Volume 16, Number 4 February 2012
Zsigmondy’s Theorem
Andy Loo (St. Paul’s Co-educational College)
Olympiad Corner
Below are the problems of the 2011-2012 British Math Olympiad Round 1 held on 2 December 2011.
Problem 1. Find all (positive or
negative) integers n for which
n2+20n+11 is a perfect square.
Problem 2.Consider the numbers 1, 2, ⋯, n. Find, in terms of n, the largest integer t such that these numbers can be arranged in a row so that all consecutive terms differ by at least t.
Problem 3. Consider a circle S. The
point P lies outside S and a line is drawn through P, cutting S at distinct points X and Y. Circles S1 and S2 are drawn through P which are tangent to S at X and Y respectively. Prove that the difference of the radii of S1 and S2 is independent of the positions of P, X and Y.
Problem 4. Initially there are m balls in
one bag, and n in the other, where m, n > 0. Two different operations are allowed:
a) Remove an equal number of balls from each bag;
b) Double the number of balls in one bag.
(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 March 28, 2012.
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
In recent years, a couple of “hard” number theoretic problems in the IMO turn out to be solvable by simple applications of deep theorems. For instances, IMO 2003 Problem 6 and IMO 2008 Problem 3 are straight forward corollaries of the Chebotarev density theorem and a theorem of Deshouillers and Iwaniec respectively. In this article we look at yet another mighty theorem, which was discovered by the Austro-Hungarian mathematician Karl Zsigmondy in 1882 and which can be used to tackle many Olympiad problems at ease.
Zsigmondy’s theorem
First part: If a, b and n are positive
integers with a>b, gcd(a, b)=1 and n≥2, then an–bn has at least one prime factor
that does not divide ak–bk for all positive
integers k<n, with the exceptions of: i) 26–16 and
ii) n=2 and a+b is a power of 2.
Second part: If a, b and n are positive
integers with a>b and n≥2, then an+bn
has at least one prime factor that does not divide ak+bk for all positive integers
k<n, with the exception of 23+13. The proof of this theorem is omitted due to limited space. Interested readers may refer to [2].
To see its power, let us look at how short solutions can be obtained using Zsigmondy’s theorem to problems of various types.
Example 1 (Japanese MO 2011). Find
all quintuples of positive integers (a,n,p,q,r) such that
an–1 = (ap–1)(aq–1)(ar–1).
Solution. If a≥3 and n≥3, then by
Zsigmondy’s theorem, an–1 has a prime
factor that does not divide ap–1, aq–1
and ar–1 (plainly n>p,q,r), so there is no
solution. The remaining cases (a < 3 or
n < 3) are easy exercises for the readers.
Example 2 (IMO Shortlist 2000). Find
all triplets of positive integers (a,m,n) such that am+1|(a+1)n.
Solution. Note that (a,m,n)=(2,3,n) with n ≥ 2 are solutions. For a > 1, m ≥ 2 and
(a,m) ≠ (2,3), by Zsigmondy’s theorem,
am+1 has a prime factor that does not
divide a+1, and hence does not divide (a+1)n, so there is no solution. The cases
(a =1 or m =1) lead to easy solutions.
Example 3 (Math Olympiad Summer Program 2001) Find all quadruples of
positive integers (x,r,p,n) such that p is a prime, n,r>1 and xr–1 = pn.
Solution. If xr–1 has a prime factor that
does not divide x–1, then since xr–1 is
divisible by x–1, we deduce that xr–1
has at least two distinct prime factors, a contradiction unless (by Zsigmondy’s theorem) we have the exceptional cases
x=2, r=6 and r=2, x+1 is a power of 2.
The former does not work. For the latter, obviously p=2 since it must be even. Let
x+1=2y. Then
2n = x2–1 = (x+1)(x–1) = 2y(2y–2). It follows that y=2 (hence x=3) and n=3.
