Volume 8, Number 1 February 2003 – March 2003
Functional Equations
Kin Y. Li
Olympiad Corner
The Fifth Hong Kong (China) Mathematical Olympiad was held on December 21, 2002. The problems are as follow.
Problem 1. Two circles intersect at points A and B. Through the point B a straight
line is drawn, intersecting the first circle at
K and the second circle at M. A line
parallel to AM is tangent to the first circle at Q. The line AQ intersects the second circle again at R.
(a) Prove that the tangent to the second circle at R is parallel to AK.
(b) Prove that these two tangents are concurrent with KM.
Problem 2. Let n ≥ 3 be an integer. In a conference there are n mathematicians. Every pair of mathematicians communicate in one of the n official languages of the conference. For any three different official languages, there exist three mathematicians who communicate with each other in these three languages. Determine all n for which this is possible. Justify your claim.
(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
February 28, 2003.
For individual subscription for the next five issues for the 02-03 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]
A functional equation is an equation whose variables are ranging over functions. Hence, we are seeking all possible functions satisfying the equation. We will let ℤ denote the set of all integers, ℤ+ or ℕ denote the positive integers, ℕ0 denote the nonnegative integers, ℚ denote the rational numbers, ℝ denote the real numbers, ℝ+ denote the positive real numbers and ℂ denote the complex numbers.
In simple cases, a functional equation can be solved by introducing some substitutions to yield more information or additional equations.
Example 1. Find all functions f : ℝ → ℝ
such that
x2 f (x) + f (1 – x) = 2 x – x4 for all x ∊ℝ.
Solution. Replacing x by 1 – x, we have
(1– x)2 f (1– x) + f ( x ) =2 (1–x) – (1–x)4. Since f (1 – x) =2 x – x4– x2 f (x) by the given equation, substituting this into the last equation and solving for f (x), we get f (x) = 1– x2.
Check: For f (x) = 1 – x2,
x2 f (x) + f (1–x) = x2 (1– x2 )+(1– (1– x)2 ) = 2 x – x4.
For certain types of functional equations, a standard approach to solving the problem is to determine some special values (such as f ( 0 ) or f ( 1 ) ), then inductively determine f ( n ) for n ∊ ℕ0, follow by the values f ( 1 / n ) and use density to find f ( x ) for all x ∊ ℝ. The following are examples of such approach.
Example 2. Find all functions f : ℚ → ℚ
such that the Cauchy equation
f ( x + y ) = f ( x ) + f ( y )
holds for all x, y ∊ℚ.
Solution. Step 1 Taking x = 0 = y, we get
f (0) = f (0) + f (0) + f (0) , which implies f (0) = 0.
Step 2 We will prove f (kx) = k f (x) for k∊ ℕ, x∊ℚ by induction. This is true for k = 1. Assume this is true for k. Taking y
= kx, we get
f ((k+1) x) = f (x + kx) = f (x) + f (kx)
= f (x) + k f (x) = (k+1) f (x).
Step 3 Taking y = –x, we get
0 = f (0) = f (x+ (–x)) = f (x) + f (–x), which implies f (–x) = – f (x). So
f (–kx) = – f (kx) = – k f (x) for k∊ℕ.
Therefore, f (kx) = k f (x) for k ∊ℤ, x∊ℚ.
Step 4 Taking x = 1/ k, we get f (1) = f (k (1/ k)) = k f (1/ k),
which implies f (1/ k) = (1/ k ) f (1).
Step 5 For m∊ℤ, n∊ℕ,
f (m/ n) = m f (1/ n) = (m/ n) f (1).
Therefore, f (x) = cx with c = f (1).
Check: For f (x) = cx with c∊ℚ , f (x+y) = c(x+y) = cx + cy = f (x) + f (y).
In dealing with functions on ℝ, after finding the function on ℚ, we can often finish the problem by using the following fact.
Density of Rational Numbers For every
real number x, there are rational numbers p1, p2, p3, … increase to x and
there are rational numbers q1, q2, q3, …
decrease to x.
This can be easily seen from the decimal representation of real numbers. For example, the number π = 3.1415… is the limits of 3, 31/10, 314/100, 3141/1000, 31415/10000, … and also 4, 32/10, 315/100, 3142/1000, 31416/10000, …. (In passing, we remark that there is a similar fact with rational numbers replaced by irrational numbers.)
