Mathematical Excalibur, Volume 18, Number 2

Volume 18, Number 2 September-October 2013

Olympiad Corner

Below are the problems of the 2013 International Mathematical Olympiad. Problem 1. Prove that for any pair of positive integers k and n, there exist k positive integers m1, m2, …, mk (not necessarily different) such that

1 2 2 1 1 1 1 1 1 1 ... 1 . k k n m m m                     Problem 2. A configuration of 4027 points in the plane is called Colombian if it consists of 2013 red points and 2014 blue points, and no three of the points of the configuration are collinear. By drawing some lines, the plane is divided into several regions. An arrangement of lines is good for a Colombian configuration if the following two conditions are satisfied: • no line passes through any point of the configuration;

• no region contains points of both colors.

Find the least value of k such that for any Colombian configuration of 4027 points, there is a good arrangement of k lines.

(continued on page 4)

IMO 2013 – Leader Report (I)

Leung Tat-Wing

The 54th _{International Mathematical} Olympiad (IMO) was held in Santa Marta, Colombia from July 18th _{to July} 28th_{, 2013. It took me 40 hours of flight} and waiting time to travel from Hong Kong to Amsterdam, then to Panama City, and then to Barranquilla, Colombia (where the leaders stayed before they met the contestants in Santa Marta after two days of 4½-hour contests held on the mornings of 23rd and 24th _{of July). Tired and exhausted, I} were picked up in the airport of Barranquilla and delivered to Hotel El Prada. We managed to settle down and be prepared for the next two days’ Jury meetings. Our team arrived at Santa Marta, three days later, safe and intact, luckily. The next day they still had to travel two hours from Santa Marta to Barranquilla, participating in another opening ceremony, then another two hours back to Santa Marta. It was tough for them. Accommodation was fine though. Contestants stayed in a nice seaside resort hotel (Iratoma), while leaders stayed in a hotel in Barranquilla. They would join the contestants after the two day contests.

Jury meetings were chaired by Maria Losada, a long time veteran of IMO activities. She was very experienced and chaired the meetings well. Interesting to note, she kept on reminding us (leaders) that we should try to form the best possible paper, a paper that can provide intellectual challenge to contestants, that has some aesthetic sense and that allows every contestant to achieve the most. We were also supposed to work out as many possible solutions as possible. We should be able to tell whether a problem is easy, medium and/or hard. Really sometimes I did not know how the goals may be attained or even verified. She also reminded us ethically we should keep the problems with strict security, not to disclose any information to any contestant beforehand, etc. Indeed the Jury meetings were very educational.

After the two days’ contests students enjoyed a break. Leaders and deputy leaders had to check the solutions of the contestants, discussed or argued with coordinators and sorted out how many points should be award to contestants. (This process is called coordination). Luckily this year many coordinators were again very experienced. Many of them are old time leaders from Europe and are experienced problem solvers. They were able to discern mistakes made by the contestants (trivial, small or big) and were able to award points accurately. Personally I recognized many of them and I think I have known many of them for at least more than 10 years. That is why little trouble was observed during the coordination process.

The awards (closing) ceremony was held near a historical site, 45 minute drive from the hotel. We were delivered to the site around 7:00 pm. Then the ceremony lasted for more than two hours. Participants were than sent back to the resort for the banquet. That night was surely hectic. The next day we started our trip home. When we arrived at Bogota, we found that the flight from Bogota to Paris was overbooked. Eventually two of us (deputy leader and a member of the team) had to take another flight from Bogota to Frankfurt, then back to Hong Kong, about 10 hours late. Air France is famous (notorious) in terms of scheduling, here is another example. All in all, we did not get delayed too much and we eventually returned home safely. Lucky! Lucky! Talking about organization of the event, personally I have no problem with the Jury meeting and/or coordination. Accommodation was very nice. However anything concerned a coach (transportation) was simply not good enough. Say, what is the point of waiting for several hours for a bus, then (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 November 8, 2013.

