Volume 13, Number 5 January-February, 2009
Generating Functions
Kin Yin Li
Olympiad Corner
The following were the problems of the Final Round (Part 2) of the Austrian Mathematical Olympiad 2008.
First Day: June 6th, 2008
Problem 1. Prove the inequality 3 1 1 1 1−ab−bc −c ≤ a
holds for all positive real numbers a, b and c with a+b+c = 1.
Problem 2. (a) Does there exist a polynomial P(x) with coefficients in integers, such that P(d) = 2008/d holds for all positive divisors of 2008? (b) For which positive integers n does a polynomial P(x) with coefficients in integers exists, such that P(d) = n/d holds for all positive divisors of n? Problem 3. We are given a line g with four successive points P, Q, R, S, reading from left to right. Describe a straight- edge and compass construction yielding a square ABCD such that P lies on the line AD, Q on the line BC, R on the line AB and S on the line CD.
(continued on page 4) Editors: 張 百 康(CHEUNG Pak-Hong), Munsang College, HK
高 子 眉 (KO Tsz-Mei) 梁 達 榮 (LEUNG Tat-Wing)
李 健 賢 (LI Kin-Yin), Dept. of Math., HKUST 吳 鏡 波 (NG Keng-Po Roger), ITC, HKPU Artist: 楊 秀 英 (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 7, 2009.
For individual subscription for the next five issues for the 09-10 academic year, send us five stamped self-addressed envelopes. Send all correspondence to:
Dr. Kin-Yin LI, Math Dept., Hong Kong Univ. of Science and Technology, Clear Water Bay, Kowloon, Hong Kong
Fax: (852) 2358 1643 Email: [email protected]
© Department of Mathematics, Hong Kong University of Science and Technology.
In some combinatorial problems, we may be asked to determine a certain sequence of numbers a0, a1, a2, a3, ….
We can associate such a sequence with the following series
f(x) = a0 + a1x + a2x2 + a3x3 + ⋯.
This is called the generating function of the sequence. Often the geometric series
L + + + + = − ) 1 2 3 1 /( 1 t t t t for |t| < 1 and its square
2 3 2 2 (1 ) ) 1 /( 1 −t = +t+t +t +L =1+2t+3t2+4t3+5t4+L will be involved in our discussions. Below we will provide examples to illustrate how generating functions can solve some combinatorial problems.
Example 1. Let a0=1, a1=1and
an = 4an−1 − 4an for n ≥ 2. Find a formula for an in terms of n.
Solution. Let f(x) = a0 + a1x + a2x2 + ⋯.
Then we have
f(x) − 1 − x = a2x2 + a3x3 + a4x4 + ⋯
= (4a1−4a0)x2 + (4a2−4a1)x3 + ⋯
= (4a1x2+4a2x3+⋯)−(4a0x2+4a1x3+⋯)
= 4x( f(x) − 1) − 4x2 f(x).
Solving for f(x) and taking |x| < ½, f (x) = (1−3x)/(1−2x)2 = 1/(1−2x)−x/(1−2x)2
∑
∑
∞ = − ∞ = − = 1 1 0 ) 2 ( ) 2 ( n n n n x n x x (2 2 ) . 0 1∑
∞ = − − = n n n n n x Therefore, an = 2n − n 2n−1.Example 2. Find the number an of ways n dollars can be changed into 1 or 2 dollar coins (regardless of order). For example, when n = 3, there are 2 ways, namely three 1 dollar coins or one 1 dollar coin and one 2 dollar coin.
Solution. Let f(x) = a0 + a1x + a2x2 + ⋯.
To study this infinite series, let |x| < 1.
For each way of changing n dollars into r 1 dollar and s 2 dollar coins, we can record it as xr x2s = xn. Now r and s may be any nonnegative integers. Adding all the recorded terms for all nonnegative integers n, then factoring, we get
.) ( 0 0 0 2 0 2 0
∑
∑∑
∑
∑
∞ = ∞ = ∞ = + ∞ = ∞ = = = = n n n r s s r s s r r x x a x f x xOn the other hand,
) 1 ( ) 1 ( 1 1 1 1 1 2 2 0 2 0 x x x x x x s s r r − − = − ⋅ − =
∑
∑
∞ = ∞ = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − + − = 2 2 1 1 ) 1 ( 1 2 1 x x(
(1 2 32 ) (1 2 4 ))
2 1 + + +L+ + + +L = x x x x 1 2 2 2 3 3 4 3 5 . L + + + + + + = x x x x x ([ /2] 1) . 0∑
∞ = + = n n x n Therefore, an = [n/2] + 1.Example 3. Let n be a positive integer.
