Mathematical Excalibur, Volume 1, Number 2

March -April, 1995

Volume 1, Number 2

Olympiad Corner

The following are the six problems from the two-day Final Selection Examfor .the 1994 Hong Kong Mathematical Olympiad Team. Would you like to try these problems to see if you could have qualified to be a Hong Kong team

member? -Editors

Fractal Game of Escape

RogerNg

Consider the following scenario. John, a secret agent, is being held captive in terrorists' headquarters. He has found an escape route, and knows it follows the quadratic equation Zn+1 = Zn2 + c if the floor map is encoded as a complex z-plane (i.e., each point (x,y) is represented by a complex number X+YI). However, John does not know the value of the complex cOnstant c. John only knows that he ~hould start from the origin with Zo = 0 -+- Oi. For which values of c, will John have not even a chance for a successful escape?

values for c (the black area) that would keep Zn bounded, i.e., the Mandelbrot set. N ow if we modify our story slightly-assume that John knows the constant c but not the starting point Zo, this will lead us to the definition of Julia sets-named after the mathematician Gaston Julia (1893-1978). For any given complex number c, some initial points Zo generate divergent sequences Zn+1 = Zn2 + c while others

generate nondivergent sequences.

The

Julia set is the boundary that separates the set of "diverging" starting points from the set of "nondiverging" starting points. Insb"uctions (the same insb"uctions were

given on both days): Answer all three questions. Each question carries 35 points. Time allowed is 411z hours.

First DaI

Question 1. In a triangle ~ABC, L C=2LB. P is a point in the interior of ~ABC satisfying thatAP= AC and PB = PC. Show that AP trisects the angle LA.

Question 2. In a table-tennis tournament of 10 contestants, any two contestants meet only once. We say that there is a winning triangle if the following s~tuation occurs: ith contestant defeated jth contestant, jth contestant defeated kth contestant, and kth contestant defeated ith contestant. Let Wi and Li be respectively the number of games won and lost by the ith contestant. Suppose Li + Wj ? 8 whenever the ith contestant wins

{continued on page 4)

To help John to answer the above question, it is natural to first try c = 0 and see what will happen. The recursion becomes Zn+l = zn2 and thus Zn = 0 for all n. That is, John will be going nowhere but staying at the origin!

If we try other values of c, there are three possible outcomes: (1) the sequence zn converges to a fIXed 'point; (2) the sequence Zn repeats in a finite cycle of points and thus becomes a periodic sequence; or (3) the sequence .zn diverges from the origin, i.e., John may have a chance to escape successfully.

Here is a simple example. For c = 0, the equation is Zn+l = Zn2. If the starting point lies within a distance of 1 from the origin, the subsequent points will get closer and closer to the origin. If the intial point is more than a distance of 1 from the origin, the subsequent points will get farther and farther away from the origin. The unit circle separate~ these two sets of starting points. This boundary is the Julia set correspOnding to c = O.

(continued on page 2)

Editors: Cheung, Pat-Hong, Curro Studies, HKU Ko, Tsz-Mei; EEE Dept, HKUST ~g, Tat-Wing, Appl. Math Dept, HKPU Li, Kin- Yin, Math Dept, HKUST Ng, Keng Po Roger, lTC, HKPU Artist: Yeung, Sau- Ying Camille, Fine Arts Dept, CU Acknowledgment: Thanks to Martha A. Dahlen,

Technical Writer, HKUST, for her comments.

The above story is a dramatization for the definition of a fractal called the Mandelbrot set. (The word "fractal" was coined by Benoit Mandelbrot to describe sets with self-similarity, i.e., they look the same if you magnify a portion of them.) The Mandelbrot set can be defined as the set of complex numbers c for which the sequence Z.+I = Z,,2 + c is bounded (i.e., does not diverge) when the starting point Zo is the origin (0,0). Figure I shbWS the asymptotic behaviour of Z. for real c's that generate bounded sequences (i;e., outcomes I and 2). The number of points on a vertical line indicates the period of the asymptotic sequence. Figure 2 shows the

The editors welcome contributions from all students. With your submission, please include your name, address, school, email address, telephone and fax numbers (if available). Electronic submissions, especially in TeX, MS Word and WordPerfect, are encouraged. The deadline for receiving material for the next issue is March 31, 1995. Send all correspondence to:

Dr. Li, Kin-Yin Department of Mathematics

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

Hong Kong

Fax: 2358-1643

Mathematical Excalibur Vol. 1, No.2, Mar-Apr, 95

_{Page 2}

Fractal Game of Escape

{continued

from page 1)

Pythagorean Triples

Kin- Yin Li

By varying c, we will obtain an infinite number of different pictures of Julia sets. Sorne,examples are shown in the figures on this page. However, no matter what cis, we observe that there are basically two major types of Julia sets. Either all the points Zo are connected in one piece, or these points are broken into a number of pieces (in fact, an infinite number of pieces to fomi something called a Cantor set).

In geometry, we often encounter

triangles whose sides are integers. Have you ever thought about how to produce many nonsimilar triangles of this kind without guessing? For this, we first define Pythagorean triples to be triples (a, b, c) of positive integers satisfying a2 + b2 = ~. For example, (3,4, 5) and (5, 12, 13) are Pythagorean triples. Clearly, if a2 + b2 = ~ , then (ad)2 + (bd)2 = (cd)2 for any positive integer d. So, solutions of a2 + b2 = C2 with a, b, c relatively pri~e (i.e., having no common prime divisors) are important. These are called primitive solutions. Below we will establish a famous theorem giving all primitive solutions.

a common prime divisor p, then the

equation will imply all three have p as a

common divisor and p ., 2. It will also

follow that(c- a)/2 =U2 and (c + a)/2 = v

are integers with p as a common

divisor.

This will contradict u, v being relatively

prime. So a, b, c must be relatively prime.

We may ask ourselves an interesting

question. For which values of c, will the corresponding Julia set be connected? This seems to be a very hard problem. It seems that we need to look at all Julia sets to find out which one is connected, and it would take an eternity to compile this huge amount of data. But mathematicians John Hubbard and Adrien Douady found a quick way to carry out this task. They proved that a Julia set is connected if the sequence Zn+1 = Zn2 + C is bounded when the starting point ?o is the origin (0,0). That is, if c belongs to the Mandelbrot set, then its corresponding Julia set will be connected! Thus the Mandelbrot set is known as the table of contents for all Julia sets.

For the second statement, we introduce modulQ arithmetic. If r, s are integers having die same remainder upon division by a positive integer m, then we say r is congruent to s modulo m and let us denote this by r = s (mod m). For example, r = 0 or 1 {mod 2) depending on whether r is even or odd. From the definition, we see that congruence is an equivalence relation between rand s. Also, if r = s (mod m) and r' = s' (mod m), then r + r' = s + s' (mod m), r- j' = s -s' (mod m), rr' = ss' (mod m) and r!c = I (mod m) for any positive integer k.

Theorem. If u, v are relatively prime positive integers, u > v and one is odd, the other even, then a = u2- vl, b = 2uv, e = U2 + v2 give a primitive solution of a2 + b2 = c2. Conversely, every primitive solution is of this form, with a possible permutation of

a and b. _{In working with squares, modulo 4 is}

often considered. This comes from the observation that r2 = 0 or 1 (mod 4) depending on ris even or odd. Now, if a2 + b2 == C2, then a2 + b2 = 0 or 1 (mod 4). For example, u = 2, v = I corresponds

toa=3, b=4, C = 5. Now let us try to see why the theorem is true. For the flfst statement, simple algebra shows a2 + b2 = u4 + 2U2v2 + v4 = cZ. If two of a, b, C have Besides this interesting relationship and

the fascinating pictures, the Julia set and many other fractals provide us insight into many physical phenomenon. As an example, the JuJia Set is directly related to the equipotential field lines of an electrostatic circular metal rod. The interested reader may refer to the book "Chaos and Fractals: New Frontiers of Science," written by H.G. Peitgen, H. JUrgens, and D. Saupe (Springer Verlag, 1992).

