Volume 8, Number 1 February 2003 – March 2003

Olympiad Corner Functional Equations The Fifth Hong Kong (China) Kin Y. Li Mathematical Olympiad was held on December 21, 2002. The problems are as follow. A functional equation is an equation Solution. Step 1 Taking x = 0 = y, we get whose variables are ranging over f (0) = f (0) + f (0) + f (0) , which implies

functions. Hence, we are seeking all f (0) = 0. Problem 1. Two circles intersect at points possible functions satisfying the A and B. Through the point B a straight Step 2 We will prove f (kx) = k f (x) for equation. We will let denote the set of line is drawn, intersecting the first circle at ℤ k , x by induction. This is true for all integers, + or denote the positive ∊ ℕ ∊ℚ K and the second circle at M. A line ℤ ℕ k = 1. Assume this is true for k. Taking y integers, denote the nonnegative parallel to AM is tangent to the first circle ℕ0 = kx, we get integers, denote the rational numbers, at Q. The line AQ intersects the second ℚ denote the real numbers, + denote f ((k+1) x) = f (x + kx) = f (x) + f (kx) circle again at R. ℝ ℝ the positive real numbers and ℂ denote = f (x) + k f (x) = (k+1) f (x). the complex numbers. Step 3 Taking y = –x, we get (a) Prove that the tangent to the second circle at R is parallel to AK. In simple cases, a functional equation 0 = f (0) = f (x+ (–x)) = f (x) + f (–x), (b) Prove that these two tangents are can be solved by introducing some which implies f (–x) = – f (x). So concurrent with KM. substitutions to yield more information

or additional equations. f (–kx) = – f (kx) = – k f (x) for k∊ℕ. Problem 2. Let n ≥ 3 be an integer. In a Therefore, f (kx) = k f (x) for k ∊ℤ, x∊ℚ. conference there are n mathematicians. Example 1. Find all functions f : ℝ → ℝ Every pair of mathematicians such that Step 4 Taking x = 1/ k, we get communicate in one of the n official x2 f (x) + f (1 – x) = 2 x – x4 f (1) = f (k (1/ k)) = k f (1/ k), languages of the conference. For any three different official languages, there for all x ∊ℝ. which implies f (1/ k) = (1/ k ) f (1). exist three mathematicians who Step 5 For m , n , communicate with each other in these Solution. Replacing x by 1 – x, we have ∊ℤ ∊ℕ

three languages. Determine all n for 2 4 f (m/ n) = m f (1/ n) = (m/ n) f (1). which this is possible. Justify your (1– x) f (1– x) + f ( x ) =2 (1–x) – (1–x) . Therefore, f (x) = cx with c = f (1). claim. Since f (1 – x) =2 x – x4– x2 f (x) by the

(continued on page 4) given equation, substituting this into the Check: For f (x) = cx with c∊ℚ , last equation and solving for f (x), we 2 f (x+y) = c(x+y) = cx + cy = f (x) + f (y). Editors: 張 百 康 (CHEUNG Pak-Hong), Munsang College, HK get f (x) = 1– x .

高 子 眉 (KO Tsz-Mei) 2 梁 達 榮 (LEUNG Tat-Wing) Check: For f (x) = 1 – x , In dealing with functions on ℝ, after finding the function on , we can often 李 健 賢 (LI Kin-Yin), Dept. of Math., HKUST x2 f (x) + f (1–x) = x2 (1– x2 )+(1– (1– x)2 ) ℚ 吳 鏡 波 (NG Keng-Po Roger), ITC, HKPU finish the problem by using the = 2 x – x4. Artist: 楊 秀 英 (YEUNG Sau-Ying Camille), MFA, CU following fact.

