Volume 14, Number 5 March - April, 2010
Ramsey Numbers and Generalizations
Law Ka Ho
Olympiad Corner
Here are the Asia Pacific Math Olympiad problems on March 2010. Problem 1. Let ABC be a triangle with ∠BAC≠90°. Let O be the circumcenter of triangle ABC and let Γ be the circumcircle of triangle BOC. Suppose that Γ intersects the line segment AB at P different from B, and the line segment AC at Q different from C. Let ON be a diameter of the circle Γ. Prove that the quadrilateral APNQ is a parallelogram.
Problem 2.
For a positive integer k, call an integer a pure k-th power if it can be represented as mk for some integer m. Show that for every positive integer n there exist n distinct positive integers such that their sum is a pure 2009-th power, and their product is a pure 2010-th power.
Problem 3. Let n be a positive integer. n people take part in a certain party. For any pair of the participants, either the two are acquainted with each other or they are not. What is the maximum possible number of the pairs for which the two are not acquainted but have a common acquaintance among the participants?
(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 May 21, 2010.
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, The Hong Kong University of Science and Technology
The following problem is classical: among any 6 people, there exist 3 who know each other or 3 who don’t know each other (we assume if A knows B, then B knows A). When 6 is replaced by 5, this is no longer true, as can be seen by constructing a counterexample. We write R(3,3) 6= and this is called a Ramsey number. In general, R m n( , ) denotes the smallest positive integer k such that, among any k people, there exist m who know each other or n who don’t know each other.
How do we know that R m n( , ) exists for all m, n? A key result is the following.
Theorem 1. For any m, n > 1, we have
( , ) ( 1, ) ( , 1)
R m n ≤R m− n +R m n− .
Proof. Take R m( −1, )n +R m n( , −1) people. We need to show that there exist m people who know each other or n people who don’t know each other. If a person X knows R m( −1, )n others, then among the people X knows, there exist either m−1 who know each other (so that together with m, there are m people who know each other) or n people who don’t know each other, so we are done. Similarly, if X doesn’t know R m n( , −1) others, we are also done. But one of these two cases must occur because the total number of ‘others’ is R m( −1, )n +R m n( , − −1) 1.
Using Theorem 1, one can easily show (by induction on m n+ ) that
2 1 ( , ) m n m R m n C + − − ≤ . This establishes an upper bound on R m n( , ). To establish a lower bound, we need a counter-example. While construction of counter- examples is in general very difficult, the probabilistic method (see Vol. 14, No. 3) may be able to help us in getting a non-constructive proof. Yet to get the exact value of a Ramsey number, the lower and upper bounds must match, which is extremely difficult. For m, n > 3, fewer than 10 values of R m n( , ) are known:
(3, 4) 9, (3,5) 14, (3,6) 18 (3,7) 23, (3,8) 28, (3,9) 36 R R R R R R = = = = = = (4, 4) 18, (4,5) 25 R = R =
Even R(5,5) is unknown at present. The best lower and upper bounds obtained so far are respectively 43 and 49. Paul Erdös once made the following remark.
Suppose an evil alien would tell mankind “Either you tell me [the value of R(5,5)] or I will exterminate the human race.”… It would be best in this case to try to compute it, both by mathematics and with a computer.
If he would ask [for the value of R(6,6)], the best thing would be to destroy him before he destroys us, because we couldn’t [determine R(6,6)].
Problems related to the Ramsey numbers occur often in mathematical competitions.
Example 2. (CWMO 2005) There are n new students. Among any three of them there exist two who know each other, and among any four of them there exist two who do not know each other. Find the greatest possible value of n.
Solution. The answer is 8. First, n can be 8 if the 8 students are numbered 1 to 8 and student i knows student j if and only if |i−j| 1, 4≡/ (mod 8). Next, suppose n=9 is possible. Then no student may know 6 others, for among the 6 either 3 don’t know each other or 3 know each other (so together with the original student there exist 4 who know each other). Similarly, it cannot happen that a student doesn’t know 4 others. Hence each student knows exactly 5 others. But this is impossible, because if we sum the number of others whom each student know, we get 9 5 45× = , which is odd, yet each pair of students who know each other is counted twice.
Mathematical Excalibur, Vol. 14, No. 5, Mar. - Apr. 10 Page 2
Remark. The answer to the above problem is R(3, 4) 1− , as can be seen by comparing with the definition of R(3,4).
