The Problem Corner · 2019-02-27 · Consider a 5x5 grid. There are 25 13 §· ¨¸ ©¹ ways to fill the grid with exactly 13 X’s and 12 O’s. How many of these have at least

The Problem Corner

Edited by Pat Costello

The Problem Corner invites questions of interest to undergraduate students. As a rule,

the solution should not demand any tools beyond calculus and linear algebra. Although

new problems are preferred, old ones of particular interest or charm are welcome,

provided the source is given. Solutions should accompany problems submitted for

publication. Solutions of the following new problems should be submitted on separate

sheets before March 15, 2016. Solutions received after this will be considered up to the

time when copy is prepared for publication. The solutions received will be published in

the Spring 2016 issue of The Pentagon. Preference will be given to correct student

solutions. Affirmation of student status and school should be included with solutions.

New problems and solutions to problems in this issue should be sent to Pat Costello,

Department of Mathematics and Statistics, Eastern Kentucky University, 521 Lancaster

Avenue, Richmond, KY 40475-3102 (e-mail: [email protected], fax: (859) 622-3051)

NEW PROBLEMS 760-768

Problem 760. Proposed by Mathew Cropper, Eastern Kentucky University, Richmond,

KY.

Consider a 5x5 grid. There are 25

13

ways to fill the grid with exactly 13 X’s and

12 O’s. How many of these have at least one row, one column, or one diagonal with

5 X’s?

Problem 761. Proposed by Ovidiu Furdiu, Technical University of Cluj-Napoca, Cluj,

Romania.

Let n ≥0 be an integer and let T2n denote the 2nth Taylor polynomial corresponding to

the cosine function at 0.

2 21

1

2

1

( ) ( 1)(2 2)!

knk

n

k

xT x

k

Let 2

2 2

0

( ) cosnn n

T x xI dx

x

.

a) Prove that In = - 1

1

(2 1)(2 )nI

n n

, n ≥1.

b) Calculate In.

mailto:[email protected],

Problem 762. Proposed by Mohammad Azarian, University of Evansville, Evansville,

IN.

If x ≠ 0, x ≠ 1, y > 0, and y ≠ 1, then find y as a function of x provided

2 1' (ln ) ln 0

( 1)y y y y y

x x

Problem 763. Proposed by D.M. Batinetu-Giurgiu, “Matei Basarab” National College,

Bucharest, Romania, Neculai Stanciu, “George Emil Palade”, Buzau, Romania.

Let x and A(x) =

1 1 1 1

1 1 1 1

1 1 1 1

1 1 1 1

x

x

x

x

.

Compute the matrix product A(0)*A(1)*A(2)*A(3).



Calculate 1/2( 1) 1/2(( 1)!) (n!)limn n

n

n n

Problem 765. Proposed by Marcel Chirita, Bucharest, Romania.

Let a, b, c be the lengths of the sides of a triangle in which 2 2 2b c a . Prove that

3 3 33 (2 2 1)b c abc a


Let x be an integer. Prove that if x5 +5x3 + 15x2 > 21x, then x5 +5x3 + 15x2 – 21x ≥ 30.

Problem 767. Proposed by the editor.

Prove that the number 1317 cannot be written as the sum of a square and a fifth power of

another integer.


Calculate the value of the series 0

(1/ 2)

( 2)( 4)

n

n n n

The Problem Corner







sheets before September 1, 2016. Solutions received after this will be considered up to

the time when copy is prepared for publication. The solutions received will be published

in the Fall 2016 issue of The Pentagon. Preference will be given to correct student






Problem 769. Proposed by the Northwest Missouri State University Problem Solving

Group, Maryville, MO.

Let Tk = ( 1)

2

k k be the kth triangular number.

(1) Under what condition(s) on n does 13 divide 3

2( 1)nT ?

(2) Under what condition(s) on n does 13 divide 3

2 1nT ?

Problem 770. Proposed by Jose Luis Diaz-Barrero, School of Civil Engineering,

Barcelona Tech - UPC, Barcelona, Spain.