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

Mathematical Excalibur, Vol. 18, No. 2, Sep.-Oct. 13 Page 2 visit an old town or take a short walk

for less than an hour, and then heading back? ? I do not mean to blame the host country. Indeed I want only to illustrate the point that it is such a gigantic and complicated task to host an IMO! Our team brought home 1 silver and 5 bronze medals. Among 97 teams, we ranked 31. I cannot say that our team did badly. Indeed all our team members managed to get medals, indicating they achieved certain standard. However in these few years, we trailed behind teams like Singapore, Canada, Australia and other teams, not to say the even stronger teams such as China, USA, Korea and Russia, etc. Do we want to do better? Can we recruit better team members? Can we afford time and energy to do that? We have to think about these problems. I can identify some weak points for our team. For example, our team members simply don’t like to do geometry and/or combinatorics problems. Our team members usually get stuck in harder problems, presentations and other things. Or perhaps our team members are too much occupied also by other contests? I know for sure IMO team members of teams such as USA, Australia and Canada would not be allowed to compete in other contests such as IOI or IPhO in the same year. Another suggestion is that we do not train our team enough, we have no intensive camp before IMO (compared with China, USA or UK), and perhaps we should start an intensive camp that will also used as a selection criterion of our team. This idea comes from none other than our old team members! We should pause to think about all these for a while, I suppose.

On the other hand, in this IMO, we confirmed that we will host IMO2016, so in 2016, IMO will be held in Hong Kong. Now we just have to do it, and do it right.

I shall discuss the problems of this IMO. First let us see how they were selected. Indeed the host country (Problem Selection Committee) shortlisted about 30 problems from hundred or so problems submitted by various countries. In the last few years, the Jury first chose an easy pair (problem 1 and 4), then a hard pair (problem 3 and 6), then a medium pair (problem 2 and 5). The 6 selected problems will be then juggled to form

the papers. However this year, it was proposed (and accepted) 4 easy problems in algebra, combinatorics, geometry and number theory were selected. Likewise 4 medium problems again from the different topics were selected. Then two easy problems were selected from the 4 easy problems, say problems of algebra and conbinatorics were selected. The medium problems of other topics (geometry and number theory) were automatically selected as the medium pair. The idea is to guarantee problems of all topics be selected either as an easy problem or a medium problem. After that it doesn’t matter what problems were selected as the hard pair. However, perhaps the end result was not as ideal as we wanted. Eventually in this IMO there are two synthetic geometry problems (Problem 3 and 4). Problem 2, which was supposed to be a combinatorics problem, is actually a problem of combinatorial geometry. Problem 6, which is a combinatorics problem, also has some geometry favor. Problem 1, which was supposed to be a number theory problem, is more like an algebra problem (no prime numbers, no factorization of integers, merely algebraic manipulation and some induction). And finally of course problem 5 is a problem of functional inequalities. So this paper is very much skewed to geometry and with no number theory. Can we say it is balanced? Really at the very beginning, the problems selected were not quite balanced. The problem selection committee suggested there were no easy combinatorics problems and no hard geometry problems! In short, Jury members tended to select problems that demand “ad hoc” considerations, no need to resort to more advanced techniques and/or theorems.

(For the statement of the problems, please see the Olympiad Corner on page 1-Ed.) Problem 1: Problem 1 and 4 (easy pair) turned out to be too easy. Many strong teams get full score in these two problems. For k = 1, we have 1 2 1 1 1 1 , n n    

and it is already of the required form. Hence it is natural to solve the problem using some kind of induction procedure. Essentially all of us did the problem using iterations. One of our team members did the problem as follows. Denote the statement that 1+(2k_{–1)/n is of the form}

1 2 1 1 1 1 1 ... 1 k m m m        _  _         by S(n,k). Note that , 1 2 1 2 2 2 1 1 2 1 2 1 1 _{1 }      _               _ n n n k k k hence if S(n,k) is valid, so is S(2n,k+1). Likewise . 1 2 1 1 2 1 1 1 2 1 2 1 1       _               n n n k k Hence if S(n,k) is valid, so is S(2n–1,k+1). Clearly the cases S(n,1) or S(1,k) are valid. Hence by reducing the cases S(2n,k) to S(n,k–1), or S(2n–1,k) to S(n,k–1), (odd or even cases), one can always obtain the cases S(p,1) or S(1,q), and we are done. Problem 2: All our members guessed the correct answer. The trouble is how to present a proof that is complete (no missing cases). Jury members also worried students didn’t realize the minimum value of k should work for all possible configurations. Thus they defined the term “Colombian”. (Another definition is the “beautiful” labeling in problem 6. In my opinion it was quite unnecessary.) First we show k ≥ 2013. Indeed we mark 2013 red points and 2013 blue points alternately on a circle, (and another blue point elsewhere), then there are 4026 arcs formed. All these arcs have two endpoints of different colors and there must be a line passing through an arc to separate the two points, also each line passing through an arc will meet another arc only once, so we see at least 4026/2=2013 lines are needed.

