Volume 15, Number 1 May 2010-June, 2010
Primitive Roots Modulo Primes
Kin Y. Li
Olympiad Corner
Below are the First Round problems of the 26th Iranian Math Olympiad.
Problem 1. In how many ways can
one choose n−3 diagonals of a regular n-gon, so that no two have an intersection strictly inside the n-gon, and no three form a triangle?
Problem 2. Let ABC be a triangle. Let Ia be the center of its A-excircle. Assume that the A-excircle touches AB and AC in B’ and C’, respectively. Let IaB and IaC intersect B’C’ in P and Q, respectively. Let M be the intersection of CP and BQ. Prove that the distance between M and the line BC is equal to the inradius of ∆ABC.
Problem 3. Let a, b, c and d be real
numbers, and at least one of c or d is not zero. Let f:ℝ→ℝ be the function defined by . ) ( d cx b ax x f + + =
Assume that f(x) ≠ x for every x∊ℝ. Prove that there exists at least one p such that f1387(p) = p, then for every x, for which f1387(x)is defined, we have f1387 (x) = x.
(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 July 10, 2010.
For individual subscription for the next five issues for the 09-10 academic year, send us five stamped self-addressed envelopes. Send all correspondence to:
Dr. Kin-Yin LI, Math Dept., Hong Kong Univ. of Science and Technology, Clear Water Bay, Kowloon, Hong Kong
Fax: (852) 2358 1643 Email: [email protected]
© Department of Mathematics, The Hong Kong University of Science and Technology
The well-known Fermat’s little theorem asserts that if p is a prime number and x is an integer not divisible by p, then
xp−1≡1(mod p).
For positive integer n>1 and integer x, if there exists a least positive integer d such that xd ≡1 (mod n), then we say d is the order of x (mod n). We denote this by ordn(x) = d. It is natural to ask for a prime p, if there exists x such that ordp(x) = p−1. Such x is called a primitive root (mod p). Indeed, we have the following
Theorem. For every prime number p,
there exists a primitive root (mod p). (We will comment on the proof at the end of the article.)
As a consequence, if x is a primitive root (mod p), then 1, x, x2, …, xp−2 (mod p) are distinct and they form a permutation of 1, 2, …, p−1 (mod p). This is useful in solving some problems in math competitions. The following are some examples. (Below, we will use the common notation a∣b to denote a is a divisor of b.)
Example 1. (2009 Hungary-Israel Math
Competition) Let p ≥ 2 be a prime number. Determine all positive integers k such that Sk = 1k + 2k + ⋯ + (p−1)k is divisible by p.
Solution. Let x be a primitive root (mod p). Then
Sk ≡ 1+xk+⋯+x(p−2)k (mod p). If p−1∣k, then Sk ≡1+⋯+1= p−1 (mod p). If p−1∤k, then since xk≢1 (mod p) and x(p−1)k ≡ 1(mod p), we have ). (mod 0 1 1 ) 1 ( p x x S k k p k − ≡ − ≡ −
Therefore, all the k’s that satisfy the requirement are precisely those integers that are not divisible by p−1.
Example 2. Prove that if p is a prime
number, then (p−1)! ≡ −1 (mod p). This is Wilson’s theorem.
Solution. The case p = 2 is easy. For p > 2, let x be a primitive root (mod p). Then
(p−1)! ≡ x1x2⋯xp−1 = x(p−1)p/2 (mod p). By the property of x, w=x(p−1)/2 satisfies w≢1(mod p) and w2 ≡1(mod p). So w ≡ −1(mod p). Then
(p−1)! ≡ x(p−1)p/2 = wp = −1 (mod p).
Example 3. (1993 Chinese IMO Team
Selection Test) For every prime number p ≥ 3, define , ) ( ( 1)/2 1 120
∑
− = = p k k p F ( ) , 2 1 ) ( ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ − = p p F p fwhere {x}=x−[x] is the fractional part of x. Find the value of f(p).
