Mathematical Excalibur, Volume 11, Number 5

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Volume 11, Number 5 January 2007 – February 2007

Difference Operator

Kin Y. Li

Olympiad Corner

Below are the 2006 British Math lympiad (Round 2) problems. O

Problem 1. Find the minimum possible value of x2 _{+ y}2_{given that x and y are real}

numbers satisfying xy(x2_{− y}2_{) = x}2_{+ y}2

nd x ≠ 0. a

Problem 2.

Let x and y be positive integers with no prime factors larger than 5. Find all such x and y which satisfy x2_{− y}2_{= 2}k_{for some}

non-negative integer k.

Problem 3. Let ABC be a triangle with

AC > AB. The point X lies on the side BA extended through A, and the point Y

lies on the side CA in such a way that BX

= CA and CY = BA. The line XY meets

the perpendicular bisector of side BC at

P. Show that _∠_BPC₊_∠_BAC₌180o. Problem 4. An exam consisting of six questions is sat by 2006 children. Each question is marked right or wrong. Any three children have right answers to at least five of the six questions between them. Let N be the total number of right answers achieved by all the children (i.e. the total number of questions solved by child 1 + the total solved by child 2 + ⋯ + the total solved by child 2006). Find the least possible value of N.

Editors: ஻ Ի ஶ(CHEUNG Pak-Hong), Munsang College, HK

ଽ υ ࣻ (KO Tsz-Mei)

గ ႀ ᄸ (LEUNG Tat-Wing)

؃ ୊ ፱ (LI Kin-Yin), Dept. of Math., HKUST

֔ ᜢ ݰ (NG Keng-Po Roger), ITC, HKPU

Artist: ྆ ؾ ़ (YEUNG Sau-Ying Camille), MFA, CU

Acknowledgment: Thanks to Elina Chiu, Math. Dept.,

HKUST for general assistance.

On-line:

http://www.math.ust.hk/mathematical_excalibur/ The editors welcome contributions from all teachers and students. With your submission, please include your name, address, school, email, telephone and fax numbers (if available). Electronic submissions, especially in MS Word, are encouraged. The deadline for receiving material for the next issue is March 25, 2007.

For individual subscription for the next five issues for the 05-06 academic year, send us five stamped self-addressed envelopes. Send all correspondence to:

Dr. Kin-Yin LI Department of Mathematics

The Hong Kong University of Science and Technology Clear Water Bay, Kowloon, Hong Kong

Fax: (852) 2358 1643 Email: [email protected]

Let h be a nonzero real number and

f(x) be a function. When f(x + h) and f(x) are real numbers, we call

) ( ) ( ) (x f x h f x f h = + − ∆

the first difference of f at x with step h. For functions f, g and real number c, we have ) ( ) ( ) )( (f g x _hf x _hg x h + =∆ +∆ ∆ and ). ( ) )( (cf x c _hf x h = ∆ ∆

Also, 0 _f(_x)or I f(x) stands for f(x).

h

∆

For any integer n ≥ 1, we define the n-th

difference by For example, ). )( ( ) (_x 1_f _x f n h h n h − ∆ ∆ = ∆ ), ( ) ( 2 ) 2 ( ) ( 2_f _x _f _x _h _f _x _h _f _x h = + − + + ∆ ). ( ) ( 3 ) 2 ( 3 ) 3 ( ) ( 3_f _x _f _x _h _f _x _h _f _x _h _f _x h = + − + + + − ∆

By induction, we can check that ( ) ( 1) ( ), (α) 0 ∑ − + = ∆ = − n k k n k n n hf x C f x kh

where 0₌1 and for k > 0, n C . ! ) 1 ( ) 1 ( k k n n n k n Ck n + − − = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = L

(Note: for these formulas, we may even et n be a real number!!!)

l

If h=1, we simply write ∆ and omit the subscript h. For example, in case of a sequence {xn}, we have ∆xn=xn+1 − xn. Facts.(1) For function f (x), n=0,1,2,…,

; ) ( ) ( 0

∑

= ∆ = + n k k k n f x C n x f in particular, if is a nonzero constant for every positive integer n,

then _; ) (n f m ∆ ) 0 ( ) ( 0

∑

= ∆ = m k k k n f C n f (2) if P(x) = axn ₊_{⋯ is a polynomial of}

degree n, then for all x, n

n

hP(x)=an!h

∆ and ∆mP(x)=0for m>n. h

Let k be a positive integer. As a function of x, k has the properties:

x C (a) k (so ); x k x k x C C C 1 ₁ + − ₊ ₌ _∆ ₌ k−1 x k x C C (b) for 0 ≤ r ≤ k, k r; x k x r_C ₌_C − ∆ for r > k, ∆ k =0; x r_C (c) 1 1(just add 2 1 + + = + + + k n k n k k _C _C _C C L 11=0 + k C to the left and apply (a) repeatedly).

