Volume 8, Number 2 April 2003 – May 2003
Countability
Kin Y. Li
Olympiad Corner
The Final Round of the 51st Czech and Slovak Mathematical Olympiad was held on April 7-10, 2002. Here are the problems.
Problem 1. Solve the system (4x)5 + 7y = 14,
(2y)5 – (3x)7 = 74,
in the domain of the integers, where (n)k stands for the multiple of the number k closest to the number n.
Problem 2. Consider an arbitrary equi- lateral triangle KLM, whose vertices K, L and M lie on the sides AB, BC and CD, respectively, of a given square ABCD. Find the locus of the midpoints of the sides KL of all such triangles KLM. Problem 3. Show that a given natural number A is the square of a natural number if and only if for any natural number n, at least one of the differences (A + 1)2 – A, (A + 2)2 – A,
(A + 3)2 – A, …, (A + n)2 – A
is divisible by n.
(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 April 26,
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]
Consider the following two questions: (1) Is there a nonconstant polynomial
with integer coefficients which has every prime number as a root? (2) Is every real number a root of some
nonconstant polynomial with integer coefficients?
The first question can be solved easily. Since the set of roots of a nonconstant polynomial is finite and the set of prime numbers is infinite, the roots cannot contain all the primes. So the first question has a negative answer.
However, for the second question, both the set of real numbers and the set of roots of nonconstant polynomials with integer coefficients are infinite. So we cannot answer this question as quickly as the first one.
In number theory, a number is said to be algebraic if it is a root of a nonconstant polynomial with integer coefficients, otherwise it is said to be transcendental. So the second question asks if every real number is algebraic.
Let’s think about the second question. For every rational number a/b, it is clearly the root of the polynomial P (x) = bx – a. How about irrational numbers? For numbers of the form n a/b, it is a
root of the polynomial P (x) = b xn – a. To some young readers, at this point they may think, perhaps the second question has a positive answer. We should do more checking before coming to any conclusion. How about π and e? Well, they are hard to check. Are there any other irrational number we can check? Recall cos(3θ)=4cos3θ-3cos θ. So setting
θ=20˚, we get 1/2 = 4cos3 20˚-3cos 20˚.
It follows that cos 20˚ is a root of the polynomial P (x) = 8x3 – 6x – 1. With
this, we seem to have one more piece of evidence to think the second question has a positive answer.
So it is somewhat surprising to learn that the second question turns out to have a negative answer. In fact, it is known that π and e are not roots of nonconstant polynomials with integer coefficients, i.e. they are transcendental. Historically, the second question was answered before knowing π and e were transcendental. In 1844, Joseph Liouville proved for the first time that transcendental numbers exist, using continued fractions. In 1873, Charles Hermite showed e was transcendental. In 1882, Ferdinand von Lindemann generalized Hermite’s argument to show π was also transcendental. Nowadays we know almost all real numbers are transcendental. This was proved by Georg Cantor in 1874. We would like to present Cantor’s countability theory used to answer the question as it can be applied to many similar questions. Let ℕ denote the set of all positive integers, ℤ the set of all integers, ℚ the set of all rational numbers and ℝ the set of all real numbers.
Recall a bijection is a function f: A→B such that for every b in B, there is exactly one a in A satisfying f (a) = b. Thus, f provides a way to correspond the elements of A with those of B in a one-to-one manner.