(continued on page 4)

Julia Sets for Various Values ofc

Due to the self-similarity of fractals, one usually needs only a few lines of computer programming to generate a fractal image. (Would you like to try?) There is also a free computer software FRACfINT (developed by the Stone Soup Group) that can generate many popular fractal images. If you would like to get a copy of this computer software, send a stamped self-addressed envelope and a PC-formatted high-density diskette to the author at the following address: Roger Ng, Institute of Textile and Clothing, Hong Kong Polytechnic University, Hung Horn, Kowloon. There are over a hundred fractal images for your investigation.

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~ .:~- ~~

"'2.

~~~

,,-' +~.J.

..~ -Ti. ""

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Mathematical Excalibur Vol. 1, No.2, Mar-Apr, 95

Page 3

Sum (HKUST), W. H. FOK (Homantin

Government

Secondary

School), and HO

Wing Yip (Clementi

Secondary

School).

Problem Corner

We welcome readers to submit solutions to the problems posed below for publication consideration. Solutions should be preceded by the solver's name, address and school affiliation. Please send submissions to Dr. Kin Y. Li, Department o/Mathematics, Hong Kong University of Science and Technology, Clear Water Bay, Kowloon. Solutions to the following problems should be submitted by March 31,1995.

= 2310. So x2 -2310x:f- 2310n = O. It follows the discriminant ~ = 231O:Z -4(2310n)=22x3x5x7x 11 x(1155-2n) must be a perfect square. Then for some positive integer k, 1155 -2n == 3 x 5 x 7 x 11 x ~ = 1155~ ~ 1155, which is a contradiction. So xy is not divisible by

2310.

Comments: A similar problem appeared in the magazine Quantum, Sept.fOct. 1993, p. 54, published by Springer- Verlag. Other commended solvers: AU Kwok Nin (Tsung Tsin College), HO Wing Yip (Clementi Secondary School), POON Wai Hoi Bobby (St. Paul's College) and SZE Hoi Wing (St. Paul's Co-ed College). Problem 6. For quadratic polynomials

P(x) = ar + bx .f. c with real coefficients satisfying I P(x) I s 1 for -'I s x ~ 1, find the maximum possible values of b and give a polynomial attaining the maximal b coefficient.

Problem 7. If positive integers a, b, C satisfy a2 + b2 = C2, show that there are at least three noncongruent right triangles with integer sides having hypotenuses all equal to d.

Problem 8. (1963 Moscow Mathematical Olympiad) Let a, = a2 = 1 and a" = (a".12 + 2)/an-2 for n = 3. 4. Show that an is an integer for n = 3, 4.

Problem 3. Show that for every positive integer n, there are polynomials P(x) of degree n and Q( x) of degree n-1 such that (P(x))2-1 = (r-1)(Q(x))2.

Solution: POON Wai Hoi Bobby, St. Paul's College.

For k = 1,2, .'., define Pt(x), Qt(x) by P1(x) =x, QJx) = 1, Pt+I(X) =xPt(x) + (r-1) Qt(x) and Qt+l(X) = Pt(x) + xQ/x). We can check that the degree of P n is n and the degree of Qn is n-1 by showing inductively thatP n(x) = 2n-lx" + ...and Qn(x) = 2n-lx"-1 + For the problem, when n = 1, P1(X)2-1 = r- P1(X)2-1 = (K-P1(X)2-1)QJx)2. Suppose the case n = k holds. Then Pt+l(X)2 -1= [xP/x) + (r-1)Q/x)f-1 = (r-1)[P/x)2 + 2xP/x)Q/x) + (r-1)Q/x)2] + P/X)2- 1 = (r-1)[P/x)2 + 2xP/x)Q/x) + (r-1)Q/x)2] + (r-l)Qk(x)2 = (r-1)Qt+J(x)2.

Comments: The solvers mainly observed that if we substitute x = cos 8, then P / cos 8) = cos kOand Q/cos8) = sin kOf sinO. The recurrence relations for Pt+1 and Qt+1 are just the usual identities for cos(kO + 8) and sin(kO + 8). The polynomials P t, Qk are (continued on page 4) Problem 9. On sides AD and BC of a

convex quadrilateral ABCD with AB < CD, locate points F and E, respectively, such that AF/FD = BE/EC =AB/CD. Suppose EF when extended beyond F meets line BA at P and meets line CD at Q. Show that LBPE = L CQE.