Find the number an of polynomials P(x) with coefficients in {0,1,2,3} such that P(2) = n.
Solution. Let f(t) be the generating
function of the sequence a0, a1, a2,
a3, …. Let P(x) = c0 + c1x + ⋯ + ckxk
with ci∊{0,1,2,3}. Now P(2) = n if and only if c0 + 2c1 + ⋯ + 2kck = n. Taking
t∊(−1,1), we can record this as . 2 21 0 c kck c n t t t t = L
Note 2ici is one of the four numbers 0, 2i, 2i+1, 3·2i. Adding all the recorded terms for all nonnegative integers n and all possible c0, c1, …, ck ∊{0,1,2,3}, then factoring on the right, we have
.) 1 ( ) ( 0 2 3 2 2 0 1
∏
∑
∞ = ⋅ ∞ = + + + = = + i n n n i i i t t t t a t f Using 1+s+s2+s3=(1−s4)/(1−s), we see L ⋅ − − ⋅ − − ⋅ − − ⋅ − − = 4 82 164 328 1 1 1 1 1 1 1 1 ) ( t t t t t t t t t f . 1 1 1 1 2 t t⋅ − − = As in example 2, we get an = [n/2] + 1.Mathematical Excalibur, Vol. 13, No. 5, Jan.-Feb. 09 Page 2 For certain problems, instead of
using the generating function of a0, a1,
a2, a3, …, we may consider the series
. 3 2 1 0 + a + a + a +L a x x x x
Example 4. (1998 IMO Shortlisted
Problem) Let a0, a1, a2, … be an
increasing sequence of nonnegative integers such that every nonnegative integer can be expressed uniquely in the form ai+2aj+4ak, where i, j and k are not necessarily distinct. Determine a1998.
Solution. For |x| < 1, let ( ) .
0
∑
∞ = = i ai x x fThe given condition implies . 1 1 ) ( ) ( ) ( 0 4 2 x x x f x f x f n n − = =
∑
∞ = Replacing x by x2, we get . 1 1 ) ( ) ( ) ( 2 4 8 2 x x f x f x f − =From these two equations, we get f(x) = (1+x) f(x8). Repeating this recursively,
we get . ) 1 )( 1 )( 1 )( 1 ( ) ( 8 82 83 L x x x x x f = + + + +
In expanding the right side, we see the exponents a0, a1, a2, … are precisely
the nonnegative integers whose base 8 representations have only digit 0 or 1. Since 1998=2+22+23+26+27+28+29+210,
so a1998=8+82+83+86+87+88+89+810.
For our next examples, we need some identities involving p-th roots of unity, where p is a positive integer. These are complex numbers λ, which are all the solutions of the equation
. 1 = p
z For a real θ, we will use the common notation eiθ = cos θ +i sin θ. Since the equation is of degree p, there are exactly p p-th roots of unity. We can easily check that they are eiθ with θ = 0, 2π/p, 4π/p, …, 2(p−1)π/p.
Below let λ be any p-th root of unity, other than 1. When we have a series
B(z) = b0 + b1z + b2z2 + b3z3 + ⋯,
sometimes we need to find the value of L
+ +
+ p p
p b b
b 2 3 . We can use the fact
0 1 1 1 2 ( 1) = − − = + + + + − j pj j p j j λ λ λ λ λ L
(for any j not divisible by p) to get
∑
− = = 1 0 ) ( 1 p j j B p λ bp+b2p+b3p+L. (*)For p odd, we have the factorization
) 1 ( ) 1 )( 1 ( 1+tp= +t +λt +λp−1t L (**)
since both sides have −1/λi (i=0,1,…, p−1) as roots and are monic of degree p.
Example 5. Can the set ℕ of all positive integers be partitioned into more than one, but still a finite number of arithmetic progressions with no two having the same common differences?