Let f : [0,1] →ℝ be a continuous concave function. Prove that 1/7 2/7 3/14

0 0 0

3 1 2( ) ( ) ( )

4 12 3f t dt f t dt f t dt



Let a < b be positive real numbers and let f : [a,b] →ℝ be a continuous function. Prove that there exists c ∈ (a, b) such that


1

2 ( ) ( )

c

a

a c b cf c f t dt

a c b cc


Solve in positive integers the equation x2 – 97y! = 2015.


Let a, b, c be real numbers greater than or equal to 3. Prove that 2 2 2 2 2 2 2 2 2

2 2 2

3 3 3min( , , )

27 27 27 9

a b b b c c c a a abc

b c a

Problem 774. Proposed by Mohammad K. Azarian, University of Evansville, Evansville,

Indiana.

If both x and y are positive real numbers, then find y as a function of x, provided

y ʹ + (y+1) ln(y+1)[1-(ln(y+1))-2((1/4)x-1 + x1/2)]x1/2 = 0.

Problem 775. Proposed by Mohammad K. Azarian, University of Evansville, Evansville,

Indiana.

Determine y explicitly as a function of x provided

x(1 + sin x)yʹ + [ (x2 + y2 + 4) – ( -3 + sin x) y – 2(1 + sin x)] = 0

y≠ -2 and x ≠ kπ.

Problem 776. Proposed by Natanael Karjanto, University College, Suwon, Republic of

Korea.

Show that for α > 0 and n , the harmonic number Hn can be represented by the

following integral:

| | 1 ( 1)| |

1 1

1 1 ( 1)sech sech

2

n nx k x k

n

k k

kH e xdx e xdx

k n

Problem. 777. Proposed by Robert Gardner and William Ty Frazier (graduate student),

East Tennessee State University, Johnson City, TN.

Let [x] represent the floor (or greatest integer) function. Let n, m ∈ ℕ with 2≤m≤n-1 and k ∈ { 0, 1, 2, …, m -1}. Use the floor function to express the smallest integer N greater than or equal to n which is congruent to k modulo m.

Problem 778. Proposed by Thomas Chu (graduate student), Western Illinois University,

Macomb, IL.

Let p1 and p2 be distinct odd primes both congruent to 1 or 3 mod 4. Prove that

1 2 1 2| |gcd , 1

2 4

p p p p


Use all the digits 1,2,3,…,9 without repeats to create two primes such that their product is

a maximum. Each digit should be used in only one of the two numbers.

The Problem Corner















Problem 780. Proposed by Daniel Sitaru, Colegiul National Economic Theodor

Costescu, Drobeta Turnu – Severin, Mehedinti, Romania.

Prove that if a,b,c ∈ [1,∞), then ab + bc + ca ≥ 3 + 2 ln(ab bc ca).

Problem 781. Proposed by Daniel Sitaru, Colegiul National Economic Theodor


Prove that if a,b,c ∈ (0,∞), then

4 4 2 2 2( ) / 2 ( ) ( ) ( ) 3a b c a b c b a c c a b abc .

Problem 782. Proposed by Jose Luis Diaz-Barrero, Barcelona Tech-UPC, Barcelona,

Spain.

Let a, b, c be the lengths of the sides of triangle ABC and ma, mb, and mc the lengths of its

medians. Prove that

2 2 2

1.2 2 2

a b cm m m

a b c


Spain.


Find all real solutions of the following system of equations

x3 + 2x + y = 9 + 3x2

3y2 + 6y + z = 21 + 9y2

5z3 + 10z + x = 33 + 15z2.



Prove that in any triangle ABC with BC = a, CA = b, AB = c and area F, the following

inequalities are true.

2 2 2 2 2 2( )sin ( )sin ( )sin 4 3 ,

2 2 2

A B Cb c c a a b F

2 2 2(1 sin ) (1 sin ) (1 sin ) 4 3 .

2 2 2

C A Bab bc ca F

Problem 785. Proposed by Iuliana Trasca, Scornicesti, Romania.

Show that x, y, z > 0, then

6 3 6 3 6 3 3 3 3

2 2 2

3

2

x z y x z y x y z xyz

x y z

.

Problem 786. Proposed by Thomas Chu, Macomb, Illinois.

Prove that if x,y,z > 1, then

2 2 2 3 3 3( )( ) 4 4 4x y z x y z x y z xy xz yz .