Solution. Let x be a primitive root (mod p). If p−1∤ 120, then x120≢ 1(mod p) and x120(p−1)≡1(mod p). So
∑
− = ≡ 1 1 120 2 1 ) ( p i i x p F 0(mod ). ) 1 ( 2 ) 1 ( 120 ) 1 ( 120 120 p x x x p ≡ − − = − Then f(p) = 1/2. If p−1∣ 120, then p∊{3, 5, 7, 11, 13, 31, 41, 61} and x120 ≡1(mod p). So ). (mod 2 1 2 1 ) ( 1 1 120 p p x p F p i i = − ≡∑
− = Then . 2 1 2 1 2 1 ) ( p p p p f = − − =Example 4. If a and b are nonnegative
integers such that 2a ≡ 2b (mod 101), then prove that a ≡ b (mod 100).
Mathematical Excalibur, Vol. 15, No. 1, May 10-Jun. 10 Page 2 Solution. We first check 2 is a
primitive root of (mod 101). If d is the least positive integer such that 2d ≡ 1 (mod 101), then dividing 100 by d, we get 100 = qd + r for some integers q, r, where 0 ≤ r < d. By Fermat’s little theorem,
1≡ 2100 =(2d)q2r ≡ 2r (mod 101), which implies the remainder r = 0. So d∣100.
Assume d < 100. Then d∣50 or d∣20, which implies 220 or 250≡1 (mod 101). But 210 = 1024 ≡ 14(mod 101) implies 220 ≡ 142 ≡ −6 (mod 101) and 250 ≡ 14(−6)2 ≡ −1 (mod 101). So d = 100. Finally, 2a ≡ 2b (mod 101) implies 2|a−b| ≡ 1 (mod 101). Then as above, dividing |a-b| by 100, we will see the remainder is 0. Therefore, a ≡ b (mod 100). Comments: The division argument in the solution above shows if ordn(x) = d, then xk ≡1 (mod n) if and only if d ∣ k. This is useful.
Example 5. (1994 Putnam Exam) For
any integer a, set
na = 101a −100×2a. Show that for 0 ≤ a,b,c,d ≤ 99,
na+nb ≡ nc+nd (mod 10100) implies {a,b}={c,d}.
Solution. Since 100 and 101 are relatively prime, na+nb ≡ nc+nd (mod 10100) is equivalent to
na+nb ≡ nc+nd (mod 100) and
na+nb ≡ nc+nd (mod 101). As na ≡ a (mod 100) and na ≡ 2a (mod 101). These can be simplified to
a+b ≡ c+d (mod 100) (*) and
2a+2b ≡2c+2d (mod 101). Using 2100≡ 1(mod 101) and (*), we get
2a2b = 2a+b ≡ 2c+d = 2c2d (mod 101). Since 2b ≡ 2c+2d −2a (mod 101), we get 2a(2c+2d −2a) ≡ 2c2d (mod 101). This can be rearranged as
(2a−2c)(2a−2d) ≡ 0 (mod 101). Then 2a ≡ 2c (mod 101) or 2a ≡ 2d (mod 101). By the last example, we get a ≡ c or d (mod 100). Finally, using a+b ≡ c+d (mod 100), we get {a,b}={c,d}.
Example 6. Find all two digit numbers n
(i.e. n = 10a + b, where a, b ∊ {0,1,…,9} and a ≠ 0) such that for all integers k, we have n | ka − kb.
Solution. Clearly, n = 11, 22, …, 99 work. Suppose n is such an integer with a ≠ b. Let p be a prime divisor of n. Let x be a primitive root (mod p). Then p ∣ xa − xb, which implies x|a−b| ≡ 1(mod p). By the comment at the end of example 4, we have p−1 ∣ |a-b| ≤ 9. Hence, p = 2, 3, 5 or 7. If p = 7 ∣ n, then 6 ∣ |a-b| implies n = 28. Now k2 ≡ k8 (mod 4) and (mod 7) hold by property of (mod 4) and Fermat’s little theorem respectively. So n = 28 works. Similarly the p = 5 case will lead to n = 15 or 40. Checking shows n = 15 works. The p = 3 case will lead to n = 24 or 48. Checking shows n = 48 works. The p = 2 case will lead to n = 16, 32 or 64, but checking shows none of them works. Therefore, the only answers are 11, 22, …, 99, 28, 15, 48.
