Volume 19, Number 5 April 2015 – June 2015
Olympiad Corner
Below are the problems of the 2015 Canadian Mathematical Olympiad held in January 28, 2015.
Notation: If V and W are two points, then VW denotes the line segment with endpoints V and W as well as the length of this segment.
Problem 1. Let ℕ = {1,2,3,…} denote the set of positive integers. Find all functions f, defined on ℕ and taking values in ℕ, such that (n−1)2 <
f(n)f(f(n)) < n2+n for every positive integer n.
Problem 2. Let ABC be an acute-angled triangle with altitudes
AD, BE and CF. Let H be the
orthocenter, that is, the point where the altitudes meet. Prove that
. 2 CF CH BE BH AD AH CB CA BA BC AC AB Problem 3. On a (4n+2)×(4n+2) square grid, a turtle can move between squares sharing a side. The turtle begins in a corner square of the grid and enters each square exactly once, ending in the square where she started. (continued on page 4)
Tournament of the Towns
Kin Yin Li
In 1980, Kiev, Moscow and Riga participated in a mathematical problem solving contest for high school students, later called the Tournament of the
Towns. At present thousands of high
school students from dozens of cities all over the world participate in this contest. In this article, we present some very interesting math problems from this contest. At the end of the article, there are some information on where interested readers can find past problems and solutions of this contest. Here are some examples we enjoy.
Example 1. (Junior Questions, Spring 1981, proposed by A. Andjans) Each of
64 friends simultaneously learns one different item of news. They begin to phone one another to tell them their news. Each conversation last exactly one hour, during which time it is possible for two friends to tell each other all of their news. What is the minimum number of hours needed in order for all of the friends to know all the news?
Solution. More generally, suppose there
are 2n friends. After n rounds, the most anyone can learn are 2npieces of gossip. Hence n rounds are necessary. We now prove by induction on n that n rounds are also sufficient. For n=1, the result is trivial. Suppose the result holds up to
n−1 for some n≥2. Consider the next
case with 2n friends. Have them call each other impairs in the first round. After this, divide them into two groups, each containing one member from each pair who had exchanged gossip. Each group has 2n−1 friends who know all the gossip among them. By the induction hypothesis, n−1 rounds are sufficient for everyone within each group to learn everything. This completes the induction argument. In particular, with 64 friends, 6 rounds are both necessary and sufficient.
Example 2. (Senior Questions, Spring 1983, proposed by A. Andjans) There
are K boys placed around a circle. Each of them has an even number of sweets. At a command each boy gives half of his sweets to the boy on his right. If, after that, any boy has an odd number of sweets, someone outside the circle gives him one more sweet to make the number even. This procedure can be repeated indefinitely. Prove that there will be a time at which all boys have the same number of sweets.
Solution. Suppose initially the
maximum number of sweets a boy has is 2m, and the minimum is 2n. We may as well assume that m>n. After a round of exchange and possible augmentation, we claim that the most any boy can have is 2m sweets. This is because he could have kept at most m sweets, and received m more in the exchange, but will not be augmented if he already has 2m sweets.
On the other hand, at least one boy who had 2n sweets will have more than that, because as long as m>n, one of these boys will get more than he gives away. It follows that while the maximum cannot increase, the minimum must increase until all have the same number of sweets.
Example 3. (Junior Questions, Autumn 1984) Six musicians gathered at a
chamber music festival. At each scheduled concert some of these musicians played while the others listened as members of the audience. What is the least number of such concerts which would need to be scheduled in order to enable each musician to listen, as a member of the audience, to all the other musicians?
(continued on page 2) 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 August 27, 2015.
For individual subscription for the next five issues for the 14-15 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
Mathematical Excalibur, Vol. 19, No. 5, Apr. 15 – Jun. 15 Page 2
Solution. Let the musicians be A, B, C, D, E and F. Suppose there are only three
concerts. Since each of the six must perform at least once, at least one concert must feature two or more musicians. Say both A and B perform in the first concert. They must still perform for each other. Say A performs in the second concert for B and B in the third for A. Now C, D, E and F must all perform in the second concert, since it is the only time B is in the audience. Similarly, they must all perform in the third. The first concert alone is not enough to allow C, D, E and F to perform for one another. Hence we need at least four concerts. This is sufficient, as we may have A, B and C in the first, A,
D and E in the second, B, D and F in the
third and C, E and F in the fourth.