The Ramsey number can be generalised in many different directions. One is to increase the number of statuses from 2 (know or don’t know) to more than 2, as the following example shows.
Example 3. (IMO 1964) Seventeen people correspond by mail with one another — each one with all the rest. In their letters only three different topics are discussed. Each pair of correspondents deals with only one of these topics. Prove that there are at least three people who write to each other about the same topic.
Solution. Suppose the three topics are A, B and C. Pick any person; he writes to 16 others. By the pigeonhole principle, he writes to 6 others on the same topic, say A. If any two of the 6 people write to each other on A, then we are done. If not, then these 6 people write to each other on B or C. Since
(3,3) 6
R = , either 3 of them write to each other on B, or 3 of them write to each other on C. In any case there exist 3 people who write to each other about the same topic.
Remark. The above problem proves (3,3,3) 17
R ≤ , where R m n p( , , ) is defined analogously as R m n( , ) except that there are now three possible statuses instead of two. It can be shown that R(3,3,3) 17= by constructing a counterexample when there are only 16 people.
Another direction of generalization is to generalise ‘m people who know each other’ or ‘n people who don’t know each other’ to other structures. (Technically, the graph Ramsey number R G H( , ) is the smallest positive integer k such that when every two of k points are joined together by a red or blue edge, there must exist a red copy of G or a blue copy of H. Hence
( , ) ( m, n)
R m n =R K K , where Km
denotes the complete graph on m vertices, i.e. m points among which every two are joined by an edge).
Example 4. N people attend a meeting, and some of them shake hands with each other. Suppose that each person shakes hands with at most 100 other people, and among any 50 people there
exist at least two who have shaken hands with each other. Find the greatest possible value of N.
Solution. The answer is 4949. We first show that N = 4949 is possible: suppose there are 49 groups of 101 people each, and two people shake hands if and only if they are in the same group. It is easy to check that the requirements of the question are satisfied. Now suppose N = 4950 and each person shakes hands with at most 100 others. We will show that there exist 50 people who have not shaken hands with each other, thus contradicting the given condition. To do this, pick a first person P1 and cross out all those who have shaken hands with him. Then pick
2
P from the rest and again cross out those who have shaken hands with him, and so on. In this way, at most 100 people are crossed out each time. After P49 is chosen, at least 4950 49 49 100 1− − × = person remains, so we will be able to choose P50. Because of the ‘crossing out’ algorithm, we see that no two of P1, P2, …, P50 have shaken hands with each other.
Remark. By identifying each person with a point and joining two points by a red line if two people have shaken hands and a blue line otherwise, we see that the above problem proves R K( 1,100,K50) 4950= . Here K1,100 is the graph on 101 points by joining 1 point to the other 100 points.
The Van der Waerden number ( , )W r k is the smallest positive integer N such that if each of 1, 2, …, N is assigned one of r colours, then there exist a monochromatic k-term arithmetic progression. The following example shows that we have
(2,3) 325
W ≤ .
Example 5. If each of the integers 1, 2, …, 325 is assigned red or blue colour, there exist three integers p, q, r which are assigned the same colour and which form an arithmetic progression.
Solution. Divide the 325 integers into 65 groups G1 = {1, 2, 3, 4, 5}, G2 = {6, 7, 8, 9,
10}, …, G65 = {321, 322, 323, 324, 325}.
There are 25 = 32 possible colour patterns
for each group. Hence there exist three groups Ga and Gb, 1≤ < ≤a b 33, whose colour patterns are the same. We note that 2b a− ≤65 and that a, b, 2b a− form an arithmetic progression. Now two of the first three numbers of Ga are of the same colour, say, the first and third are red (it can be seen that the proof goes exactly the
same way if it is the first and second, or second and third). If the fifth is also red, then we are done. Otherwise, the first and third numbers of both Ga and Gb (recall that they have identical colour patterns) are red while the fifth is blue. If the fifth number of G2b-a is
red, then it together with the first number of Ga and the third number of Gb form a red arithmetic progression; if it is blue, then it together with the fifth numbers of Ga and Gb form a blue arithmetic progression.
It can be shown via a two-dimensional inductive argument that ( , )W r k exists for all r, k. We see that the existence of Ramsey numbers and van der Waerden numbers are very similar: both say that the desired structure exists in a sufficiently large population.