Example 4 (Czech-Slovak Match 1996). Find all positive integral
solutions to px–yp=1, where p is a prime.
Solution. The equation can be rewritten
as px = yp+1. Now y = 1 leads to (p,x) =
(2,1) and (y,p) = (2,3) leads to x = 2. For
y > 1 and p≠3, by Zsigmondy’s theorem, yp+1 has a prime factor that does not
divide y+1. Since yp+1 is divisible by
y+1, it follows that yp+1 has at least two
prime factors, a contradiction.
Remark. Alternatively, the results of Examples 3 and 4 follow from Catalan’s conjecture (proven in 2002), which guarantees that the only positive integral solution to the equation xa–yb=1
Mathematical Excalibur, Vol. 16, No. 4, Feb. 12 Page 2
Example 5 (Polish MO 2010 Round 1). Let p and q be prime numbers with
q>p>2. Prove that 2pq–1 has at least
three distinct prime factors.
Solution. Note that 2p–1 and 2q–1
divide 2pq–1. By Zsigmondy’s theorem,
2pq–1 has a prime factor p
1 that does not divide 2p–1 and 2q–1. Moreover,
2q–1 has a prime factor p
2 that does not divide 2p–1. Finally, 2p–1 has a prime
factor p3.
The next example illustrates a more involved technique of applying Zsigmondy’s theorem to solve a class of Diophantine equations.
Example 6 (Balkan MO 2009). Solve
the equation 5x – 3y = z2 in positive integers.
Solution. By considering (mod 3), we
see that x must be even. Let x=2w. Then 3y = 52w–z2 = (5w–z)(5w+z). Note
that
(5w–z,5w+z) = (5w–z,2z)
= (5w–z,z)
= (5w,z) = 1,
so 5w – z = 1 and 5w + z = 3a for some
positive integer a ≥ 2. Adding, 2(5w) =
3a + 1. For a = 2, we have w = 1,
corresponding to the solution x = 2, y = 2 and z = 4. For a ≥ 3, by Zsigmondy’s theorem, 3a+1 has a prime factor p that
does not divide 32 + 1 = 10, which implies p≠2 or 5, so there is no solution in this case.
Example 7. Find all positive integral
solutions to pa–1=2n(p–1), where p is a
prime.
Solution. The case p=2 is trivial.
Assume p is odd. If a is not a prime, let
a=uv. Then pu–1 has a prime factor that
does not divide p–1. Since pu–1
divides pa–1=2n(p–1), this prime factor
of pu–1 must be 2. But by Zsigmondy’s
theorem, pa–1 has a prime factor that
does not divide pu–1 and p–1, a
contradiction to the equation. So a is a prime. The case a=2 yields p=2n–1, i.e.
the Mersenne primes. If a is an odd number, then by Zsigmondy’s theorem again, pa–1=2n(p–1) has a prime factor
that does not divide p–1; this prime factor must be 2. However, 2 divides
p–1, a contradiction.
(continued on page 4)
A Geometry Theorem
Kin Y. Li
The following is a not so well known, but useful theorem.
Solution. Let ∠BAC=∠DAC=θ and
G’ be on segment BC such that
∠G’AC =∠EAC=α. We will show G’, F, D are collinear, which implies G’=G.
Applying the subtended angle theorem to △ABE, △ABC and △ACD respectively, we get (1 α β A B D C ) AE AB AF θ α α θ ) sin sin sin( + = + , (2
Subtended Angle Theorem. D is a point
inside ∠BAC (<180˚). Let α =∠BAD and
β =∠CAD. D is on side BC if and only if
) AC AB AG ) sin( sin ' sinθ = α + θ−α and (3) AC AD AE ) sin( sin sinθ α θ−α + = . (*). sin sin ) sin( AB AC AD β α β α+ = + Doing (1)−(2)+(3), we get
Proof. Note D is on segment BC if and
only if the area of ΔABC is the sum of the areas of ΔABD and ΔACD. This is
' sin sin ) sin( AG AD AF θ α α θ+ = + . . 2 sin 2 sin 2 ) sin(α β ABAD α ACAD β AC
AB⋅ + = ⋅ + ⋅ By the subtended angle theorem, G’, F,
D are collinear. Therefore, G = G’.