Mike buys some pants and shorts at the Great Pants Store. Mike buys shorts that cost $11

each and pants that cost $14 each. His total before taxes is $283. How many shorts and

how many pants did Mike buy?

Problem 788. Proposed by George Heineman, Worcester Polytechnic Institute,

Worcester, MA.

A Sujiken™ puzzle has a triangular grid of cells containing digits from 1 to 9. You must

place a digit in each of the empty cells with the constraint that no digit can repeat in any

row, column, or diagonal. Additionally, no digit can repeat in the 3x3 large squares with

thick borders or the three triangular regions with thick borders. The puzzle below is of

intermediate difficulty.

The Problem Corner







sheets before October 1, 2017. Solutions received after this will be considered up to the


the Fall 2017 issue of The Pentagon. Preference will be given to correct student






Problem 789. Proposed by Daniel Sitaru, “Thodor Costescu” National Economic

College, Traian National College, Drobeta Turnu – Severin, Mehedinti, Romania.

In triangle ABC, let I = the incenter, O = the circumcenter, G = the centroid, and a,b,c

the lengths of the sides. Prove that ( )( )( )IA OA GA a b b c c a

Problem 790. Proposed by Daniel Sitaru, “Thodor Costescu” National Economic

College, Traian National College, Drobeta Turnu – Severin, Mehedinti, Romania.

Prove that if a, b ϵ ℝ with a < b, then

2 sin 2 2 3

ln2 sin 2 3

bb a

a



Determine whether the real number

ln(11 5 2)

ln(5 11 2)

is rational or not.




Let x1 , x2, …, xn be real numbers lying in the interval (0,π/2). Prove that

1 1

1 1 1sin cos

2

n n

k k

k k

x xn n



If a [0,/4], compute 2 2

0( )(ln(1 tan tan )

a

x ax a x a dx



Let a, b, c be positive real numbers. Prove that

2 2 2(1 )(1 )(1 ) 1 1 1

ab bc caa b c

a b b c c a

Problem 795. Proposed by Michal Kremzer, Glicice, Silesia, Poland.

Let Q be the set of rational numbers. Does there exist a function f: (Q-{0}) -> (Q-{0}) so

that f(x) < f(3x) < f(2x) for all x in the set (Q-{0})?

Problem 796. Proposed by Kadir Altintas, Turkey and Leonard Giugiuc, Romania.

If A, B, and C are the angles of a triangle, prove that

6(1 cos cos cos 4cos cos cos2 2 2

A B CA B C


Let integer n be called a consecutives concatenated number when n is formed by

concatenating two consecutive integers. For example, 67 and 1314 are consecutives

concatenated numbers. 67 is prime, but 1314 is composite. It turns out there are lots of

consecutives concatenated primes. Find a consecutives concatenated prime of 20 digits

where the first integer of the two concatenated is divisible by 23*33.

The Problem Corner
















In 2002, Britney Gallivan (high school junior) found a formula for paper folding and

managed to do 12 folds of a long sheet of toilet paper. She found that

2 4 2 16

n ntL

where t represents the thickness of the material to be folded, L is the length of the paper

to be folded and n is the number of folds desired (in only one direction). Suppose you

tape together sheets of standard 8.5” x 11” copier paper (thickness .0035”) end to end,

how many sheets would be needed to be able to fold the long taped sheet 14 times?

Problem 799. Proposed by Daniel Sitaru, “Theodor Costescu” National Economic

College, Drobeta Turnu – Severin, Mehedinti, Romania.

Prove that if a,b,c ∈ (0,2] then

( 2

3 2 2b a a a b c

c b c a




Prove that if a ∈ R, then

5 8 2 11

3 6 9

ln(1 ) ln(1 ) ln(1 ) ln(1 )

a a a a

x x x x

a a a a

e dx e dx e dx e dx

.


Spain.

Compute

2 2

3 2 2 31

3

1lim

1 2

n

nk

k n

n k n n

n


Spain.