Example 7. Let p be an odd prime number.
Determine all functions f : ℤ→ℤ such that for all m,n∊ℤ,
(i) if m ≡ n (mod p), then f(m) = f(n) and (ii) f(mn) = f(m)f(n).
Solution. For such functions, taking m = n = 0, we have f(0) = f(0)2, so f(0) = 0 or 1. If f(0) = 1, then taking m = 0, we have 1 = f(0) = f(0) f(n) = f(n) for all n∊ℤ, which is clearly a solution.
If f(0) = 0, then n ≡ 0 (mod p) implies f(n) = 0. For n≢0 (mod p), let x be a primitive root (mod p). Then n ≡ xk(mod p) for some k∊{1,2,…,p−1}. So f(n) = f(xk) = f(x)k. By Fermat’s little theorem, xp ≡ x (mod p). This implies f(x)p = f(x). So f(x) = 0, 1 or −1. If f(x) = 0, then f(n) = 0 for all n∊ℤ. If f(x) = 1, then f(n) = 1 for all n≢0 (mod p). If f(x) = −1, then for n congruent to a nonzero square number (mod p), f(n) = 1, otherwise f(n)= −1.
After seeing how primitive roots can solve problem, it is time to examine the proof of the theorem more closely. We will divide the proofs into a few observations. For a polynomial f(x) of degree n with coefficients in (mod p), the congruence
f(x) ≡ 0 (mod p)
has at most n solutions (mod p). This can be proved by doing induction on n and imitating the proof for real coefficient polynomials having at most n roots.
If d∣p−1, then xd−1 ≡ 0(mod p) has exactly d solutions (mod p). To see this, let n = (p−1)/d, then
xp−1−1= (xd−1)(x(n−1)d+x(n−2)d+⋯+1). Since xp−1−1≡ 0 (mod p) has p−1 solutions by Fermat’s little theorem, so if xd−1 ≡ 0 (mod p) has less than d solutions, then
(xd−1)(x(n−1)d+x(n−2)d+⋯+1) ≡ 0 (mod p) would have less than d + (n−1)d = p−1 solutions, which is a contradiction. Suppose the prime factorization of p −1
is ek
k
e p
p L1
1 , where pi’s are distinct primes and ei ≥ 1 . For i = 1, 2, …, k, let
.
i e i i p
m = Using the observation in the last paragraph, we see there exist mi−mi/pi > 1 solutions xi of equation 0 1≡ − i m
x (mod p), which are not solutions of xmi/pi −1≡0(mod p). It follows that the least positive integer d such that xid − 1 ≡ 0 (mod p) is
.
i e i i p
m = That means xi has order
i e i i p
m = in (mod p).
Let r be the order of xixj in (mod p). By the comment at the end of example 4, we have | i ej. j e i p p r Now ), (mod 1 ) ( ) (x x xx p x rd j i rd j r d i rd j ≡ = ≡
which by the comment again, we get
. | rd pej j Since j e j p and ei i p d = are relatively prime, we get pej| r.
j Interchanging the roles of pi and pj, we also get pej| r. j So p p | r. j i e j e i Then . j i e j e i p p r= So x = x1x2⋯xk will have order 1 1, 1 p = p− p ek k e L which implies x
is a primitive root (mod p).
For n > 1, Euler’s theorem asserts that if x and n are relatively prime integers, then xφ(n) ≡1(mod n), where φ(n) is the number of positive integers among 1,2,…,n that are relatively prime to n. Similarly, we can define x to be a primitive root (mod n) if and only if the least positive integer d satisfying xd ≡ 1(mod n) is φ(n). For the inquisitive mind who wants to know for which n, there exists primitive roots (mod n), the answers are n = 2, 4, pk and 2pk, where p is an odd prime. This is much harder to prove. The important thing is for such a primitive root x (mod n), the numbers xi (mod n) for i = 1 to φ(n) is a permutation of the φ(n) numbers among 1,2,…,n that are relatively prime to n.