We say a set S is countable if and only if S is a finite set or there exists a bijection f : ℕ→S. For an infinite set, since such a bijection is a one-to-one correspondence between the positive integers and the elements of S, we have
1↔s1, 2↔s2, 3↔s3, 4↔s4, …
and so the elements of S can be listed orderly as s1, s2, s3, … without repetition
or omission. Conversely, any such list of the elements of a set is equivalent to showing the set is countable since assigning f (1) = s1, f (2) = s2, f (3) =
Mathematical Excalibur, Vol. 8, No. 2, Apr 03- May 03 Page 2 Certainly, ℕ is countable as the
identity function f : ℕ→ℕ defined by f (n)= n is a bijection. This provides the usual listing of ℕ as 1, 2, 3, 4, 5, 6, …. Next, for ℤ, the usual listing would be
…, –4, –3, –2, –1, 0, 1, 2, 3, 4, …. However, to be in a one-to-one correspondence with ℕ, there should be a first element, followed by a second element, etc. So we can try listing ℤ as
0, 1, –1, 2, –2, 3, –3, 4, –4, …. From this we can construct a bijection g : ℕ→ℤ, namely define g as follow:
g (n) = (1 – n) / 2 if n is odd and
g (n) = n / 2 if n is even. For ℚ, there is no usual listing. So how do we proceed? Well, let’s consider listing the set of all positive rational numbers ℚ+ first.
Here is a table of ℚ+.
O
M
M
M
M
M
M
L
L
L
L
L
L
6 6 5 6 4 6 3 6 2 6 1 6 6 5 5 5 4 5 3 5 2 5 1 5 6 4 5 4 4 4 3 4 2 4 1 4 6 3 5 3 4 3 3 3 2 3 1 3 6 2 5 2 4 2 3 2 2 2 1 2 6 1 5 1 4 1 3 1 2 1 1 1In the m-th row, the numerator is m and in the n-th column, the denominator is n.
Consider the southwest-to-northeast diagonals. The first one has 1/1, the second one has 2/1 and 1/2, the third one has 3/1, 2/2, 1/3, etc. We can list ℚ+ by writing down the numbers on
these diagonals one after the other. However, this will repeat numbers, for example, 1/1 and 2/2 are the same. So to avoid repetitions, we will write down only numbers whose numerators and denominators are relatively prime. This will not omit any positive rational numbers because we can cancel common factors in the numerator and denominator of a positive rational number to arrive at a number in the table that we will not skip. Here is the list we will get for ℚ+:
1/1, 2/1, 1/2, 3/1, 1/3, 4/1, 3/2, 2/3,
1/4, 5/1, 1/5, 6/1, 5/2, 4/3, 3/4, … . Once we have a listing of ℚ+, we can list
ℚ as we did for ℤ from ℕ, i.e.
0, 1/1, –1/1, 2/1, –2/1, 1/2, –1/2, 3/1, –3/1, 1/3, –1/3, 4/1, –4/1, 3/2, …. This shows ℚ is countable, although the bijection behind this listing is difficult to write down.
If a bijection h : ℕ → ℚ is desired, then we can do the following. Define h (1) = 0. For an integer n > 1, write down the prime factorization of g(n), where g is the function above. Suppose
g(n) = ± 2a 3b 5c 7d…. Then we define
h (n) = ± 2g(a+1) 3g(b+1) 5g(c+1) 7g(d+1)… with g (n), h (n) taking the same sign. Next, how about ℝ? This is interesting. It turns out ℝ is uncountable (i.e. not countable). To explain this, consider the function u : (0,1) →ℝ defined by u(x) = tan π(x–1/2). It has an inverse function v(x) = 1/2 + (Arctan x)/ π . So both u and v are bijections. Now assume there is a bijection f : ℕ→ ℝ. Then F = v◦f : ℕ→ (0,1) is also a bijection. Now we write the decimal representations of F(1), F(2), F(3), F(4), F(5), … in a table. F (1) = 0.a11a12a13a14… F (2) = 0.a21a22a23a24… F (3) = 0.a31a32a33a34… F (4) = 0.a41a42a43a44… F (5) = 0.a51a52a53a54… F (6) = 0.a61a62a63a64…
Consider the number
r = 0. b1b2b3b4b5b6…,
where the digit bn = 2 if ann = 1 and bn = 1 if ann ≠ 1. Then F (n) ≠ r for all n because ann ≠ bn. This contadicts F is a bijection. Thus, no bijection f : ℕ→ℝ can exist. Therefore, (0,1) and ℝ are both uncountable.