Solution. (Due to Donald J. Newman) Assume the set ℕ can be partitioned into sets S1, S2,…,Sk, where Si={ai+ndi: n∊ℕ} with d1 > d2 > ⋯> dk. Then for |z| < 1,
. 1 1 1 1 2 2 1 1
∑
∑
∑
∑
∞ = + ∞ = + ∞ = + ∞ = + + + = n nd a n nd a n nd a n n z z zk k z LSumming the geometric series, this gives . 1 1 1 1 2 2 1 1 k k d a d a d a z z z z z z z z − + + − + − = − L
Letting z tend to e2πi/d1, we see the left side has a finite limit, but the right side goes to infinity. That gives a contradiction.
Example 6. (1995 IMO) Let p be an odd prime number. Find the number of subsets A of the set {1,2,…,2p} such that (i) A has exactly p elements, and (ii) the sum of all the elements in A is divisible by p.
Solution. Consider the polynomial
Fa(x) = (1+ax)(1+a2x)(1+a3x)⋯(1+a2px)
When the right side is expanded, let cn,k count the number of terms of the form
) ( ) )(
(ai1x ai2x L aikx , where i1, i2, …, ik are
integers such that 1≤ i1< i2 <⋯< ik ≤ 2p and i1+i2 +⋯+ik = n. Then . 1 ) ( 2 1 1 , k p k n n k n a x c a x F
∑ ∑
= ∞ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + =Now, in terms of cn,k, the answer to the problem is C=cp,p +c2p,p+c3p,p+L. To get C, note the coefficient of
x
pin Fa(x) is∑
∞ =1 , . n n p n a c By (*) above, we see∑∑
− = ∞ = = 1 0 1 , . 1 p j n nj p n c p C ωNow the right side is the coefficient of xp in 1 1 ( ), 0
∑
− = p j x F p ωj which equals∑
− = + + + 1 0 2 2 ) (1 ). 1 )( 1 ( 1 p j pj j jx x x p ω ω L ωFor j = 0, the term is (1+x)2p. For 1 ≤ j
≤ p−1, using (**) with λ = ωj and t = λx, we see the j-th term is (1+xp)2. Using these, we have
]. ) 1 )( 1 ( ) 1 [( 1 ) ( 1 1 2 2 0 p p p j x p x p x F p
∑
j = + + − + − = ωTherefore, the coefficient of
x
pis ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − + ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = 1 2 2(p 1) p p p C .So far all generating functions were in one variable. For the curious mind, next we will look at an example involving a two variable generating function
∑∑
∞ = ∞ = = 0 1 , ) , ( i j j i j i xy a y x fof the simplest kind.
Example 7. An a×b rectangle can be tiled by a number of p×1 and 1×q types of rectangles, where a, b, p, q are fixed positive integers. Prove that a is divisible by p or b is divisible by q. (Here a k×1 and a 1×k rectangles are considered to be different types.)
Solution. Inside the (i, j) cell of the a×b rectangle, let us put the term xiyj for i=1,2,…,a and j=1,2,…,b. The sum of the terms inside a p×1 rectangle is xiyj+⋯+ xi+p−1y j=(1+ x + ⋯ + xp−1) xiyj, if the top cell is at (i, j), while the sum of the terms inside a 1×q rectangle is xiy j+⋯+ xiy j+q−1= xiyj (1+ y + ⋯ + y q−1), if the leftmost cell is at (i, j). Now take
p i e
x = 2π / and y = e2πi/q. Then both sums become 0. If the desired tiling is possible, then the total sum of all terms in the a×b rectangle would be
. ) 1 )( 1 ( ) 1 )( 1 ( 0 1 1 x y y x xy y x b a a i b j j i − − − − = =
∑∑
= =This implies that a is divisible by p or b is divisible by q.
For the readers who like to know more about generating functions, w
e
recommend two excellent references: T. Andreescu and Z. Feng, A Path to Combinatorics for Undergraduates, Birkhäuser, Boston, 2004.
M. Novaković, Generating Functions, The IMO Compendium Group, 2007 (www.imomath.com)
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 sending solutions is March 7, 2009.