Let n ≥ 1 be an integer. Prove that

1

1 1 1

1

2

n n n

n n nk k k

k k k

F F L

where Fn and Ln are the nth Fibonacci and Lucas numbers defined by F1 = F2 = 1 and

Fn = Fn-1 + Fn-2 for n ≥ 3 and by L1 = 1, L2 = 3 and Ln = Ln-1 + Ln-2 for n ≥ 3.

Problem 803. Proposed by Ovidiu Furdui and Alina Sintamarian, Technical University

of Cluj-Napoca, Cluj-Napoca, Romania.

Calculate

2

1 1 ( )

n m

n m

H

n n m

where Hn = 1 + ½ + … + 1/n denotes the nth harmonic number.


Bucharest, Romania and Neculai Stanciu, “George Emil Palade” School, Buzau,

Romania.

Compute the following limit

3

1

2! 3! ... !lim

(2 1)!!

n n

n n

n

n


Bucharest, Romania and Neculai Stanciu, “George Emil Palade” School, Buzau,

Romania.

Let zk = xk + i yk be a complex number where k ∈ { 1,2, …, n }. Prove that

4 4 2

1

1 1

2| |

2

n n

k n k k

k k

x y z

Problem 806. Proposed by Marius Dragan, Bucharest, Romania and Neculai Stanciu,

“George Emil Palade” School, Buzau, Romania.

If a1, a2, …, an > 0 such that 1

1n

k

k

a

, then prove that

2 2 2 21 2 1

2 3 11 1/ 1 1/ ... 1 1/ 1 1/ 1n nna na na na

na a a a n

Problem 807. Proposed by Titu Zvonaru, Comanesti, Romania.

If A, B, and C are the angles of a triangle and = A/2, = B/2, = C/2, prove that

6(1 cos cos cos ) 2sin sin sin (1 8sin sin sin ) 4cos cos cosA B C

SOLUTIONS TO PROBLEMS 780-788

Problem 780. Proposed by Daniel Sitaru, Colegiul National Economic College Theodor


Prove that if a,b,c ∈ [1,∞), then ab + bc + ca ≥ 3 + 2 ln(ab bc ca).

Solution by Richdad Phuc, University of Sciences, Hanoi, Vietnam.

We have LHS – RHS = b(a – 2 ln a) + c(b – 2 ln b) + a(c – 2 ln c) – 3 or

LHS – RHS = (b/a)a(a – 2 ln a) + (c/b)b(b – 2 ln b) + (a/c)c(c – 2 ln c) - 3

Let f(x) = x(x – 2 ln x) for x ≥ 1. The derivative is f’(x) = 2x – 2 ln x – 2 and

f″(x) = 2 – 2/x ≥ 0 for all x ≥ 1 f’(x) ≥ f’(1) = 0 for all x ≥ 1. This means f(x) is strictly increasing on [1, ∞). Then f(x) ≥ f(1) for all x ≥ 1. Hence a(a – 2 ln a) ≥ 1, b(b – 2 ln b) ≥ 1, c(c – 2 ln c) ≥ 1. Then

LHS – RHS ≥ (b/a) + (c/b) + (a/c) – 3 which is ≥ 0 by the AM-GM inequality. Equality holds if a = b = c = 1.

Also solved by Angel Plaza, Universidad de Las Palmas de Gran Canaria, Spain;

Henry Ricardo, New York Math Circle, NY; and the proposer.

Problem 781. Proposed by Daniel Sitaru, Colegiul National Economic College Theodor


Prove that if a,b,c ∈ (0,∞), then

4 4 2 2 2( ) / 2 ( ) ( ) ( ) 3a b c a b c b a c c a b abc .

Solution by the proposer.

We prove that if x,y ∈ (0,∞), then 2 2

2

x yx y xy

(*).

We denote 2 2

2

x yu

which means 2u2 = x2 + y2 and let v xy

2v xy

With these notations, we have 2u2 + 2v2 = x2 +2xy + y2 = (x+y)2

We can rewrite (*) as x + y – v ≥ u or (x + y)2 ≥ (u + v)2

⇔ 2u2 + 2v2 ≥ (u + v)2

⇔ 2u2 + 2v2 – u2 – v2 – 2uv ≥ 0

⇔ (u – v)2 ≥ 0

Now replace x with x/y and y with y/x in (*) to get

2 2( / ) ( / )

2

x y x y y x x y

y x y x

⇔ 2 2 4 41

12

x y x y

xy xy

⇔4 4

2 2

2

x yx y xy

For x = a and y = b and multiplying by c we have

4 42 2

2

a ba c b c c abc

Analogously,

4 4

2 2

2

b cb a c a a abc

and 4 4

2 2

2

c ac b a b b abc

Adding the last three inequalities gives the desired result.