We remark that the above argument shows no matter how the elements of (0,1) are listed, there will always be numbers omitted. The number r above is one such number.
So some sets are countable and some sets are uncountable.
For more complicated sets, we will use the following theorems to determine if they are countable or not.
Theorem 1. Let A be a subset of B. If
B is countable, then A is countable.
Theorem 2. If for every integer n, Sn is a countable set, then their union is countable.
For the next theorem, we introduce some terminologies first. An object of the form (x1,…,xn) is called an
ordered n-tuple. For sets T1, T2, …, Tn,
the Cartesian product T1×⋯×Tn of
these sets is the set of all ordered n-tuples (x1,…,xn), where each xi is
an element of Ti for i = 1,…, n.
Theorem 3. If T1, T2, …, Tn are
countable sets, then their Cartesian product is also countable.
We will give some brief explanations for these theorems. For theorem 1, if A is finite, then A is countable. So suppose A is infinite, then B is infinite. Since B is countable, we can list B as b1, b2, b3, … without
repetition or omission. Removing the elements bi that are not in A, we get a list for A without repetition or omission.
For theorem 2, let us list the elements of Sn without repetition or omission in the n-th row of a table. (If Sn is finite, then the row contains finitely many elements.) Now we can list the union of these sets by writing down the diagonal elements as we have done for the positive rational numbers. To avoid repetition, we will not write the element if it has appeared before. Also, if some rows are finite, it is possible that as we go diagonally, we may get to a “hole”. Then we simply skip over the hole and go on.
For theorem 3, we use mathematical induction. The case n = 1 is trivial. For the case n = 2, let a1, a2, a3, … be
a list of the elements of T1 and b1, b2,
b3, …be a list of the elements of T2
without repetition or omission. Draw a table with (ai, bj) in the i-th row and j-th column. Listing the diagonal elements as for the positive rational numbers, we get a list for T1×T2
without repetition or omission. This takes care the case n = 2.
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 April 26, 2003.
Problem 176. (Proposed by Achilleas PavlosPorfyriadis,AmericanCollege of Thessaloniki “Anatolia”, Thessaloniki, Greece) Prove that the fraction n n m n m − + + + ) 1 ( 1 ) 1 ( is irreducible for all positive integers m and n.
Problem 177. A locust, a grasshopper and a cricket are sitting in a long, straight ditch, the locust on the left and the cricket on the right side of the grasshopper. From time to time one of them leaps over one of its neighbors in the ditch. Is it possible that they will be sitting in their original order in the ditch after 1999 jumps?
Problem 178. Prove that if x < y, then there exist integers m and n such that x < m + n 2 < y.
Problem 179. Prove that in any triangle, a line passing through the incenter cuts the perimeter of the triangle in half if and only if it cuts the area of the triangle in half.
Problem 180. There are n ≥ 4 points in the plane such that the distance between any two of them is an integer. Prove that at least 1/6 of the distances between them are divisible by 3. *****************
Solutions
****************
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
Solution. LEE Man Fui (STFA Leung
Kau Kui College, Form 6) and TANG Ming Tak (STFA Leung Kau Kui College, Form 6).
Since the inequality is symmetric in a and b, without loss of generality, we may assume a ≥ b. For every x, bx ≥ b[x]. Since b[x] is an integer, we get [bx] ≥ b[x]. Let x = c/(ab). We have
c[c/(ab)] – [c/a][c/b] = c[x] – [bx][c/b] ≤ (c/b)[bx] – [bx][c/b] = [bx]( (c/b) – [c/b] )
< bx· 1 = c/a = c min{1/a,1/b}. Other commended solvers: CHEUNG Yun Kuen (Hong Kong Chinese Women’s Club College, Form5), Antonio LEI (Colchester Royal Grammar School, UK, Year 13), SIU Tsz Hang (STFA Leung Kau Kui College, Form 7), Rooney TANG Chong Man (Hong Kong Chinese Women’s Club College, Form 5).