Problem 316. For every positive integer n > 6, prove that in every n-sided convex polygon A1A2…An, there exist i ≠ j such that
. ) 6 ( 2 1 | cos cos | − < ∠ − ∠ n A Ai j
Problem 317. Find all polynomial P(x) with integer coefficients such that for every positive integer n, 2n−1 is divisible by P(n).
Problem 318. In ΔABC, side BC has length equal to the average of the two other sides. Draw a circle passing through A and the midpoints of AB, AC. Draw the tangent lines from the centroid of the triangle to the circle. Prove that one of the points of tangency is the incenter of ΔABC. (Source: 2000 Chinese Team Training Test)
Problem 319. For a positive integer n, let S be the set of all integers m such that |m| < 2n. Prove that whenever 2n+1 elements are chosen from S, there exist three of them whose sum is 0. (Source: 1990 Chinese Team Training Test)
Problem 320. For every positive integer k > 1, prove that there exists a positive integer m such that among the rightmost k digits of 2m in base 10, at least half of them are 9’s.
(Source: 2005 Chinese Team Training Test)
*****************
Solutions
****************Problem 311. Let S = {1,2,…,2008}. Prove that there exists a function f : S → {red, white, blue, green} such that there does not exist a 10-term arithmetic progression a1,a2,…,a10 in S
satisfying f(a1) = f(a2) = ⋯ = f(a10).
Solution 1. Kipp JOHNSON (Valley
Catholic School, teacher, Beaverton, Oregon, USA) and PUN Ying Anna (HKU Math, Year 3).
The number of 10-term arithmetic progressions in S is the same as the number of ordered pairs (a,d) such that a, d are in S and a+9d ≤ 2008. Since d ≤ 2007/9=223 and for each such d, a goes from 1 to 2008−9d, so there are at most
∑
= − × ×223 − 1 ) 10 2008 ( 4 (2008 9 ) 4 d d = 41999×223000functions f :S→{red, white, blue, green} such that there exists a 10-term arithmetic progression a1,a2,…,a10 in S satisfying
f(a1) = f(a2) = ⋯ = f(a10), while there are
more (namely 42008) functions from S to
{red, white, blue, green}. So the desired function exists.
Solution 2. G.R.A. 20 Problem Solving Group (Roma, Italy
)
.Replace red, white, blue, green by 0, 1, 2, 3 respectively. It can be seen by a direct checking that f:{1,2,…,2048}→ {0,1,2,3} given by 2 mod 2 mod 128 1 2 8 1 ) (n =⎢⎣⎡ −n ⎥⎦⎤ + ⎢⎣⎡ −n ⎥⎦⎤ f
avoids any 9-term arithmetic progression having the same value (where kmod 2 is 0 if
k is even and 1 if k is odd). The range of f is ((0818)8(2838)8)8, where for any string x,
x8 denotes the string obtained by putting
eight copies of the string x one after another in a row and f(n) is the n-th digit in the specified string.
Commended solvers: LKL Problem Solving Group (Madam Lau Kam Lung Secondary School of Miu Fat Buddhist Monastery).
Problem 312. Let x, y, z > 1. Prove that . 48 ) 1 ( ) 1 ( ) 1 ( 2 4 2 4 2 4 ≥ − + − + − x z z y y x
Solution. Glenier L. BELLO-BURGUET
(I.E.S. Hermanos D`Elhuyar, Spain), Kipp JOHNSON (Valley Catholic School, teacher, Beaverton, Oregon, USA), Kelvin LEE (Trinity College, University of Cambridge, Year 2), LEUNG Kai Chung (HKUST Math, Year 2), LKL Problem Solving Group (Madam Lau Kam Lung Secondary School of Miu Fat Buddhist Monastery), MA Ka Hei (Wah Yan College, Kowloon), NGUYEN Van Thien (Luong The Vinh High School, Dong
Nai, Vietnam) and PUN Ying Anna (HKU Math, Year 3).