Example 3. Multiplying by 2/(AB·AC·AD) yields (*).
Below, we will write PQ∩RS=X to mean lines PQ and RS intersect at point X. Example 1.
(Butterfly Theorem) Let
A,C,E,B,D,F be points in cyclic order
on a circle and CD∩EF=P is the midpoint of AB. Let M = AB∩DE and
N = AB∩CF. Prove that MP = NP.
Let AD∩BC=K, AB∩CD=L,
BD∩KL=F and AC∩KL=G. Prove that
1/KL = ½(1/KF +1/KG). β βα α Ο D Α B P C F E R Q N Μ α β A B C D K L G F
Solution. Applying the subtended angle theorem to △KAL, △KDL, △KDF and
△KAG, we get Solution.theorem, PC·PD=PE·PF, call this x. By the intersecting chord
Applying the subtended angle theorem to △PDE and △PCF, we get
KD KL KC KA KL KB β α β α β α β
α sin( ) sin sin
, sin sin ) sin( + = + + = + . sin sin ) sin( , sin sin ) sin( PF PC PN PE PD PM β α β α β α β α + = + + = + KA KG KC KD KF KB β α β α β α β
α sin( ) sin sin
, sin sin ) sin( + = + + = +
Call these (1), (2), (3), (4) respectively. Doing (1)+(2)−(3)−(4), we get
Subtracting these equations, we get
, sin sin sin 2 0 KG KF KL α α α − − = . sin sin 1 1 ) sin( x PC PD x PE PF PN PM − − − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − + α β β α (*)
Let Q and R be the midpoints of EF and
CD respectively. Since OP⊥AB, we
have PF−PE = 2PQ = 2OP cos(90°−α) = 2OP sin α. Then similarly we have
PD−PC = 2OP sin β. Hence, the right
side of (*) is zero. So the left side of (*) is also zero. Since 0 < α+β < 180°, we get sin (α+β) ≠ 0. Then PM = PN. which implies the desired equation.
Example 2. (1999 Chinese National Math
Competition) In the convex quadrilateral ABCD, diagonal AC bisects∠BAD. Let E
be on side CD such that BE∩AC=F and
DF∩BC=G. Prove that ∠GAC =∠EAC.
α
α
Α
Β
C DΕ
F GG'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 March 28, 2012.
Problem 386. Observe that 7+1=23 and 77+1= 23×113×911. Prove that for n = 2, 3, 4,…, in the prime factorization
of , the sum of the
n+3.
Problem 387. Determine (with proof) all functions f : [0,+∞) →[0,+∞) such that for every x ≥ 0, we have 4f(x) ≥ 3x and f (4f(x) − 3x) = x.
Problem 388. In ΔABC, ∠BAC=30° and ∠ABC=70°. There is a point M lying inside ΔABC such that ∠MAB= ∠MCA=20°. Determine ∠MBA (with
proof).
Problem 389. There are 80 cities. An airline designed flights so that for each of these cities, there are flights going in both directions between that city and at least 7 other cities. Also, passengers from any city may fly to any other city by a sequence of these flights. Determine the least k such that no matter how the flights are designed subject to the conditions above, passengers from one city can fly to another city by a sequence of at most k flights.
Problem 390. Determine (with proof) all ordered triples (x, y, z) of positive integers satisfying the equation x2y2 = z2(z2 − x2 − y2).
*****************
Solutions
****************
Problem 381. Let k be a positive integer. There are 2k balls divided into
a number of piles. For every two piles
A and B with p and q balls respectively,
if p ≥ q, then we may transfer q balls from pile A to pile B. Prove that it is always possible to make finitely many such transfers so as to have all the balls end up in one pile.