Similar to fact (1), if f(x) is a degree m polynomial, then ₍ ₎ ₍₀₎_. (β) 0

∑

= ∆ = m k k k x f C x f

(This is because both sides are degree

m polynomials and from property (b),

the k-th differences at 0 are the same for k = 0 to m, which implies the values of both sides at 0, 1, 2, …, m are the same.)

Example 1. Sum Sn = 14 + 24 + ⋯ + n4

n terms of n. i

Solution. Let f(x) = x4. By (β) and (c),

∑∑

∑

= = = ∆ = = n j k k k j n j n f j C f S 1 4 0 1 ) 0 ( ) (

_{∑ ∑}

_∑

= = = + +∆ = ∆ = 4 0 1 4 0 1 1 (0). ) 0 ( ) ( k n j k k k n k k j f C f C x : 0 1 2 3 4 f(x) : 0 1 16 81 256 : ) (x f ∆ 1 15 65 175 : 14 50 110 ) ( 2_f _x ∆ : ) ( 3_f _x ∆ 36 60 : 24 ) ( 4_f _x ∆ Therefore, ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + + ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + + ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + + ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + = 5 1 24 4 1 36 3 1 14 2 1 n n n n Sn = n(n+1)(2n+1)(3n2_+3n−1)/30.

Example 2. (2000 Chinese IMO Team

Selection Test) Given positive integers k, m, n satisfying 1 ≤ k ≤ m ≤ n. Find

∑

= − + + + + + − n i i i m i n i i n m i k n 0 . )! ( )! ( ! )! ( 1 ) 1 ( Solution. Define . ) ( ) 2 )( 1 ( ) ( k n x n m x m x m x x g + + + + + + + + = L From 1 ≤ k ≤ m ≤ n, we see m + 1 ≤ n + k ≤ m + n. So g(x) is a polynomial of degree n−1. By fact (2) and formula (α),

∑ − = ∆ − = = n i i n i n n _g _C _g_i 0( 1) () ) 0 ( ) 1 ( 0 i k n i m i n m i n i n n i i + + ∑ + + + − − = = 1 )! ( )! ( )! ( ! ! ) 1 ( 0 .

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Mathematical Excalibur, Vol. 11, No. 5, Jan. 07 – Feb. 07 Page 2

Example 3. (1949 Putnam Exam) The

sequence x0, x1, x2, … is defined by the

onditions x c 0 = a, x1 = b and for n ≥ 1, , 2 ) 1 2 ( 1 1 n x n x x n n n − + = − +

where a and b are given numbers. Express limn→∞ xn in terms of a and b.

Solution. The recurrence relation can

be written as . 2 1 n x x n n − ∆ − = ∆

Repeating this n − 2 times, we get ). ( ! 1 2 1 ! 1 2 1 1 b a n x n x n n n ⎟ − ⎠ ⎞ ⎜ ⎝ ⎛− = ∆ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛− = ∆ Then . ! 1 2 1 ) ( 1 0 1 0 0 i a b a x x x i n i n i i n

∑

− = − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛− − + = ∆ + =

Using the fact

, ! 0

∑

∞ = = i i x i x e

we get limn→∞ xn = a + (b − a)e−1/2.

Example 4. (2004 Chinese Math

Olympiad) Given a positive integer c,

let x1, x2, x3, … satisfy x1 = c and

1 ) 2 ( 2 1 1 ⎥⎦+ ⎤ ⎢⎣ ⎡ − + + = − − n n x x x n n n

for n = 2,3,…, where [x] is the greatest integer less than or equal to x. Find a

eneral formula of x

g n in terms of n.