An analogy to this (though not mathematically rigorous) is that when there are sufficiently many stars in the sky, one can form from them whatever picture one wishes. (This is one of the lines in the movie A Beautiful Mind!)
Yet another generalization of the van der Waerden Theorem (which says that ( , )W r k exists for all r, k) is the Hales-Jewett Theorem. The exact statement of the theorem is rather technical, but we can look at an informal version here. We are familiar with the two-person tic-tac-toe game played on a 3 3× square in two dimensions. We also have the two- person tic-tac-toe game played on a 4 4 4× × cube in three dimensions (try it out at http://www.mathdb.org/fun/ games/tie_toe/e_tie_toe.htm!). Both games can end in a draw. However, it is easy to see that a two-person tic-tac- toe game played on a 2 2× square in two dimensions cannot end in a draw. The Hales-Jewett Theorem says that for any n and k, the k-person tic-tac-toe game played on an n n× × ×L n (D factors of n, where D is the dimension) hypercube cannot end in a draw when D is large enough! (For instance, we have just seen that when n = 2 and k = 2, then D = 2 is large enough, while when n = 3 and k = 2, then D = 2 is not large enough.) In case k = 2 (i.e. a two- person game) and when D is large enough so that a draw is impossible, it can be shown (via a so-called strategy stealing argument) that the first player has a winning strategy!
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 May 21, 2010.
Problem 341. Show that there exists an infinite set S of points in the 3-dimensional space such that every plane contains at least one, but not infinitely many points of S.
Problem 342. Let f(x)=anxn+⋯+a
1x+p
be a polynomial with coefficients in the integers and degree n>1, where p is a prime number and
|an|+|an−1|+⋯+|a1| < p.
Then prove that f(x) is not the product of two polynomials with coefficients in the integers and degrees less than n. Problem 343. Determine all ordered pairs (a,b) of positive integers such that a≠b, b2+a=pm (where p is a prime number, m is a positive integer) and a2+b is divisible by b2+a.
Problem 344. ABCD is a cyclic quadrilateral. Let M, N be midpoints of diagonals AC, BD respectively. Lines BA, CD intersect at E and lines AD, BC intersect at F. Prove that
. 2 EF MN BD AC AC BD− = Problem 345. Let a1, a2, a3, ⋯ be a
sequence of integers such that there are infinitely many positive terms and also infinitely many negative terms. For every positive integer n, the remainders of a1, a2, ⋯, an upon divisions by n are
all distinct. Prove that every integer appears exactly one time in the sequence.
*****************
Solutions
****************
Problem 336. (Due to Ozgur Kircak, Yahya Kemal College, Skopje, Macedonia) Find all distinct pairs (x,y) of integers satisfying the equation
. 2009 2009 3 3 y y x x + = +
Solution. CHOW Tseung Man (True Light Girls’ College), CHUNG Ping Ngai (La Salle College, Form 6), HUNG Ka Kin Kenneth (Diocesan Boys’ School), D. Kipp JOHNSON (Valley Catholic School, Beaverton, Oregon, USA), LI Pak Hin (PLK Vicwood K. T. Chong Sixth Form College), Emanuele NATALE (Università di Roma “Tor Vergata”, Roma, Italy), Pedro Henrique O. PANTOJA (UFRN, Natal, Brazil), PUN Ying Anna (HKU), TSOI Kwok Wing (PLK Centenary Li Shiu Chung Memorial College), Simon YAU Chi- Keung and Fai YUNG.
All pairs (x,x) satisfy the equation. If (x,y) satisfies the equation and x≠y, then
). 3 (mod 2 2009 3 3 2 2 = ≡ − − = + + y x y x y xy x
However, x2+xy+y2≡x2−2xy+y2=(x−y)2≡ 0
or 1 (mod 3). So there are no solutions with x≠y.
Problem 337. In triangle ABC, ∠ABC = ∠ACB=40°. P and Q are two points inside the triangle such that ∠PAB = ∠ QAC =20° and ∠PCB =∠QCA =10°. Determine whether B, P, Q are collinear or not.
Solution 1. CHUNG Ping Ngai (La Salle College, Form 6) and HUNG Ka Kin Kenneth (Diocesan Boys’ School).