Example 4. (Junior Questions, Autumn 1984, proposed by V. G. Ilichev) On the
Island of Camelot live 13 grey, 15 brown and 17 crimson chameleons. If two different chameleons of different colours meet, they both simultaneously change colour to the third colour (eg. If a grey and a brown chameleon meet each other they both change to crimson). Is it possible that they will eventually all be the same colour?
Solution. In this case the numbers of
chameleons of each colour at the start have remainders of 0, 1 and 2 when divided by three. Each “meeting” maintains such a situation (not necessarily in any order) as two of these remainders must either be reduced by 1 (or increased by 2) while the other must be increased by 2 (or reduced by 1). Thus at least two colours are present at any stage, guaranteeing the possibility of obtaining all of the three colours in fact by future meetings.
Note. The only way of getting
chameleons to be of the same colour would be getting an equal number of two colours first. This would mean getting two with the same remainder on division by three. This would have been possible if we had started with, say 15 of each colour. From this position we can obtain sets with remainders equal to {0,0,0}, {1,1,1} and {2,2,2}.
Example 5. (Junior Questions, Spring 1985, proposed by S. Fomin) There are
68 coins, each coin having a different
weight that that of each other. Show how to find the heaviest and lightest coin in 100 weighings on a balance beam.
Solution 1. First divide into 34 pairs and
perform 34 weighings, each time identifying the heavier and lighter coins. Put all the heavier coins into one group and the lighter coins into another. Divide the group with heavier coins into 17 pairs, and perform 17 weighings on these to identify the 17 heavier coins. Continue this process with the group of heavier coins each time. If there is an odd number of coins at any stage, the odd coin out must be carried over to the following stage. There will be a total of 17+8+4+2+1+1=3 such weighings required for identifying the heaviest coin.
A similar 33 weighings of the lighter group will identify the lightest coin. The total number of weighing is thus 34+33+33=100, as required.
Solution 2. More generally, we show that
3n−2 weighings are sufficient for 2n coins. We first divide the coins into n pairs, and use n weighings to sort them out into a “heavy” pile and a “light” pile. The heaviest coin is among the n coins in the “heavy” pile. Each weighing eliminates 1 coin. Since there are n coins, n−1 weighings are necessary and sufficient. Similarly, n−1 weighings will locate the lightest coin in the “light” pile. Thus the task can be accomplished in 3n−2 weighings.
Example 6. (Junior Questions, Spring 1987, proposed by D. Fomin) A certain
number of cubes are painted in six colours, each cube having six faces of different colours (the colours in different cubes may be arranged differently). The cubes are placed on a table so as to form a rectangle. We are allowed to take out any column of cubes, rotate it (as a whole) along its long axis and place it in a rectangle. A similar operation with rows is also allowed. Can we always make the rectangle monochromatic (i.e. such that the top faces of all the cubes are the same colour) by means of such operations?
Solution. The task can always be
accomplished, and we can select the top colour in advance, say red. By fixing a cube, we mean bringing its red face to the top. Given a rectangular block, we fix one cube at a time, from left to right, and from front to back.
Suppose that the cube in the i-th row and the j-th column is the next to be fixed. Suppose that we need to rotate the i-th row. In order not to unfix the first j−1 cubes of this row, we rotate each of the first j−1 columns so that all red faces are to the left. They remain to the left when the i-th row is rotated. We can now refix the first j−1 columns. Similarly, if we need to rotate the j-th column, we can go through an analogous three-step process.
Example 7. (Senior Questions, Autumn 1987, proposed by A. Andjans) A
certain town is represented as an infinite plane, which is divided by straight lines into squares. The lines are streets, while the squares are blocks. Along a certain street there stands a policeman on each 100th intersection. Somewhere in the town there is a bandit, whose position and speed are unknown, but he can move only along the streets. The aim of the police is to see the bandit. Does there exist an algorithm available to the police to enable them to achieve their aim?
Solution. We assume that (a) there is no
limit to how far a policeman can see along the street he is on; (b) there is no overall time limit, and (c) if the bandit is ever on the same street as a policeman he will be seen.
Let i, j and k denote integers, let the North-South streets be x=i for all i, the East-West streets y=j for all j and suppose the k-th policeman is at (100k,0).