Solution. First tabulate some values.

x x x x x x1 2 3 4 5 6 c=1 1 1 1 1 1 1 c=2 2 3 4 5 6 7 c=3 3 5 7 10 13 17 c=4 4 7 11 16 22 29 c=5 5 9 14 20 27 35 c=6 6 11 17 25 34 45 c=7 7 13 21 31 43 57 Next tabulate first differences in each

olumn. c column 1: 1,1,1,1,1,1,… column 2: 2,2,2,2,2,2,… column 3: 3,3,4,3,3,4,… column 4: 4,5,6,4,5,6,… column 5: 5,7,9,5,7,9,… column 6: 6,10,12,6,10,12,…

We suspect they are periodic with period 3. Let x(c,n) be the value of xn

for the sequence with x1 = c. For rows

1 and 2, the first differences seem to be constant and for row 4, the second

differences seem to be constant. Using act (1) and induction, we get

f

x(1,n) = 1, x(2 n) = n + 1 (i) , and x(4,n) = (n2_{+3n+4)/2 for all n. Now}

. 2 ) 2 )( 1 ( ) , 1 ( ) , 4 ( n −x n = n+ n+ x

To check the column difference eriodicity, we claim that for a fixed c, p

x(c + 3, n) = x(c, n) + (n + 1)(n + 2)/2.

If n = 1, then x(c + 3,1) = c + 3 = x(c,1) + 3 and so case n = 1 is true. Suppose the case

n−1 is true. By the recurrence relation, x(c+3,n) equals . 1 ) 2 ( ) 1 , 3 ( 2 ) 1 , 3 ( + − +_⎢⎣⎡ + − − + _⎥⎦⎤+ n n n c x n c x

From the case n − 1, we get x(c + 3, n − 1) = x(c, n − 1) + n(n + 1)/2. Using this, the displayed expression simplifies to

, 2 4 3 ) 2 ( ) 1 , ( 2 ) 1 , ( + 2+ + ⎥⎦ ⎤ ⎢⎣ ⎡ − − + + − n n n n n c x n c x which is x(c, n) + (n + 1)(n + 2)/2 by the recurrence relation. This completes the induction for the claim.

Now the claim implies

, 2 ) 2 )( 1 ( 3 ) , ( ) , ( ⎟ + + ⎠ ⎞ ⎜ ⎝ ⎛ − + =xdn c d n n n c x (ii)

where d = 1,2 or 3 subject to c ≡ d(mod 3). Since x(1,n) and x(2,n) are known, all we need to find is x(3,n).

For the case c = 3, studying x1,x3,x5,… and x2,x4,x6,… separately, we can see that the

second differences of these sequences seem to be constant. Using fact (1) and

nduction, we get i

x(3,n) = (n2_{+4n+7)/4 if n is odd and} x(3,n) = (n2_{+4n+8)/4 if n is even. (iii)}

Formula (ii) along with formulas (i) and (iii) provided the required answer for the problem.

Example 5. Let g(x) be a polynomial of

degree n with real coefficients. If a ≥ 3, then prove that one of the numbers |1 − g(0)|, |a − g(1)|, |a

2_{− g(2)|, … , |a}n+1 ₋ g(n+1)| is at least 1.

Solution. Let f(x) = a − g(x). We have x

∆ax _{= a}x+1_−ax _{= (a−1)a}x_,

_∆2ax _{= (a−1)∆a}x _{= (a−1)}2_ax_,

…,

_∆n+1ax _{= (a−1)}n+1_ax_.

In particular, _∆n+1_a0 _{= (a−1)}n+1_{. Now}

_∆n+1f(0)=_∆n+1_a0₋_∆n+1_{g(0) = (a−1)}n+1_. Since a ≥ 3, we get 2n+1 _≥_∆n+1_f(0). Assume |ak_{−g(k)| < 1 for k = 0,1,…,} n+1. Then

(

)

∑

+ = + − + + ₌ ₋ ₋ ∆ 1 0 1 1 1 ₍₀₎ n ₍ ₁₎ ₍ ₎ k k k n k n n _f _C _a _g _k _< _∑+ ₌ = + + 1 0 1 1 2 , n k n k n C which is a contradiction.

Example 6. (1984 USAMO) Let P(x)

e a polynomial of degree 3n such that b

P(0) = P(3) = _{⋯ = P(3n) = 2,} P(1) = P(4) = _{⋯ = P(3n−2) = 1,} P(2) = P(5) = _{⋯ = P(3n−1) = 0.} If P(3n + 1), then find n.