Let ∠PBA=a, ∠PBC=b, ∠QBA=a’ and ∠QBC=b’. By the trigonometric form of Ceva’s theorem, we have
, 20 sin 30 sin sin 80 sin 10 sin sin sin sin sin sin sin sin 1 o o o o b a PAB PAC PCA PCB PBC PBA = ∠ ∠ ∠ ∠ ∠ ∠ = . 80 sin 10 sin ' sin 20 sin 30 sin ' sin sin sin sin sin sin sin 1 o o o o b a QAB QAC QCA QCB QBC QBA = ∠ ∠ ∠ ∠ ∠ ∠ =
As sin10°sin80° = sin10°cos10° =½sin20° = sin30°sin20°, we obtain sin a= sin b and sin a’ = sin b’. Since 0<a,b,a’,b’<90° and a+b=40°=a’+b’, we get a=b=a’=b’=20°, i.e. ∠ PBA= ∠ PBC= ∠ QBA= ∠ QBC. Therefore, B, P, Q are collinear.
Solution 2. LEE Kai Seng.
We will show B,P,Q collinear by proving lines BQ and BP bisect ∠ABC.
Draw an equilateral triangle BDC with D on the same side of BC as A. Since ∠ABC =∠ACB=40°, AB=AC. Then both D and A are equal distance from B and C. So DA
bisects ∠BDC. We have A B C D Q ∠QCD = 60°−∠BCQ = 30° = ∠ADC. Also, ∠ DCA =∠QCD −∠QCA = 20° = ∠QAC, which implies QA||CD. Then AQCD is an isosceles trapezoid, so AD = QC. This with BD=BC and ∠ BDA = 30° = ∠QCB imply ΔBDA ≅ ΔQCB. Then BA=BQ. Since ∠BAQ = ∠BAC −∠QAC = 100°−20° = 80°, we get ∠ABQ = 20° = ½∠ABC. So BQ bisects ∠ABC.
A
B Q C
P
E
Extend BA to a point E so that BE=BC. Then ∠BCE = ½(180°−∠ABC) = 70°. Next, we will show ΔEPC is equilateral.
We have ∠PCE=∠BCE−∠PCB=60°, ∠ACE=∠BCE−∠BCA=30°=½∠PCE. So CA bisects ∠PCE. Next, ∠CAE = 180°−∠BAC = 80° = ∠BAC−∠BAP = ∠CAP. Then ΔCAE ≅ ΔCAP. So CE = CP and ΔEPC is equilateral. Then B, P are equal distance from E and C. Hence BP bisects ∠ABC.
Other commended solvers: CHAN Chun Wai (St. Paul’s College), LI Pak Hin (PLK Vicwood K. T. Chong Sixth Form College), PUN Ying Anna (HKU).
Problem 338. Sequences {an} and {bn} satisfy a0=1, b0=0 and for n=0,1,2,…,
. 4 7 8 , 3 6 7 1 1 − + = − + = + + n n n n n n b a b b a a
Prove that an is a perfect square for all n=0,1,2,…
Solution 1. CHUNG Ping Ngai (La Salle College, Form 6), HUNG Ka Kin Kenneth (Diocesan Boys’ School), D. Kipp JOHNSON (Valley Catholic School, Beaverton, Oregon, USA), LI Pak Hin (PLK Vicwood K. T. Chong Sixth Form College), NGUYEN Van Thien (Luong The Vinh High School, Dong Nai, Vietnam), O Kin Chit Alex (G.T. (Ellen Yeung) College), Ercole SUPPA (Teramo, Italy) and YEUNG Chun Wing (St. Paul’s College).
Mathematical Excalibur, Vol. 14, No. 5, Mar. - Apr. 10 Page 4 Solving for bn in the first equation and
putting it into the second equation, we have
an+2=14an+1−an−6 for n=0,1,2, … (*) with a0=1and a1=4. Let dn=an−½. Then
(*) becomes dn+2 =14dn+1−dn. Since the roots of x2 − 14x + 1 = 0 are 7±4 3,
we get dn is of the form α(7−4 3)n + β(7+4 3)n. Using d 0=½ and d1=3½, we get α=¼ and β=¼. So . 4 ) 3 4 7 ( ) 3 4 7 ( 2 2 1 n n n n d a = + = + − + +
Now, consider the sequence {cn} of positive integers, defined by c0=1, c1=2
and
cn+2=4cn+1−cn for n=0,1,2,…. (**) Since the roots of x2−4x+1= 0 are
, 3 2± as above we get . 2 ) 3 2 ( ) 3 2 ( n n n c = − + +
Squaring cn, we see an=cn2.