For all even k the k-th policeman remains stationary throughout. This traps the bandit in the infinite strip between x=200k and x=200(k+1) for some k, say k*.
All other policemen first travel along
y=0 towards (0,0) until they reach the
first cross street x=s for which there is a policeman on every street x=i for i between 0 and s. Police are to travel at regulation speed, say one block per minute, but nevertheless there will come a time, dependent only on k*, when every street x=i on the k* strip will be policed.
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 August 27, 2015.
Problem 466. Let k be an integer greater than 1. If k+2 integers are chosen among 1,2,3,…,3k, then there exist two of these integers m,n such that k<|m−n|<2k.
Problem 467. Let p be a prime number and q be a positive integer. Take any pq consecutive integers. Among these integers, remove all multiples of p. Let
M be the product of the remaining
integers. Determine the remainder when M is divided by p in terms of q. Problem 468. Let ABCD be a cyclic quadrilateral satisfying BC>AD and
CD>AB. E, F are points on chords BC, CD respectively and M is the midpoint
of EF. If BE=AD and DF=AB, then prove that BM⊥DM.
Problem 469. Let m be an integer greater than 4. On the plane, if m points satisfy no three of them are collinear and every four of them are the vertices of a convex quadrilateral, then prove that all m of the points are the vertices of a m-sided convex polygon.
Problem 470. If a, b, c>0, then prove that . 3 ) 2 ( ) 2 ( ) 2 ( 2 2 2 2 2 2 ca bc ab a c a c c b c b b a b a *****************
Solutions
****************Problem 461. Inside rectangle ABCD, there is a circle. Points W, X, Y, Z are on the circle such that lines AW, BX, CY,
DZ are tangent to the circle. If AW=3, BX=4, CY=5, then find DZ with proof. Solution. Adnan ALI (Atomic Energy Central School 4, Mumbai, India), Adithya BHASKAR (Atomic Energy School 2, Mumbai, India), Andrea
FANCHINI (Cantú, Italy), William FUNG, KWOK Man Yi (Baptist Lui Ming Choi Secondary School, S4), Jon GLIMMS, LKL Excalibur (Madam Lau Kam Lung Secondary School of MFBM), Corneliu MĂNESCU-AVRAM (“Henri Mathias Berthelot” Secondary School, Ploieşti, Romania), MANOLOUDIS Apostolos (4 High School of Korydallos, Piraeus, Greece), Vijaya Prasad NALLURI (Retired Principal, AP Educational Service, India), Alex Kin-Chit O (G.T. (Ellen Yeung) College), Toshihiro SHIMIZU (Kawasaki, Japan), Titu ZVONARU (Comăneşti, Romania) and Neculai STANCIU (“George Emil Palade’’ Secondary School, Buzău, Romania).
A
B
C
D
O
P
Q
W
X
Z
Y
Let r be the radius of the circle. By Pythagoras’ theorem, we have
r2 = AW2−AO2 =BX2−BO2 = CY2−CO2 = DZ2−DO2. (*)
Let P,Q be the feet of perpendiculars from
O to AB, CD respectively. Then
AO2−BO2 = (AP2 +PO2)− (BP2+PO2) = (DQ2+QO2) − (CQ2+QO2)= DO2−CO2. Using (*), we get AW2−BX2 =AO2−BO2 =
DO2−CO2 =DZ2−CY2. Then . 2 3 2 2 2 AW BX CY DZ
Problem 462. For all x1, x2, …, xn ≥ 0, let
xn+1 = x1, then prove that
n k k k k n x x x 1 1 2 2 1 2 ( 1) 2. ) 1 ( 1Solution. Adnan ALI (Atomic Energy
Central School 4, Mumbai, India), Adithya BHASKAR (Atomic Energy School 2, Mumbai, India), Ioan Viorel CODREANU (Secondary School Satulung, Maramures, Romania), DHRUV Nevatia (10th Standard, Ramanujan Academy, Nashik, India), KWOK Man Yi (Baptist Lui Ming Choi Secondary School, S4), LKL Excalibur (Madam Lau Kam Lung Secondary School
of MFBM), MAMEDOV Shatlyk (School of Young Physics and Maths N 21, Dashogus, Turkmenistan), Corneliu MĂNESCU- AVRAM (“Henri Mathias Berthelot” Secondary School, Ploieşti, Romania), Paolo PERFETTI (Math Dept, Università degli studi di Tor Vergata Roma, via della ricerca scientifica, Roma, Italy), Ángel PLAZA (Universidad de Las Palmas de Gran Canaria, Spain), Toshihiro SHIMIZU (Kawasaki, Japan), WADAH Ali (Ben Badis College, Algeria), Nicuşor ZLOTA (“Traian Vuia” Technical College, Focşani,
Romania), Titu ZVONARU
(Comăneşti, Romania) and Neculai STANCIU (“George Emil Palade’’ Secondary School, Buzău, Romania). By squaring both sides or RMS-AM inequality, we have for all a,b ≥ 0,
. 2 2
2 b a b
a
Applying this, we get
n k k k n k k k k n x x x x x 1 1 1 1 2 2 1 2 . 2 ) 1 1 1 ( 1 1 2 1 ) 1 ( ) 1 ( 1Problem 463. Let S be a set with 20 elements. N 2-element subsets of S are chosen with no two of these subsets equal. Find the least number N such that among any 3 elements in S, there exist 2 of them belong to one of the N chosen subsets.