Now we have to show k = 2013 is indeed enough. The official solution goes as follows. First if there are two points of the same color, say A and B, then one can draw two lines parallel to AB, and are sufficiently close and there are only two points between these lines, namely A and B. This statement is intuitively clear. Draw the convex hull P of the points, and there are two cases. (continued on page 4)

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 November 8, 2013. Problem 426. Real numbers a, b, x, y satisfy the property that for all positive integers n, axn_+byn ₌₁₊₂n+1_{. Determine} (with proof) the value of xa_+yb_.

Problem 427. Determine all (m,n,k), where m, n, k are integers greater than 1, such that 1! + 2! + _{⋯ + m! = n}k_.

Problem 428. Let A1A2A3A4 be a convex quadrilateral. Prove that the nine point circles of _∆A1A2A3, ∆A2A3A4, ∆A3A4A1 and ∆A4A1A2 pass through a common point.

Problem 429. Inside _{∆ABC, there is} a point P such that ∠APB =∠BPC = ∠CPA. Let PA = u, PB = v, PC = w, BC = a, CA = b and AB = c. Prove that



( ) ( )



. ) ( 2 2 b a c b a c b a ca bc ab w v u           

Problem 430. Prove that among any 2n+2 people, there exist two of them, say A and B, such that there exist n of the remaining 2n people, each either knows both A and B or does not know A nor B. Here, x knows y does not necessarily imply y knows x.

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

Solutions

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

Problem 421. For every acute triangle ABC, prove that there exists a point P inside the circumcircle ω of _∆ABC such that if rays AP, BP, CP intersect ω at D, E, F, then DE: EF: FD = 4:5:6.

Solution. Jon GLIMMS (Vancouver,

Canada), Jeffrey HUI Pak Nam (La Salle College, Form 6) and William PENG.

For such a point P, let us apply the exterior angle theorem to ΔABP and ΔACP. Then we have

∠BPC =∠BAC+∠ABE +∠ACF =∠BAC +∠FDE. Similarly, _{∠CPA =∠CBA+∠DEF.}

L₁ L₂ C B A P D E F Q R

To get such a point P, we first draw ΔXYZ with XY = 4, YZ = 5 and ZX = 6. Let α =_{∠ZXY and β =∠XYZ. Next we consider} the locus L1 of point P such that _{∠BPC =} ∠BAC + α, which is a circle through B and C. Also, let L2 be the locus of point P such that _{∠CPA = ∠CBA + β, which is a} circle through C and A.

Let the tangents to L1 and L2 at C intersect ω at Q and R. Then

∠QCB +∠RCA

= 180_{°−(∠BAC+α)+180°−(∠CBA+β)} = _{∠ACB +∠YZX > ∠ACB.}

This implies L1 and L2 intersect at a point P inside ω. Define D, E, F as in the statement of the problem. From the last two paragraphs, we get _{∠ZXY = α} =_{∠FDE and ∠XYZ = β =∠DEF. These} imply Δ DEF and Δ XYZ are similar. Therefore, DE: EF: FD = 4:5:6.

Problem 422. Real numbers a1, a2, a3, … satisfy the relations

an+1an + 3an+1 + an + 4 = 0 and a2013 ≤ an for all positive integer n. Determine (with proof) all the possible values of a1.

Solution. CHEUNG Wai Lam (Queen

Elizabeth School, Form 4), Jon GLIMMS (Vancouver, Canada), William PENG and TAM Pok Man (Sing Yin Secondary School, Form 6).

The recurrence relation can be written as (an+1+2)(an+2) = (an+2)−(an+1+2). If ai =_{−2 for some i, then all a}n = −2 by induction. So a1 =_{−2 is a possible value.} Suppose no ai =−2. Then . 2 1 1 2 1 1      n n a a

Letting bn = 1/(an+2), we easily get bn=

n_−1+b1≠0 for all positive integer n. Then b1≠0,−1,−2,… and an=−2+1/(n−1+b1). Now for positive integer n, an is least when n_−1+b1< 0 and nearest 0, i.e.

n−1+b1 < 0 < n+b1.

Setting n = 2013 and b1 = 1/(a1+2), we can solve the inequality to get

. 2013 4027 2012 4025 1   a

Other commended solvers: Jeffrey HUI Pak Nam (La Salle College, Form 6) and LKL Excalibur (Madam Lau Kam Lung Secondary School of MFBM).

