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The Geometric Problems in the Bulgarian National Competitions – Tendency to Development

Sava Grozdev, Svetlozar Doichev, Svetoslav Savchev

Bulgarian Academy of Sciences, Acad. G. Bonchev Str., BI.8 1113 Sofia,Bulgaria

51, Knjaz Batenberg Str., E, apt.125

6003 Stara Zagora,Bulgaria

58 A, Chehov Str. 1113 Sofia,Bulgaria

E-mail: savagroz@banmatpc.math.bas.bg

The Bulgaria National Olympiad is one of the oldest national Olympiads in the world. It started during the academic year 1949-50. Some of its main goals are the following: to popularize Mathematics among rising generations and to simulate its study; to active teachers; to discover talented students and give them possibility of expression; to ensure conditions for future training of skilled scientists in the field of Mathematics and its applications, but also in near scientific domains; to select and prepare a strong and competitive national team for the International Mathematical Olympiad. In 1971 the first Spring National Mathematical Competition took place with similar goals. It was followed by the Winter National Mathematical Competition with its first edition in 1982. Now about 30 national and regional competitions in Mathematics are organized in Bulgaria annually. The National Olympiad and both the Winter and the Spring Competitions are approved as the most important, the most qualitative and with the most numerous

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participation. Moreover, they are used to select the national teams for both the International and Balkan Mathematical Olympiads. This article examines the geometric problems from these three competitions. The following features are discovered: the proportion between synthetic and analytic character is changing through the years and the analytic one is leading at present; the genre variety decreases gradually and the problem formulation becomes more classical on account of it; problem difficulty increases gradually; aspiration appears to approach the ideas of the International Mathematical Olympiad . To prove this idea about 35 problems are discussed by dividing them into two groups. The first one contains 18 problems from the 80-ies, which accent synthetic character, genre variety and some comparable easiness of the solutions. The second group contains 17 problems from the 90-ies with the accent of analytic character, classical sounding and comparable difficulty of the solutions. Examples are given using the idea of Ivan Ganchev to present problems in systems which turns out to be quite useful in simplifying solutions and developing skills in problems solving.

Founded in the academic year 1949/1950, the Bulgarian national Olympiad is among the oldest mathematical contests in the word. Some of its original goals are popularizing mathematics at high-school level, discovering and encouraging students of high mathematical ability, motivating teachers and, later on, selecting and coaching a competitive team for the International Mathematical Olympiad. The Spring National Mathematical Competition, whose goals are similar, was organized for the first time in 1971. It was followed by the Winter National Mathematical Competition, with its first edition in 1982. There are about 30 national and regional mathematical contests in Bulgaria at present. However, the three mentioned here remain the most important ones, of highest quality and with most numerous participation. They are also used in the process of team selection for both the International and the Balkan Mathematical Olympiad. The subject of this article is the geometry problems posed on these three contests. Several dozens of problems are considered. We found it best to classify them mainly with respect to their chronological order by the following

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reason. It turns out that each stage of development of the Bulgarian mathematical contests heavily influences their style and mathematical contents.

• Whereas synthetic solutions were dominant in the 1980ies, analytical approaches prevail after 1990.

• The diversity of geometric settings, so easy to spot in the years 1980--

89, decreases more and more.

• The recent geometry posed on Bulgarian competitions is now of a flavor much more classical than some 15 years ago.

• In return, the problems are much harder as mathematical challenges.

Following the IMO geometry tradition is the major concern of the proposers.

Some reasons for and consequences of this state of matters are briefly touched upon below. Our emphasis is on the most recent editions of the contests which display some new trends of development. Examples are given following Ivan Ganchev's idea to present problems in systems. This approach turns out to be quite useful in simplifying solutions and developing problem solving skills. 1 The 1980ies: The Fruits of a Strategy The Bulgarian national olympiad went through substantial changes during the years of its existence. Mass participation was the main objective in the 1950ies and 1960ies. The style of problems proposed was basically the one present at the university entrance exams. This started changing in the late 1960ies and the early 1970ies. A co-founder of the International Mathematical Olympiad, Bulgaria was lacking the traditions of participating countries like Hungary and the former Soviet Union, for instance. So a number of renowned mathematicians was concerned with establishing a different kind of mathematical culture in the overall approach to high-school students. Without changing the olympiad structure,