Problem Corner
We welcome readers to submit their solutions to the problems posed below for publication consideration. The solutions should be preceded by the solver’s name, home (or email) address and school affiliation. Please send submissions to Dr. Kin Y. Li, Department of Mathematics, The Hong Kong University of Science & Technology, Clear Water Bay, Kowloon, Hong Kong. The deadline for sending solutions is July 10, 2010.
Problem 346. Let k be a positive integer. Divide 3k pebbles into five piles (with possibly unequal number of pebbles). Operate on the five piles by selecting three of them and removing one pebble from each of the three piles. If it is possible to remove all pebbles after k operations, then we say it is a harmonious ending.
Determine a necessary and sufficient condition for a harmonious ending to exist in terms of the number k and the distribution of pebbles in the five piles. (Source: 2008 Zhejiang Province High School Math Competition)
Problem 347. P(x) is a polynomial of degree n such that for all w∊{1, 2, 22, …, 2n}, we have P(w) = 1/w. Determine P(0) with proof.
Problem 348. In ∆ABC, we have
∠BAC = 90° and AB < AC. Let D be the foot of the perpendicular from A to side BC. Let I1 and I2 be the incenters of ∆ABD and ∆ACD respectively. The circumcircle of ∆AI1I2 (with center O) intersects sides AB and AC at E and F respectively. Let M be the intersection of lines EF and BC. Prove that I1 or I2 is the incenter of the ∆ODM, while the other one is an excenter of ∆ODM.
(Source: 2008 Jiangxi Province Math Competition)
Problem 349. Let a1, a2, …, an be rational numbers such that for every positive integer m, m n m m a a a1 + 2 +L+
is an integer. Prove that a1, a2, …, an are integers.
Problem 350. Prove that there exists a
positive constant c such that for all positive integer n and all real numbers a1, a2, …, an, if P(x) = (x − a1)(x − a2) ⋯ (x − an), then . ) ( max ) ( max ] 1 , 0 [ ] 2 , 0 [ P x c x P x n x∈ ≤ ∈ *****************
Solutions
****************Problem 341. Show that there exists an infinite set S of points in the 3-dimensional space such that every plane contains at least one, but not infinitely many points of S.
Solution. Emanuele NATALE and
Carlo PAGANO (Università di Roma
“Tor Vergata”, Roma, Italy).
Consider the curve σ : ℝ→ℝ3 defined by σ(x) = (x, x3, x5). Let S be the graph of σ. If ax+by+cz=d is the equation of a plane in ℝ3, then the intersection of the plane and the curve is determined by the equation
ax + bx3+ cx5 = d,
which has at least one and at most five solutions.
Other commended solvers: HUNG Ka
Kin Kenneth (Diocesan Boys’ School), D.
Kipp JOHNSON (Valley Catholic
School, Beaverton, Oregon, USA) and LI
Pak Hin (PLK Vicwood K. T. Chong
Sixth Form College).
Problem 342. Let f(x)=anxn+⋯+a1x+p be a polynomial with coefficients in the integers and degree n>1, where p is a prime number and
|an|+|an−1|+⋯+|a1| < p.
Then prove that f(x) is not the product of two polynomials with coefficients in the integers and degrees less than n.
Solution. The 6B Mathematics Group
(Carmel Alison Lam Foundation Secondary School), CHUNG Ping Ngai (La Salle College, Form 6), LEE Kai Seng (HKUST), LI Pak Hin (PLK Vicwood K. T. Chong Sixth Form College), Emanuele
NATALE (Università di Roma “Tor
Vergata”, Roma, Italy), Pedro Henrique
O. PANTOJA (University of Lisbon,
Portugal).
Let w be a root of f(x) in ℂ. Assume |w|≤1. Using anwn+⋯+a1w+p=0 and the triangle inequality, we have ,| | | || | 1 1 1
∑
∑
∑
= = = ≤ ≤ = n i i n i i i n i i iw a w a a pwhich contradicts the given inequality. So all roots of f(x) have absolute values greater than 1.