Acknowledgment: Thanks to Elina Chiu, Math. Dept., HKUST For certain types of functional equations, Density of Rational Numbers For every for general assistance. a standard approach to solving the real number x, there are rational On-line: http://www.math.ust.hk/mathematical_excalibur/ problem is to determine some special numbers p1, p2, p3, … increase to x and The editors welcome contributions from all teachers and values (such as f ( 0 ) or f ( 1 ) ), then there are rational numbers q1, q2, q3, … students. With your submission, please include your name, inductively determine f ( n ) for n , decrease to x. address, school, email, telephone and fax numbers (if available). ∊ ℕ0 Electronic submissions, especially in MS Word, are encouraged. follow by the values f ( 1 / n ) and use This can be easily seen from the decimal The deadline for receiving material for the next issue is density to find f ( x ) for all x . The February 28, 2003. ∊ ℝ representation of real numbers. For following are examples of such For individual subscription for the next five issues for the 02-03 example, the number π = 3.1415… is the approach. academic year, send us five stamped self-addressed envelopes. limits of 3, 31/10, 314/100, 3141/1000, Send all correspondence to: 31415/10000, … and also 4, 32/10, Example 2. Find all functions f : → Dr. Kin-Yin LI ℚ ℚ 315/100, 3142/1000, 31416/10000, …. Department of Mathematics such that the Cauchy equation The Hong Kong University of Science and Technology Clear Water Bay, Kowloon, Hong Kong f ( x + y ) = f ( x ) + f ( y ) (In passing, we remark that there is a similar fact with rational numbers Fax: (852) 2358 1643 holds for all x, y . Email: [email protected] ∊ℚ replaced by irrational numbers.) Mathematical Excalibur, Vol. 8, No. 1, Feb 03- Mar 03 Page 2