Solution. AN-anduud Problem Solving
Group (Ulaanbaatar, Mongolia), CHAN
Chun Wai and LEE Chi Man (Statistics and Actuarial Science Society SS HKUSU
),
Andrew KIRK (Mearns Castle High School, Glasgow, Scotland),Kevin LAU Chun Ting (St. Paul’s
Co-educational College, S.3), LO Shing
Fung (F3E, Carmel Alison Lam
Foundation Secondary School) and Andy
LOO (St. Paul’s Co-educational College). We induct on k. For k=1, we can merge the 2 balls in at most 1 transfer.
Suppose the case k=n is true. For k=n+1, since 2k is even, considering (odd-even)
parity of the number of balls in each pile, we see the number of piles with odd numbers of balls is even. Pair up these piles. In each pair, after 1 transfer, both piles will result in even number of balls. So we need to consider only the situation when all piles have even number of balls. Then in each pile, pair up the balls. This gives altogether 2n pairs. Applying the
case k=n with the paired balls, we solve the case k=n+1.
Problem 382. Let v0 = 0, v1 = 1 and vn+1 = 8vn−vn−1 for n = 1,2,3,….
Prove that vn is divisible by 3 if and only if
vn is divisible by 7.
Solution. Alumni 2011 (Carmel Alison Lam Foundation Secondary School) and
AN-anduud Problem Solving Group
(Ulaanbaatar, Mongolia) and Mihai
STOENESCU (Bischwiller, France). For n = 1,2,3,…, vn+2 = 8(8vn−vn−1) −vn =
63vn− 8vn−1. Then vn+2 ≡ vn−1 (mod 3) and
vn+2 ≡ −vn−1 (mod 7). Since v0= 0, v1=1, v2 = 8, so v3k+1, v3k+2 ≢ 0 (mod 3) and (mod 7) and v3k ≡ 0 (mod 3) and (mod 7).
Other commended solvers: CHAN Chun
Wai and LEE Chi Man (Statistics and Actuarial Science Society SS HKUSU
),
CHAN Long Tin (Diocesan Boys’
School), CHAN Yin Hong (St. Paul’s Co-educational College), Andrew KIRK (Mearns Castle High School, Glasgow, Scotland), Kevin LAU Chun Ting (St. Paul’s Co-educational College, S.3), LKL
Excalibur (Madam Lau Kam Lung
Secondary School of MFBM), Andy
LOO (St. Paul’s Co-educational College),
NGUYEN van Thien (Luong The Vinh
High School, Dong Nai, Vietnam), O Kin
Chit Alex (G.T.(Ellen Yeung) College)
,
Ángel PLAZA (Universidad de Las
Palmas de Gran Canaria, Spain), Yan Yin
WANG (City University of Hong Kong, Computing Math, Year 2), ZOLBAYAR
Shagdar (Orchlon School, Ulaanbaatar, Mongolia),Titu ZVONARU (Comăneşti, Romania) and Neculai STANCIU (“George Emil Palade’’ Secondary School, Buzău, Romania).
Problem 383. Let O and I be the circumcenter and incenter of ΔABC respectively. If AB≠AC, points D, E are midpoints of AB, AC respectively and
BC=(AB+AC)/2, then prove that the
line OI and the bisector of ∠CAB are perpendicular.
Solution 1. Kevin LAU Chun Ting (St. Paul’s Co-educational College, S.3).
B C A I D E F O 1 77 + = n n A exponents is at least 2
From BC = (AB+AC)/2 = BD+CE, we see there exists a point F be on side BC such that BF=BD and CF=CE. Since BI bisects ∠FBD, by SAS, ΔIBD≅ΔIBF. Then ∠BDI=∠BFI. Similarly, ∠CEI = ∠CFI. Then
∠ADI +∠AEI
= (180°−∠BDI ) + (180°−∠CEI ) = 360° −∠BFI −∠CFI = 180°.