Problem 423. Determine (with proof) the largest positive integer m such that a mm square can be divided into seven rectangles with no two having any common interior point and the lengths and widths of these rectangles form the sequence 1,2,3,4,5,6,7,8,9,10, 11,12,13,14.

Solution. Jon GLIMMS (Vancouver,

Canada), William PENG and ZOLBAYAR Shagdar (Orchlon International School, Ulaanbaatar, Mongolia).

Let a1, a2, a3 , a4, …, a2n−1, a2n be a permutation of 1, 2, 3, 4, …, 2n_{−1, 2n.} We claim the maximum of a1a2 + a3a4 + ⋯ + a2n−1a2n is Sn =1×2 + 3×4 + ⋯ + (2n−1)×2n. The cases n = 1 or 2 can be checked. Suppose cases 1 to n are true. For the case n+1, if _(2n+1)(2n+2) is one of the term, then we can switch it with the last term and apply the case n to get

a1a2+a3a4+⋯+ a2n−1a2n+(2n+1)(2n+2) ≤ Sn+ (2n+1)×(2n+2) = Sn+1. Otherwise, 2n + 1 and 2n + 2 are in different terms. We can switch terms so that a2n−1=2n+1 and a2n+1=2n+2. If we try switching (2n+1)a2n+(2n+2)a2n+2 to a2na2n+2+(2n+1)(2n+2), then since a2n and a2n+2 are at most 2n, we have

[(2n+2)_−a2n][(2n+1)_−a2n+2] > 0. Expanding, we see

a2na2n+2+(2n+1)(2n+2) > (2n+1)a2n_+(2n+2)a2n+2 = a2n−1a2n + a2n+1 a2n+2.

Adding a1a2+a3a4+_{⋯+ a}2n−3a2n−2 and using case n_{−1, we see S}n+1 again is the maximum.

For the problem, the claim implies m2 _≤ S7=1_{×2+3×4+⋯+13×14= 504. Then} m ≤ 22. To finish, we show a 22_×22 square which can be so divided.

Mathematical Excalibur, Vol. 18, No. 2, Sep.-Oct. 13 Page 4 12 10 5 3 13 14 4 2 6 7 11 8 9 1

Other commended solvers: LKL Excalibur (Madam Lau Kam Lung Secondary School of MFBM).

Problem 424. (Due to Prof. Marcel Chirita, Bucuresti, Romania) In _∆ABC, let a=BC, b=CA, c=AB and R be the circumradius of _{∆ABC. Prove that}

. 3 3 2 ) , , max( 2 2 2 R abc ab c ca b bc a    

Solution. Jeffrey HUI Pak Nam (La Salle College, Form 6), TAM Pok Man (Sing Yin Secondary School, Form 6), Alex TUNG Kam Chuen (La Salle College), ZOLBAYAR Shagdar (Orchlon International School, Ulaanbaatar, Mongolia) and Titu ZVONARU (Comăneşti, Romania) and Neculai STANCIU (“George Emil Palade’’ Secondary School, Buzău, Romania).

By the extended sine law, c/sin C = 2R. Let [ABC] denote the area of ΔABC. Then [ABC] = ½ab sin C = abc/(4R). So ab = 2[ABC]/sin C. Using these below, we have , 3 2 ) 3 / ) sin(( 3 ] [ 4 sin 1 sin 1 sin 1 ] [ 4 ) ( 2 ) , , ( max 3 2 2 2 2 2 2 R abc C B A ABC B A C ABC ca bc ab ab c ca b bc a ab c ca b bc a           _{ }             

where the second inequality is by expanding (a−b)2_+(b−c)2_+(c−a)2_≥0 and the third inequality is by applying Jensen’s inequality to f(x)=1/sin x. Other commended solvers: Ioan Viorel CODREANU (Secondary School Satulung, Maramures, Romania) and KWOK Man Yi (Baptist Lui Ming Choi Secondary School, Form 2).

Problem 425. Let p be a prime number greater than 10. Prove that there exist distinct positive integers a1, a2, …, an such that n ≤ (p+1)/4 and

n n a a a a p a p a p   2 1 2 1)( ) ( ) (   

is a positive integral power of 2.

Solution. Jeffrey HUI Pak Nam (La

Salle College, Form 6), LKL Excalibur (Madam Lau Kam Lung Secondary School of MFBM), Alex TUNG Kam Chuen (La Salle College) and ZOLBAYAR Shagdar (Orchlon International School, Ulaanbaatar, Mongolia).