Mathematical Excalibur, Vol. 8, No. 1, Feb 03- Mar 03 Page 2
Example 3. Find all functions
f :ℝ→ℝ such that
f ( x + y) = f ( x ) + f ( y ) for all x, y ∊ ℝ and f (x) ≥ 0 for x ≥ 0.
Solution. Step 1 By example 2, we
have f (x) = x f (1) for x∊ℚ.
Step 2 If x ≥ y, then x – y ≥ 0. So f (x) = f ((x–y)+y) = f (x–y)+f (y )≥ f (y).
Hence, f is increasing.
Step 3 If x ∊ℝ, then by the density of
rational numbers, there are rational pn,
qn such that pn ≤ x ≤ qn, the pn’s
increase to x and the qn’s decrease to x.
So by step 2, pn f (1) = f (pn) ≤ f (x) ≤
f (qn) = qn f (1). Taking limits, the
sandwich theorem gives f (x) = x f (1) for all x. Therefore, f (x) = cx with c ≥ 0. The checking is as in example 2.
Remarks. (1) In example 3, if we
replace the condition that “f (x) ≥ 0 for
x ≥ 0” by “f is monotone”, then the
answer is essentially the same, namely
f (x) = cx with c = f (1). Also if the
condition that “f (x) ≥ 0 for x ≥ 0” is replaced by “f is continuous at 0”, then steps 2 and 3 in example 3 are not necessary. We can take rational pn’s
increase to x and take limit of pn f (1) =
f (pn) = f (pn–x) + f (x) to get x f (1) = f (x)
since pn–x increases to 0.
(2) The Cauchy equation f ( x + y ) =
f ( x ) + f ( y ) for all x, y ∊ ℝ has
noncontinuous solutions (in particular, solutions not of the form f (x) = cx). This requires the concept of a Hamel
basis of the vector space ℝ over ℚ
from linear algebra.
The following are some useful facts related to the Cauchy equation.
Fact 1. Let A = ℝ, [0, ∞) or (0, ∞). If
f :A→ℝ satisfies f ( x + y ) = f ( x ) + f (y) and f (xy) = f (x) f (y) for all x, y ∊ A, then either f (x) = 0 for all x
∊ A or f (x) = x for all x ∊ A.
Proof. By example 2, we have f (x) =
f (1) x for all x∊ℚ. If f (1) = 0, then f (x) = f (x·1) = f (x) f (1)=0 for all x∊A.
Otherwise, we have f (1) ≠ 0. Since
f (1) = f (1) f (1), we get f (1) = 1.
Then f (x) = x for all x ∊ A ∩ ℚ. If y ≥ 0, then f (y) = f ( y1/2 )2 ≥ 0 and
f (x + y) = f (x) + f (y) ≥ f (x),
which implies f is increasing. Now for any x∊A∖ℚ, by the density of rational numbers, there are pn, qn∊ℚ such that pn
< x < qn, the pn’s increase to x and the
qn’s decrease to x. As f is increasing, we
have pn = f (pn) ≤ f (x) ≤ f (qn) = qn.
Taking limits, the sandwich theorem gives f (x) = x for all x∊A.
Fact 2. If a function f : ( 0, ∞ ) → ℝ
satisfies f (xy) = f (x) f ( y) for all x, y > 0 and f is monotone, then either f(x)=0 for all x > 0 or there exists c such that f (x) = xc for all x > 0.
Proof. For x > 0, f (x) = f (x1/2)2 ≥ 0. Also
f (1) = f (1) f (1) implies f (1) = 0 or 1. If f (1) = 0, then f (x) = f (x) f (1) = 0 for all x > 0. If f (1) = 1, then f (x) > 0 for all x >
0 (since f (x) = 0 implies f (1) = f (x(1/x)) = f (x) f (1/x) = 0, which would lead to a contradiction). Define g: ℝ→ℝ by g (w) = ln f (ew ). Then g (x+y) = ln f (ex+y) = ln f (ex ey) =ln f (ex) f (ey) = ln f (ex) + ln f (ey) = g(x) + g(y).
Since f is monotone, it follows that g is also monotone. Then g (w) = cw for all w. Therefore, f (x) = xc for all x > 0.
As an application of these facts, we look at the following example.
Example 4. (2002 IMO) Find all
functions f from the set ℝ of real numbers to itself such that
( f (x) + f (z))( f (y) + f (t)) = f ( xy − zt ) + f ( xt + yz )
for all x, y, z, t in ℝ.