Let x = a + 1, y = b + 1 and z = c + 1. Applying the AM-GM inequality twice, we have 2 4 2 4 2 4 ) 1 ( ) 1 ( ) 1 ( − + − + x− z z y y x 2 4 2 4 2 4 ( 1) ( 1) ) 1 ( a c c b b a + + + + + = 3 / 1 2 2 2 4 4 4( 1) ( 1) ) 1 ( 3 ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + + + ≥ c b a c b a 3 (2 ) (2 ) (2 ) 48. 3 / 1 2 2 2 4 4 4 = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ≥ c b a c b a
Commended solvers: CHUNG Ping Ngai (La Salle College, Form 5), G.R.A. 20 Problem Solving Group (Roma, Italy), NG Ngai Fung (STFA Leung Kau Kui College, Form 6), Paolo PERFETTI (Dipartimento di Matematica, Università degli studi di Tor Vergata Roma, via della ricerca scientifica, Roma, Italy), Dimitar TRENEVSKI (Yahya Kemal College, Skopje, Macedonia) and TSOI Kwok Wing (PLK Centenary Li Shiu Chung Memorial College, Form 6).
Problem 313. In
Δ
ABC, AB < AC and O is its circumcenter. Let the tangent at A to the circumcircle cut line BC at D. Let the perpendicular lines to line BC at B and C cut the perpendicular bisectors of sides AB and AC at E and F respectively. Prove that D, E, F are collinear.O
C
B
A
D
N
F
E
M
Solution. Glenier L. BELLO-
BURGUET (I.E.S. Hermanos D`Elhuyar, Spain), CHUNG Ping Ngai (La Salle College, Form 5), Kelvin LEE (Trinity College, University of Cambridge, Year 2), NG Ngai Fung (STFA Leung Kau Kui College, Form 6) and PUN Ying Anna (HKU Math, Year 3).
Let M be the midpoint of AB and N be the midpoint of AC. Using
∠
ABE =∠
ABC − 90°,∠
FCA = 90° −∠
ABC and the sine law, we haveMathematical Excalibur, Vol. 13, No. 5, Jan.-Feb. 09 Page 4 FCA CN ABE BM CF BE ∠ ∠ = cos / cos / . sin / sin / 2 2 2 1 2 1 AC AB BCA AC ABC AB = ∠ ∠ =
From ΔDCA~ΔDAB, we see
. sin sin sin sin AC AB ABC ACB DBA DAB DA DB DC DA = ∠ ∠ = ∠ ∠ = = So . 2 2 DC DB DA DB DC DA AC AB CF BE= = ⋅ =
Then ∠BDE =∠CDF. Therefore D,E,F are collinear.
Commended solvers: Stefan LOZANOVSKI and Bojan JOVESKI (Private Yahya Kemal College, Skopje, Macedonia).
Problem 314. Determine all positive integers x, y, z satisfying x3 − y3 = z2,
where y is a prime, z is not divisible by 3 and z is not divisible by y.
Solution. CHUNG Ping Ngai (La
Salle College, Form 5) and PUN Ying Anna (HKU Math, Year 3).
Suppose there is such a solution. Then z2 = x3 − y3 =(x−y)(x2+xy+y2)
= (x−y) ((x−y)2+3xy). (*)
Since y is a prime, z is not divisible by 3 and z is not divisible by y, (*) implies (x,y)=1 and (x−y,3)=1. Then
(x2+xy+y2, x−y)=(3xy, x−y)=1. (**)
Now (*) and (**) imply
x−y=m2, x2+xy+y2=n2 and z=mn
for some positive integers m and n. Consequently,
4n2= 4x2+4xy+4y2=(2x+y)2+3y2.
Then 3y2=(2n+2x+y)(2n−2x−y). Since
y is prime, there are 3 possibilities: (1) 2n+2x+y = 3y2, 2n−2x−y = 1
(2) 2n+2x+y = 3y, 2n−2x−y = y (3) 2n+2x+y = y2, 2n−2x−y = 3.
In (1), subtracting the equations leads to 3y2−1 = 2(2x+y) = 2(2m2+3y). Then
m2 + 1 = 3y2 − 6y − 3m2 ≡ 0
(
mod 3)
.However, m2 + 1 ≡ 1 or 2 (mod 3). We
get a contradiction.
In (2), subtracting the equations leads to x = 0, contradiction.
In (3), subtracting the equations leads
to y2−3 = 2(2x+y) = 2(2m2+3y), which can
be rearranged as (y−3)2−4m2=12. This
leads to y =
7
and m =1
. Then x = 8 and z = 13. Since 83−73=132, this gives the onlysolution.