Solution. Jon GLIMMS, KWOK Man Yi (Baptist Lui Ming Choi Secondary School, S4), LKL Excalibur (Madam Lau Kam Lung Secondary School of MFBM) and Toshihiro SHIMIZU (Kawasaki, Japan).
Let x∈S be contained in k of the N 2-elements subsets of S, where k is least among the elements of S.
Let x1, x2, …, xk be the other elements in
k of the N 2-element subsets with x. As k
is least, so each of the xi’s is also contained in at least k of the N 2-element subsets of S.
Also, there are m=19−k elements w1,
w2, …, wm∈S not in any of the N 2-element subsets of S with x. For
Mathematical Excalibur, Vol. 19, No. 5, Apr. 15 – Jun. 15 Page 4 every pair wr, ws of these, {wr, ws} is
one of these N 2-element subsets of S (otherwise, no two of x, wr, ws form one of the N 2-element subsets). Then
. 90 90 ) 9 ( 2 19 2 1 2 k k k N
To get the least case of N=90, we divide the 20 elements into two groups of 10 elements. Then take all 2-element subsets of each of the two groups to get 45+45=90 2-element subsets of S. Problem 464. Determine all positive integers n such that for n, there exists an integer m with 2n−1 divides m2+289. Solution. Adnan ALI (Atomic Energy
Central School 4, Mumbai, India), Adithya BHASKAR (Atomic Energy School 2, Mumbai, India), KWOK Man Yi (Baptist Lui Ming Choi Secondary School, S4), LKL Excalibur (Madam Lau Kam Lung Secondary School of MFBM), Corneliu MĂNESCU-AVRAM (“Henri Mathias Berthelot” Secondary School, Ploieşti, Romania), PANG Lok Wing and Toshihiro SHIMIZU (Kawasaki, Japan).
The case n=1 is a solution. For n>1, we first show if a prime q of the form 4k+3 divides a2+b2, then q divides a and b. Assume gcd(q,a)=1. Let c=aq−2. Then by Fermat’s little theorem, ac=aq−1≡1 (mod q). As q|a2+b2, so b2≡−a2 (mod
q). Then (bc)2≡−(ac)2≡ −1 (mod q) and (bc)q−1 = (bc)2(2k+1) ≡ = −1 (mod q), contradicting Fermat’s little theorem. So q divides a (and b similarly). If n>1, then 2n−1≡3 (mod 4). Hence 2n−1 has a prime divisor q≡3 (mod 4). By the fact above, q divides m2+289 implies q divides m and 17. Then q=17 ≢3 (mod 4), contradiction.
Problem 465. Points A, E, D, C, F, B lie on a circle Γ in clockwise order. Rays AD, BC, the tangents to Γ at E and at F pass through P. Chord EF meets chords AD and BC at M and N respectively. Prove that lines AB, CD,
EF are concurrent.
Comments. A number of solvers
pointed out if lines AB, CD are parallel, then by symmetry lines AB, CD, EF are all parallel. So below, we present solutions for the case when lines AB and CD intersect at a point.
Solution 1. Jon GLIMMS.