Example 3. Find all functions f (x + y) = f (x) + f (y) ≥ f (x), = f (x). Since f (x) is not the zero f : → such that function, f (1) = 1. Setting z = 0, t = 0, ℝ ℝ which implies f is increasing. Now for we get f (x) f (y) = f (xy) for all x,y. In f ( x + y) = f ( x ) + f ( y ) any x A , by the density of rational ∊ ∖ℚ particular, f (w) = f (w1/2)2 ≥ 0 for numbers, there are p , q such that p for all x, y ∊ ℝ and f (x) ≥ 0 for x ≥ 0. n n∊ℚ n w > 0. < x < q , the p ’s increase to x and the n n q ’s decrease to x. As f is increasing, we Solution. Step 1 By example 2, we n Setting x = 0, y = 1 and t = 1, we have have p = f (p ) ≤ f (x) ≤ f (q ) = q . have f (x) = x f (1) for x∊ℚ. n n n n 2 f (1) f (z) = f (−z) + f (z), which Taking limits, the sandwich theorem implies f (z) = f (−z) for all z. So f is Step 2 If x ≥ y, then x – y ≥ 0. So gives f (x) = x for all x A. ∊ even. f (x) = f ((x–y)+y) = f (x–y)+f (y )≥ f (y). Fact 2. If a function f : ( 0, ∞ ) → ℝ Define the function g: (0, ∞) → by Hence, f is increasing. satisfies f (xy) = f (x) f ( y) for all x, y > ℝ g(w)= f (w1/2) ≥ 0. Then for all x,y>0, 0 and f is monotone, then either f(x)=0 Step 3 If x ∊ℝ, then by the density of for all x > 0 or there exists c such that g (xy) = f ((xy)1/2) = f (x1/2 y1/2) rational numbers, there are rational pn, f (x) = xc for all x > 0. = f (x1/2) f (y1/2) = g (x) g (y). qn such that pn ≤ x ≤ qn, the pn’s increase to x and the qn’s decrease to x. Proof. For x > 0, f (x) = f (x1/2)2 ≥ 0. Also Next f is even implies g (x2) = f (x) for So by step 2, pn f (1) = f (pn) ≤ f (x) ≤ f (1) = f (1) f (1) implies f (1) = 0 or 1. If all x. Setting z = y, t = x in the given f (qn) = qn f (1). Taking limits, the sandwich theorem gives f (x) = x f (1) f (1) = 0, then f (x) = f (x) f (1) = 0 for all equation, we get x > 0. If f (1) = 1, then f (x) > 0 for all x > for all x. Therefore, f (x) = cx with c ≥ 0. ( g (x2) + g (y2) )2 = g ( (x2 + y2)2 ) 0 (since f (x) = 0 implies f (1) = f (x(1/x)) The checking is as in example 2. = g ( x2 + y2 )2 = f (x) f (1/x) = 0, which would lead to a Remarks. (1) In example 3, if we contradiction). for all x,y. Taking square roots and letting a = x2, b = y2, we get g(a)+g (b) replace the condition that “f (x) ≥ 0 for w x ≥ 0” by “f is monotone”, then the Define g: ℝ→ℝ by g (w) = ln f (e ). = g(a+ b) for all a, b > 0. answer is essentially the same, namely Then By fact 1, we have g (w) = w for all w f (x) = cx with c = f (1). Also if the g (x+y) = ln f (ex+y) = ln f (ex ey) > 0. Since f (0) = 0 and f is even, it condition that “f (x) ≥ 0 for x ≥ 0” is =ln f (ex) f (ey) follows f (x) = g (x2) = x2 for all x. replaced by “f is continuous at 0”, then = ln f (ex) + ln f (ey) steps 2 and 3 in example 3 are not = g(x) + g(y). Check: If f (x) = x2, then the equation necessary. We can take rational pn’s Since f is monotone, it follows that g is reduces to increase to x and take limit of pn f (1) = also monotone. Then g (w) = cw for all w. 2 2 2 2 2 2 f (pn) = f (pn–x) + f (x) to get x f (1) = f (x) (x + z )(y + t ) = (xy−zt) + (xt+yz) , Therefore, f (x) = xc for all x > 0. since pn–x increases to 0. which is a well known identity and (2) The Cauchy equation f ( x + y ) = As an application of these facts, we look can easily be checked by expansion f ( x ) + f ( y ) for all x, y ∊ ℝ has at the following example. or seen from | p |2 | q |2 = | pq |2, where noncontinuous solutions (in particular, p = x + iz, q = y + it ∊ℂ. solutions not of the form f (x) = cx). Example 4. (2002 IMO) Find all This requires the concept of a Hamel functions f from the set ℝ of real The concept of fixed point of a basis of the vector space ℝ over ℚ numbers to itself such that function is another useful idea in from linear algebra. solving some functional equations. ( f (x) + f (z))( f (y) + f (t)) Its definition is very simple. We say = f ( xy − zt ) + f ( xt + yz ) The following are some useful facts w is a fixed point of a function f if and related to the Cauchy equation. for all x, y, z, t in ℝ. only if w is in the domain of f and f (w) = w. Having information on the Fact 1. Let A = ℝ, [0, ∞) or (0, ∞). If Solution. (Due to Yu Hok Pun, 2002 fixed points of functions often help to f :A→ℝ satisfies f ( x + y ) = f ( x ) Hong Kong IMO team member, gold solve certain types of functional + f (y) and f (xy) = f (x) f (y) for all medalist) Suppose f (x) = c for all x. equations as the following examples x, y ∊ A, then either f (x) = 0 for all x Then the equation implies 4c2 = 2c. So c will show. ∊ A or f (x) = x for all x ∊ A. can only be 0 or 1/2. Reversing steps, we can also check f (x) = 0 for all x or f (x) = Example 5. (1983 IMO) Determine + + Proof. By example 2, we have f (x) = 1/2 for all x are solutions. all functions f : ℝ → ℝ such that + f (1) x for all x∊ℚ. If f (1) = 0, then f ( x f (y) ) = y f (x) for all x, y ∊ ℝ f (x) = f (x·1) = f (x) f (1)=0 for all Suppose the equation is satisfied by a and as x → + ∞ , f (x) → 0. x∊A. nonconstant function f. Setting x = 0 and z = 0, we get 2 f (0) (f (y) + f(t)) = 2 f (0), Solution. Step 1 Taking x = 1 = y, we Otherwise, we have f (1) ≠ 0. Since which implies f (0) = 0 or f (y) + f (t) = 1 get f ( f (1)) = f (1). Taking x = 1 and y f (1) = f (1) f (1), we get f (1) = 1. for all y, t. In the latter case, setting y = t, = f (1), we get f ( f ( f (1))) = f (1)2. Then f (x) = x for all x ∊ A ∩ ℚ. we get the constant function f (y) = 1/2 Then f (1)2 = f ( f ( f (1))) = f ( f (1)) = for all y. Hence we may assume f (0) = 0. f (1), which implies f (1) = 1. So 1 is a 1/2 2 If y ≥ 0, then f (y) = f ( y ) ≥ 0 and fixed point of f.