Also solved by Titu Zvonaru, Comanesti, Romania and Neculai Stanciu,“George Emil

Palade” School, Buzau, Romania; Ioan Viorel Codreanu, Satulung, Maramures, Romania; Soumitra Moukherjee, Scottish Church College, Chandar Nagore, India; Ravi Prakash, Oxford University Press, New Delhi, India.


Spain.

Let a, b, c be the lengths of the sides of triangle ABC and ma, mb, and mc the lengths of its

medians. Prove that

2 2 2

1.2 2 2

a b cm m m

a b c

Solution by Titu Zvonaru, Comanesti, Romania and Neculai Stanciu,“George Emil

Palade” School, Buzau, Romania.

Since 2

c

a bm

, by the AM-GM inequality we have

22 2 2 2 2 2 2 2 2 c

a bma b a b

Writing the other two similar inequalities and adding all three gives the desired result.

Also solved by Madison Estabrook, Missouri State University, Springfield, MO; Rovsen

Pirkulyev, Baku State University, Sumgait, Azerbaidjian; and the proposer.


Spain.

Find all real solutions of the following system of equations

x3 + 2x + y = 9 + 3x2

3y2 + 6y + z = 21 + 9y2

5z3 + 10z + x = 33 + 15z2.

Solution by the proposer.

We can rewrite the system as

3 – y = x3 - 3x2 + 2x – 6

3 – z = 3(y3 – 3y2 + 2y - 6)

3 – x = 5(z3 – 3z2 +2z – 6)

Since t3 – 3t2 +2 t - 6 = (t – 3)(t2 +2), we have

3 – y = (x – 3)(x2 +2)

3 – z = 3(y – 3)(y2 +2)

3 – x =5(z – 3)(z2 +2).

Multiplying these together gives

-(x-3)(y-3)(z-3) = 15(x-3)(y-3)(z-3)(x2 +2)(y2 +2)(z2 +2).

From this we get

0 = (x-3)(y-3)(z-3)(15(x2 +2)(y2 +2)(z2 +2) + 1)

Since the last factor above is positive, either x = 3, y = 3 or z = 3.

If we assume that x = 3, the first equation says y = 3. Substituting this into the second

equation implies that z=3. The same occurs for starting with y=3 or z=3. So x = y = z = 3

is the only real solution.

Also solved by Soumava Chakraborty, Softweb Technologies, Kolkota, India.


Bucharest, Romania, Neculai Stanciu, “George Emil Palade” School, Buzau, Romania.

Prove that in any triangle ABC with BC = a, CA = b, AB = c and area F, the following

inequalities are true.

2 2 2 2 2 2( )sin ( )sin ( )sin 4 3 ,

2 2 2

A B Cb c c a a b F

2 2 2(1 sin ) (1 sin ) (1 sin ) 4 3 .

2 2 2

C A Bab bc ca F

Solution by Ioan Viorel Codreanu, Satulung, Maramures. Romania.

We have 2 2 sin 2( )sin 2 sin 2 sec

2 2 2cos cos

2 2

A A bc A F Ab c bc F

A A

Similarly, 2 2( )sin 2 sec2 2

B Bc a F and 2 2( )sin 2 sec

2 2

C Ca b F

Then 2 2( )sin 2 sec2 2

A Ab c F .

Using Jensen’s Inequality and that f(x) = sec x on (0, π/2) is a convex function, we get

sec 3sec 2 3.2 6

AA

Thus 2 2( )sin 4 3 .

2

Ab c F

Next 2 sin(1 sin ) 2 sin 2 sec .

2 2 2cos

2

C C ab C Cab ab F

C

Similarly, 2(1 sin ) 2 sec2 2

A Abc F and 2(1 sin ) 2 sec .