Solution. By fact (2) and (α),

∑ − + − = ∆ − = + = + + + 3 1 0 3 1 1 3 1 ₍₀₎ ₍₁₎ ₍₃ ₁ ₎ ) 1 ( 0 n k k n k n n _P _C _P _n _k

_∑

= = + + + + ₊ ₋ − + = n j n j j n j j n j _C _C 0 1 3 1 3 3 1 3 1 3 1 3 ₍ ₁₎ _. ) 1 ( 2 730

We can write this as 2a+b = −36_{, where}

, . ) 1 ( 0 1 3 1 3 1 3

∑

= + + + − = n j j n j _C a

∑

= + − = n j j n j_C b 0 3 1 3 3 ) 1 (

To find a and b, we consider the cube root of unity ω=e2πi/3_{, the binomial}

expansion of f(x) = (1−x)3n+1_{and let}

₍ ₁₎ _. 1 1 3 1 3 1 3

∑

= − + − − = n j j n j _C c

Now 0 = f(1)=b−a−c, f(ω)= b−aω−cω2

and f(ω2_{)= b−aω}2_{−cω. Solving, we see} a = − ( f(1) + ω2_{f(ω) + ωf(ω}2_{) )/3}

( )

_π 6 1 3 cos 3 2 3 1 − = n− n and b = ( f(1) + f(ω) + f(ω2_{) )/3}

( )

_. 6 1 3 cos 3 2 3 1 + π = n− n

Studying the equation 2a + b = −36_{, we}

find that it has no solution when n is odd and one solution when n is even, namely when n = 4.

Example 7. (1980 Putnam Exam) For

which real numbers a does the sequence defined by the initial condition u0=a and the recursion un+1=2un−n2 have un>0 for all n ≥ 0?

Solution. Among all sequences

satisfying un+1 = 2un−n2 for all n ≥ 0, the

difference vn of any two such

sequences will satisfy vn+1 = 2vn for all n ≥ 0. Then vn=2nv0 for all n ≥ 0.

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h

P

? Give a proof.

Problem Corner

We welcome readers to submit their solutions to the problems posed below for publication consideration. The solutions should be preceded by the solver’s name, home (or email) address and school affiliation. Please send submissions to Dr. Kin Y. Li,

Department of Mathematics, The Hong Kong University of Science & Technology, Clear Water Bay, Kowloon, Hong Kong. The deadline for

submitting solutions is March 25,

2007.

Problem 266. Let

N = 1+10+102₊_⋯+101997_.

Determine the 1000th_{digit after the}

decimal point of the square root of N in base 10.

Problem 267. For any integer a, set

na = 101a − 100·2a.

Show that for 0 ≤ a, b, c, d ≤ 99, if

na + nb ≡ nc+ nd (mod 10100),

then {a,b} = {c,d}.

Problem 268. In triangle ABC,

ABC = ACB = 40˚. Points P and

Q are inside the triangle suc that

∠ AB = ∠ QAC = 20˚ and ∠ PCB =∠ QCA = 10˚. Must B, P, Q be collinear

∠ ∠

Problem 269. Let f(x) be a polynomial with integer coefficients. Define a sequence a0, a1, … of integers such that a0 = 0, an+1 = f (an) for all n ≥ 0. Prove

that if there exists a positive integer m for which am = 0, then either a1 = 0 or a2 = 0.

Problem 270. The distance between any two of the points A, B, C, D on a plane is at most 1. Find the minimum of the radius of a circle that can cover these four points.

*****************

Solutions

****************

Problem 261. Prove that among any 13 consecutive positive integers, one of them has sum of its digits (in base 10) divisible by 7.

Solution. Jeff CHEN (Virginia, USA), CHEUNG Wang Chi (Raffles Junior

College, Singapore), G.R.A. 20 Math Problem Group (Roma, Italy), Naoki S. D. LING, Anna Ying PUN (HKU, Math, Year 1), Simon YAU Chi Keung, YIM Wing Yin (HKU, Year 1) and Fai YUNG. Consider the tens digits of the 13 consecutive positive integers. By the pigeonhole principle, there are at least [13/2] + 1 = 7 of them with the same tens digit. The sums of digits for these 7 numbers are consecutive. Hence, one of the sums of digits is divisible by 7.