Solution 2. William CHAN and Invisible MAK (Carmel Alison Lam Foundation Secondary School).
The equations imply
an+2=14an+1−an−6 for n=0,1,2, … (*) We will prove anan+2=(an+1+3)2 by math
induction. The case n=0 is 1×49=(4+3)2.
Suppose an−1an+1=(an+3)2. Then
2 1 2−( + +3) + n n na a a . 0 ) 3 ( ) 3 ( ) 6 14 ( 9 6 6 14 ) 3 ( ) 6 14 ( 2 1 1 2 1 1 1 2 1 2 1 2 1 1 = + − = + − − − = − − − − − = + − − − = + − + + + + + + + n n n n n n n n n n n n n n n n n a a a a a a a a a a a a a a a a a
This completes the induction.
Next, we will show all an’s are perfect squares. Now a0=12 and a1=22. Suppose
an−1=r2 and an=s2, we get an+1=(an+3)2/r2
and an+2=(an+1+3)2/s2. Since the square
root of a positive integer is an integer or an irrational number, an+1 and an+2 are perfect squares. By mathematical induction, the result follows.
Other commended solvers: PUN Ying Anna (HKU), TSOI Kwok Wing (PLK Centenary Li Shiu Chung Memorial College).
Problem 339. In triangle ABC,∠ACB =90°. For every n points inside the
triangle, prove that there exists a labeling of these points as P1, P2, …, Pn such that
. 2 2 1 2 3 2 2 2 1P PP P P AB P + +L+ n− n ≤
Solution. Federico BUONERBA
(Università di Roma “Tor Vergata”, Roma, Italy), HUNG Ka Kin Kenneth (Diocesan Boys’ School) and PUN Ying Anna (HKU).
We will prove the following more general result:
Let ABC be a triangle with ∠ACB =90°. For every n points inside or on the sides of the triangle, there exists a labeling of these points as P1, P2, …, Pn such that
. 2 2 2 1 2 2 1 2 1 PP P P PB AB AP + +L+ n− n + n ≤ We prove this by induction on n. For the case n=1, since ∠AP1B ≥ 90°, the cosine
law gives AP12+P1B 2 ≤ AB 2.
Next we assume all cases less than n are true. For the case n, we can divide the original right triangle into two right triangles by taking the altitude from C to H on the hypotenuse AB. We can assume that the two smaller right triangles AHC and BHC contain m > 0 and n−m > 0 points respectively (otherwise, one of these two smaller triangles contains all the points and we keep dividing in the same way the smaller right triangle which contains all the points). Since m < n and n−m < n, by the induction hypothesis, there exist a labeling of points in triangle AHC as P1, P2, …, Pm such that
2 2 2 1 2 2 1 2 1 PP P P PC AC AP + +L+ m− m+ m ≤ and a labeling of points in triangle BHC as Pm+1, Pm+2, …, Pm such that . 2 2 2 2 1 2 1 P P PB CB CPm+ + m+ m+ +L+ n ≤
Since ∠PmCPm+1 ≤ 90°, the cosine law gives PmPm+12≤ PmC2+CPm+12. Then 2 2 1 2 2 1 2 1 PP P P PB AP + +L+ n− n + n ≤ AC2+CB2 = AB2.
Problem 340. Let k be a given positive integer. Find the least positive integer N such that there exists a set of 2k+1 distinct positive integers, the sum of all its elements is greater than N and the sum of any k elements is at most N/2.
Solution. CHAN Chun Wai (St. Paul’s College), CHOW Tseung Man (True Light Girls’ College), CHUNG Ping Ngai (La Salle College, Form 6), HUNG Ka Kin Kenneth (Diocesan Boys’ School),
LI Pak Hin (PLK Vicwood K. T. Chong Sixth Form College), PUN Ying Anna (HKU).