2 2

B Bca F

Then 2(1 sin ) 2 sec 4 3 .2 2

C Aab F F

Also solved by Soumava Chakraborty, Softweb Technologies, Kolkota, India; Ravi Prakash, Oxford University Press, New Delhi, India; Soumitra Moukherjee, Scottish Church College, Chandar Nagore, India; and the proposer.

Problem 785. Proposed by Iuliana Trasca, Scornicesti, Romania.

Show that x, y, z > 0, then

6 3 6 3 6 3 3 3 3

2 2 2

3

2

x z y x z y x y z xyz

x y z

.

Solution by Soumava Chakraborty, Softweb Technologies, Kolkota, India. The inequality is equivalent to 2(x6z3 + y6x3 +z6y3) ≥ x5y2z2 +y5z2x2 +z5x2y2 + 3x3y3z3. Using the AM-GM inequality, we have x6z3 + x6z3 + x3y6 ≥ 3x5y2z2 y6x3 + y6x3 + y3z6 ≥ 3y5z2x2 z6y3 + z6y3 + z3x6 ≥ 3z5x2y2 Adding these gives 3(x6z3 + y6x3 + z6y3) ≥ 3(x5y2z2 + y5z2x2 + z5x2y2). Dividing by 3 says x6z3 + y6x3 + z6y3 ≥ x5y2z2 +y5z2x2 + z5x2y2

The AM-GM inequality also says x6z3 + y6x3 + z6y3 ≥ 3x3y3z3. Summing the previous two inequalities gives the inequality that is equivalent to the one of the problem. Also solved by Soumitra Moukherjee, Scottish Church College, Chandar Nagore, India; Ioan Viorel Codreanu, Satulung, Maramures, Romania; Titu Zvonaru, Comanesti, Romania; and the proposer.

Problem 786. Proposed by Thomas Chu, Macomb, Illinois.

Prove that if x,y,z > 1, then

2 2 2 3 3 3( )( ) 4 4 4x y z x y z x y z xy xz yz .

Solution by Angel Plaza, Universidad de Las Palmas de Gran Canaria, Spain.

By changing variables x = 1+a, y = 1+b and z = 1+c, the problem reads as: Prove that if a,b,c > 0, then

2 3(1 ) 3 (1 ) 4(1 )(1 ) 4(1 )(1 ) 4(1 )(1 )a a a a b b c c a

Expanding the right-hand side and left-hand sides, we get

LHS = 12 + 12(a+b+c) + 4(ab+ac+bc) + 2 2 38 2a a b a

RHS = 12 + 8(a+b+c) + + 4(ab+ac+bc) We can clearly see that the LHS > RHS when a,b,c > 0. Also solved by Anas Adlany (student), Omar Ben Abdelaziz University, El Jadida, Morroco; Myagmarsuren Yadamsuren, Ulanbataar University, Ulanbataar, Mongolia; Soumava Chakraborty, Softweb Technologies, Kolkota, India; and the proposer.


Mike buys some pants and shorts at the Great Pants Store. Mike buys shorts that cost $11

each and pants that cost $14 each. His total before taxes is $283. How many shorts and

how many pants did Mike buy?

Solution by Robert Bailey (former KME national President 2001-2005), Niagara

University, NY.

Let x = number of shorts and y = number of pants. We have 11x + 14y = 283 which is a

linear Diophantine equation in two variables. Then 14y = 283 – 11x which is equivalent

to 14y ≡ 283 (mod 11) or 3y ≡ 8 (mod 11) or 3y ≡-3 (mod 11). Since 3 is relatively

prime to 11, we get y ≡-1 (mod 11). This means y = 10, 21, 32, …. The only value for y

that causes x to be positive in the equation 11x + 14y = 283 is y = 10 in which case

x = 13.

Also solved by Michael Bhujel, Bobbie Legg, Katie Tyson (students), and Bill Yankosky,

North Carolina Wesleyan College, Rocky Mount, NC; Jeremiah Bartz, University of

North Dakota, Grand Forks, ND; and the proposer.

Problem 788. Proposed by George Heineman, Worcester Polytechnic Institute,

Worcester, MA.