Problem 262. Let O be the center of the circumcircle of _{∆ABC and let AD be a} diameter. Let the tangent at D to the circumcircle intersect line BC at P. Let line PO intersect lines AC, AB at M, N respectively. Prove that OM=ON.

Solution 1. Jeff CHEN (Virginia, USA).

O D A B C P M N Q R L

We may assume B is between P and C (otherwise interchange B and C, then N and M). Through C, draw a line parallel to line MN and intersect line AN at Q. Let line AO intersect line CQ at R. Since

MN||CQ, triangles AMN and ACQ are

similar. To show OM = ON, it suffices to show RC = RQ.

Let L be the midpoint of BC. We will show LR || BQ (which implies RC = RQ). Now ∠ OLP = ∠ OLB = 90˚ = ∠ ODP, which implies O, P, D, L are concyclic. Then ∠ ODL = ∠ OPL. From OP || RC, we get ∠ RDL = ∠ RCL, which implies

L,R,D,C are concyclic. Then

∠ RLC =180˚− ∠ RDC = 180˚− ∠ ADC = 180˚−∠ ABC = ∠ QBC. Therefore, LR || BQ as claimed.

Solution 2. CHEUNG Wang Chi

(Raffles Junior College, Singapore). Set O as the origin and line MN as the

x-axis.

Let P’ be the reflection of P with respect to O. Then the coordinates of P and P’ are of the form (p,0) and

(

−p,0).

O A D P P' C B N _M

The equation of the circumcircle as a conic section is of the form x2_+y2_−r2_=0.

The equation of the pair of lines AP’ and BC as a (degenerate) conic section is

(

y−m(x+p)

)(

y−n(x− p)

)

=0, where m is the slope of line AP’ and n is the slope of line BC. Since these two conic sections intersect at A, B, C, so the equation of the pair of lines AB and

AC as a (degenerate) conic section is of

he form t

(

( )

)(

( )

)

, 2 2 2 _y _r _y _m_x _p _y _n_x _p x + − =λ − + − − for some real number λ. When we set y = 0, we see the x-coordinates of M and

N satisfies x2 _−r2_{= λmn(x}2_{− p}2_{), whose}

roots are some positive number and its negative. Therefore, OM = ON.

Commended solvers: Courtis G.

CHRYSSOSTOMOS (Larissa, Greece, teacher) and Anna Ying PUN (HKU, Math, Year 1).

Problem 263. For positive integers m,

n, consider a (2m+1)×(2n+1) table,

where in each cell, there is exactly one ant. At a certain moment, every ant moves to a horizontal or vertical neighboring cell. Prove that after that moment, there exists a cell with no ant.

Solution. Jeff CHEN (Virginia, USA), CHEUNG Wang Chi (Raffles Junior College, Singapore), G.R.A. 20 Math Problem Group (Roma, Italy), Naoki S. D. LING, Anna Ying PUN (HKU, Math, Year 1), YIM Wing Yin (HKU, Year 1) and Fai YUNG.

Assign the value (−1)i+j_{to the cell in}

the i-th row, j-th column of the table. Then two horizontal or vertical neighboring cells will have values of opposite sign. Since 2m+1 and 2n+1 are odd, there is exactly one more cell with negative values than cells with positive values. Before the moment, there is one more ant in cells with negative values than ants in cells with positive values. After the moment, two of the ants from cells with negative values will occupy a common cell with a positive value. Then there exists a cell with no ant.

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Mathematical Excalibur, Vol. 11, No. 5, Jan. 07 – Feb. 07 Page 4

Problem 264. For a prime number p > 3 and arbitrary integers a, b, prove that

is divisible by 6p. p

p _ba ab −

Solution. Samuel Liló ABDALLA

(São Paulo, Brazil), Claudio ARCONCHER (Jundiaí, Brazil), Jeff CHEN (Virginia, USA), CHEUNG Wang Chi (Raffles Junior College, Singapore), G.R.A. 20 Math Problem Group (Roma, Italy), HO Ha Fai (Carmel Divine Grace Foundation Secondary School, Form 6), D. Kipp JOHNSON (Valley Catholic School, Teacher, Beaverton, Oregon, USA), Anna Ying PUN (HKU, Math, Year 1), Simon YAU Chi Keung, YIM Wing Yin (HKU, Year 1) and Fai YUNG. Observe that )]. 1 ( ) 1 [( 1₋ ₋ 1₋ = − p p− p− p _ba _ab _b _a ab For q = 2, 3 or p, if a or b is divisible by

q, then the right side is divisible by q.