So A,D,I,E are concyclic.
Since OD ⊥ AD and OE ⊥ AE, so
A,D,O,E are also conyclic. Then A,D,I,O
are concyclic. So ∠OIA =∠ODA=90°. Solution 2. AN-anduud Problem
Solving Group (Ulaanbaatar,
Mongolia), Ercole SUPPA (Teramo, Italy), Titu ZVONARU (Comăneşti, Romania) and Neculai STANCIU (“George Emil Palade’’ Secondary School, Buzău, Romania).
Let a=BC, b=CA, c=AB and let R, r, s be the circumradius, the inradius and the semiperimeter of ΔABC respectively. By the famous formulas OI2 = R2−2Rr, s−a = AI cos(A/2), Rr = abc/(4s) and
cos2(A/2) = s(s−a)/(bc), we get
, ) ( ) 2 / ( cos ) ( 2 2 2 s a s bc A a s AI = − = − . 2 2 2 2 2 s abc R Rr R OI = − = −
If a = (b+c)/2, then we get 2s = 3a and
bc(s−a)/s = abc/(2s). So AI2+OI2 = R2 = OA2. By the converse of Pythagoras’ Theorem, we get OI⊥AI.
Comment: In the last paragraph, all
steps may be reversed so that OI⊥AI if and only if a = (b+c)/2.
Mathematical Excalibur, Vol. 16, No. 4, Feb. 12 Page 4
Other commended solvers: Alumni
2011 (Carmel Alison Lam Foundation Secondary School), CHAN Chun Wai and LEE Chi Man (Statistics and Actuarial Science Society SS HKUSU
),
Andrew KIRK (Mearns Castle High School, Glasgow, Scotland), AndyLOO (St. Paul’s Co-educational College), MANOLOUDIS Apostolos (4° Lyk. Korydallos, Piraeus, Greece),
NGUYEN van Thien (Luong The
Vinh High School, Dong Nai, Vietnam),
Mihai STOENESCU (Bischwiller,
France) and ZOLBAYAR Shagdar (Orchlon School, Ulaanbaatar, Mongolia).
Problem 384. For all positive real numbers a,b,c satisfying a + b + c = 3, prove that . 4 ) 4 ( 3 ) 4 ( 3 ) 4 ( 3 2 2 2 2 2 2 2 2 2 ≥ − + + − + + − + ca ca a c bc bc c b ab ab b a
Solution. William PENG. Let , ) 4 ( ) 4 ( ) 4 ( 2 2 2 a ca c bc c b ab b a A − + − + − = , ) 4 ( 1 ) 4 ( 1 ) 4 ( 1 ca c bc b ab a B − + − + − = c ca b bc a ab C=4− +4− +4− and 1 1 1. c b a
D= + + Then A+3B is the left side of the desired inequality. Now since a + b + c = 3, we have C = 4D −3. By the Cauchy-Schwarz inequality, we have (a+b+c)D ≥ 32, AC ≥ D2 and BC ≥ D2. The first of these gives us D ≥ 3 so that (D−3)(D−1)≥0, which implies
D2 ≥ 4D−3. The second and third imply
. 4 3 4 4 4 3 2 2 ≥ − = ≥ + D D C D B A
Other commended solvers: Alumni
2011 (Carmel Alison Lam Foundation Secondary School), AN-anduud
Problem Solving Group (Ulaanbaatar, Mongolia), Andrew KIRK (Mearns Castle High School, Glasgow, Scotland), LKL Excalibur (Madam Lau Kam Lung Secondary School of MFBM), Andy LOO (St. Paul’s Co-educational College), NGUYEN
van Thien (Luong The Vinh High
School, Dong Nai, Vietnam) and Paolo
PERFETTI (Math Dept, Università degli studi di Tor Vergata Roma, via della ricerca scientifica, Roma, Italy).