Solution. (Due to Yu Hok Pun, 2002
Hong Kong IMO team member, gold medalist) Suppose f (x) = c for all x.
Then the equation implies 4c2 = 2c. So c can only be 0 or 1/2. Reversing steps, we can also check f (x) = 0 for all x or f (x) = 1/2 for all x are solutions.
Suppose the equation is satisfied by a nonconstant function f. Setting x = 0 and
z = 0, we get 2 f (0) (f (y) + f(t)) = 2 f (0),
which implies f (0) = 0 or f (y) + f (t) = 1 for all y, t. In the latter case, setting y = t, we get the constant function f (y) = 1/2 for all y. Hence we may assume f (0) = 0. Setting y = 1, z = 0, t = 0, we get f (x) f (1)
= f (x). Since f (x) is not the zero function, f (1) = 1. Setting z = 0, t = 0, we get f (x) f (y) = f (xy) for all x,y. In particular, f (w) = f (w1/2)2 ≥ 0 for
w > 0.
Setting x = 0, y = 1 and t = 1, we have 2 f (1) f (z) = f (−z) + f (z), which implies f (z) = f (−z) for all z. So f is even.
Define the function g: (0, ∞) →ℝ by
g(w)= f (w1/2) ≥ 0. Then for all x,y>0,
g (xy) = f ((xy)1/2) = f (x1/2 y1/2) = f (x1/2) f (y1/2) = g (x) g (y). Next f is even implies g (x2) = f (x) for all x. Setting z = y, t = x in the given equation, we get
( g (x2) + g (y2) )2 = g ( (x2 + y2)2 ) = g ( x2 + y2 )2 for all x,y. Taking square roots and letting a = x2, b = y2, we get g(a)+g (b) = g(a+ b) for all a, b > 0.
By fact 1, we have g (w) = w for all w > 0. Since f (0) = 0 and f is even, it follows f (x) = g (x2) = x2 for all x.
Check: If f (x) = x2, then the equation reduces to
(x2 + z2)(y2 + t2) = (xy−zt)2 + (xt+yz)2, which is a well known identity and can easily be checked by expansion or seen from | p |2 | q |2 = | pq |2, where
p = x + iz, q = y + it ∊ℂ.
The concept of fixed point of a function is another useful idea in solving some functional equations. Its definition is very simple. We say
w is a fixed point of a function f if and
only if w is in the domain of f and
f (w) = w. Having information on the
fixed points of functions often help to solve certain types of functional equations as the following examples will show.
Example 5. (1983 IMO) Determine
all functions f : ℝ+ → ℝ+ such that
f ( x f (y) ) = y f (x) for all x, y ∊ ℝ+ and as x → + ∞ , f (x) → 0.
Solution. Step 1 Taking x = 1 = y, we
get f ( f (1)) = f (1). Taking x = 1 and y = f (1), we get f ( f ( f (1))) = f (1)2. Then f (1)2 = f ( f ( f (1))) = f ( f (1)) =
f (1), which implies f (1) = 1. So 1 is a
fixed point of f.
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.
The deadline for submitting solutions is February 28, 2003.
Problem 171. (Proposed by Ha Duy Hung, Hanoi University of Education, Hanoi City, Vietnam) Let a, b, c be
positive integers, [x] denote the greatest integer less than or equal to x and min{x,y} denote the minimum of x and y. Prove or disprove that
. 1 , 1 min ≤ − b a c b c a c ab c c
Problem 172. (Proposed by José Luis Díaz-Barrero, Universitat Politècnica de Catalunya, Barcelona, Spain) Find
all positive integers such that they are equal to the square of the sum of their digits in base 10 representation. Problem 173. 300 apples are given, no one of which weighs more than 3 times any other. Show that the apples may be divided into groups of 4 such that no group weighs more than 3/2 times any other group.
Problem 174. Let M be a point inside acute triangle ABC. Let A′, B′, C′ be the mirror images of M with respect to BC,
CA, AB, respectively. Determine (with
proof) all points M such that A, B, C, A′,
B′, C′ are concyclic.
Problem 175. A regular polygon with n sides is divided into n isosceles
triangles by segments joining its center to the vertices. Initially, n + 1 frogs are placed inside the triangles. At every second, there are two frogs in some common triangle jumping into the interior of the two neighboring triangles (one frog into each neighbor). Prove that after some time, at every second, there are at least [ (n + 1) / 2 ] triangles, each containing at least one frog.