Otherwise, a and b are relatively prime to q. Now p − 1 is divisible by q − 1, which is 1, 2 or p − 1. By Fermat’s little theorem, both ≡ 1 (mod

q). So ≡ 1 (mod q). Hence, the

bracket factor above is divisible by q. Thus is divisible by 2, 3 and

p. Therefore, it is divisible by 6p. 1 1_, − − q q _b a 1 1_, − − p p _b a p p _ba ab −

Problem 265. Determine (with proof) the maximum of

∑

= − n j j j x x 1 5 4 ₎ ( ,

where x1, x2, …, xn are nonnegative real

numbers whose sum is 1.

(Source: 1999 Chinese IMO Team

Selection Test)

Solution. Jeff CHEN (Virginia, USA), D. Kipp JOHNSON (Valley Catholic School, Teacher, Beaverton, Oregon, USA), Anna Ying PUN (HKU, Math, Year 1) and YIM Wing Yin (HKU, Year 1).

Let f ( x ) = x4 _{− x}5 _{= x}4 _{(1 − x). Since}

, we see that f(x) is strictly convex on [0, 3/5]. Suppose

n ≥ 3. Without loss of generality, we

may assume x ) 5 3 ( 4 ) (_x _x2 _x f ′′ = − 1≥ x2≥ ⋯ ≥ xn. If x1 ≤ 3/5, then since ), , , , ( 0 , , 0 , 5 2 , 5 3 2 1 x xn x _L f L ⎟ ⎠ ⎞ ⎜ ⎝ ⎛

by the majorization inequality (see

Math Excalibur, vol. 5, no. 5, pp. 2,4),

). 5 2 ( ) 5 3 ( ) ( 1 f f x f n i i + ≤

∑

= If x1 > 3/5, then 1 − x1, x2, …, xn are in [0, 2/5]. Since

(

1−x₁,0,_K,0

)

_f(x₂,_K,x_n), by the majorization inequality,

). 1 ( ) ( ) ( ₁ ₁ 1 x f x f x f n i i − + ≤

∑

=

Thus the problem is reduced to the case n = 2. So now consider nonnegative a, b with a + b = 1. We have f(a) + f(b) = a4_{(1−a) + b}4_(1−b) = a4_b+b4_{a = ab(a}3_+b3₎ = ab[(a+b)3_−3ab(a+b)] = 3ab(1−3ab)/3 ≤ 1/12

by the AM-GM inequality. Equality case holds when ab = 1/6 in addition to a + b = 1, for example when

). 6 3 3 , 6 3 3 ( ) , (ab = + −

Therefore, the maximum is 1/12.

Difference Operator

(continued from page 2)

Next we look for a particular solution of

un+1 = 2un−n2 for all n ≥ 0. Observe that n2

= un − (un+1 − un) = (I −

∆

)un. From the

sum of geometric series, we guess

2 2 2 1 ₍ ₎ ) (I n I n u_n= −∆ − = +∆+∆ +_L = n2_{+(2n+1)+2= n}2_+2n+3

should wor

k.

Indeed, this is true since (n + 1)2 _{+ 2(n + 1) + 3 = 2(n}2 _{+ 2n + 3) − n}2_.

Combining, we see that the general solution to un+1 = 2un − n2 for all n ≥ 0 is un = n2 + 2n + 3 + 2nv0 for any real v0.

Finally, to have u0 = a, we must choose v0

= a − 3. Hence, the sequence we seek is un = n2+2n+3+2n(a−3) for all n ≥ 0.

Since , 3 2 2 lim ₂ =+∞ + + ∞ → _n _n n n

un will be negative for large n if a − 3 < 0.

Conversely, if a − 3 ≥ 0, then all un > 0.

Therefore, the answer is a ≥ 3.

Example 8. (1971 Putnam Exam) Let c

be a real number such that nc_{is an integer}

for every positive integer n. Show that c is a non-negative integer.