Assume f(x) is the product of two integral coefficient polynomials g(x) and h(x) with degrees less than n. Let b and c be the nonzero coefficients of the highest degree terms of g(x) and h(x) respectively. Then |b| and |c| ≥ 1. By Vieta’s theorem, |g(0)/b| and |h(0)/c| are the products of the absolute values of their roots respectively. Since their roots are also roots of f(x), we have |g(0)/b| > 1 and |h(0)/c| > 1. Now p = |f(0)| = |g(0)h(0)|, but g(0), h(0) are integers and |g(0)| > |b| ≥ 1 and |h(0)| > |c| ≥ 1, which contradicts p is prime.
Problem 343. Determine all ordered pairs (a,b) of positive integers such that a≠b, b2+a=pm (where p is a prime number, m is a positive integer) and a2+b is divisible by b2+a.
Solution. CHUNG Ping Ngai (La
Salle College, Form 6), HUNG Ka
Kin Kenneth (Diocesan Boys’ School) and LI Pak Hin (PLK Vicwood K. T. Chong Sixth Form College).
For such (a,b),
2 4 2 2 2 b a b b b a b a b a + + + − = + + implies pm = a + b2 | b4 + b = b(b3+1). From a ≠ b, we get b < 1+b < a+b2. As gcd(b, b3+1) = 1, so pm divides b3+1 = (b+1)(b2−b+1).
Next, by the Euclidean algorithm, we have gcd(b+1,b2−b+1) = gcd(b+1,3) | 3. Assume we have gcd(b+1,b2−b+1)=1. Then b2+a=pm divides only one of b+1 or b2−b+1. However, both b+1, b2−b+1< b2+a=pm. Hence, b+1 and b2−b+1 must be divisible by p. Then the assumption is false and
p = gcd(b+1,b2−b+1) = 3. (*) If m = 1, then b2+a = 3 has no solution. If m = 2, then b2+a = 9 yields (a,b) = (5,2). For m ≥ 3, by (*), one of b+1 or b2−b+1 is divisible by 3, while the other one is divisible by 3m−1. Since , 3 1 3 1 1< 2+ + = /2+ < −1 + b a m m b so 3m−1 | b2−b+1. Since m ≥ 3, we have b2 − b + 1 ≡ 0 (mod 9). Checking b ≡ −4, −3, −2, −1, 0, 1, 2, 3, 4 (mod 9) shows there cannot be any solution.
Mathematical Excalibur, Vol. 15, No. 1, May 10-Jun. 10 Page 4 Problem 344. ABCD is a cyclic
quadrilateral. Let M, N be midpoints of diagonals AC, BD respectively. Lines BA, CD intersect at E and lines AD, BC intersect at F. Prove that
. 2 EF MN BD AC AC BD− =
Solution 1. LEE Kai Seng (HKUST).
Without loss of generality, let the circumcircle of ABCD be the unit circle in the complex plane. We have
M = (A+C)/2 and N = (B+D)/2. The equations of lines AB and CD are
B A Z AB Z+ = + and D C Z CD Z+ = +
respectively. Solving for Z, we get
. CD AB D C B A Z E − − − + = = Similarly, . BC AD D C B A F − + − − = In terms of A, B, C, D, we have 2MN = |A+C−B−D|, EF = E−F . ) )( ( ) )( )( ( BC AD CD AB D B C A A C D B BC AD D C B A CD AB D C B A − − − − + − − = − + − − − − − − + =
The left and right hand sides of the equation become , ) )( ( | | | | 2 2 D B C A C A D B BD AC AC BD − − − − − = − . ) )( ( ) )( ( 2 A C D B BC AD CD AB EF MN − − − − =
It suffices to show the numerators of the right sides are equal. We have |B−D|2−|A−C|2 B D D B A C C A C A C A D B D B − − + = − − − − − = ( )( ) ( )( ) and (AB−CD)(AD−BC) . ) )( ( B D C A A C D B BC AD CD AB + − − = − − =
Comments: For complex method of solving geometry problems, please see Math Excalibur, vol. 9, no. 1.