*****************
Solutions
****************In the last issue, problems 166, 167 and 169 were stated incorrectly. They are revised as problems 171, 172, 173, respectively. As the problems became easy due to the mistakes, we received many solutions. Regretfully we do not have the space to print the names and affiliations of all solvers. We would like to apologize for this.
Problem 166. Let a, b, c be positive integers, [x] denote the greatest integer less than or equal to x and min{x,y} denote the minimum of x and y. Prove or disprove that
c [a/b] – [c/a] [c/b] ≤ c min{1/a, 1/b}.
Solution. Over 30 solvers disproved the
inequality by providing different counter- examples, such as (a, b, c) = (3, 2, 1). Problem 167. Find all positive integers such that they are equal to the sum of their digits in base 10 representation.
Solution. Over 30 solvers sent in solutions
similar to the following. For a positive integer N with digits an, … , a0 (from left to right), we have
N = an 10n + an−1 10n−1 + ⋯ + a0 ≥ an + an−1 + ⋯ + a0
because10k
> 1 for k> 0. So equality holds if and only if an=an−1=⋯=a1=0. Hence,
N=1, 2, …, 9 are the only solutions. Problem 168. Let AB and CD be nonintersecting chords of a circle and let
K be a point on CD. Construct (with
straightedge and compass) a point P on the circle such that K is the midpoint of the part of segment CD lying inside triangle
ABP. (Source: 1997 Hungarian Math Olympiad)
Solution. SIU Tsz Hang (STFA Leung Kau
Kui College, Form 7)
Draw the midpoint M of AB. If AB || CD, then draw ray MK to intersect the circle at P. Let AP, BP intersect CD at Q,R, respectively. Since AB || QR, ∆ABP ~ ∆QRP. Then M being the midpoint of AB will imply K is the midpoint of QR.
If AB intersects CD at E, then draw the circumcircle of EMK meeting the original circle at S and S′. Draw the circumcircle of
BES meeting CD at R. Draw the
circumcircle of AES meeting CD at Q. Let
AQ, BR meet at P. Since ∠PBS = ∠RBS =
∠RES = ∠QES = ∠QAS = ∠PAS, P is on the original circle.
Next, ∠SMB = ∠SME = ∠SKE = ∠SKR and ∠SBM = 180° − ∠SBE = 180° − ∠SRE
= ∠SRK imply ∆SMB ~ ∆SKR and
MB/KR = BS/RS. Replacing M by A and K by Q, similarly ∆SAB ~ ∆SQR and AB/QR = BS/RS. Since AB = 2MB, we
get QR = 2KR. So K is the midpoint of
QR.
Problem 169. 300 apples are given, no one of which weighs more than 3 times any other. Show that the apples may be divided into groups of 4 such that no group weighs more than 11/2 times any other group.
Solution. Almost all solvers used the
following argument. Let m and M be the weights of the lightest and heaviest apple(s). Then 3m≥ M. If the problem is false, then there are two groups A and B with weights wA and wB such that
(11/2) wB < wA. Since 4m≤ wB and wA ≤
4M, we get (11/2)4m < 4M implying 3m≤ (11/2)m < M , a contradiction. Problem 170.
(Proposed by
Abderrahim Ouardini, Nice, France)
For any (nondegenerate) triangle with sides a, b, c, let ∑’ h (a, b, c) denote the sum h (a, b, c) + h (b, c, a )+ h (c, a, b). Let f (a, b, c) = ∑’ ﴾a / (b + c – a)﴿2 and
g (a, b, c) =∑’ j(a, b, c), where j(a,b,c)=
(b + c – a) / (c+a −b)(a +b−c) . Show that f (a, b, c)≥ max{3,g(a, b, c)} and determine when equality occurs. (Here max{x,y} denotes the maximum of x and y.)
Solution. CHUNG Ho Yin (STFA
Leung Kau Kui College, Form 6), CHUNG Tat Chi (Queen Elizabeth School, Form 6), D. Kipp JOHNSON (Valley Catholic High School, Beaverton, Oregon, USA), LEE Man Fui (STFA Leung Kau Kui College, Form 6), Antonio LEI (Colchester Royal Grammar School, UK, Year 13), SIU Tsz Hang (STFA Leung Kau Kui College, Form 7), TAM Choi Nang Julian (SKH Lam Kau Mow Secondary School) and WONG Wing Hong (La Salle College, Form 5).