Problem 385. To prepare for the IMO, in everyday of the next 11 weeks, Jack will solve at least one problem. If every
week he can solve at most 12 problems, then prove that for some positive integer n, there are n consecutive days in which he can solve a total of 21 problems.
Solution. AN-anduud Problem Solving
Group (Ulaanbaatar, Mongolia), CHAN
Chun Wai and LEE Chi Man (Statistics and Actuarial Science Society SS HKUSU
),
Andrew KIRK (Mearns Castle High School, Glasgow, Scotland),Andy LOO (St. Paul’s Co-educational College) and Yan Yin WANG (City University of Hong Kong, Computing Math, Year 2).
Let Si be the total number of problems
Jack solved from the first day to the end of the i-th day. Since he solves at least one problem everyday, we have 0 < S1 < S2 < S3 < ⋯ < S77. Since he can solve at most 12 problems every week, we have
S77≤12×11=132.
Consider the two strictly increasing sequences S1, S2, ⋯ , S77 and S1+21, S2+21, ⋯, S77+21. Now these 154 integers are at least 1 and at most 132+21=153. By the pigeonhole principle, since the two sequences are strictly increasing, there must be m < k such that Sk=Sm+21.
Therefore, Jack solved a total of 21 problems from the (m+1)-st day to the end of the k-th day.
Olympiad Corner
(continued from page 1)
Problem 4 (cont). Is it possible to empty both bags after a finite sequence of operations?
Operation b) is now replaced with b') Triple the number of balls in one bag. Is it now possible to empty both bags after a finite sequence of operations?
Problem 5. Prove that the product of four consecutive positive integers cannot be equal to the product of two consecutive positive integers.
Problem 6. Let ABC be an acute-angled triangle. The feet of all the altitudes from
A, B and C are D, E and F respectively.
Prove that DE+DF ≤ BC and determine the triangles for which equality holds.
Zsigmondy’s Theorem
(continued from page 2)
Example 8. Find all positive integral solutions to
(a+1)(a2+a+1)⋯(an+an–1+⋯+1)
= am+am-1+⋯+1.
Solution. Note that n = m = 1 is a trivial
solution. Other than that, we must have
m>n. Write the equation as
2 1 3 1 1 1 1 ... 1 1 1 1 n m a a a a a a a a + + − ⋅ − ⋅ ⋅ − = − − − − − , 1
then rearranging we get
)
By Zsigmondy’s theorem, we must have a = 2 and m + 1 = 6, i.e. m = 5 (otherwise, am–1 has a prime factor that
does not divide a2–1, a3–1, …, an+1–1,
a contradiction), which however does not yield a solution for n.
The above examples show that Zsigmondy’s theorem can instantly reduce many number theoretic problems to a handful of small cases. We should bear in mind the exceptions stated in Zsigmondy’s theorem in order not to miss out any solutions.
Below are some exercises for the readers.
Exercise 1 (1994 Romanian Team
(
)(
) (
(
)
(
)
2 3 1 1 1 1 1 ... 1 1 1 . n n m a a a a a + − + − − − = − −Selection Test). Prove that the
sequence an=3n–2n contains no three
terms in geometric progression.
Exercise 2. Fermat’s last theorem asserts that for a positive integer n ≥ 3, the equation xn+yn=zn has no integral
solution with xyz≠0. Prove this statement when z is a prime.
Exercise 3 (1996 British Math Olympiad Round 2). Determine all sets of non-negative integers x, y and z which satisfy the equation 2x+3y=z2.