Solution 2. CHUNG Ping Ngai (La Salle
College, Form 6).
Without loss of generality, let AC > BD. Since ∠EAC =∠EDB and ∠AEC =∠DEB, we get ∆AEC~∆DEB. Then
DB MC DN AM DB AC DE AE= = =
and ∠ECA =∠EBD. So ∆AEM ~∆DEN and ∆CEM ~∆BEN. Similarly, we have ∆AFC ~ ∆BFD, ∆AFM ~ ∆BFN and ∆CFM ~ ∆DFN. Then . FM FN FA FB AC BD AE DE EM EN = = = = (*)
Define Q so that QENF is a parallelogram. Let P = MQ∩EF. Then
∠EQF =∠FNE=180°−∠ENB−∠FND =180°−∠EMC−∠FMC=180°−∠EMF. Hence, M, E, Q, F are concyclic. Then ∠MEQ=180°−∠MFQ.
By (1), EN×FM = EM×FN. Then [EMQ] = ½ EM×FN sin ∠MEQ
= ½ EN×FM sin ∠MFQ =[FMQ], where [XYZ] denotes the area of ∆XYZ. Then EP=FP, which implies M, N, P, Q are collinear. Due to M, E, Q, F concyclic, so ∆PEM ~∆PQF and ∆PEQ ~∆PMF. Then . , PF NP PF QP FM QE FM FN PF PM QF EM EN EM= = = = =
Using these relations, we have FM FN EN EM AC BD BD AC − = − , 2 / EF MN PF NP PF MP− = =
which is the desired equation.
Problem 345. Let a1, a2, a3, ⋯ be a sequence of integers such that there are infinitely many positive terms and also infinitely many negative terms. For every positive integer n, the remainders of a1, a2, ⋯, an upon divisions by n are all distinct. Prove that every integer appears exactly one time in the sequence.
Solution. CHUNG Ping Ngai (La Salle
College, Form 6), HUNG Ka Kin
Kenneth (Diocesan Boys’ School), LI
Pak Hin (PLK Vicwood K. T. Chong
Sixth Form College), Emanuele
NATALE and Carlo PAGANO
(Università di Roma “Tor Vergata”, Roma, Italy).
Assume there are i > j such that ai = aj. Then for n > i, ai ≡ aj (mod n), which is a contradiction. So any number appears at most once.
Next, for every positive integer n, let Sn={a1, a2, …, an}, max Sn = av and min Sn = aw. If k = av− aw ≥ n, then k ≥ n ≥ v, w and av ≡ aw (mod k), contradicting the given fact. So
max Sn −min Sn = av− aw ≤ n − 1. Now Sn ⊆[min Sn , max Sn] and both contain n integers. So the n numbers in Sn are the n consecutive integers from min Sn to max Sn.
Now for every integer m, since there are infinitely many positive terms and also infinitely many negative terms, there exists ap and aq such that ap < m < aq. Let r > max{p,q}, then m is in Sr. Therefore, every integer appears exactly one time in the sequence. Comment: An example of such a sequence is 0, 1, −1, 2, −2, 3, −3, ….
Olympiad Corner
(continued from page 1)
Problem 4. Let a∊ℕ be such that for every n∊ℕ, 4(an+1) is a perfect cube. Show that a = 1.
Problem 5. We want to choose some phone numbers for a new city. The phone numbers should consist of exactly ten digits, and 0 is not allowed as a digit in them. To make sure that different phone numbers are not confused with each other, we want every two phone numbers to either be different in at least two places or have digits separated by at least 2 units, in at least one of the ten places.
What is the maximum number of phone numbers that can be chosen, satisfying the constraints? In how many ways can one choose this amount of phone numbers?
Problem 6. Let ABC be a triangle and H be the foot of the altitude drawn from A. Let T, T’ be the feet of the perpendicular lines drawn from H onto AB, AC, respectively. Let O be the circumcenter of ∆ABC, and assume that AC = 2OT. Prove that AB = 2OT’.