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.
Solution. D. Kipp JOHNSON (Valley
Catholic High School, Beaverton, Oregon, USA), Antonio LEI (Colchester Royal Grammar School, UK, Year 13), SIU Tsz Hang (STFA Leung Kau Kui College, Form 7) and and WONG Wing Hong (La Salle College, Form 5).
Suppose there is such an integer n and it has k digits. Then 10k–1 ≤ n ≤ (9k)2. However,
for k ≥ 5, we have
(9k)2 = 81k2 < (54/2)2k ≤ (5k–1/2)2k = 10k–1. So k ≤ 4. Then n ≤ 362. Since n is a perfect
square, we check 12, 22, …, 362 and find
only 1 and 92 = 81 work.
Other commended solvers: CHAN Yat Fei (STFA Leung Kau Kui College, Form 6) and Rooney TANG Chong Man (Hong Kong Chinese Women’s Club College, Form 5).
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. (Source: 1997 Russian Math Olympiad)
Solution. CHEUNG Yun Kuen (Hong
Kong Chinese Women’s Club College,
Form 5) and D. Kipp JOHNSON (Valley Catholic High School, Beaverton, Oregon, USA).
Let a1, a2, … , a300 be the weights of the
apples in increasing order. For j = 1, 2, …, 75, let the j-th group consist of the apples with weights aj, a75+j, a150+j, a225+j.
Note the weights of the groups are increasing. Then the ratio of the weights of any two groups is at most
226 151 76 1 300 225 150 75 a a a a a a a a + + + + + + 226 151 76 1 1 226 151 76 3 a a a a a a a a + + + + + + ≤ 1 226 151 76 )/ ( 1 2 1 a a a a + + + + = . Since 3 ≤ (a76 + a151 + a226) / a1 ≤ 9, so the
ratio of groups is at most 1+2/(1+3)=3/2. Other commended solvers: CHAN Yat Fei (STFA Leung Kau Kui College, Form 6), Terry CHUNG Ho Yin (STFA Leung Kau Kui College, Form 6), SIU Tsz Hang (STFA Leung Kau Kui College, Form 7) and TANG Ming Tak (STFA Leung Kau Kui College, Form 6).
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.
Solution. Achilleas Pavlos
PORFYRIADIS (American College of Thessaloniki “Anatolia”, Thessaloniki, Greece).
For such M, note the points around the circle are in the order A, B′, C, A′, B, C′. Now ∠ACC′ = ∠ABC′ as they are subtended by chord AC′. Also, AB′=AC′ because they both equal to AM by symmetry. So ∠ABC′ = ∠ACB′ as they are subtended by chords AC′ and AB′ respectively. By symmetry, we also have ∠ACB′ = ∠ACM. Therefore, ∠ACC′ = ∠ACM and so C, M, C′ are collinear. Similarly, A, M, A′ are collinear. Then CM ⊥ AB and AM ⊥ BC. So M is the orthocenter of ∆ABC. Conversely, if M is the orthocenter, then ∠ACB′ = ∠ACM = 90°−∠BAC = ∠ABB′, which implies A, B, C, B′ are concyclic. Similarly, A′ and C′ are on the circumcircle of ∆ABC.
Mathematical Excalibur, Vol. 8, No. 2, Apr 03- May 03 Page 4 Yun Kuen (Hong Kong Chinese
Women’s Club College, Form 5), Antonio LEI (Colchester Royal Grammar School, UK, Year 13), SIU Tsz Hang (STFA Leung Kau Kui College, Form 7) and WONG Wing Hong (La Salle College, Form 5).
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. (Source: 1993 Jiangsu Province Math Olympiad)
Solution. (Official Solution)
By the pigeonhole principle, the process will go on forever. Suppose there is a triangle that never contains any frog. Label that triangle number 1. Then label the other triangles in the clockwise direction numbers 2 to n. For each frog in a triangle, label the frog the number of the triangle. Let S be the sum of the squares of all frog numbers. On one hand, S ≤ (n + 1) n2. On
the other hand, since triangle 1 never contains any frog, then at every second, some two terms of S will change from i2 +
i2 to ( i + 1 )2 + ( i – 1 )2 = 2 i2 + 2 with i < n.