Reference
[1] PISOLVE, The Zsigmondy
Theorem,
http://www.artofproblemsolving.com/ Forum/viewtopic.php?f=721&t=42233 0
[2] Lola Thompson, Zsigmondy’s
Theorem,
http://www.math.dartmouth.edu/~tho mpson/Zsigmondy's%20Theorem.pdf
The following are solutions by
Dalton Fung (Shatin College),
who was one of the nine contestants, achieved perfect scores on the 2012 BMO Round 1 paper.Problem
1. Let n2+20n+ll=k2 for some nE~, kENu{O}. Completing square, we get 89=(n+l0)2-k2 =(n+ 1 O+k)(n+ 1 O-k). Note that kE N implies n+ 1 O+Ien+ 1 O-k. Since 89 is a prime, we have (n+ 1 O+k = 89 and n+ 1 O-k = 1) or (n+ 1 O+k = -1 and n+ 1 O-k = -89). In each system, we add the equations and solve for n. The first system yields n=35 and k=44. The second system yields n=-55 and k=44. All steps are reversible. Therefore, the only solutions are n = 35 or -55.Problem
2. We denote the required t by tmax . We consider two cases.Case 1 Cn is even). First we show tmax
:s
nl2. (We note that iftmax > n/2, then an arrangement that worksfor tmax also works for (n/2)+ 1. However, no arrangement can work for (n/2)+ 1 because such
arrangement would contain n/2 and a neighbor x of n/2 such that (n/2)-x2:(n/2)+ 1 or x-(nl2)2:(n/2)+ 1, which leads to -12:x or x2:n+ 1, contradiction.) Next we will construct a sequence that works for n/2. We partition {l,2, ... ,n} into {l,1+(nl2)}, {2, 2+(n/2)}, ... ,{n/2, n}. Then 1, 1+(nl2), 2, 2+(n/2), ... ,n/2, n is such a sequence for n/2. So tmax= n/2.
Case 2 Cn is odd). First we show tmax
:s
(n-l )/2. (We note that if tmax > (n-l )/2, then an arrangement thatworks for tmax also works for (n+ 1 )/2. However, no arrangement can work for (n+ 1)12 because such
arrangement would contain (n+ 1)/2 and a neighbor x of (n+ 1)/2 such that (n+ l)l2-x2:(n+ 1)/2 or x-(n+ 1)/22:(n+ 1)/2, which leads to 02:x or x2:n+ 1, contradiction.) Next we will construct a sequence that works for (n-l)/2. We partition {1,2, ... ,n} into {n}, {(n-1)/2, n-l}, {(n-3)12, n-2}, ... ,{2, 2+(n-l)/2}, {I, 1+(n-1)/2}. Then n, (n-1)/2, n-l, (n-3)/2, n-2, ... ,2, 2+(n-l)/2, 1, 1+(n-l)/2 is such a sequence for (n-1 )/2. So tmax= (n-l )/2.
To conclude, tmax = Lnl2J for both n even and odd.
o
r
y
(Figure for Problem 3)
Problem
3. Let 0, 0], O2 be the centers of S, S], S2 respectively. Let 1', r],r2 be the radii of S, S], S2 respectively. Since S] and S2 are tangent to S at X and Y respectively, we see O],X,O are coilinear and also 02,O,Y are collinear. Now we have
LO]PX=LO]XP=LOXY=LOYX=L02YP=L02PY,
call this 8.By considering
6.
PXO], we get r] = (Y2PX)/(cos 8). Similarly, r2 = (YzPY)/(cos 8) and r = (YzXY)/(cos 8). NowIr]-r21
==
I(YzPX)/(cos8)-
(YzPY)/(cos8)1
=1
(YzXY)/(cos8)1
= r, which is independent of the positions ofP, X and Y.Problem
4. For the first part of the problem, we assume, without loss of generality, that m2:n. We set a=m-n. Note that aE N U {O}. We consider two cases.Case 1 Cn> [ml2] ). Then n-a =2n-m2:0. Now we remove n-a balls from each bag by operation (a). Then we have m-(n-a)=2a balls in the first bag and n-(n-a)= a balls in .the second bag. We use operation (b) to double the balls in the second 'bag. Now we have 2a balls in each bag. By operation (a), we can empty both bags.