A Sujiken™ puzzle has a triangular grid of cells containing digits from 1 to 9. You must

place a digit in each of the empty cells with the constraint that no digit can repeat in any

row, column, or diagonal. Additionally, no digit can repeat in the 3x3 large squares with

thick borders or the three triangular regions with thick borders. The puzzle below is of

intermediate difficulty.

Solution

Solved by Jamie Farrar, Destinee Fisher, Nicole Kettle, Courtney Lush (students), Ed

Wilson (retired faculty), Eastern Kentucky University, Richmond, KY; Jeremiah Bartz,

University of North Dakota, Grand Forks, ND; Katie Tyson (student), Gail Stafford,

Carol Lawrence, Bill Yankosky, North Carolina Wesleyan College, Rocky Mount, NC.

The Problem Corner







sheets before November 1, 2018. Solutions received after this will be considered up to

the time when copy is prepared for publication. The solutions received will be published

in the Fall 2018 issue of The Pentagon. Preference will be given to correct student








Prove that if a, b, c ∈ [1, ∞) then

/ / /

a b c ab bc cab c a

b a c b a c a b c

e ea b c

e e



Prove that if a, b, c ∈ (2, ∞) then

2 ( ( 2) ( 2)) 3a b b a abc



Compute 4 3 2

5 51

1 4 12 9lim

( 3) 243

n

n

x x x xL dx

n x x



Let a, b, c be three positive integers. Prove that

3ab bc ca bc ca aba b c abc




Let S be a finite set. Consider three partitions of it each one with n elements:

A1,…, An; B1,…, Bn; C1,…, Cn. If for all 1≤ i, j, k ≤ n, it holds that

,i j j k k iA B B C C A n then prove that 3

.3

nS When does equality hold?



Prove that if a, b, c ∈ (1, ∞) then

2 2 2 2 2 2

3log log log .

5ab c a bc a b ca b c



Let the sequence (an)n≥1, defined by a1= 1, an+1 = (n+1)! an for any positive integer n.

Compute 2

2 !lim .

n

n nn

n

a

Problem 815. Proposed byMarius Dragan, National College Mircea cel Batran,

Bucharest, Romania, and Neculai Stanciu, “George Emil Palade” School, Buzau,

Romania.

Let a, b, c be positive real number such that 4 4 4 3a b c . Show that

2 2 2 2 2 2 2 2 2 2 2 2 22( )( )( ) ( )a b b c c a a b b c c a abc

Problem 816. Proposed by Stanescu Florin, Serban Cioclescu School, Gaesti, Romania.

Determine the largest real number with the property that for any function f :[0,1] [0,∞) where the following hold:

i) f is convex and f(0) = 0;

ii) there exists ε > 0 with f differentiable on [0,ε) and (0) 0f ;

the following inequality holds:

1 12

21

0 0

00

( )( )

( ) ( )

x

x f xdx x dx

f t dt f t dt

Problem 817. Proposed by Stanescu Florin, Serban Cioclescu School, Gaesti, Romania.

Consider the complex numbers a, b, c, d all of modulus one which have the following properties:

a) arg a < arg b < arg c < arg d;

b)

2( )( )

2 ( )( )( )( ) 4.a c b d

b ai c bi d ci a diabcd

Show that 2 24 4 4

max 1 ; 18 ( ) 4 ( ) 17

i a i b

i a b d i b a c

Problem 818. Proposed by Titu Zvonaru, Comanesti, Romania and Neculai Stanciu,

“George Emil Palade”, Buzau, Romania.

Determine all positive integers a,b,c,d,x,y,z,t such that a ≠ b ≠ c≠ d and

a + b + c = td, b + c + d = xa, c + d + a = yb, d + a + b = zc.


Billy has created a snowman with a base, torso, and a head all of which are spheres where

the torso is smaller than the base and the head is smaller than the torso. The radius of

each piece is a positive integer with the base having a 12 inch radius. He decides that he

wants to use the snow to make 3 similar and identical snowmen for his younger siblings.

He discovers that he can do this. What are the radii of the original and smaller snowmen?