Problem 2. Given N objects and B(?2) boxes, find an inequality involving Nand B such that if the inequality is satisfied, then at least two of the boxes have the same number o! objects. .

Solution: POON Wai Hoi Bobby, St. Paul's College.

Denote the number of objects in the kth box by Nt" Suppose no two boxes have the same number of objects. ThenN = NJ + N2 + ...+ NB ? 0 + 1+ 2 + ...+ 1) = B (B-I )/2. So if N < B (B-(B-I)/2, then at least two of the boxes have the same number of objects.

Other commended solvers: CHAN Wing

Construction Without Words:

Inscribe a Regular Pentagon in a Unit Circle

Problem 10. Show that every integer k >

1 has a multiple which is less than k4 and can be written in base 10 with at most four different digits. [Hint: First consider numbers with digits 0 and 1 only.] (This was a problem proposed by Poland in a past IMO.)

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

Solutions

Problem 1. The sum of two positive

integers is 2310. Show that their product

is not divisible by 2310.

Solution: W. H. FOK, Homantin

Government

Secondary

School.

Let x, y be two positive integers such

that x + y = 2310. Suppose

xy is divisible

by 2310, then xy = 2310n for some

positive integer n. We get x + (2310n/x)

Mathematical Excalibur Vol. 1, No.2, Mar-Apr, 95

Page 4

that AP = 2(u2-v)/(U2+V) and BP = 4UV/(U2+V), where u, v are as in the theorem. Since Ap2 + Bp2 = AB2, all such P's are on the unit circle. Using similar triangles, we find the coordinates of P is (x,y), where x = (AP2/2) -1 and y = ;tAP.BP/2 are both rational. Let 0= LBOP = 2LBAP. Then cos( 0/2) = (1 +x)/AP and sin(0/2) = lyl/AP are rational. Finally, for two such points P and P', pp' = 21 sin( 8 tl)/21 = 21 sin( 0/2)cos( 0'/2) -cos( 0/2) sin( 0'/2) 1 is rational.

Problem Corner

(continued

from page 3)

Chung Ming Secondary School), W. H. FOK (Homantin Government Secondary

School) and HO Wing Yip (Clementi Secondary School).

Without loss of generality, suppose a( < ~ < :.. < an. For k = 2, 3, .~., n, since the differences are distinct, at = at + (a2 -au + ...+ (at -at.u ~ 1 +(2 + 4 + ...+ 2(k-l)) = 1 + t2 -k. Summing from k = 1 to n, we get a, + a2 + ...+ an ? n (n2 + 2)/3. Comments: Ho Wing Yip proved the result by induction on n, which did not require the formula for summing t2 in the last step.

Pythagorean Triples

{continued

from page 2)

Example 3. Find all positive integral solutions of3x + 4)1 = 5'. (cf. W. Sierpinski, On the Equation 3X + 4)1 = 5' (polish), Wiadom. Mat.(1955/56), pp. 194-5.) Solution. We will show there is exactly one solution set, namely x = y = z = 2. To simpl~ the equation, we consider modulo 3. Wehavel=O+I)1= 3x+4)1= 5' = (-1)' (mod 3). It follows that z must be even, say z = 2w. Then 3X = 5' -4)1 = (5W + 2Y)(5W- 2Y). Now 5w + 2)1 and 5W -2)1 are not both divisible by 3, since their sum is not divisible by 3. So, 5w + 2)1 = 3x and 5w-2Y = 1. Then, (-I)W + (~I))I = 0 (mod 3) and (-l)W -(-1))1 = 1 (mod 3). Consequently,

w is odd and y is even. If y > 2, then 5= 5w + 2Y = 3x = 1 or 3 (mod 8), a contradiction. So y = 2. Then 5W -2)1 = 1 implies w = 1 and z = 2. Finally, we get x = 2.

So, if a, b, c are also relatively prime, then one of a or b is odd and the other is even. Let us say a is odd and b is even. Then c is odd and it follows m : (c -a)/2 and n = (c + a)/2 are positive integers. Note a (: m-n) and c (: m+n) relatively prime implies m, n cannot have a common' prime divisor. Now considering the prime factorization of (b/2f, which equals mn, it follows that both m and n are perfect squares with no common prime divisors. Let us say m = U2 and n = y2. Then a = U2-y2, b = 2uv and c = U2 + y2.