Solution. 2 is an integer implies c ≥ 0. c

Next we will do the case c is between 0 and 1 using the mean value theorem. This

motivates and clarifies the general argument for the case c ≥ 1. Assume 0 ≤ c < 1. Then choose a positive integer n > c1/(1−c)_{. Applying the mean}

value theorem to f(x) = xc_{on [n, n + 1],}

we know there exists a number w between n and n + 1 such that = (n+1)

c n

∆

c _{− n}c _{= cw}c−1_{. On the left side, we}

have an integer, but on the right side, since w > n > c1/(1−c)_{, we have 0 ≤ cw}c−1

< 1. Hence, c = 0.

For c ≥ 1, let us mention there is an extension of the mean value theorem, which asserts that if f is continuous on [a,b], k-times differentiable on (a,b), 0<h ≤ (b−a)/k and x + kh ≤ b, then there exists a number v such that a < v < b and ). ( ) ( _f()_v h x f k k k h ₌ ∆

Taking this for the moment, we will finish the argument as follows. Let k be the integer such that k−1≤ c < k. Choose an integer n so large that

c(c−1)(c−2)⋯(c−k+1)nc−k_{< 1.}

Applying the extension of the mean value theorem mentioned above to f(x) = xc_{on [n, n + k], there is a number v}

between n and n + k such that . ) 1 ( ) 2 )( 1 ( ck c k_n ₌_c _c₋ _c₋ _c₋_k₊ _v − ∆ L

Again, the left side is an integer, but the right side is in the interval [0, 1). Therefore, both sides are 0 and c = k−1. Now the extension of the mean value theorem can be proved by doing math induction on k. The case k = 1 is the mean value theorem. Next, suppose the extension is true for the case k−1.

et 0<h ≤(b−a)/k. On [a,b−h], define L . ) ( ) ( ) ( ) ( h x f h x f h x f x g ₌∆h ₌ + −

Applying the case k−1 to g(x), we know there exists a number v0 such that a < v0 < b−h and ). ( ) ( ) ( 0 ) 1 ( 1 1 v g h x g h x f k k k h k k h − − − = ∆ = ∆

gJ2) = $2

+

3': with z'y' = :. -zy

2

0.

. .

The

question

tells

us

that

. . . .

3

..

-

-. c . ..

and a& usto firid %e aii+rnm Mlue of A.

By

the%M-GM

inequality,

. . 2 . _ . .. Thus we have 7 . - - . . . . . . .

so

B

2

2 (smce.A =.O not aliowed). _{. .}

Reairanging-equatian

(I),

we find that . . . .

u2

=

B2(A2

-@) . ' . . . . _. _. . . . i . -

.

_- . . _. . . . . '. . ' @. . . , . - and thus . .

-

_.-...

_-.

A2 =! ==4' . . .

(We can divide through by B2

-

4 since B =

2

isnot a solution

6

the previ6us . . . '.

. . - .

equation.) _. _-.

-

. .

NOW (B2

y2

=

4

2

4.

Finally,

we note that x2

-kg

cah indeed take

the

value 4,. fbr example when x = J r & a n d y =

G Z .

-

k(cos 8

+

isin

e),

where i2 = -1. Then

and I

.

ksih8

₌

3 ( a )

_{= 2zy.}

H&e R(a)

is the real part of a,

and

Q(Q) is the

imaginary

part.

Now we

are told that z2

+

y2 = xy(z2

-

y2), and SO IayI =

+q,)~(~).

"

In other words, k = i 2 k 2 cos6sin8, and so, since k

>

0 as x

#

0, 4/k = 2cosBsin8 =sin28

5

1. Thus

k

3

4. .

So x2

+

y2

2

4, &d,'as'above, we check that equdjty can be obtained when z = J c d a n d y =

d a :

'

. .

.

.-

@

-Solution

by

Jona

than

Lee, Loughborough Grammar School,

slightly

edited

. It is easy to check that there are no solutions for k = 0 or k = 1, since differ- Now,. it suffices to look,for solutions where x, and 'y

are

not both even, ences.behveen positive squares are at least 3.

2"-'. Since not

both

are

even, n = 1.

Now,

we

may

write x'= 321 5=2, y

=

39' 592. Since x- y = 2, x and y canriot

both be divisible by either 3 or: 5. ,Thus either'x =

3a,

y = 5b or

x

= Sa, y = 3 4 for some a and b.

If?