Hence, S will keep on increasing , which contradicts S ≤ (n + 1) n2. Thus, after some
time T, every triangle will eventually contain some frog at least once.
By the jumping rule, for any pair of triangles sharing a common side, if one of them contains a frog at some second, then at least one of them will contain a frog from then on. If n is even, then after time T, the n triangles can be divided into n / 2 = [ (n + 1) / 2] pairs, each pair shares a common side and at least one of the triangles in the pair has a frog. If n is odd, then after time T, we may remove one of the triangles with a frog and divide the rest into (n – 1)/2 pairs. Then there will exist 1 + ( n – 1) / 2 = [ (n + 1) / 2] triangles, each contains at least one frog.
Other commended solvers: SIU Tsz Hang (STFA Leung Kau Kui College, Form 7).
Olympiad Corner
(continued from page 1)
Problem 4. Find all pairs of real numbers a, b for which the equation in the domain of the real numbers
x x b x ax = − + − 1 24 2 2
has two solutions and the sum of them equals 12.
Problem 5. A triangle KLM is given in the plane together with a point A lying on the half-line opposite to KL. Construct a rectangle ABCD whose vertices B, C and D lie on the lines KM, KL and LM, respectively. (We allow the rectangle to be a square.)
Problem 6. Let ℝ+ denote the set of
positive real numbers. Find all functions f : ℝ+→ℝ+ satisfying for all x, y ∊ℝ+ the
equality
f (x f (y) ) = f ( xy ) + x.
Countability
(continued from page 2)
Assume the case n = k is true. For k+1 countable sets T1, …, Tk, Tk+1, we apply
the case n = k to conclude T1×⋯×Tk is countable. Then (T1×⋯×Tk)×Tk+1 is
countable by the case n = 2.
We should remark that for theorem 1, if C is uncountable and B is countable, then C cannot be a subset of B. As for theorem 2, it is also true for finitely many set S1, …, Sn because we can set
Sn+1, Sn+2, … all equal to S1, then the
union of S1, …, Sn is the same as the
union of S1, …, Sn, Sn+1, Sn+2, ….
However, for theorem 3, it only works for finitely many sets. Although it is possible to define ordered infinite tuples, the statement is not true for the case of infinitely many sets.
Now we go back to answer question 2 stated in the beginning of this article. We have already seen that C = ℝ is uncountable. To see question 2 has a negative answer, it is enough to show the set B of all algebraic numbers is countable. By the remark for theorem 1, we can conclude that C = ℝ cannot be a subset of B. Hence, there exists at least
one real number which is not a root of any nonconstant polynomial with integer coefficients.
To show B is countable, we will first show the set D of all nonconstant polynomials with integer coefficients is countable.
Observe that every nonconstant polynomial is of degree n for some positive integer n. Let Dn be the set of all polynomials of degree n with integer coefficients. Let ℤ′ denote the set of all nonzero integers. Since ℤ′ is a subset of ℤ, ℤ′ is countable by theorem 1 (or simply deleting 0 from a list of ℤ without repetition or omission).
Note every polynomial of degree n is of the form
anxn + an–1xn–1 + ⋯ + a
0 (with an ≠ 0),
which is uniquely determined by its coefficients. Hence, if we define the function w : ℤ′×ℤ×⋯ ×ℤ→ Dn by w(an, an-1, …, a0)=anxn+an–1xn–1 +⋯+a0 ,
then w is a bijection. By theorem 3, ℤ′×ℤ×⋯ ×ℤ is countable. So there is a bijection q : ℕ → ℤ′×ℤ×⋯ ×ℤ. Then w◦ q : ℕ → Dn is also a bijection. Hence, Dn is countable for every positive integer n. Since D is the union of D1, D2, D3, …, by theorem 2, D is
countable.