called Chebychev polynomials and have many interesting properties.

We thank Professor Andy Liu (University of Alberta, Canada) for informing us that his colleague Professor Murray Klamkin located this problem in Goursat-Hedrick's "A Course in Mathematical Anaysis", vol. 1, p. 32, published by Ginn and Company in 1904. Professor Klamkin has a calculus solution, first showing Q divides p', then obtaining Q = nP' and solving a differential equation in P to get P(x) = cos(n arccos x). Professor liu also forwarded an alternative recurrence approach by Byung-Kyu Chun, a Korean-Canadian secondary school student He observed that P n(x) = 2xP n-'(x) -P.-z{x) and Q.(x) = 2xQn-I(X) -Qn-2(X) and showed by simultaneous induction that P n(X)P n~I(X) -X = (~- l)Qn(X)Q.-l(X) and P n(X)2 -1 = (~ -1 )Qn(X)2.

Other commended solver: 80 Wing Yip (Clementi Secondary School).

Example 1. Show that there are exactly three right triangles whose sides are integers while the area is twice the perimeter as numbers. (This was a problem on the 1965 Putnam Exam, a North American Collegiate Competition.) Solution: For such a triangle, the sides are of the fonna = (U2- y2)d, b = 2uvd and c = (U2 + y2)d, where u, v are relatively prime, u > v, one is odd, the other even and d is the greatest common divisor of the three sides. The condition ab/2 = 2(a+b+c) expressed in tenDS of u, v, d can be simplified to (u-v)vd = 4. It follows that u -v being odd must be I. Then v = 1,2

or 4; u = 2, 3 or 5; d = 4, 2 or I

corresponding to the 12-16-20, 10-24-26 and 9-40-41 triangles.

Example 2. Show that there are infinitely many points on the unit circle such that the distance between any two of them is rational. (This was essentially a problem in the 1975 International Mathematical Olympiad).

Solution: Let A = (-1, 0), B = (1, 0) and 0 be the origin. Consider all points P such

Olympiad Corner

(continued

from page 1)

the jth contestant. Prove that there are

exactly 40 winning triangles in this

tournament.

Question 3. Find all the non-negative

integers x, y. and z satisfying that 7x + 1 =

3Y + 5z.

Problem 4. If the diagonals of a quadrilateral in the plane are perpendicular, show that the midpoints of its sides and the feet of the perpendiculars dropped from the midpoints to the opposite sides lie on a circle.

Solution: Independent solution by W. H. FOK (Homantin Government Secondary School) and POON Wai Hoi Bobby (St. Paul's College).

Let ABCD be a quadrilateral such that AC is perpendicular to BD. Let E, F, G, H

be the midpoints of AB, BC, CD, DA, respectively. By the midpoint theorem, EH, BD, FG are parallel to each other and so are EF,AC, HG. Since AC and BD are perpendicular, EFGH is a rectangle. Hence E.. F, G, Hare concyclic.

Let M be the foot of the perpendicular fromEto CD, then LEMG = LEFG= 90°. SoH, F, M, G, H lie on a circle. Similarly, the other feet of perpendiculars are on the same circle.

Problem 5. (1979 British Mathematical Olympiad) Let ai, a2' ..., an be n distinct positive odd integers. Suppose all the differences I a(ajl are distinct, 1 oS i <j oS n. Prove that a, + a2 + ...+ an ~ n(n2+2)/3. Solution: Independent solution by Julian CHAN Chun Sang (Lok Sin Tong Wong

Second Da):

Question 4. Suppose that yz + zx + xy = 1 andx,y, and z?; O. Prove thatx(l-f)(I-i) + y(l-thatx(l-f)(I-i)(I-r) + z(l-r)(I-f) .s 4,fj/9. Question 5. Given that a function f<n) defined on natural numbers satisfies the conditions:f<n) = n -12 if n > 2000, and f<n) =fij{n+16)) ifn.s 2000.

(a) Findf<n).

(b) Find all solutions tof<n) = n. Question 6. Let m and n be positive integers where m has d digits in base ten and d s n. Find the sum of all the digits