D A B Q C P E F
Let lines AB, CD meet at Q. We have (1) ∠AFE =∠ADE=180°−∠PDE, (2) ∠EFD =∠PED, (3) ∠FDQ =∠PFC, (4) ∠QAF =∠FCB=180°−∠PCF. (5) ∠DAQ =∠DCP, (6) ∠QDA =180°−∠PDC. Then . sin sin sin sin , sin sin sin sin , sin sin sin sin PC PD PDC DCP QDA DAQ PF PC PCF PFC QAF FDQ PD PE PED PDE EFD AFE
Multiplying these and using PE=PF, we have . 1 sin sin sin sin sin sin PC PD PF PC PD PE QDA DAQ QAF FDQ EFD AFE
Applying the converse of the trigonometric form of Ceva’s theorem to ΔADF and point Q, we get lines AB, CD, EF are concurrent at Q.
Solution 2. Adnan ALI (Atomic Energy
Central School 4, Mumbai, India), Adithya BHASKAR (Atomic Energy School 2, Mumbai, India) and William FUNG.
Since the tangents to Γ at E and at F intersect at P, line EF is the polar of P. Since lines AD, BC intersect at P, the polar of P (that is, line EF) passes through the intersection of lines AB and CD.
Other commended solvers: KWOK Man
Yi (Baptist Lui Ming Choi Secondary School, S4), MANOLOUDIS Apostolos (4 High School of Korydallos, Piraeus, Greece) and Toshihiro SHIMIZU (Kawasaki, Japan).
Olympiad Corner
(Continued from page 1) Problem 3 (Cont’d). In terms of n, what is the largest positive integer k such that there must be a row or a column that the turtle has entered at least k distinct times?
Problem 4. Let ABC be an acute-angled triangle with circum- center O. Let Γ be a circle with centre on the altitude from A in ABC, passing through vertex A and points P and Q on sides AB and AC. Assume that
BP·CQ=AP·AQ. Prove that Γ is
tangent to the circumcircle of triangle
BOC.
Problem 5. Let p be a prime number for which (p−1)/2 is also prime, and let
a, b, c be integers not divisible by p.
Prove that there are at most 1 2p positive integers n such that n<p and p divides an+bn+cn.
Tournament of the Towns
(Continued from page 2) When this happens the bandit will be trapped on some street y=j*, on a single block between x=i* and x=i*+1 for some i*.For each k, as soon as all streets on the
k-th strip are policed, one of the
policemen travels north and another travels south. For k=k* this will inevitably reveal the bandit.
After reading these examples, should anyone want to read more, below are websites, which books on this contest can be ordered or problems and solutions of the recent Tournament of the Towns can be found.
www.amtt.com.au/ProductList.php?pa ger=1&startpage=1 www.artofproblemsolving.com/comm unity/c3239_tournament_of_towns www.math.toronto.edu/oz/turgor/archi ves.php
Pf"o
bl?M
1
- Solution; The only such function is J(n) = n.
Assume that f satisfies the given condition. It will be shown by inductio~ that f(n) = n for all n E N. Substituting-n = 1 yields that O < /(1)/(/(1)) < 2 which implies the bMe cage /(:1) = 1. Now assume that f(k) = k for all. k < n and assume for contradiction that
f(n) :/, n. ·. . . . .
· On the one hand, if f(n) ::; n-1 then f (!(n)) = f(n) and J(n)J(f(n)) = f (n)2 ::; (n-1)2
which is a contradiction, ·.On the other hand,·if /(n) ~ n+ 1 then there are several ways to proceed.
Method 1: ·Assume f(n)
=
M ~ n+
1. Then (n + 1)/(M) ::; /(n)J(/(Tf,)) < n2 + n.Therefore /(M) < n, and hence J(f(M))
=
/(M.) and /(M)/(/(M))=
J(M)2 < n2 ::;(M -1)2, which is a contradiction. This completes the induction. 0
Method 2: First note that if la-bl > 11 then the intervals ((a-1)2, a2+a) and ((b-1)2, b2+b)
are disjoint which implies that /(a) and /(b) cannot be equal. ·
Assuming f(n) ~ n
+
1, it follows that /(f(n)) <iJ{t;.f ::;
n. This implies that for somea :Sn - 1, /(a) = J(f(n)) which is a contradiction since 1/(n) - al ~ n
+
1 - a 2: 2. Thiscompletes the induction. · 0 .