Case 2 (n<[m/2]). In this case, we can use operation (b) finitely many times to get the number of balls, n', in the second bag satisfies m2:n'2:fm/21. (This is always possible because assuming for some tEN, t < f m/21 and 2t> m, we get t < f m/21 =? 2t < 2 f m/21
:s m+ 1 (so 2t :s
m), which contradicts 2t > m).Hence, the first part of the problem has an affirmative answer. Next we will show the second part of the problem has a negative answer. This follows from the
Lemma.
The total number of balls in both bags modulo 2 is an invariant.Proof. For operation (a), we have to remove an even number of balls in total. For operation (b'), if a bag has an odd (resp. even) number of balls before operation (b'), then it still has an odd (resp. even) number of balls after operation (b ').
By the lemma, if m+n is odd in the beginning, then it is not possible to empty both bags after a finite sequence of operations.
Problem
5. Suppose (n-1)n(n+ 1)(n+2)=i(i+ 1) for some nE{2,3,4, ... } and iEN. Then (n2+n-2)(n2+n) = (n-1)(n+2)n(n+ 1) = i(i+ 1). Note that n2+n-3 > O.Case 1 (i>n2+n). Then i(i+1) > (n2+n)(n2+n+1) > (n2+n)(n2+n-2). Case 2 0=n2+n). Then i(i+ 1) = (n2+n)(n2+n+ 1) > (n2+n)(n2+n-2). Case 3 0=n2+n-1). Then i(i+ 1) = (n2+n-1)(n2+n) > (n2+n-2)(n2+n). Case 4 0=n2+n-2). Then i(i+ 1) = (n2+n-2)(n2+n-1) < (n2+n-2)(n2+n). Case
5
(i=n2+n-3). Then i(i+ 1) =(n2+n-3)(n2+n-2) < (n2+n)(n2+n-2). Case 6 0<n2+n-3). Then i(i+ 1) < (n2+n-3)(n2+n-2) < (n2+n)(n2+n-2).B
Therefore there are no solutions.
A
D
(Figure for Problem 6)
Problem
6. LetL.A,
L.B, L.C denote L.BAC, L.ABC, L.BCA respectively. Without loss of generality, we.assume L.B2:L.C.
Lemma
1. L.B2:L.C
¢::} AF 2: AE.Proof. L.B
2:L.C
¢::} AC 2: AB by sine rule. NowAF/AC
= cosL.A=AE/AB.
So AC 2: AB ¢::} AF 2: AE.Lemma
2. DExDF=DCxBD.Proof. By the property of orthocenter, we get D, H, E, Care concyclic; F, C H, D, Bare concyclic; E, H, F, A are concyclic; A, F, D, Care concyclic; C, E, F, Bare concyclic; B, D, E, A are concyclic. Now applying sine rule to 6.ADE and the fact D, H, E, Care concyclic, we have
DE = AE(sin L.DAE/sin L.ADE)=AE(sin
L.DAC/sin L.ACF)=AE[(DC/AC)/(AF/AC)]
= DCCAE/AF) .... (1) .
Similarly, DF=
BD(AF/AE)BD ...
(2). Then DExDF=DC(AE/AF)xBD(AF/AE) = DCxBD.Now DE+DF ::; BC=BD+DC ¢::} DE2+2DExDF+DF2::; BD2+2BCxDC+DC2 ¢::} DE2+DF2::; BD2+DC2
(by Lemma 2) ¢::} BD2+DC22:
DC2(AE2/AF2)+BD2(AF2/AE2)
(by (1) and (2)) ¢::}DC2(1-AE2/AF2)
2:BD2(1-AF2/AE2)
¢::}DC2/AF2
2:BD2/AE2
¢::}DC/BD
2:AF/AE
¢::}(AD/tanL.C)/(AD/tanL.B)
2:sin