+

3', y = 5', then 5! =

2m-'

-

2t. . .. . = 5 .2t, y = ' . , . .

Alternative method by Lee Zhao, Nottingbam High School We can

2m-1

-

1 by stating Mih5iIes'm's theorem, which was conjecked by Catalan, and was proved by MihZlescu

The theorem says that the only solution to the equation'mp

-

nq = 1 with

rn, n positive integers and p , Q

1

2 is 32

-

23 = 1. This

quickly recovers

the solutions above.

/' proceed as above, but we catt sdve the equations 5b = .2m-1.- 1 and 3b =

(6)

.@

Solution by Julia Robson, PerSc School for Girls Dr$w in the bisector of

A,

and let this meet BC at

R,

adthe circumcjrcle of AABC-at

Q.

-kt X Y

m&t

BC

at

TB

a i d k t

M

be the &point of BC.

Then

weknow.that Q . t b on

the perpendicular bisector of

BC

fwhich is the line

PM),

since BQ = CQ-

+-

.

.BAQ = CAQ. _c _.

A

Since

AB

=

CY

and BX = CAB we have AX = Y A and so AXAY is line XY is pardel to line AX his shows that

AACR

is similar to AYCT

(angle-angleangle). Thus

iso~eles. Then X P A = Y Z A = (18OO

-

X A Y ) / 2 =,CAB/2 = CAR, and SO

. .

by similar triangles and. the definition pf the point,

Y.

By the angle b k o r .

theorem, . .

- . BR . B A

. - = -

CR C A : . . .

and so from the last &o equations, C T = BR.'So,'since M is the midp&t of

BC;

we must havqMR =

MT.

So since PdIT = Q i i R , and pT is parallel to

RQ, we have APMT is congruqt to

A Q M R

Thus P M = Q M . %.now, by

sideangleside, we know that A.@ltfP is congruent to ABMQ. So

. .

PB'M = Q B M . .

. .

= .

,CAQ

= .'BAC/2. .

c:

AQ bisects BAC).

But BPC = 2BpM = 180" -.2PBM (since P M B = QO"), i d

&."

- -

: ' .

Bl%

+

diAC = 188O

-

_(by_angles

-

_{in the same segment)}

. . . .

.. .

'

@

Solution by Dominic Rowland,'Winchester College Consider all the students who scored.four or less. If any students scored less than four, giye

them extra correct solutions arbitmily ~mtil each has four.

There

&

15 ways to choose 4 from 6, so if there are more than 30 studenti with exactly four solutions, some three will have &e same by the pi.geonhole pri+iple. But then they have only four between them.

Therefoe at kast 2006

-

30 students scbred five or more.

!3gwesee that a possible candidate for a miniin# total number of solutions

1976 students scored exactly five;and

:

is as fonows: . . .

. . /

. . .

. .. 30 sjudents scored exactly four, in such a way that each combination of

. fouq pmbleiw was iolved by exactly two Students.

. ?

This gives

N

= 19?6 * 5.+ 30 * 4 = 10,000.

.

To

provethat'this is indeed the minimum, we must show that there is no configuration containing 30 students who still satisfy the conditions but who have fewer than 120 solutions be*een &em. We ~ & l l prove it by contradic- . tiox suppose

there

is.

'Consider

the 30 worst students.

Clearly

none scored six, or: else

N

would be far greater than 10,OOa.

Let A be the number who scored five,

B

be the number who scored four, and so on, with F.being the number who scored none.

Now, we.know ._

A + B.

+

C

+

D

+

E

+

F = 30 ₍₁₎

.and .

(4

5 A + 4 B + 3 C + 2 D + E

<

120.

We

know

that we cannot have more thgn 30 students who scored four.

Also, a student who scored three c& be given a fourth correct solution in three ways. If we replaced that student with three more scoring four. one with each possible extra solt~tion, then we still get a valid configuration, . (which must.stil1 be under 30 in humber).

Similarly, a score of tivo can be increased to'a score of four in six possible ways, so a.student scoring two Can be replaced by six students scoring fou''

Likewise, a student scoring me qin be replaced by 10 'scoring four, and a therefore

(3)

i'aking

(3)-(1)

gives us

-A

0, but C, D , E ind F are all nonnegative

integers. This

3

a contradiction; so

N

,= 10,000 is minimal. ' ' .

. . -. -