Setting y = 1, z = 0, t = 0, we get f (x) f (1) (continued on page 4) Mathematical Excalibur, Vol. 8, No. 1, Feb 03- Mar 03 Page 3

Problem Corner = SRK imply ∆SMB ~ ∆SKR and In the last issue, problems 166, 167 and ∠ We welcome readers to submit their 169 were stated incorrectly. They are MB/KR = BS/RS. Replacing M by A and solutions to the problems posed below revised as problems 171, 172, 173, K by Q, similarly ∆SAB ~ ∆SQR and for publication consideration. The respectively. As the problems became AB/QR = BS/RS. Since AB = 2MB, we solutions should be preceded by the easy due to the mistakes, we received get QR = 2KR. So K is the midpoint of solver’s name, home (or email) address many solutions. Regretfully we do not QR. and school affiliation. Please send have the space to print the names and submissions to Dr. Kin Y. Li, affiliations of all solvers. We would like to Problem 169. 300 apples are given, no Department of Mathematics, The Hong apologize for this. one of which weighs more than 3 times Kong University of Science & any other. Show that the apples may be Technology, Clear Water Bay, Kowloon. Problem 166. Let a, b, c be positive The deadline for submitting solutions integers, [x] denote the greatest integer divided into groups of 4 such that no is February 28, 2003. less than or equal to x and min{x,y} group weighs more than 11/2 times any denote the minimum of x and y. Prove or other group. Problem 171. (Proposed by Ha Duy disprove that Solution. Almost all solvers used the Hung, Hanoi University of Education, c [a/b] – [c/a] [c/b] ≤ c min{1/a, 1/b}. following argument. Let m and M be Hanoi City, Vietnam) Let a, b, c be the weights of the lightest and heaviest positive integers, [x] denote the Solution. Over 30 solvers disproved the apple(s). Then 3m≥ M. If the problem greatest integer less than or equal to x inequality by providing different counter- is false, then there are two groups A and min{x,y} denote the minimum of x examples, such as (a, b, c) = (3, 2, 1). and B with weights w and w such that and y. Prove or disprove that A B (11/2) w < w . Since 4m≤ w and w ≤ Problem 167. Find all positive integers B A B A  c   c   c   1 1  4M, we get (11/2)4m < 4M implying c   −     ≤ c min  , . such that they are equal to the sum of their  ab   a   b   a b  digits in base 10 representation. 3m≤ (11/2)m < M , a contradiction.