Let x = b + c − a, y = c + a − b and z = a +
b − c. Then a = (y + z)/2, b = (z + x)/2 and c = (x + y)/2.
Substituting these and using the AM-GM inequality, the rearrangement inequality and the AM-GM inequality again, we find
f ( a, b, c ) 2 2 2 2 2 2 + + + + + = z y x y x z x z y 2 2 2 + + ≥ z xy y zx x yz
Mathematical Excalibur, Vol. 8, No. 1, Feb 03- Mar 03 Page 4 zx yz xy yz xy zx xy zx yz + + ≥ ) , , (a b c g xy z zx y yz x + + = = . 3 33 = ≥ xy zx yz xyz
So f (a,b,c)≥ g(a,b,c) = max{3,g(a,b,c)} with equality if and only if x = y = z, which is the same as a = b = c.
Olympiad Corner
(continued from page 1)
Problem 3. If a ≥ b ≥ c ≥ 0 and a + b + c =3, then prove that ab2 + bc2 + ca2 ≤ 27/8 and determine the equality case(s).
Problem 4. Let p be an odd prime such that p ≡ 1 (mod 4). Evaluate (with reason) 1 2 2 1 , p k k p − =
∑
where {x} = x − [x], [x] being the greatest integer not exceeding x.
Functional Equations
(continued from page 2) Step 2 Taking y = x, we get f ( x f ( x)) = x f (x). So w = x f (x) is a fixed point of f
for every x ∊ ℝ+.
Step 3 Suppose f has a fixed point x > 1.
By step 2, x f (x) = x2 is also a fixed point, x2 f (x2) = x4 is also a fixed point and so on. So the xm’s are fixed points
for every m that is a power of 2. Since x > 1, for m ranging over the powers of 2, we have xm → ∞, but f (xm) = xm → ∞ ,
not to 0. This contradicts the given property. Hence, f cannot have any fixed point x > 1.
Step 4 Suppose f has a fixed point x in
the interval (0,1). Then
1 = f ((1/x) x) = f ((1/x) f (x)) = x f (1/ x), which implies f (1 / x) = 1 / x. This will lead to f having a fixed point 1 / x > 1, contradicting step 3. Hence, f cannot
have any fixed point x in (0,1).
Step 5 Steps 1, 3, 4 showed the only fixed
point of f is 1. By step 2, we get x f (x) = 1 for all x ∊ ℝ+. Therefore, f (x) = 1 / x for all x ∊ ℝ+.
Check: For f (x) = 1/x, f (x f (y)) = f (x/y) = y/x =y f (x). As x →∞ , f (x) = 1/x → 0.
Example 6. (1996 IMO) Find all functions f :
ℕ0 → ℕ0 such that
f ( m + f (n) ) = f ( f (m) ) + f (n)
for all m, n∊ℕ0.
Solution. Step 1 Taking m = 0 = n, we get
f ( f (0)) = f ( f (0) ) + f (0), which implies f (0) = 0. Taking m = 0, we get f ( f ( n )) = f (n), i.e. f (n) is a fixed point of f for every n ∊ℕ0. Also the equation becomes f ( m + f (n) ) = f (m) + f (n).
Step 2 If w is a fixed point of f, then we
will show kw is a fixed point of f for all k ∊ℕ0. The cases k = 0, 1 are known. If kw is a fixed point, then f ((k + 1) w) = f ( kw +
w ) = f ( kw ) + f (w) = kw + w = (k + 1) w
and so (k + 1) w is also a fixed point.
Step 3 If 0 is the only fixed point of f, then f (n) = 0 for all n ∊ℕ0 by step 1. Obviously, the zero function is a solution.
Otherwise, f has a least fixed point w > 0. We will show the only fixed points are kw,
k∊ℕ0. Suppose x is a fixed point. By the division algorithm, x = kw + r, where 0≤ r <w. We have
x = f (x) = f (r + kw) = f (r + f (kw))
= f (r) + f (kw) = f (r) + kw.
So f (r) = x − kw = r. Since w is the least positive fixed point, r = 0 and x = kw. Since f (n) is a fixed point for all n ∊ℕ0 by step 1, f (n) = cnw for some cn ∊ ℕ0. We have c0 = 0.