Let a1, a2, …, a2k+1 be such a set of
2k+1 of positive integers arranged in increasing order. We have
. 1 2 1 2 1 2 1 2 1 + ≥ + ≥
∑
∑
+ + = + = k k i i k i i a N a Then 1 1 1 2 2 1≥∑
−∑
+ = + + = + k i i k k i i k a a a . 1 1 ) 1 ( 1 ) ( 2 1 1 1 + + = + + ≥ + − =∑
∑
= = + + k k k a a k i k i i k i Also,∑
∑
+∑
+ = + + = + + + + = + − = ≥ 2 1 2 1 2 2 1 1 1 2 2 ) ( 2 k k i k k i k k i k k i i a a a a N . 2 3 3 2 ) 1 ( ) 1 ( 2 3 2 1 2 2 k k k k k k k i k k i + + = + + + − − ≥∑
+ + =Now all inequalities above become equality if we take ai=k2+i for i=1,
2, …, 2k+1. So the least positive value of N is 2k3+3k2+3k.
Olympiad Corner
(continued from page 1)
Problem 4. Let ABC be an acute triangle satisfying the condition AB>BC and AC>BC. Denote by O and H the circumcenter and orthocenter, respectively, of the triangle ABC. Suppose that the circumcircle of the triangle AHC intersects the line AB at M different from A, and that the circumcircle of the triangle AHB intersects the line AC at N different from A. Prove that the circumcenter of the triangle MNH lies on the line OH. Problem 5. Find all functions f from the set R of real numbers into R which satisfy for all x, y, z ∊ R the identity f(f(x)+f(y)+f(z))
Solution: From the assumption that the circle
r
intersects both of the line segments ABand AC, it follows that the 4 points N, C, Q, 0 are iocated on
r
in the order of N, C, Q, 0 or in the .order of N, C, 0, Q. The following argument for the proof of the assertion, of the problem is valid in either case. Since, LNQC and LNOC are subtended by the same arcNC of
r
at the points Q,and.o,~espectively,
onr,
we have LNQC:=;LNOC,~We
also have LBOC=
2LBAC, since LBOC and LBAC are subtended by the same arc BC of the circum-circle of the triangle ABC at the center 0 of the circle and at the point A on the circle, respectively. From OB=
OC and the fact that ON is a diameter ofr,
it follows that the tri'angles OBN and OCN are congruent, and th,erefore we obtain 2LNOC=
L.BOC.Consequently, we have LNQO
= ~LBOC
=
L.BAC,:which shows that the 2 lines AP, QN~are parallel. '
In the same manner , we can show tl\at the 2 lines AQ, P N are also parallel. Thus, the quadrilateral APNQ is a parallelogram. '
f2:I
Solution: For the sake of simplicity, let us set k = 2009. , . '\.7'
,First of all, choose n distinct positive integers bl, ... , bn SUItably so that their product '~~~pure k+ 1-th power (for example, let bi = ik+1 for i
=
1,·· . , n). Then we have bl ... bn=
t . for some positive integer t. Set bl+ .,. +
bn=
s. . ' . 'Now we set ai
=
biSk2-1 for i = 1,· .. , n, and show that al,'" ,an satisfy the reqUIred; conditions. Since bl,'" , bn are ~istin_c_t·positive integers, it is clear that so are ~1!'" , an- ,From
al
+ ... +
an = sk2-1(bl+ ... +
bn )=
sk2 = (sk?009,( k2-1)nb b _' (, k2-1)ntk+.l
=
(s(k-l)nt)2010 al ... an = s 1 . .'. n - sth t a satisfy the conditions on the sum and th~ product as well. This we can see a al,''', n
ends the proof of the assertion. "
~\ Solution: When 1 participant, say the person A, is mutually.acquainted ;;ith e.ach of the
GV
. .
1 ticipants and if there are no other acquamtance relatIOnshIps among remammg n - par , ' . . . ' t . l ' A the two are not mutual the participants, then for any paIr of partlclp~ts no mvo v~ng A ~o any such pair satisfies acq1J,aintances, but they have a commo~ac~ual~tance, ~a~~~(n~Z) = n2-3n+2.the requirement. Thus, tp.e number desITed m thiS case IS 2 Z
Suppose, then, n participants are separated into k(k ;:::: 2) groups, and t~e number. of pe~ple in each group is given by ai, i
=
1,··· , k. In such a case, the nu:nber of,raIrs for WhI%h paired people are not mutually acquainted but have a common acquamtance IS at most ~i=l a,.C2,where we set 1C2 = 0 for convenience. Since aOZ
+
b02:::; a+bOZ holds for any paIr of positive integers a, b, we have L:~=1 ai 02:::; al02+
n-al Oz. Fro~ , 'nZ - n ' n Z n2 - 2n
alC2
+
n-al C2=
at -
nal+
--2-=
(al -'2)
+
4-'t f 11 th t C'
+
Oz takes its maXimum value when a1=
I, r/;. - 1. Therefore, we1 0 ows a aI' 2 n-al
h ave ",k wi-l ai 0 Z -
<
n-l 0 2, which shows that in the case where the number Qf groups ,are 2 all . t d b tor more -the number of the pairs for which paired people are not mutu y acquam e u
, 2-3n±Z h h d . d .