Solution. Over 30 solvers sent in solutions Problem 170. (Proposed by Problem 172. (Proposed by José Luis similar to the following. For a positive Abderrahim Ouardini, Nice, France) Díaz-Barrero, Universitat Politècnica integer N with digits an, … , a0 (from left For any (nondegenerate) triangle with to right), we have de Catalunya, Barcelona, Spain) Find sides a, b, c, let ∑’ h (a, b, c) denote the n n−1 all positive integers such that they are N = an 10 + an−1 10 + + a0 sum h (a, b, c) + h (b, c, a )+ h (c, a, b). ⋯ 2 equal to the square of the sum of their ≥ a + a + + a Let f (a, b, c) = ∑’ ﴾a / (b + c – a)﴿ and n n−1 ⋯ 0 digits in base 10 representation. k g (a, b, c) =∑’ j(a, b, c), where j(a,b,c)= because 10 > 1 for k> 0. So equality holds (b + c – a) / . if and only if a =a = =a =0. Hence, (c + a − b )( a + b − c ) n n−1 ⋯ 1 Show that f (a, b, c)≥ max{3,g(a, b, c)} Problem 173. 300 apples are given, N=1, 2, …, 9 are the only solutions. and determine when equality occurs. no one of which weighs more than 3 times any other. Show that the apples (Here max{x,y} denotes the maximum Problem 168. Let AB and CD be of x and y.) may be divided into groups of 4 such nonintersecting chords of a circle and let that no group weighs more than 3/2 K be a point on CD. Construct (with Solution. CHUNG Ho Yin (STFA times any other group. straightedge and compass) a point P on Leung Kau Kui College, Form 6), CHUNG Tat Chi (Queen Elizabeth the circle such that K is the midpoint of the School, Form 6), D. Kipp JOHNSON Problem 174. Let M be a point inside part of segment CD lying inside triangle (Valley Catholic High School, acute triangle ABC. Let A′, B′, C′ be the ABP. (Source: 1997 Hungarian Math Beaverton, Oregon, USA), LEE Man mirror images of M with respect to BC, Olympiad) Fui (STFA Leung Kau Kui College, CA, AB, respectively. Determine (with Form 6), Antonio LEI (Colchester Solution. SIU Tsz Hang (STFA Leung Kau Royal Grammar School, UK, Year 13), proof) all points M such that A, B, C, A′, Kui College, Form 7) SIU Tsz Hang (STFA Leung Kau Kui B′, C′ are concyclic. College, Form 7), TAM Choi Nang Draw the midpoint M of AB. If AB || CD, Julian (SKH Lam Kau Mow Secondary Problem 175. A regular polygon with then draw ray MK to intersect the circle at P. School) and WONG Wing Hong (La Salle College, Form 5). n sides is divided into n isosceles Let AP, BP intersect CD at Q,R, respectively. triangles by segments joining its center Since AB || QR, ∆ABP ~ ∆QRP. Then M Let x = b + c − a, y = c + a − b and z = a + to the vertices. Initially, n + 1 frogs are being the midpoint of AB will imply K is the b − c. Then a = (y + z)/2, b = (z + x)/2 and placed inside the triangles. At every c = (x + y)/2. midpoint of QR. second, there are two frogs in some If AB intersects CD at E, then draw the Substituting these and using the AM-GM common triangle jumping into the inequality, the rearrangement inequality interior of the two neighboring circumcircle of EMK meeting the original and the AM-GM inequality again, we find triangles (one frog into each neighbor). circle at S and S′. Draw the circumcircle of Prove that after some time, at every BES meeting CD at R. Draw the f ( a, b, c ) second, there are at least [ (n + 1) / 2 ] circumcircle of AES meeting CD at Q. Let triangles, each containing at least one 2 2 2 AQ, BR meet at P. Since PBS RBS  y + z   z + x   x + y  frog. ∠ = ∠ = =   +   +   RES = QES = QAS PAS , P is on  2 x   2 y   2 z  ∠ ∠ ∠ = ∠ the original circle. 2 2 2 *****************  yz   zx   xy  ≥   +   +   Solutions       Next, ∠SMB = ∠SME = ∠SKE = ∠SKR  x   y   z  **************** and ∠SBM = 180° − ∠SBE = 180° − ∠SRE Mathematical Excalibur, Vol. 8, No. 1, Feb 03- Mar 03 Page 4

have any fixed point x in (0,1). = f ( f (m) ) + f (n) . yz zx zx xy xy yz ≥ + + Other than the fixed point concept, in xy yz zx Step 5 Steps 1, 3, 4 showed the only fixed solving functional equations, the point of f is 1. By step 2, we get x f (x) = 1 x y z injectivity and surjectivity of the = + + = g (a, b, c) for all x +. Therefore, f (x) = 1 / x for yz zx xy ∊ ℝ functions also provide crucial all x +. ∊ ℝ informations.