Commended solvers: LKL Problem Solving Group (Madam Lau Kam Lung Secondary School of Miu Fat Buddhist Monastery).
Problem 315. Each face of 8 unit cubes is painted white or black. Let n be the total number of black faces. Determine the values of n such that in every way of coloring n faces of the 8 unit cubes black, there always exists a way of stacking the 8 unit cubes into a 2×2×2 cube C so the numbers of black squares and white squares on the surface of C are the same.
Solution. CHUNG Ping Ngai (La Salle
College, Form 5) and PUN Ying Anna (HKU Math, Year 3).
The answer is n = 23 or 24 or 25. First notice that if n is a possible value, then so is 48−n. This is because we can interchange all the black and white coloring and the condition can still be met by symmetry. Hence, without loss of generality, we may assume n ≤ 24. For the 8 unit cubes, there are totally 24 pairs of opposite faces. In each pair, no matter how the cubes are stacked, there is one face on the surface of C and one face hidden.
If n ≤ 22, there is a coloring that has [n/2] pairs with both opposite faces black. Then at least [n/2] black faces will be hidden so that there can be at most n−[n/2] ≤ 11 black faces on the surface of C. This contradicts the existence of a stacking with 12 black and 12 white squares on the surface of C. So only n = 23 or 24 is possible.
Next, start with an arbitrary stacking. Let b be the number of black squares on the surface of C. For each of the 8 unit cubes, take an axis formed by the centers of a pair of opposite faces and rotate the cube about that axis by 90°. Then take an axis formed by the centers of another pair of opposite faces of the same cube and rotate the cube about the axis by 90° twice. These three 90° rotations switch the three exposed faces with the three hidden faces. So after doing the twenty-four 90° rotations for the 8 unit cubes, there will be n−b black squares on the surface of C.
For n = 23 or 24 and b ≤ n, the average of b
and n−b is 11.5 or 12, hence 12 is between b and n−b inclusive.
Finally, observe that after each of the twenty-four 90° rotations, one exposed square will be hidden and one hidden square will be exposed. So the number of black squares on the surface of C can only increase by one, stay the same or decrease by one.
Therefore, at a certain moment, there will be exactly 12 black squares (and 12 white squares) on the surface of C. Commended solvers: G.R.A. 20 Problem Solving Group (Roma, Italy) and LKL Problem Solving Group (Madam Lau Kam Lung Secondary School of Miu Fat Buddhist Monastery).
Olympiad Corner
(continued from page 1)
Second Day: June 7th, 2008
Problem 4. Determine all functions f mapping the set of positive integers to the set of non-negative integers satisfying the following conditions: (1) f(mn) = f(m)+f(n),
(2) f(2008) = 0, and
(3) f(n) = 0 for all n ≡ 39 (mod 2008). Problem 5. Which positive integers are missing in the sequence {an}, with
] [ ] [ n 3 n n an= + +
for all n ≥ 1? ([x] denotes the largest integer less than or equal to x, i.e. g with g ≤ x < g+1.)
Problem 6. We are given a square ABCD. Let P be a point not equal to a corner of the square or to its center M. For any such P, we let E denote the common point of the lines PD and AC, if such a point exists. Furthermore, we let F denote the common point of the lines PC and BD, if such a point exists. All such points P, for which E and F exist are called acceptable points. Determine the set of all acceptable points, for which the line EF is parallel to AD.
So
Gmec,
,
The analogous construction can be found for B. Since LQBR = 90°, B must lie on the circle with diameter QR. Since BD is the bisector of LQBR, the point 'in which BD meets the
@
Solution: Since (1 - a)+
(1-
b)+
(1-
c) = 2, the left-hand side of the given inequality is circurncircle of QBR must lie on the bisector of QR. The line BD must therefore also pass the weighted geometric mean of a, b and c with weights (1-
a)/2, (1-
b ) / 2 and (1-
c)/2 through one of the points in which the bisector of QR meets the circle with diameter QR, both respectively. By the weightedAM-GM
inequality (and applying theAM-QM
inequality), we of which can also be readily constructed. We obtain four possible lines BD, each of whichtherefore have can be intersected with the respective circle to determine B and D. If the two polnts are on
J
-( 1 - a ) . a + -( l - b ) . b + -( l - c ) . c different sides of g, the comers of the resulting square are not labelled in the correct older, and
al-"p-bc'-c
<
-
2 correct points B and D therefore only result from lines joining points on the same slde of g.