Method .3: Assumi~g /{n) 2:.n + l,.it follows that f(f(n)) <
~
:S. n and /(/(/(n)))=
/(/(n)). This impliEIB that (f(n)-1)2 '< J(f(n))J(f(f(n))) = /(/(n))2 </(n)2+
f(n) andtherefore that /(/(ri)') ·;; f(n) since /(n)2 is the unique square satisfying this constraint. This implies tl).at /(n)/(/(n)) = f (n)2 2: (n
+
1)2 which is a contradiction, completing theinduction. · · 0
PYO/o·f
ew,
7...~
Solution. ,\ Let AB.= c, AC= b, and BC= a denote the three side lengths of .
the triangle. .
As LBFH = LBDH = 90°, FHDB is a cyclic quadrilateral. By the Power-of-a-Point
Theorem, AH · AD .= AF ·°AB.
Sin~e AF =.AC. cos LA, we ·have AH· AD = AC· AB · cos LA = be cos LA.
.
~+2-~.
~+2-~
By the Cosine Law, cos LA =
2bc , which implies that AH · AD
=
2~+e-f
~+~-2
. By symmetry, we can show that BH ·BE= _2 and CH· CF= 2
Hence,
AH, AD +BH ·BE+CH · CF ~+~-~ 2
.+
~+e-~
2.+
~~f-2
2a2+b2
+2
.
.
2 (1)
. . AB ,AC+BO · BA+CA·CB . · . l tt th · 1·t·
Our desired mequality, < 2, is eqmva en o e mequa I y
.
AH·AD+BH·BE+CH,CF-cb
!+~;+~•
ba~
2, which simplifies to 2a2+
2b2+
22 2: 2ab+
2bc+ 2ca.
Blit this last inequality is easy to prove, as it is equivalent to (a.:..b)2+(a-c)2+(b-c)2
~
0.Therefore, we have established the desired inequality. The proof also shows that equality
occurs if and only if a= b ~ c, i.e., 6.ABC is equilateral. 0
Solution. We shall prove that the answer is 2n
+
2. ·Number the rows in increasing .order, from top to bottom, and number the columns from left to right. By symmetry, we may ( and shall) assume that the turtle starts .in the top right corner square.First we shall prove that some row or column must be entered at least 2n + 2 times. Let
m = 4n
+
2. First note that each time· the turtle mov!',S, she enters either .a. row or a column. Let r; denote the number of times·the turtle enters row i, and let C; be similarly defined forcolumn i. Since the turtle moves m2 times,
. . . 2
r1.
+
r2+ · "
+rm+ Ct+
Cz+···+Cm
= m ,Now note that each time the turtle enters column 1, the next column she enters .. must be column 2. Therefore Ct is equal to the number of tirp.es the turtle enters column 2 from column 1. Furthermore, the turtle must enter column: 2 from column 3 at least once, which implies that
ez
> Ct, Therefore since. the 2m terrufir,
and e; are not all equal, one must be strictly greater than m2 /(2m) .= 2n+
1 and therefore at least 2n + 2.Now we construct an example to show tha.t i~ .is possible tha.t no row or column is entered more than 2n
+
2 times. Partition the square grid into four (2n+
1) x (2n+
1) quadrants A, B, C, and D, containing the upper left, upper r,iglit, lower .left, and lower right corners, respectively. The turtle begins at the top right cornE\r square of B, moves one square down, and 'then moves left ·through the whole second row of B. She then moves one ·square. down and moves right through the whole third row of B. She continues in this pattern, moving through e~h remaining row of B in: succession and moving one square down. when each row is completed. Since 2n+
1 is odd, the turtle' ends at the bottom right corner of B. She_ then moves one square down into D and through each column of D in turn, moving. one square to the left 'when each column is completed. She ends at the lower left corner of D and moves 'left into C and through the rciws of.C, moving one square up when each row is completed, ending fo the upper left corner of G. She then enters A a.nd moves through the columns of A, moving one square right when each column is completed. This takes her to the upper . · right corner of A, whereupon she enters B and moves· right through the top row of B, which · returns her to her starting point. Each row passing through A and B is entered at most2n
+
1 times in A and once in: B, and thus at most 2n + 2 times in total, Sim,ilarly; l;lach row and. column in the grid is· entered at most 2n+
2 times by this path. (See figure below.)I+
P(l)fo~4--so1ution. Let w be the circumcircle of BOG. Let ·M be the point diametrically opposite to-0 on w and let the line AM intersect w at M and K. Since O is the circumcenter of
ABC, it follows that OB = OG and therefore that O is the midpoint ~f the arc
BOO
of w.Since M is diametrically opposite to 0, it follows that
M
is the midpoint of the arcBMC·
of w. This implies since K is on w that 'KM is the bisector ~f LBKG.Since K is on w, this impliesthat LBKM = LGKM, i.e. KM is the bisector of LBKG. .Since O is the circumcenter of ABC, it follows that LBOC = 2LBAC. Since B, K, 0 .. arid C all lie on w, it iµso follows that LBKC = lBOC = 2LBAC. Since KM bisects
LBKC, it follows that LBKM = LGKM = LBAG. The fact.that A,K and M lie.on a
.· liµe therefore implies that LAkB .= L.AKC = 180° - L.BAC. Now it follows that
LKBA= 180° - LAKE - LKAB = LBAC.- LKAB = _L.KAC.