Finally, let P1, P2, P3, … be a list of all
the elements of D. For every n, let Rn be the set of all roots of Pn, which is finite by the fundamental theorem of algebra. Hence Rn is countable. Since B is the union of R1, R2, R3, …, by
theorem 2, B is countable and we are done.
Historically, the countability concept was created by Cantor when he proved the rational numbers were countable in 1873. Then he showed algebraic numbers were also countable a little later. Finally in December 1873, he showed real numbers were uncountable and wrote up the results in a paper, which appeared in print in 1874. It was this paper of Cantor that also introduced the one-to-one correspondence concept into mathematics for the first time!
Solution. Fkom the first equation of the given system it follows that the number 7y - 14 = 7(y - 2) is divisible by five, so y = 53
+
2 for a suitable integer 3. Thus 2y = 108+4, hence (2y)s = 1Os+5. Substituting this back into the system we obtain the pair of equations (45)s+
35s = 0 and 10s - (3x)7 = 69. Multiplying the first equation by two, the second by seven, and subtracting, we eliminate the unknown s and obtain the equation 2(4x)~+7(3x), = -483 for the unknown x. Since the function F(t) = 2(4t)s +7(3t)7 is nondecreaing in the integer variable t and F(-18) = -532, F ( - 1 7 ) = -483 and F(-16) = -473, our eqcation F ( x ) = -483 has the unique solution z = -17. From the equation ( 4 s ) ~+
355 = 0 i t then follows t h a t s = 2, soy = 12. It is easily checked that (z, y) = (-17, 12) i s indeed a solution. The given system thus has a unique solution (x, y) = (-17,12).
A n o t h e r solution. For any integer t there clearly hold the inequalities t
-
2 Q( t ) s
<
t+
2 and t - 3 ,< ( t ) ,<
t+
3. Using them we obtain from the given system the following system of inequalities12
<
42+
7y<
16, 69<
2y - 3 s<
79.From these we easily eliminate, for instance, the unknown x: for the expression 3(4z
+
7y)+
4(2y-
3x), which equals 29y, we thus obtain the estimates29y
<
3 . 1 6+
4 - 7 9 = 364 and 29y 2 3 . 1 2+
4 . 6 9 = 312.From the inequalities 312
<
29y<
364 it follows that y E {11,12}. For y = 11 the first equation OE the given system gives (42)s = -63, which is not a multiple of five, while for y = 12 we get (42)s = -70, whence -72<
4 s<
-68, so x E {-18, -17). Therefore we must have y = 12; upon substituting into the second equation of the system we find that the latter is satisfied for x = -17, but not for x = -18. The only solution is thus the pair ( z , y ) = (-17,12).f d J ! ! 2&
Solution. Let S be the midpoint of the side K L of any of such triangles K L M (Fig.1). Since both angles LCM and L S M are right, the quadrangle C M S L is chordal, and therefore JLMCSI = ILMLSl = 60". The point S thus lies on the fixed segment C E , whose endpoint E E
AB
is determined by the equality ILECDl = 60". We show that the sought locus of the midpoints S is a certain segment between the points C and E , which is determined by the conditions S E C E ,(i) lASl> JBSl and (ii) ILCBSl 2 45'.
These conditions clearly imply that the segment in question is FG, where F is the vertex of the equilateral triangle C D F and G is the point of the side CF lying on the diagonal E D , Fig. 2. This is because the points of the segment C E satisfying the condition (i) are precisely those in the segment C F , and the condition (ii) precisely the points in the segment EG.
