stadler, Herrliberg, Switzerland; Toshihiro Shimizu, Kawasaki, Japan; David Stone and John Hawkins, Georgia Southern University, Statesboro, GA, and the proposer. 530: Proposed by Oleh Faynshteyn, Leipzig, Germany Let a, b, c be the side-lengths, α, β, γ the angles, and R, r the radii respectively of the circumcircle and incircle of a triangle. Show that a 3 cosβ γ) + b 3 cosγ α) + c 3 cosα β) b + c) cos α + c + a) cos β + a + b) cos γ 6Rr. Solution by Dionne Bailey, Elsie Campbell, and Charles Diminnie, Angelo State University, San Angelo, TX By the Law of Cosines, and hence, cos α b + c a bc b + c) cos α b + c) b + c a ) bc a b + c) b + c a ). abc

12 Similarly, and Therefore, c + a) cos β b c + a) c + a b ) abc a + b) cos γ c a + b) a + b c ). abc b + c) cos α + c + a) cos β + a + b) cos γ a b + c) b + c a ) + b c + a) c + a b ) + c a + b) a + b c ) abc a bc + ab c + abc abc a + b + c. ) If K is the area of the given triangle, then K ab sin γ bc sin α ca sin β and we have Thus, sin α K bc, K K sin β, and sin γ ca ab. By Heron s Formula, Hence, a 3 cos β γ) a 3 [cos β cos γ + sin β sin γ] [ c a 3 + a b ) a + b c ) ca ab [ a b c ) ] + 6K a bc a [ a b c ) ] + 6K. abc ] + K ca) ab) 6K a + b + c) a + b c) b + c a) c + a b) [a + b) c ] [ c a b) ] a b + b c + c a ) a + b + c ). a 3 cos β γ) a [a b c ) + a b + b c + c a ) a + b + c )] abc a [ b c + a b + b c + c a )] abc Similarly, a abc b 3 cos γ α) b c + a b + b c + c a ). b c a + a b + b c + c a ) abc

13 and As a result, c 3 cos α β) c a b + a b + b c + c a ). abc a 3 cos β γ) + b 3 cos γ α) + c 3 cos α β) a b c + a b + b c + c a ) + b c a + a b + b c + c a ) abc abc + c a b + a b + b c + c a ) abc abc 6a b c 3abc. ) By ) and ), a 3 cos β γ) + b 3 cos γ α) + c 3 cos α β) b + c) cos α + c + a) cos β + a + b) cos γ 3abc a + b + c. 3) Finally, if s a + b + c, then and we get R abc K 6Rr 6 and K rs ) ) abc K K s 3abc s 3abc a + b + c. ) Conditions 3) and ) yield the desired result. Solution by Moti Levy, Rehovot, Israel After substituting Rr becomes abc a+b+c) in the right hand side of the original inequality, it cyc a3 cos β γ) 3abc b + c) cos α a + b + c. cyc Thus, we actually need to prove two identities which appeared many times before in the literature): b + c) cos α a + b + c, ) cyc a 3 cos β γ) 3abc. ) cyc 3

14 Dropping a perpendicular from C to side c, it divides the triangle into two right triangles, and c into two pieces c a cos β + b cos α, and similarly for all sides: c a cos β + b cos α, a b cos γ + c cos β, b c cos α + a cos γ. To prove ), we add the three equations, and get immediately: a + b + c a cos β + b cos α + b cos γ + c cos β + c cos α + a cos γ cyc b + c) cos α. To prove ), we use the following trigonometric identity cos y) sin cos + sin y cos y, sin + y) and the triangle identity a sin α b sin β c sin γ. a 3 3 sin β cos β + sin γ cos γ cos β γ) a sin β + γ) 3 sin β cos β + sin γ cos γ a sin α 3 b cos β + c cos γ a a b cos β + a c cos γ a a b cos β + a c cos γ ) cyc a 3 cos β γ) cyc ab a cos β + b cos α) + ac c cos α + a cos γ) + bc b cos γ + c cos β) 3abc. Also solved by Arkady Alt, San Jose, CA; Hatef I. Arshagi, Guilford Technical Community College, Jamestown, NC; Bruno Salgueiro Fanego, Viveiro, Spain; Kee-Wai Lau, Hong Kong, China; Kevin Soto Palacios, Huarmey, Peru; Neculai Stanciu, Geroge Emil Palade School Buzău, Romania and Titu Zvonaru, Comănesti, Romania; Nicusor Zlota, Traian Vuia Technical College, Focsani, Romania, and the proposer. 53: Proposed by José Luis Díaz-Barrero, Barcelona Tech, Barcelona, Spain Let F n be the n th Fibonacci number defined by F, F and for all n 3, F n F n + F n. Prove that ) FnFn+ n is an irrational number and determine it *). The asterisk ) indicates that neither the author of the problem nor the editor are aware of a closed form for the irrational number. Solution by Moti Levy, Rehovot, Israel