More generally, we prove this is true for all odd integers p≥3. Let

n n a a a a p a p a p X   2 1 2 1)( ) ( ) (    

If p≡1 (mod 4), then let n=(p_{−1)/4 and for} i=1,2,…,n, let ai =2i−1. We have

. 2 ) 2 ( 4 2 ) 2 ( 4 2 ) 1 2 ( 3 1 ) 2 2 ( ) 2 4 ( 4 ) 1 2 ( 3 1 ) 2 2 ( ) 2 4 ( 4 2n n n n n n n n n n n X                    

If p≡3 (mod 4), then let n=(p_{+1)/4 and for} i=1,2,…,n, let ai =2i−1. We have

. 2 ) 2 2 ( 4 2 ) 2 2 ( 4 2 ) 1 2 ( 3 1 ) 2 ( ) 4 4 )( 2 4 ( ) 1 2 ( 3 1 ) 2 ( ) 4 4 )( 2 4 ( 1 2                 n n n n n n n n n n n X      

Olympiad Corner

(continued from page 1) Problem 3. Let the excircle of triangle ABC opposite the vertex A be tangent to the side BC at the point A1. Define the points B1 on CA and C1 on AB analogously, using the excircles opposite B and C, respectively. Suppose that the circum- centre of triangle A1B1C1lies on the circumcircle of triangle ABC. Prove that triangle ABC is right-angled.

Problem 4. Let ABC be an acute-angled triangle with orthocenter H, and let W be a point on the side BC, lying strictly between B and C. The points M and N are the feet of the altitudes from B and C, respectively. Denote by ω1 the circumcircle of BWN, and let X be the point on ω1 such that WX is a diameter of ω1. Analogously, denote by ω2 the circumcircle of CWM, and let Y be the

point on ω2 such that WY is a diameter of ω2. Prove that X, Y, H are collinear. Problem 5. Let ℚ>0 be the set of positive rational numbers. Let f: ℚ>0→ℝ be a function satisfying the following three conditions:

(i) for all x,y ∈ ℚ>0, we have f (x) f (y) ≥ f (xy);

(ii) for all x, y ∈ ℚ>0, we have f(x+y) ≥ f(x) + f(y);

(iii) there exists a rational number a > 1 such that f (a) = a.

Prove that f (x) = x for all x _{∈ ℚ>0.} Problem 6. Let n ≥ 3 be an integer, and consider a circle with n+1 equally spaced points marked on it. Consider all labellings of these points with the numbers 0,1,…, n such that each label is used exactly once; two such labellings are considered to be the same if one can be obtained from the other by a rotation of the circle. A labelling is called beautiful if, for any four labels a < b < c < d with a+d = b+c, the chord joining the points labelled a and d does not intersect the chord joining the points labelled b and c.

Let M be the number of beautiful labellings, and let N be the number of ordered pairs (x,y) of positive integers such that x+y ≤ n and gcd(x,y)=1. Prove that M = N+1.

IMO 2013–Leader Report (I)

(continued from page 2) Case 1. If there is a red point A on the convex hull P, we can draw a line separating A draw all other points. Then we pair up the remaining 2012 red points into 1006 pairs, and as remarked, draw 1006 pairs of parallel lines (2012 lines), separating each pair of red points from all other points. Thus 2012+1=2013 lines are needed. Case 2. All vertices of the convex hull P are blue. Take any pair of consecutive blue points A and B, separating them from all other points by a line (one line) parallel to AB. Then pair up the remaining 2012 blue points into 1006 pairs as before, separating each pair from all other points by 1006 pairs of parallel lines (2012 lines). Thus again 2013 lines are used.

Soc~-tQJ'Y\.<;

-to

Ot

₁

)"\,r'~J.

Gvvte-< )

v-o-C

l

g

Ylo~

·z

: (}) We proceed by induction on k. For k

=

1 the statement is trivial. Assuming we have proved it for k

=

j - 1, W!3 now prove it for k

=

j.

Case 1. n = 2t - 1 for some positive integer

t.

0 bserve that

1 +

zi

- l

= 2(t

+

zi-l

-.1).

~

= ( 1 +

zi-1

-1) ( 1 + _1_).

2t - 1 . 2t 2t - 1 t 2t - 1 By the induction hypothesis we can find m 1 , ... , mj-l such that

1

+

zi-1

1

=

(1

+

_!_)

(1

+ _1 ) ...