have a common acquaintance is at most n-102
=
n 2 ' and ence t e esne maximum, ' . . h . t ' n2-3n±2
number of the pairs satlsfymg t e reqmremen IS 2 •
@
S.olution: In thes~quel,
we denote LBAC = a,~CBA
=' (3, LACE!= /.
Let 0' be the Clrcumcenter of the trIangle M N H. The lengths of hne segments startmg from the point Hwill be treated as. signed quantities:
Let us denot~ by M', N' the point of intersection of CH, BH, respectively, with the cir-cumcircle of the triangle ABC (distinct from C, B, respectively.)_ From the fact that 4 points A, M, H, C lie on the same circle, we see· that LMH M', = ex holds. Furthermore,
L.BM'C, LBN'O and a are al1$ubtended by the same arc B~O of the circumcircle of the triangle ABC at points on the circle, and therefore, we have LBM'O = ex, and LB}!'C = a as well. We also have LABH= LA ON' as they are subtended by the same arc AN' of the circumcircle of the triangle ABC' at points on the circle. Since H M' 1. B M, H N' 1. AC, we
conclude that ' - '
LM'HB ='= 90° - LABH
=
90° - LACN'=
a 'is valid as well. Putting these facts together, we obtain the fact that the quadrilateral
HBM'M is a rhombus. In a similar mariner, we can conclude that the quadrilateral HCN'N
is also a rhombus. Since both ofithese rhombuses are made up of 4 right triangles with an angle of magnitude ex, we also see that these rhombuses are similar. ,
, Let us denote by P, Q the feet, of the perpendicular lines on HM and HN, respectively, drawn from the point 0'. Since 0' is the circutncenter of the triangle M N H, P, Q are re-spectively, the midpoints of the line segments H M, H N. Furthermore,' if we denote by R, S the feet of the perpendicular lines oIv'HM and HN, respectively, drawn from the point 0, then since 0 is the circumcenter oYboth the triangle M' BC and the triangle N' BO, we see that R is the intersection point pff HM and the perpendicular bisector of BM', and S is the intersection point of H N and the iperpendicular bisector of ON'.
We note that the similarity map ifJ between tlie rhombuses H B M' M and H C N' N carries the perpendicular bisector of BMI onto the perpendicular bisector of Cff', and straight line
Ii M onto. the straight line H N, and hence ifJ maps R onto S, and Ponto Q. Therefore, we get
IIP: HR = HQ : HS. If we now denote by X, Y the intersection points of the line HO' with the line through R and perpendicular to H P, and with the line through Sand perpendicu1ar to HQ, respectively, then we get
HO': HX = HP: HR= HQ :H8
=
HO': HYso that we must have HX = HY,' and therefore, X = Y; But it is obvious that the point of intersection of the line through! R and perpendicular to HP with the line through Sand perpendicular to HQ must be 0, and therefore,'we conclude that X
=
Y=
0 and that the points H, 0', 0 are collinear;®
Solution: It, is clear that iff.
is a constantfunctio~
which satisfies the given equation, then the constant must beo.
Conversely, f(x)=
0 cle:arly satisfies the given equation, so, ,the identicallyb
function is a solution. In the sequel, we consider the case where f is not aconstant f u n c t i o n . , '
Let t E R and substitute (x, y, z) = (t, 0, 0) and
(c,
y, z) = (0, t, 0) into the given functional equation. Then, we obtain, respectively,f(l(t)
+
2f(0))=
f(l(t) - f(O))+
f(l(O))*
2f(0), f(f(t)+
2f(0))=
f(l(O) - f(t))+
fU(O))+
2f(0),from which we conclude that f(f(t) 0:::.1(0)).