xyz ≥ 3 3 = 3. Check: For f (x) = 1/x, f (x f (y)) = f (x/y) = yz zx xy Example 7. (1987 IMO) Prove that y/x =y f (x). As x →∞ , f (x) = 1/x → 0. there is no function f: → such ℕ0 ℕ0 So f (a,b,c)≥ g(a,b,c) = max{3,g(a,b,c)} that f ( f (n)) = n + 1987. Example 6. (1996 IMO) Find all functions f : with equality if and only if x = y = z, → such that which is the same as a = b = c. ℕ0 ℕ0 Solution. Suppose there is such a f ( m + f (n) ) = f ( f (m) ) + f (n) function f. Then f is injective because f (a) = f (b) implies for all m, n∊ℕ0. a = f ( f (a))−1987 = f ( f (b))−1987 = b. Olympiad Corner Solution. Step 1 Taking m = 0 = n, we get f ( f (0)) = f ( f (0) ) + f (0), which implies Suppose f (n) misses exactly k distinct (continued from page 1) f (0) = 0. Taking m = 0, we get f ( f ( n )) = values c , … , c in , i.e. f (n)≠ c , …, 1 k ℕ0 1 f (n), i.e. f (n) is a fixed point of f for every c for all n . Then f ( f ( n )) misses Problem 3. If a ≥ b ≥ c ≥ 0 and a + b + k ∊ ℕ0 n ∊ℕ0. Also the equation becomes the 2k distinct values c1, …, ck and c =3, then prove that ab2 + bc2 + ca2 ≤ f ( m + f (n) ) = f (m) + f (n). f (c1), …, f (ck) in ℕ0. (The f (cj)’s are 27/8 and determine the equality distinct because f is injective.) Now if case(s). Step 2 If w is a fixed point of f, then we w≠ c1, … , ck, f (c1), … , f (ck), then will show kw is a fixed point of f for all k there is m ∊ ℕ0 such that f (m) = w. Problem 4. Let p be an odd prime ∊ℕ0. The cases k = 0, 1 are known. If kw Since w≠ f (cj), m≠ cj, so there is n ∊ is a fixed point, then f ((k + 1) w) = f ( kw + such that f (n) = m, then f ( f (n)) = w. such that p ≡ 1 (mod 4). Evaluate ℕ0 w ) = f ( kw ) + f (w) = kw + w = (k + 1) w This shows f ( f (n)) misses only the 2k (with reason) and so (k + 1) w is also a fixed point. values c1, … , ck, f (c1), … , f (ck) and no p − 1 others. Since n + 1987 misses the 1987 2 k 2 Step 3 If 0 is the only fixed point of f, then values 0, 1, …, 1986 and 2k ≠ 1987, ∑ , k = 1 p f (n) = 0 for all n ∊ℕ0 by step 1. Obviously, this is a contradiction. the zero function is a solution. where {x} = x − [x], [x] being the Example 8. (1999 IMO) Determine all greatest integer not exceeding x. Otherwise, f has a least fixed point w > 0. functions f : ℝ → ℝ such that We will show the only fixed points are kw, f (x − f (y)) = f (f (y)) + x f (y) + f (x) − 1 k . Suppose x is a fixed point. By the ∊ℕ0 division algorithm, x = kw + r, where 0≤ r for all x, y ∊ ℝ.

Since f (n) is a fixed point for all n ∊ℕ0 by Next, if we set y = 0, we get Step 3 Suppose f has a fixed point x > 1. step 1, f (n) = cnw for some cn ∊ ℕ0. We By step 2, x f (x) = x2 is also a fixed { f (x − c) − f (x) : x ∊ ℝ } have c0 = 0. 2 2 4 = { cx + f ( c ) − 1 : x } = point, x f (x ) = x is also a fixed point ∊ ℝ ℝ and so on. So the xm’s are fixed points Step 4 For n∊ℕ0, by the division because c≠ 0. Then A − A = { y1 − y2 : for every m that is a power of 2. Since x algorithm, n = kw + r, 0 ≤ r < w. We have y1, y2 ∊ A} = ℝ. > 1, for m ranging over the powers of 2, we have xm → ∞, but f (xm) = xm → ∞ , f (n) = f (r + kw) = f (r + f (kw)) Now for an arbitrary x∊ℝ, let y1, y2∊A not to 0. This contradicts the given = f (r) + f (kw) = crw + kw be such that y1 − y2 = x. Then = (cr+ k) w = (cr + [n/w]) w. property. Hence, f cannot have any fixed point x > 1. f (x)= f (y1−y2) = f (y2) + y1y2 + f (y1) − 1 Check: For each w > 0, let c0 = 0 and let 2 2 = (c+1−y2 )/2+y1y2+(c+1−y1 )/2 −1 c1. …, cw−1 ∊ℕ0 be arbitrary. The function 2 2 Step 4 Suppose f has a fixed point x in = c − ( y1−y2) /2 = c − x /2. f(n)=(cr+[n/w])w, where r is the remainder the interval (0,1). Then 2 of n divided by w, (and the zero function) However, for x∊A, f (x) = (c + 1 − x )/2. 2 1 = f ((1/x) x) = f ((1/x) f (x)) = x f (1/ x), are all the solutions. Write m = kw + r and So c = 1. Therefore, f (x) = 1 − x /2 for n = lw + s with 0≤ r, s < w. Then all x ∊ℝ. which implies f (1 / x) = 1 / x. This will 2 lead to f having a fixed point 1 / x > 1, f (m + f (n)) = f (r + kw + (cs + l) w) Check: For f (x) = 1 − x /2, both sides 2 4 2 2 contradicting step 3. Hence, f cannot = crw + kw + csw + lw equal 1/2 + y /2 − y /8 + x−xy /2 − x /2.