(1

+

_1_.) ,

t

m1 m2 IDj-1

so setting IDj

=

2t-1 gives the desired expression.

Case 2. n

=

2t for some positive integer

t.

Now we have

2j-1

2t+2j-·1

2t+2i·-2 _ (.

1

) (

zi-

1

_-1)

1

+

~

=

2t

+

2j -

2 •

2t

.-

1

+

2t

+

2j - 2

1

+

t

I noting that 2t

+

2i - 2

>

0. Again, we use that

1

+

zi-1

~

=:=

(1

+

_!_)

(1

+

__!__) ...

(1

+

_1_) .

t m1 m2 IDj-1

Settirig IDj = 2t

+

zi -

2 then gives the desired expression.

Firstly, let .us present an example showing that k ;??: 2013. Mark 2013 red and 2013

blue points on some drcle alternately, and mark one more blue point somewhere in the plane. The circle is thus split into 4026 ~res, each .arc having endpoints of different colors. Thus, if the goal is reached, then each arc shouid intersect some of the drawn lines. Since any line contains at most two points of the circle, one needs at least

4026/2

=

2013 lines.

It remains to prove that one .can reach the goal using 2013 lines. First of all, let us mention that for .every two points 'A and B having the same color, one can draw two lines separating these points from all other ones. Namely, it suffices to take two lines parallel to AB and lying on different sides of

AB

sufficiently close to it: the only two points between these lines will be A and B.

Now, let P be the convex hull of all marked points. Two cases are possible.

Case

1.

Assume that P has· a red vertex

A,.

Then one rnay draw a line separating A from all the other points, pair up the other 2012 red points into 1006 pairs, and separate each pair from the other points by two lines. Thus, 2013 lines will be used.

Case· 2. Assume now that all the vertices of P are blue. Consider any two consecutive vertices of P, say A 'and B. One may separate these two points from the others by a line parallel to AB.

Then, as in the previous case, one pairs up all the other 2012 blue points into 1006 pairs, and

sepa~ates each pair from the other points by two lines. Again, 2013 lines will be used.

~

· .

De~ote

the

circumcircle~

of the triangles ABC and A1B1 C1 by D and

r,

respectively.

Denote the m1dpomt of the arc 0 B of D containing A by A_{0 ,} and define Bo as well as Oo analogously.

By

our hypothesis the centre Q of

r

lies on D.

Lemm~.

One has

Ao~1

=

Ao01. Moreover, the points A, A0 , B1 , and 01 are concyclic. Finally,

the pomts A and Ao he on the same side of B1 C1 . Similar. statements hqld forB and C.

Proof. Let us consi~er the case A ==: A0 first. Then the triangle ABC is isosceles at A, which

implies AB1

=

AC1 while the remaining assertions of the Lemma are obvious, So Jet us suppose

A # Ao from now on.

By the definition of Ao, we have A0B

=

A0C. It is also well known and easy to show that B0₁.

=

CB1. Next, we have L.C1BAo

=

L.ABAo = L.ACAo

=

_L.B10A0 • Hence the triangles AoBC₁

and AoC B1 are congruent.' This implies AoC1

=

A0B1 , establishing the first· part of the Lemma.

It also follows that L.AoC1A

=

L.AoB1A, as these are exterior angles at the corresponding vertices

01 and B1 of the congruent triangles A0B01 and A00B1 • For that reason the points A, A0, B1,

and 01 are indeed the vertices of some cyclic quadrilateral two opposite sides of which are AAo

· Now we

tur~

to the solution. Evidently the points A1, B1 , aria

C1

lie:interior to some semicircle

arc of

r,

so the triangle A_{1B 1C1 is obtuse-angled. Without los~ of generalit~, we will assume that} its angle

at

B₁ is obtuse. Thus Q and B1 lie on different sides of A1.C1i obviously, the same holds

for the points Band B1. So, the points

Q

and Bare on the same stde of A.tCt. . . .

Notice that the perpendicular bisector of AI C

₁

intersects

n

at two pomts lymg on. different sides of A₁

c

₁. By the first statement from the Lemma, both

poi~ts

.Bo. and

Q

ar.e among these

points of intersection; since they share the sarrie side of AI C1, they 'comctde (see Figure 1) ·

Bo(= Q)

r

Co

Figure 1

Now, by the first part of the Lemma again, the lines QAo and QCo are the perpendicular

bisectors of B1C1 and

A1B11

respectively. Thus

LCIBoA1

=

LCIBoB1

+

LBtBoAl

=

2LA0BoBt

+

2LB1BoCo = 2LAoBoCo = 180o - LABC,

recalling that A0 and C0 are the midpoints of the arcs CB and BA, respectively.