The Problem Corner

Edited by Pat Costello The Problem Corner invites questions of interest to undergraduate students. As a rule, the solution should not demand any tools beyond calculus and linear algebra. Although new problems are preferred, old ones of particular interest or charm are welcome, provided the source is given. Solutions should accompany problems submitted for publication. Solutions of the following new problems should be submitted on separate sheets before March 15, 2019. Solutions received after this will be considered up to the time when copy is prepared for publication. The solutions received will be published in the Spring 2019 issue of The Pentagon. Preference will be given to correct student solutions. Affirmation of student status and school should be included with solutions. New problems and solutions to problems in this issue should be sent to Pat Costello, Department of Mathematics and Statistics, Eastern Kentucky University, 521 Lancaster Avenue, Richmond, KY 40475-3102 (e-mail: [email protected], fax: (859) 622-3051) NEW PROBLEMS 820-828 Problem 820. Proposed by the editor.

Find a 4-digit positive integer N = abcd which is divisible by 11 and N/11 = b2 + c2 + d2. Problem 821. Proposed by Daniel Sitaru, “Theodor Costescu” National Economic College, Drobeta Turnu – Severin, Mehedinti, Romania. Prove that if a,,b,c ∈ R then 2 2 24 | (1 ) | (1 )

cyclic cyclica b b a b− ≤ +∑ ∑

Problem 822. Proposed by Daniel Sitaru, “Theodor Costescu” National Economic College, Drobeta Turnu – Severin, Mehedinti, Romania. Prove that in any acute-angled ABC∆ you have

6 6

3 tan tan2 tan 3(tan tan tan )2cyclic cyclic

A BA A B C+≥ + + +∑ ∑

.

Problem 823. Proposed by Pedro H.O. Pantoja, University of Campina Grande, Brazil. Let x,y,z be positive real numbers. Prove that

3 3 3

3 3 3 2

1 3 3 3( 2 ) ( 2 ) ( 2 ) (3 )

x y z x y y z z xxy yz zx y z z x x y xyz

+ +≤ + + ≤

+ + + + +


Problem 824 Proposed by Pedro H.O. Pantoja, University of Campina Grande, Brazil. Find all positive integers a, b, c where a and b are prime numbers with a ≠ 0 (mod c) such that 51a + 7ab + bc2 = abc2. Problem 825. Proposed by Ovidiu Furdui and Alina Sintamarian, Technical University of Cluj-Napoca, Cluj-Napoca, Romania. Let k ≥ 0 be an integer. Calculate

2 21

1 1 1( 1)n n n n k

∞

=

+ + − + +

∑

Problem 826. Proposed by D.M. Batinetu-Giurgiu, “Matei Basarab” National College, Bucharest, Romania and Neculai Stanciu, “George Emil Palade” School, Buzau, Romania. Let Fn and Ln be the nth Fibonacci and Lucas numbers defined by F1 = F2 = 1 and Fn = Fn-1 + Fn-2 for n ≥ 3 and by L1 = 1, L2 = 3 and Ln = Ln-1 + Ln-2 for n ≥ 3.

Let k be a positive integer and F(k) = 2 2

112 2

11

k kk k

k kk k

L LF FL LF F++

++

.

Evaluate1

( )n

k

F k=∏ as a multiple of the matrix

1 11 1

.

Problem 827. Proposed by D.M. Batinetu-Giurgiu, “Matei Basarab” National College, Bucharest, Romania and Neculai Stanciu, “George Emil Palade” School, Buzau, Romania.

Let (an) be a sequence of positive real numbers such that lim!n

n

a an→∞

= > 0. Find

2 2

11

( 1)limn n n

n n

n na a→∞ +

+

+−

Problem 828. Proposed by D.M. Batinetu-Giurgiu, “Matei Basarab” National College, Bucharest, Romania and Neculai Stanciu, “George Emil Palade” School, Buzau, Romania.

Determine all injective functions f: R → R with f(0) ≠ 1/b and

f(f(x)y3) + ax9y9= bf(x3)f(y3) for all x,y ∈ R where a>0, b>0.

Documents

The Problem Corner · 2019-02-27 · Consider a 5x5 grid. There are 25 13 §· ¨¸ ©¹ ways to fill the grid with exactly 13 X’s and 12 O’s. How many of these have at least