( a + b + c ) - ( a 2 + @ + $ ) Once B and D have been found, the segment AC is found by rotating BD by 90" around the
-
-
2 mid-point of BD, and we are finished.
1 - 3 -
(d*l2
-
-2 -
@
Solution: Since 2008 = 23-
251, combining 2) and I) yields 3.
f(2)+
f (251) = 0. Since allvalues off (m) are non-negative, we therefore obtain f (2) = f (251) = 0. For any n = 2".251b 7n 1 - 3 . ( 9 ) ~
.
5 2 with gcd(m, 2008) = 1 we therefore also have f (n) = a . f ( 2 )
+
b f (251)+
f (m) = f (rn)1 For any m with gcd(m, 2008) = 1 there exists a positive integer k such that km r 1 (mod 2008),
- -
-
3' and therefore 39km r 39 (mod 2008). By 3) we therefore have f (39km) = f (39)
+
f (k)+
-- - f (m) = 0, and we therefore also have f (m) = 0 for all such m. In summary, we see that
- - - - - - -- -- f(m) = 0 for all positive integers m, and the only function with the required properties is
Solution: a) Assuming that such a polynomial exists, we note that P(2) = 1004 must hold. f (m) E 0.
0
C e d a a P(2) modulo 2, we see that the absolute d d m t at^ must be even. On theother hand, the analogous argument for P(8) = 251 shows us that a0 must be odd. This is a -- -
contradiction, and we see that there can be no polynomial with the rquired properties. solution: The three partial sequences b, = n, c, =
f i
and dn =f i
are all increasing) andb) If n contains any prime factor more than once, the argument used in part a) can be used in ,
.
therefore so is %. In each step (from n-
1 to n) bn increases by one. G only Increases byan analogous way to show that no such polynomial can exist. If n contains any two different one if the step is for n = k2 and d, only if n = k3. Both c, and dn increase by one only for prime factors p < q, we see that P(n) = 1 would yield % = k6. hi^ means that there are only certain specific numbers missing in the sequaenrr,
of which we name specifically. If n = k2 or n = k3 but not n = k6 exactly one number will yield q p (mod q), which is a contradiction. We see that n must be either equal to ,l or a
be missing to the sequence of d positive integers. If n = k6, two such numbers prime number.
be missing. iqoting also that the numbers 1 and 2 are also missing since the sequence beg1ns In either case, we can take Pn(x) = n
+
1-
x, getting Pn(l) = n and Pn(n) = 1, fdfdling therequirements of the problem. We se that such a polynomial exists only for n = 1 or n prime. with a1 = 3, numbers of the following types are missing:
if k is not a perfect cube
---
Solution: S i c e LPDS = 90°, D must lie on the circle with diameter PS. hrthermore, since k 3 + [ @ ] + k - 1 if k is not a perfect square
@
B D is the bisector of LPDS, the point in which BD meets the circumcircle of P D S must liek6
+
k3 -I- k2-
1 for any k on the b i i r of PS. We see that the line BD must pass through one of the points in whichthe bisector of PS meets the circle with diameter PS, both of which can be readily constructed k6
+
k3+
k2-
2 for any k. by straight-edge and compass methods.- -- -.
---
@
Solution: We asUme that P is given in such a way that EF is parallel to AD. sins p ,.mot be ill a Corner of the square or in M , neither E nor F is in any of these points.we
consider the triangle DEC. Since EF is parallel to AD, it is also perpendicular to C D , and since F-
liesOn BD, DF is perpendicular to EC. We see that F is the orthocenter of DEC, and FC is perpendicular to DE. We see that LDPC is a right angle, and P therefore lies on the circle with diameter CD.
Taking P anywhere on this circle except in C, Dl M or in the point symmetric to M with respect to CD, we can construct E and F. The resulting triangles DEC are always non-degenerate and not right-angled with orthocenter in F. The required set of points is therefore the set of points on the circle with diameter CD, not including C , Dl M or in the point symmetric to M