Step 4 For n∊ℕ0, by the division algorithm, n = kw + r, 0 ≤ r < w. We have
f (n) = f (r + kw) = f (r + f (kw))
= f (r) + f (kw) = crw + kw
= (cr+ k) w = (cr + [n/w]) w.
Check: For each w > 0, let c0 = 0 and let
c1. …, cw−1∊ℕ0 be arbitrary. The function
f(n)=(cr+[n/w])w, where r is the remainder
of n divided by w, (and the zero function) are all the solutions. Write m = kw + r and
n = lw + s with 0≤ r, s < w. Then f (m + f (n)) = f (r + kw + (cs + l) w)
= crw + kw + csw + lw
= f ( f (m) ) + f (n) . Other than the fixed point concept, in solving functional equations, the injectivity and surjectivity of the functions also provide crucial informations.
Example 7. (1987 IMO) Prove that
there is no function f: ℕ0 → ℕ0 such that f ( f (n)) = n + 1987.
Solution. Suppose there is such a
function f. Then f is injective because
f (a) = f (b) implies
a = f ( f (a))−1987 = f ( f (b))−1987 = b. Suppose f (n) misses exactly k distinct values c1, … , ck in ℕ0 , i.e. f (n)≠ c1, …,
ck for all n∊ ℕ0. Then f ( f ( n )) misses the 2k distinct values c1, …, ck and
f (c1), …, f (ck) in ℕ0. (The f (cj)’s are
distinct because f is injective.) Now if
w≠ c1, … , ck, f (c1), … , f (ck), then
there is m ∊ ℕ0 such that f (m) = w. Since w≠ f (cj), m≠ cj, so there is n ∊
ℕ0 such that f (n) = m, then f ( f (n)) = w. This shows f ( f (n)) misses only the 2k values c1, … , ck, f (c1), … , f (ck) and no
others. Since n + 1987 misses the 1987 values 0, 1, …, 1986 and 2k ≠ 1987, this is a contradiction.
Example 8. (1999 IMO) Determine all
functions f : ℝ → ℝ such that
f (x − f (y)) = f (f (y)) + x f (y) + f (x) − 1
for all x, y ∊ ℝ.
Solution. Let c = f (0). Setting x = y = 0,
we get f (−c) = f (c) + c − 1. So c≠ 0. Let A be the range of f, then for x = f (y) ∊ A, we get c = f (0) = f (x) + x2 + f (x) − 1. Solving for f (x), this gives f (x) = ( c + 1 − x2 ) / 2 .
Next, if we set y = 0, we get { f (x − c) − f (x) : x ∊ ℝ } = { cx + f ( c ) − 1 : x ∊ ℝ } = ℝ because c≠ 0. Then A − A = { y1 − y2 :
y1, y2 ∊ A} = ℝ.
Now for an arbitrary x∊ℝ, let y1, y2∊A be such that y1 − y2 = x. Then
f (x)= f (y1−y2) = f (y2) + y1y2 + f (y1) − 1 = (c+1−y22)/2+y1y2+(c+1−y12)/2 −1 = c − ( y1−y2)2/2 = c − x2/2. However, for x∊A, f (x) = (c + 1 − x2)/2. So c = 1. Therefore, f (x) = 1 − x2/2 for all x ∊ℝ.
Check: For f (x) = 1 − x2/2, both sides equal 1/2 + y2/2 − y4/8 + x−xy2/2 − x2/2.
December 2 1,2002 Time allowed: 3 hours
Each problem is worth 7 marks
Two circles intersect at points A and B. Through the point B a straight line is drawn, intersecting the first circle at K and the second circle at A4. A line parallel to AM is tangent to the first circle at Q. The line AQ intersects the second circle again at R. (a) Prove that the tangent to the second circle at R is parallel to AK.
(b) Prove that these two tangents are concurrent with KM.
Let n 2 3 be an integer. In a conference there are n mathematicians. Every pair of mathematicians communicate in one of the n official languages of the conference. For any three different official languages, there exist three mathematicians who communicate with each other in these three languages. Determine all n for which this is possible. Justify your claim.
If aZb>c>O and atbtc=3,thenprove that ab2+bc2+ca2<y and determine the equalities case(s j.
Let p be an odd prime where (x) =x-[xl, [x]
such that p = l(mod4). Evaluate with reasons, being the greatest integer not exceeding x.
***End of Paper***
I ,
1.