We now prove the claim above by picking an arbitrary point S in the interior of the segment CE and trying to construct the correSponding triangle K L M whose side K L has S for its midpoint. We fmd that such triangle KLM exists if and only if the point S satisfies both conditions (i) and (ii). Let us return to Fig.1. Since the angle K B L is right, the Thaletian theorem implies that all three segments SKI SB and S L are congruent. Thus starting horn the point S the points K and L can be obtained as intersections of the'segments AB and BC , respectively, with the circle of center S and radius ISBI. Such intersection K (K # B ) exists if and only if (i) holds, and the intersection L ( L # B) exists if and only lBSl Q ICSl, that is; [LRCSI
<
ILCYRSI. Since, however, / / F C S I = 30", the last inequality is guaranteed by the stronger condition (ii), the necessity of which will become apparent shortly. Once the points K and L are known, the point M can be determined asequilateral, since the quadrangle CMSL is chordal (the angles at the vertices C and 3 are rieht), and thus ILMLSl = lLMCSl = 60". It thus remains to examine when the intersection of the segment C D with the perpendicular bisector of the segment K L exists, that is, when the points C and D lie in opposite half-planes determined by that bisector. These half-planes are characterized by the inequalities lKXl
<
(15x1 and ( K X ( 2 lLXl, respectively; since IKC(2
lBCl and lBCl2
ILCl, hence (KC( 3 [LC(, our task is to find out when the inequality IKDI Q lLD[ holds. From the right triangles K D A and LDC we deduce that the last inequality holds if and only if [AKl<
ILCl, that is, IKBl 2 ILBl, or ILBLKl2
45". But the angle BLK is congruent with the angle CBS (we know that lSBl= ISLI), thus we obtain the condition (ii). This completes the proof.7 ~ . - -. -
? y O \ i \ t h ~ ~ Solution. (i) Assume first that A = d' for some natural d. Then for any j =
1 , 2 ,
...,
n,wehave ( A + j ) ' - A = ( d ' + j ) ' - d * = (da--d+j)(d'+d+j);sinceoneof the n consecutive numbers (8
-
d+
j ) , j = 1,2,.. .
,n, is divisible by n, so must be the corresponding number (A+
j)' - A.(ii) Assume now that the number A is not a square of any natural number. Then in the factorization of A into prime factors there must be a prime p which enters in an odd power, that is, p2k-'
1
A and pzk1
A for a suitable natural k. We claim that the number n = p2k then fails to have the property described in the statement of the problem. Indeed assume that for some '-
1,2,.. .
,pZk the difference ( A+
j)' - Ais divisible by
ti.
The numbers ( A+
j$&d A then give the same remainders upon division by p 2,
hence also upon division by pzk-'. Since the number A is divisible by p2'-', but not by p2', the same must hold for the number (A+
j)'. But this is a contradiction, since ( A+
j)' is a square.T 17 A Fig. 1 i Fig. 2 ... . . 7n)t)\em
4
Solution. Upon multiplying both sides by the expression x 2 - 1 (which vanishes if and only if 3: E (-1,l)) and moving all terms over to one side, we obtain the cubic equation
(1) 'X - ax2
+
232 - b = 0.It is well known that a cubic equation with real coefficients hss either one or t h e
roots in the real domain (counting multiplicities). Since both solutions of the original equation are roots of the equation (l), the latter must thus have three real roots. These roots XI, 5 2 , 1 3 and the coefficients of the equation (1) are related by the familiar formulas of Vibta:
In order to avoid some checks later on, let us also recall that every solution of the system (2) is formed by the three roots of the equation (l), that is, all solutions to (2) are permutations of one and the same triple of numbers.
The condition about the two solutions of the original equation means that either precisely one of the roots q , x 2 , 1 3 belongs to the set { - 1 , l ) and the other two roots are different, or one of the toots 21, x p , x 3 is double and none of them belongs to the set {-I, 1). The solutions of the original equation can thus be labelled hy s and 12-5 in such a way that one of the following cases occurs: ( x ~ , x ~ , x ~ ) = (-1,s, 12 - s), ( x i , ~ i , ~ 3 ) = ( 1 , ~ , 1 2 - ~ ) , or ( ~ 1 ~ x 2 ~ ~ 3 ) = ( s , s , 1 2 - s ) ; a t thesame time, always s $ { - I , 1 , G . 11.13) WP n o w consider all these cases separately.