This implies that triangles KB A and K AC. are similar. Rearranging the condition in the problem statement yields that BP/AP= AQ/OQ which, when combined with the.fact that
KBA and KAO are similar, implies that triangles KPA and KQC are-similar. Therefore
LKPA
=
LKQG = 180° - L.KQA which implies that K lies onr.
Now let. S denote the centre of
r
and let T denote the centre of w. Note that T is the midpoint of segmentOM
and that TM and AS, which are both perpendicular to BO, are parallel. This implies that L.K MT = LK AS since A, K and M are collinear. Further, sinceKTM and KSA .are-isosceles triangles, _it follows that LTKM = LKMT and L.SKA =
. LKSA; Therefor~·LTKM= LSKA which:Ji:hplies thatp, T and Kare collinear. Therefore
r
and w intersect at a point K which ·lies on the line ST connecting the centres of the two circles. T_his implies that the circlesr
and. w are tangent .at K. 0. Pr-oMem.s::.
-Solution. First suppose b-= ±a (mod p) and c
=
:l;:b (mod p). Theµ, for any.n,.we h_avean+ bn
+
en=
±an or ±3an (mod p) .. We a.re given that p#-
3. (since 3;1 is not prime) and
p ,/'a; so it follows t~at an+ bn
+
ent
o· (mod p). The claim is trivial in this case. Otherwise, we may assume without loss of generality that b ¢;. ±a (mod·p) ===} ba-1 ¢;. ±1 (mod p).. Now let q =
1?·
By Fermat's little- theorem; we know that the order of ba-1 _mod pdivides p - 1 = 2q. However, since ba-1 ¢;. ±1 (mod p}, the order of ba-1 does not divide 2.
· Thus, the order must be either-q or 2q. · · ·
Next, let S denote the set of positive integers n < p such that an
+
bn+
en=
0 (mod p),and let St denote the number of ordered pairs (i,j} CS such that .i - j
=
t (mod p-1).Lemma: If t is a positive integer less than
2q
and not equal to q, then St . .$ 2. Proof: Consider i,j ES with j - i=
t (mod p - 1). Then we have'a• +bi +ci
=
0 (modp}===} a•d-•
+
b'd-•+
d=
0 (mod p)=> o}d~·
+
bid-• - ai - .fl= o·(mod p)=> ~i. (ct. - at) i::: bi,
W -
ct) (mod p).· If ct
=
at (mod p), then this implies ct=
bt (mod p) as well, so (ab-1)t=
1 (mod p).However, we know the order of ab-1 is.q or 2q, and q Jt, so this impossible. Thus,·we can
. write ·
. (ab-1/
=
W-
ct) . (ct - atrl (modp).For a fixea' t, the right-hand side cif this equation is fixed, so ( ab-1 )' is also fixed. Since· the
order of ab-:-1 is either q or 2q, it follows that there are at most 2 solutions for i, and the
lemma is prov(;ln. D
Now, for ea.ch element i in S, there are at least
ISi -
2 other elements that differ from i by a quantity other than.q: (mod p-'-1). Therefore, the lemma implies thatISi · (ISi -
2) .$L
St _$. 2 • (p - 2)t'Fq
===}