(a) (Siu Tsz Hang) Let PR be the tangent to the second circle at R. Join AB and MR. We have LARP = LMRP + LARM = LMAR + LABK = LLQA + LABK
= LQBA+ LABK = 180” - LK4Q. Thus AKIIPR, i.e., the tangent to the second circle at R is parallel to AK.
(b) Let the two tangents meet at J. Note that LBRP = LBAR = LBQJ (tangent property). Thus BJRQ is cyclic. On the other hand, by (a) LKBQ = LaQJ. This means BMproduced will meet PR at J.
2. (Yu Hok Pun) It is possible if and only if n is odd. The problem is equivalent to color the edges of K,, by n colors, and such that for any three colors, there exist three vertices such that their edges are of these three colors. Now notice that there are CT triples of colors and there are C; triples of vertices, that means for the edges each triple of vertices they must be colored by a unique triple of colors, and vice versa. In particular for every triangle, the edges must be of different colors. Now fix a color S, there exist exactly C,"-' triangles with one edge of color S. Yet an edge of color S is connected with n -2 vertices (hence they form n - 2 triangles). Therefore there are
c;-’ n-l
- = 2 edges of color S. Now the condition that y
n-2 is an integer cannot be
fulfilled if n is even.
Assume n is odd, denote the vertices by 1,2, e.. , n and the colors by S1, S, , . en, S, . Color the edge connecting i andj by the color St with t = i + j (mod n). Then for every triple of colors S,, , S(, , and St, the system
i+ j=ttl (rnodn)
j+k = t, (modn) k+i = t, (modn)
has a unique solution, as the determinant of the coefficient matrix
110 110 110
0 1 1 = 0 1 1 = 0 1 1 = 2 f 0, since n is odd. Therefore for each triple of 1 0 1 0 -1 1 0 0 2
colors, there exists a unique triangle with edges of these colors. 3. Let f(a, b, c) = ab2 + bc2 + ca2. Then
f(a,b, c) + f(a, c,b) = ab2 + bc2 + cu2 + UC’ + cb2 + ba2 =(a+b+c)(ab+bci-ca)-3abc
= 3(ab + bc + ca - abc)
= 3[(1- a)(1 - b)(l - c) + (a + b + c) - 11 = 3(1- a)(1 - b)(l - c) + 6.
Since a>b>c>O,wehave cllla and b<z,
If bll, then (l-a)(l-b)(l-c)_<O sothat f&b,c)+f(a,c,b)<6.
If l<b<G, then
(1 - a)(1 - b)(l - c) 5 (a - l)(b - 1) I = f so that 3 27
f (a, b, 4 + f (a, c, b) 5 6 - = -. Note that equality holds if and only if c = 0 and 4 4
Now fi,c,b)-f(a,b,c)=(a-b)(b-c)(a-c)>O,
whichimplies f(a,b,c)lf(a,c,b). Then f(a,b,c)iiCf(a,b,c)+ f(a,c,b))Sy.
Equality holds if and only if c = 0 and a = b = +.
a-b
(AlternateSolutionbyYuHokPun)Put ~=~,y=~,,z=~.Then x+y+,z=:,
b
x+y=-, x=5 3(x+y+z)
3 3 Add together get 3x + 2y + z = 1. Thus a = 3x+2y+z ’
b= 3(x+39 , c_ 3x
3x+2y+z - 3x+2y+z’ By expressing a, b, and c in terms of x, y, and z , get 3x3 + z3 + 6x2y + 4xy2 + 3x2z + xz2 + 4y2z + 6yz2 + 18~72 2 0. As x, y, z 2 0, equality holds if and only if x = z = 0, andhence c=O, a=b=S.
4. First note that (-k)’ = (p - k)2 = k2 (modp). Also if x2 = y2(tnodp), where P-l
llx,yl--
2 ’ then (x-y)(x+y) = 0 (modp). But 1~ x+y < p, hence x = y. These
imply is a reduced residue system modulo p. Now since
p = 1 (mod 4), by Euler’s criterion, = 1, hence -1 is a square. Because
($j=(;)(;).
thus b is a square if and only if -b = p-b (modp) is. Therefore the set {12,22,..., consists of pairs {a,,~-a,,a,,p -a2,...,ap_,,p-a,_l}-7 -7
modulop. As l<a,,p-a, <p, we have {4)+{p-a,)=a’+p-ai = 1, thus the sum
P P P
is P-l 4 *
*“*End*“”