Ob. the other hand, by the second part of the Lemma we have

LGtBoAt

=

LG1BA1 = LABC.

From the last two equalities, we get LABC = 90°, whereby the problem is solved.

@

Let L be the foot.'

of

the altitude from A, and let Z be the second intersection point of circles WI and w2 , other than-1)'. We show that X, Y, Z and H lie on the sameline.

Due to

LBNC

=

LBMC

=goo,

the points

B,

C, Nand Mare concyclic; denote their circle

by w3,. Observe tQ.at the line WZ is the radical axis of w1 and w2; similarly, EN is the radical axis

.of WI an:d.w_3, and CM is the radical axis Of

w

₂ and

w

_{3 •} Hence A= EN n CM is the radical center

of the three circles, and therefore W Z passes through A.

Since W X and WY ar,e diameters in w1 and w2, respectively, we have LW Z X = LW ZY

=

goo,

so the points X and Y lie on the li~e through Z, perpendicular to W Z.

A

Y ...

The quadrilateral BLH N is cyclic, because it has two opposite right angles. From the power of A with respect to the circles wi and BLH N we find AL · AH

=

AB ·AN= AW · AZ. If H lies

. on the line AW then this implies H

=

Z immediately. Otherwise, by ~~

=

;{V

the triangles AHZ

;g;

\91

Denote by Z>o the set of positive integers.

Plugging x == 1, y == a into (1) we get /(1) ~ 1. Next, by an easy induction on n we get ·from

(2)

that

f(nx) ~ nf(x) for all n e Z>o and x e Q>o·

(3)

In particular, we have

f(n) ;.:;;.: n/(1) ~ n for all n e Z>o·

(4)

From (1) again we have f(m/n)f(n) ~ f(m), so j(q)

>

0 for all q e Q!>o·

Now,

(2)

implies that

J

is strictly increasing; this fact together with

(4)

yields

. j

(X)

~ j

(l

X

j)

~

l

X

j >

X -

1 for all

X

~ 1.

' .

By"an easy induction we.get from (1) that f(x.)n ~ f(xn), so

j(xt ~ j(xn)

>

Xn- 1 ::===> j(x) ;_:;;.: :jxn -1 for all

X>

1 and n E Z>O·

This yields

f(x);; x for every x

>

1.

(5)

' (Indeed, if x > y > 1 then :;;n- yn == (x- y)(xn-l

+

xn-2_y

_{+ · · • +}

_{yn) > n(x- y), so for a large n}

we have xn-1

>

yn and thus f(x)

>

y.) .

Now,

(1)

and

(5)

give

an

=

f(a)n;,:;;.: !(an);.:;;.: an, so·f(an) ==an.

Now, for

x>

llet:us choose

n

e Z>o such that

an- x

>

1. Then by

(2)

and

(5)

we get

an= f(an) ;.:;;.: f(x)

+

f(an- x)

~

x

+(an- x) =an

and therefore f(x) = x for x

>

1. Finally, for every x E Q>o and every n E Z>o, from

(1)

and

(3)

we get

nj(x) = f(n)f(x) ;.:;;.: f(nx) ~ nf(x),

which gives f(nx)

=

nf(x). ·Therefore f(mjn)

=

f(m)jn

=

m/n for all m,n E Z>O·

. @

Given a circular arrapgement of [0, n]

=

{0, 1, ... , n}, we

d~fine

a k-chord to be a (possibly degenerate) chord whose (possibly equal)' endpoints add up to k. We say that three chords of a circle are aligned if one of them separates the other two. Say that m ~ 3 chords are aligned if any three of them are aligned. For instance, in Figure 1, A, B, and Care aligned, while B, C, and

lJ

are not.

Figure 1

n-t

Figure 2 Claim. In a beautiful arrangement, the k-chords are aligned for any integer k.