15 It is well known that F n F n+ n Fk, ) k hence : FnFn+ n ) can be epressed as or F + k F ) F ) + ) ) ) +, F F F 3 a a a k, a k F k. ) The series ) is the Engel epansion of the positive real number. See [] for definition of Engel epansion. In 93, Engel established the following result See [] page 303): Every real number has a unique representation c + k a a a k, where c is an integer and a a a 3 is a sequence of integers. Conversely, every such sequence is convergent and its sum is irrational if and only if lim k a k. Therefore, by Engel s result, n is irrational, since lim FnF n+ k F k. I do not know how to epress in closed form. However, it can be shown that it is transcendental. To this end, I rely on a result given in [] on page 35): Let fn)) n be a sequence of positive integers such that lim n fn+) for every integer d, the number n d fn) In our case, d and f n) F n F n+. We check that fn) is transcendental. µ >. Then f n + ) F n+ F n+ lim lim n f n) n F n F n+ F n+ + lim n F n+ lim n F n >. F n F n+ + F n lim n F n Then n References: FnF n+ [] Wikipedia Engel epansion. is transcendental. [] Ribenboim Paulo, My Numbers, My Friends: Popular Lectures on Number Theory, Springer 000. Solution by Ulrich Abel, Technische Hochschule Mittelhessen, Friedberg, Germany Let p be a prime. For the sake of brevity put c k F k F k+. We prove that the number ) ck s p k 5

16 is transcendental, in particular irrational. The partial sum s n n k ) ck a n p b n with positive integers a n and b n p cn satisfies ) ck 0 < s s n p kn+ ) cn+ p p p p cn ) c n+ cn ) cn+, k0 ) k p because c k+ c k F k+ F k+ F k F k+ Fk+. Since ) c n+ F n+ F n+ Fn+ lim lim lim Fn+ + ) > n c n n F n F n+ n F n F n+ By the theorem of Thue, Siegel and Roth, for any fied) algebraic number and ε > 0, the inequality 0 < a < b b +ε is satisfied only by a finite number of integers a and b. Hence, s is transcendental. Also solved by the Kee-Wai Lau, Hong Kong, China first part of the problem), and the proposer, first part of the problem) 53: Proposed by Ovidiu Furdui and Alina Sîntămărian, Technical University of Cluj-Napoca, Cluj-Napoca, Romania Find all differentiable functions f : 0, ) 0, ), with f), such that ) f, > 0. f) Solution by Arkady Alt, San Jose, CA ) First note that f f ), > 0 f ) f Then, since f ) ) f ) ) f )), f f f f f ), ) ) ) and ), > 0. we obtain f ) f )) f ) f ) f ) f )) f )) f ) f ) f )) 6

17 ) f ) f f ) ) f ) + + c f ) f ) + c +. Since f ) f ) then f ) f ) + + c + c c 0. f ) Therefore, f ) + f ) f ) using f) + ln f ) ln + ) + d and, again, we obtain ln f ) ln + ) + d ln ln + d d 0. Thus, f ) +. Solution by Albert Stadler, Hirrliberg, Switzerland The differential equation f ) f > 0. We differentiate the equation f )f f )f ) ) f )f f ) or equivalently f )f) f )). ) ) shows that f is differentiable infinitely often in ) and get f ) f ) f) 0, By assumption f) and thus f ) f) We integrate ) and apply partial integration to get. dt t ft) f t) f t)ft) f t)) dt ) d dt f ft)dt t) f t) + f t) dt f) f ) f) f ) + f) f ) +. So f) f ) + or equivalently f ) f) +. 7

18 We integrate again and get ln f) ln f) Therefore f) +. f t) ft) dt t + t dt ln + ) ln. Solution 3 by Bruno Salgueiro Fanego, Viveiro, Spain Let f : 0, + ) 0, + ) be a differentiable function that satisfies the hypothesis of the problem and let g : 0, + ) 0, + ) be the differentiable function defined by g). Since f is differentiable, and by the hypothesis f ), > 0, we f g)) conclude that ) f is also differentiable and, differentiating ) both ) side of the equality f )f, we obtain that f )f + f )f 0, and since ) f, or equivalently, f )) f )f) f )), or what is the same, ) f f ), > 0. Integrating both sides, we conclude that f) f ) + + C, > 0, for some C R. If we take at the start of the inequality, and since f), we obtain that f ) and f) f + C, from where C 0, which implies, because ) f) > 0 > 0 by hypothesis and f) f ) and f ) f), > 0. + Integrating both sides of this last equality, we conclude that ) ln f)) log + + D, > 0 for some D R. Taking in this equality and using the fact that f), we find that D 0 and therefore f) +, > 0. Since the function f : 0, + ) 0, + ) defined by f) +, > 0, is differentiable with f ) + and satisfies that f), and that f ) +, > 0, we conclude that the only differentiable function f) that satisfies the conditions of the problem is the function f) +, > 0. Solution by Toshihiro Shimizu, Kawasaki, Japan We have f ) f ). Letting to we also have f ) f d f ) f d Integrating it, we have )) f ) f. ) + ) f ) f ) ). Thus, ) 8

19 f ) f ) + + C Letting, we have + C or C 0. Therefore f ) f ) +. Multiplying f ) to ), we have + ) f ) f ) f ) f ) + Integrating again, we have d log f ) + + d + ) + d log + ) + D Thus, we can write f ) D + where D is some constant. Letting, we have D. Therefore, we have f ) +, this function actually satisfies the condition. Also solved by Abdallah El Farsi, Bechar, Algeria; Hatef I. Arshagi, Guilford Technical Community College, Jamestown, NC; Michael N. Fried, Ben-Gurion University, Beer-Sheva, Israel; Moti Levy, Rehovot, Israel; Ravi Prakash, New Delhi, India, and the proposers. 9