1. Using a simple induct.ion, one C<.lll see tha.t.whenever n-3 diagonals of a convex n-gbn are select.ed, such that they do not have any intersections iIlIlide the ri-gon, the n-goll is divided into exactly n - 2 tri,angles. Oile . can !'limply corlllider one of the. cjiagonals a.nd divide the n-gon into two . pa,rts according to the dia,go'nal, in order to complete the induction. The CHHe
n
=
:3 is easily dealt with. There is 110 dia.gQna) to choose,and hence the answer ii:l 1.
When n
>
3,n6 triangle can share 3 sides with the n-goll. Howevereach triangle must at least .::;hare one side. We have n '-2 triangles and
n
sides. Hence exactly two of the triangles will share two sides, and .therellt will share one Ride,' each. Now consider 011e of those two triangles.
it'
cau be any of then
trianglefi made fi'om two adjacent sides of then-goll. " ' .
Now consider the third Ride of that triangle, the one which is a diag-onal. The other triangle
wllich
uses
tha.t diagoilal, canbe Ct!ly
of the two tdl1llgleR which use tha.t. diagonal and share a side with the n-gon. . Just nice the picture below.' ".
~---'
--
-
---/--,.
-
..._---
--'Considering (;he 'new diagonal again, another triangle uses it. It again
has
two 'po~sihi1itieR. COlltinuing, each new (liagonai''llC'is twopossihili-tieR. The first one ha.d
n
po~sibilities~ But we are counting everything twice, because lhere are two t.riangles which sharetyvo
~ides,vit:h
the n-gon, and choosing any of them as the beginning triar~gle'would yield the Rarne resl.tlt. Hence there are n2n-4divided' by 2, or equivalently
, r,~2n-5 pos.':ln;le ways. ' ' '
2. Let D he the point wher,e the A-excircle'touches BO. Then'CD = CO',
hecallSe IaC'C a.nd I(l.DCare congruent: hoth having It right a.ngle,'
two congruent angleR, and two' congruent sides. LetQ' be the foot
of tJle perpendicular liue drawn from B auto .f(LG. Thell DGQ' ~
Hence
L.G'Q'Q
=L.DQ'G.
SinceL.EQ'IIL
andL.BDla
aTe both right angles, EDQ'Ia is cyclic. Hence L.DQ'C = L.DBII(. Sihce BIu iR the a1igle bisector of L.DB13', we haye L.PBI1L==
1:'.B:BI((. Because'L.BB'Ia
+
L.BQ'Ia = 90°+
DOC) = ,1'80°, BB'J«(Q' is cyclic. Hence L.E'EIu
= L.B' Q' IlL' Fii'lallyl, putting all thiugR together, we haveL.B'Q'Ia =L.GQ'O'. This proves tliat G', Q', B' 111'e collinear. Hence Q'
is the intersection of B',C' and G'Ia. This proves that Q,
==
Q'. HenceL.BQC = 90°. Similarly L.CPE = 90°. ' , '
Let I be the incenter of ABC. TheIl GI ..L OIa , because they are the
inner and outer bise<:xors of L.ECA. L.BQC = DOC) which shows 'that
BQ 1. CI/I' Hence EQ
/I
CI. Similarly GPII
BI. This shows that BMCI is l:l. parallelogram. Hence the diRtance betweenAi
and EC isequal to the distance,b'etween I and BO which is equal to the lnradius.
3. Since f1387(a) = a weltave j1388(a)
=
j(a), because obviously f(a) is defined (in-tactf1387(o.)
is
defined).If
we
letb::; f(o.)
thenP387(b) ==
b.
SiilCe f(a)
=f
a, a, b are two different solutions 0f t387(ic) = x. LetC
=.f(b).
Then again t381(C) ::; c. Since c::;feb),
we,luwe c=f..
b.
We want a, b, c to be three different numberfl. The only pair of them tha.t ' might possibly be e(lW1.1 is a, c. Assume that a =c.
Then f2(a) = a .$0
rea)
=
pea) =
a
and conthmillg like thiB we get p388(a)= a.