Proof We proceed by induction. For n :::;; 3 the state1llent is trivial. Now let n ~ 4, and proceed,

by contradiction. ·Consider a beautiful arrangementS where the three k-c4ords A, B₁ Care not'

aligned. If

n

is not among the endpoints of

A, B,

and

c,·

then by deleting

n

from

·s

we obtain a beautiful arrangement 8\{n} of [O,n-

1],

where A, B, and Care aligned by the induction hypothesis. Similarly, if 0 is not among these endpoints, then deleting 0 and decreasing all the

numbers by lgives a beautiful arrangement 8\{0}'where A, B, and Care aligned. Therefore

both 0 and n are among the endpoints of these segments. If x and y are their respective partners,

we have n ;.:;;.: 0

+

x = k = n

+

y ;,:;;.: n. Thus 0 and n are the endpoints of one of the chords; say it is C.

Let D be the chord formed by the numbers u and v which are adjacent to 0 and n and on the same side of 0 as A and B, as shown in Figure 2. Set t

=

u

+

v. If we had

t

== n, then-chords A, B, and D would not be aligned in the beautiful arrangement S\ { 0, n}, contradicting the induction hypothesis. If t < n,_ then the t-chord from 0 tot cannot intersect D; sb the chord C separates t

and D. The chord E from t ton-t does not intersect C, sot and n- t are on the same side of 0.

But then the chords A, B, and E are not aligned in S\{0, n}, a contradiction. Finally, the case

t > n ls equivalent to the case

t

< n via the beauty-preserving relabelling x ~ n- x for 0 ~ x ~ n,

H!iwing established the Claim, we prove the desired result by induction. The case n = 2 is

trivi~?-1. Now assume that n ~ 3. Let S be a beautiful arrangement of [0, n] and delete n to obtain

the beautiful arrangement T of

[0,

n-

1].

Then-chords ofT are aligned, and they contain every point except 0 .. Say T is of Type

i

if 0 lies between two of these n-chords, and it is of Type

2

otherwise;~ i.e., if 0 is aligned ·with these n-chords. We will show that each Type 1 arrangement of

[0,

n ~ l}"arises from a unique arrangement of

[0,

n], and each Type

2

arrangement of

[0,

n-

1]

arises from exactly two beautiful arrangell,1ents of

[0,

n].

If T is of Type 1, let 0 lie between chords A and B. Since the chord from 0 to n must be aiigned with A and B in S, n must be on the other arc between A and B. Therefore Scan be

recovered .uniquely from T. In the other direction, if T is of Typ~ 1':{jlnd we insert n .as above, then we claim the resulting arrangement S is beautiful. For

·o-<

k < n, the k-chords of S are also k-chords of T, so they are aligned. Finally, for n < k < 2n, notice that the n-chords of S are parallel by construction, so there is an antisymmetry axis

e

such that x is symmetric to n- x with respect to f for all x. If we had two k-chords which intersect, then their reflections across

e

would be. two (2n - k )~chords which intersect, where 0

<

2n-k

<

n, a contradiction.

·If T is of Type 2, there are two possible positions for n in S, on either side of 0. As above, we . check that both positions lead to beautiful arrangements of

[0,

n].

· Hence if we let Mn be the number of beautiful arrangements of

[0,

n],

and let Ln be the number of beautiful arrangements of

[0,

n - 1] qf Type

2,

we have

To· prove this, consider a Type .2 beautiful arrangement of

[0,

n -

1].

Label the positions 0, ... ,

n-

1 (mod

n)

clockwise arorlnd the circle, so that number 0 is in position 0. Let

f(i)

be the number in' position

i;

note that

f

is a permutation of

[0,

n-

1].

Let a be the position such.

that j(a)

=

n- 1. .

Since the n-chords are aligned with 0, and every point is in art n-chord, these chords are all

parallel and ·

f(i)

+

f(-i)

=

n for all i.

Similarly, since the ( n - 1 )-chords are aligned and every point is in an ( n - 1 )-chord, -these chords are also parallel and

f(i)

+

f(a-

i)

=

n-

1 for all i. Therefore

f(a-

i)

=

f(

-i)

-1 for all i; and since

f(O) =

0, we get

f(-ak)

=

k

for all k. (1)

Recall that this is an equality modulo n. Since

f

is a permutation, we must have

(a,

n) = 1. Hence

Ln-1 ~ 1p(n).

To prove equality, it remains to observe that the labeling (1) is beautiful. To see this, consider four numbers w,

x,

y,

z

on the circle .with w

+

y

=

x

+

z.

Their positions around the circle satisfy

( -aw)

+ (

-ay) = (-ax)

+ (

-az), which means that the chord from w to y and the chord from

x to z are parallel. Thus (1) is beautiful, and by construction it has Type 2. The desired result follows.