=
fU(O) - f(t)) holds for all t ER. Now, suppose for some pair Ul, U2, feUl) = f(U2) is, sat!sfied. Then by subs.tituting (x, y, z)=
(a, 0, Ul) an~ (X, y, z)=
(a, 0, U2) into the functional equation and comparing the resulting identities, we" can easily conclude thatf(aul)
=
f( au2) (*)holds for all a E R. Since f is not a constant function there ~xists an ao such that f(ao)- f(O) =1=
o. If We put Ul
=
j(ao) - f(O), U2 = -Ul, then f(Ul) = !(U2), sO we have by (*)f(aul)
=
f( au2)=
f( -~Ul)for all a E R. Since ~l =1= 0, we con~lude_,tnat
". .
f(x) =)(-x)
holds for all x E R. ,
Next, if feu)
= f(O)
for some U =1= 0, then by (*), we have f(au)=
f(aO) = f(O) for aU' a, which implies that f is a constant function, contradicting our assumption. Therefore, we must have f(a) =1= f(O) whenever a =1=o.
,
.
We
will
now show that if f(x) = fey) holds, then eithcf x= y
or x= -y
must hold. Suppose on the contrary that f(xo) ,; f(yo) holds for ~ome pair of non-zero numbers Xo, Yo for which Xo =1= Yo, Xo =1= -Yo. Since I( -Yo) = f(yo), we may assume, by replacing Yo by -Yo if necessary, that Xo and Yo have the same sign. In vieW of (*), we see that f(axo)= f(ayo)
holds for all a, and therefore, there exists some r>
O,r'#
1 such thatf(x)
=
f(rx)holds for all x. Replacing x by rx and y by ry in the given functional equation, we obtain f(f(rx)
+
f(ry)+
fez)) = f(f(rx) - f(ry))+
f(2r2d;y+
fez)) f 2f(r(x - y)z) (i),and replacing x by r2x in the functional equation, we get
f(j(r 2x)
+
fey)+
fez)) == f(l(r 2x) -fey))+
f(2r2x,y+
fez))+
2f((r2x - y)z) (ii).Since f(rx)
=
f(x) holds for all x E R, we see that except for the last term on the right-hand side, all the corresponding terms appearing in the identities (i) and (ii) above are equal, andhence ,we conclude that "
f(r(x - y)z)
=
f((r2:x;, - y)z)) (iii)must hold for arbitrary choice of x,y,z E R. Forarbitrarily fixed pair u,v ER, substitute (x, y, z) = (:2-::'~>'r2~l" 1) into the identity (iii)~Then we obtain f(v) ~ f(ru) = feu), since x - y = u, r2;i; - y = v, z = 1. But this implies ~hat the ftInction f is a constant, contradicting our assumption. Thus we conclude that if f(x) = fey) then either x = y or x
= -y
must hold.By substituting z
=
0 in the functional equation, we getf(f(x)
+
fey)+
f(O))=
f(j(x) - fey)+
f(O)) = f((I(x) - fey))+
f(2xy+
f(O))+
2f(0).-Changing y to -y in th'e identity above and using the fact that fey)=
f( -v), we see that' all the terms except the second term on the right-hand side in the identity above remain the same. Thus we conclude that f(2xy+
f(O))= f( -2xy
+
f(O)), from which we get either 2xy+
f(O)= -2xy
+
f(O) or 2xy+
f(O)=
2xy - f(O) for all x, Y E R. The first of these alternatives says that 4xy=
0, which is impossible if xy =1= O. Therefore the second alternative must be valid and we get that f(O) = O.Finally, let us show that if f satisfies the given functional equation and is n'ot ~ constant function, then f(x) = x2. Let x = y in the functional equation, then since f(O) = 0, we get
f(2f(x)
+
fez)) = f(2x2+
fez)),, from which we conclude that either 2f(x)
+
fez)=
2x2+
fez) or 2f(x)+
fez)=
_2x2 - fez) must hold. Suppose there exis;s Xo for which f(xo) =1= x6,then from the second alternative, we s,ee that fez)= -
f(xo) - Xo must hold for all z, which means that f must be a constant , function, contrary to our assumption. Therefore, the first alternative above must hold, andwe have f(o;).= x2 for all x, establishing our claim.
,It is easy to check that f(x)