But j1388(a) = f(f1387(a» = f(a). Which is a contradiction, becaufle
f(a)
=f
a. Hence q" b1 c are all different solutions to f(x)=
x.The composition of two funct~orlfl of the fOrIn ~~~~t,~, iR itself I:\. ftinction.
of the form :;t~ '(with diffe'reIlt a, b, c, d). Hence f1381 is a function like ",
em+ f. The equation I!;f,+ I
=
X is equival~mt to gx2+
(h - e)x ...:f
=
0
p~ ~~ .
which is
a
polynomial equation of degree two, and hence can have atmost. two solutions unless it is totally zero, which mea.ns 9
=
f
=
a
aude
=h
wllich mea~ls f1387(x) = x for every x. ) 4. 4(0.3+1) and 4(a9+1) are perfect. cubes. Hellce t.heir quotient 0.6
-0.
3+1'is, also a perfect cube. If a
>
1, then 0.6- 0.3
+
1<
(0.2)3, = 0.6 anda6_a3+1
>
a6-3a4+3a2-1=,(0.
2_1)3 because.3a4-a3:-3a2 +2;:=:3. 2a3 - (t3 - 3a2
+
2>
2a3+
2>.0. But (a2)3 mld (a2 ":"'1)3are two ~onsecutive perfect cubes. So a6
-:-a3
.+
1 which lies between them, can not be a perfect cube., " ':,G. Let.8
={1,.2, ... , OJ.
Each phqne numher is a point in8
10phone numbers a.re not confused with each other iff they have a dif!tance greatertha,n 1 in SlO. So the problem if! to select. the maximum number
of
ilOn-adjacentpoints
in tlie lattice S10. . .'We URe induction on 71 to show tha.t the r~1a.xil1lu1l1 number of
non-acijfi.cent point,s which call be selected from,Sn if! (IS~I
+
1)/2 and thereis only one Way to do tha,t. For n = 0,
Isnl
= 1 and the claim is'obviously true. Let 71
>
O.811.
is D copies of811.-\
those that. have 1as their first coordinate, those that ha:ve 2 and so all. Let Ai be those'
elements that have i as their first coordinate. The maximum number'
of elements which can be chosen from
AI; A2
is(IAII
+
IA2D/2 becl1)lseif we pair elelllerltR of
A
I,A
2 according to their,last 71-1 ~ligits, fro111each pair at most one element call be chosen. Use, the same 'argument
for A3, A4 , for
A
5 ,A6
and A7 • As. Finally' use the induct.ion hypothesisfor
A
g • Th~s shows tlmt the number of Relected element,1Y is at most(lsn!
+
1)/2. However we can select all elements whose sum of digitsa.re equa,l to
n
modulo 2; obviously no t.wo elements will be acijacent.Dut'these are
(Isnl
+
i)/2
elements.It remaillR' to prove th(\t there is only one Way to do fiO. From the last.
a.rgument it if> obvious~ha~ the elements of Ag are uni9uely determine~.
Dut then the elements of As aTe, because we can pair points of As, Ag.
Hence t,he selected elementR of As are exactly thofie tlJ».t are not selected
in
A
g• Now'putAs, A~ aside,
and repeat the last argument forA
7 , andtheu'
A
6 • COllt.inuing like thifi we see.that
there is only one way to select,(1811.1
+
1)/2
non":ad,i.a.c811t elements,G. Using the Stewart,'::; t,heorem in OAB with the Cevian 01' we get
If we let B = LABG then TB = BE ·cosB = AH· Got Bcof:l B and·
TA
=
AH . t:Ol:J(DOO - B) ~ AH . sine.
Hence OT2=
R2 - AH2 (~()S2 B.We have AG = 2Rl-iinB, hence (AG/2)2
=
R2 sin2e.
So '(A.G/2) = 0'11iff
R2 - AH2 cos2 B=
R2 sin2 ~ or equivalently AH = R. The sameiR true for 01~". ,9illce AG/2 - 01' we have R = AH and hence