-- ..-. .. !
___
(i) (21,22,23) = ( - l , s , l 2
-
s). Upon substituting into the system (2) and aI
simple manipulation, we obtainQ = 11, s2
-
12s-
35 = 0, b = -412-
9).The second equation has two roots s = 5 and s = 7, t o which there corresponds, by the third equation, the same value b = -35. The pair (a, b) = (11, -35) is a solution to our problem.
(ii) (21,z2,z3) = ( 1 , s , 1 2
-
s). Upon substituting into the system (2) and a simple manipulation, we obtaina = 13, s2
-
123 t 11 = 0, b = s(12-
s).The second equation has roots s = 1 and s = 11, which, however, belong to the excluded values of s (see above).
(iii) (z1,22,z3) = (9, s, 12
-
s). Upon substituting into (2) and a simple manipulation, we obtain
The second equation has roots s = 1 and s = 23. The value s = 1 L excluded, for
s = 23 the first and the third equation give Q = 35 and b = -11 . 232 = -5 819. The
pair (a, b) = (35, -5 819) is a solution to our problem. The sought pairs (a, b) are thus (11, -35) and (35, -5819).
g
w
-- ----
-
___
- _
Solution. Let ABCD be the sought rectangle, and denote by A‘B’C‘D’ its image under the translation by the vector
a
(Fig.3, B‘ = A ) . The point A’ lies on the line centrally symmetric to the line KM with respect to the center A-
let the intersections of this line with the lines LK and L M be K’ and M’, respectively. Since the diagonal AC of the sought rectangle lies on the line KL, the diagonal A’C‘ of the translated rectangle A’B‘C’D’ is parallel to KL. In the homothety with centerM’
which takes the point A’ into the point K‘ (and the point C‘ = D into the point L ) , .
the right triangle A‘AC‘ is mapped into the triangle K‘AI‘L. The point A” we can already construct, since it lies on the Thaletian circle with diameter K’L and also on the line M‘A. Now it is already easy to construct the sought rectangle ABCD: first determine the points A’ and C’ = D, which are images of the points K‘ and
L,
respectively, in the homothety with center
M’
taking A” into A , and then supply the vertices B and C as the images of thepoints B’ = A and C’ = D, respectively, under the translation by the vector A’A = AE.-
--
__
Since the point A lies in the interior of the segment K‘L and M’
#
A , the lineM’A intersects the Thaletian circle with diameter K’L always a t two points. It A’‘ is one of them and
M‘
#
A”, then the points A and A‘’ determine uniquely the above- mentioned homothety with center M’. Thus if the point M ” does not lie on the Thaletian circle with diameter K’L, the problem has two distinct solutions ABCD and AiBlClD1 (Fig. 4). Otherwise there is only one solution.I 1
D1
Fig. 3 Fig. 4 -.-
r - --- --- Q c b L \ e y I .L
Solution. Substituting f(z) for 3: in the given equation, we obtain the equation
from which we express f ( f ( z ) ~ ) = f(f(z)f(y)) - f(z). Another expression for the same quantity f(f(z)y) can be obtained by interchanging the variables 5 and y in
the original equation; this gives f ( f ( z ) y ) = f(y3:)
+
y. Comparing both expressiona, we thus arrive at the equation’
’
whose left-hand side remains unchanged upon interchanging I and y. The right-hand side must therefore enjoy the same property, that is, we must have
Upon a simple manipulation this gives f(z)
-
2 = f(y)-
y, which is to be satisfied for any r, y E R t . This means that the function z H f(z)-
z is constant on R+, sothe sought function f must be of the form f(2) = 2
+
c for a suitable real number c. Substituting this recipe into both sides of the original equation,I
I
I
we find that the latter is satisfied if and only i f c = 1. Thus the sought function f is unique and is given by f(z) = I