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more and more extracurricular elements were being introduced into the unwritten syllabus of the olympiad. The general state educational policy was also changing at that time. Studying mathematics was intensified in the regular schools, and specialized mathematical schools were opened in all regional centers (1971-75). Mathematical teachers of high ability created their own schools and circles of talented students. The main incentive was a good performance at the olympiad, but new forms of extracurricular activity came into being as well. The authority of the Central Olympiad Committee increased considerably. The results became visible already around 1976, when the Bulgarian team came out fourth at the international olympiad. However, the entire system of extracurricular activities reached maturity a little later-and the stable performance of the national IMO team in 1981-1989 is merely one of the many indications. We mention only the International Olympiad in Prague, 1984, where Bulgaria achieved the second highest combined score among all participants. The work on running high-school mathematical activities was a well-coordinated effort of the Ministry of Education and the Union of Bulgarian Mathematicians, assisted by the Mathematical Department of the Sofia University and the Institute of Mathematics and Mechanics of the Bulgarian Academy of Sciences. The social prestige of the olympiad movement was surprisingly high. The centralized higher education in Bulgaria honored unconditionally the achievements of the IMO contestants, allowing them to become university students without taking the (quite competitive) entrance exams. A new factor was the Balkan Olympiad which started out in 1984. The successes of the young Bulgarian mathematicians at the time in question were due to a carefully developed orderly system, the opportunities for a broader selection because of the numerous participants, to the enthusiasm of teachers in many regional centers. We pass on to the mathematical contents of the olympiad between 1981 and 1989. More exactly, let us take a closer look at its final rounds, the third and the fourth. For one thing, the goal of encouraging mass participation was still present as a codition sine qua non. On the other hand, the final rounds had to complete a major part of the IMO selection in an evironment where the top contestants' preliminary preparation greatly exceeded the one of a regular participant. The typical result was a contest paper of difficulty comparable to that of an IMO paper, but also reflecting the curriculum of the Bulgarian school. For instance, solid geometry is practically missing at the international olympiad but well represented at Bulgarian contests in the eighties.

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The following list of problem statements may seem excessive but it reflects important features of the whole period. Problem 1 (National Olympiad, fourth round, 1982) Determine the locus of the centroids of the equilateral triangles whose vertices lie on the sides of a given square. Problem 2 (National Olympiad, fourth round, 1983) Find a square of minimum side length which can contain 5 nonoverlapping unit circles. Problem 3 (National Olympiad, fourth round, 1984) In space, 1+n points nPPP ,...,, 21 and Q are given, 4≥n , no four of which are in the same plane. It is known that for each triple of distinct points

ji PP , and kP one can find a point lP such that Q is interior to the tetrahedron

lkji PPPP . Show that n must be even. Problem 4 (National Olympiad, third round, 1985) Five points in the plane have the following property: Among any 4 of them, one can choose 3 which are the vertices of an equilateral triangle.

a) Prove that 4 of the points are the vertices of a rhombus with acute angle 60 .

b) Find the number of equilateral triangles with vertices at the given

points. Problem 5 (National Olympiad, third round, 1985) The midpoints of the edges AB and CD of the tetrahedron ABCD and the center of its inscribed sphere are on the same straight line. Prove that the center of the circumscribed sphere of the tetrahedron lies on the same line. Problem 6 (National Olympiad, fourth round, 1985) In the triangle ABC, the angle γ=∠ ACB is acute and BCAC ≠ . A point P is taken on the medianCM with the property that the bisectors of PAC∠ and

PBC∠ intersect at a point Q on CM . Find APB∠ and AQB∠ . Problem 7 (National Olympiad, fourth round, 1985)

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A sphere with center O is inscribed in a quadrangular pyramid ABCDM , with base ABCD . The point O lies on the altitude MH of the pyramid. Each of the planes, )(ACM , )(BDM and )(ABO divides the lateral surface of the pyramid into two parts of equal area. The areas of the plane sections of the pyramid with the planes )(ACM and )(ABO are in ratio 4:)22( + . Find the angle between the planes )(ACM and )(ABO and the dihedral angle of the pyramid with edge AB . Problem 8 (National Olympiad, fourth round, 1986) Find the least positive integer n for which there exist an n -gon P and a point A in its interior such that, if a source of light is placed at A , no side of P will be completely lit by the light. For the value of n obtained, prove that one can find two points inside the n -gon P with the property that, if sources of light are placed at them, all sides of P will be completely lit. Problem 9 (National Olympiad, fourth round, 1986) A regular tetrahedron T of unit edge is given. Find the maximum volume of a cube contained in T and such that one of its diagonals lies on an altitude of the tertahedron. Problem 10 (National Olympiad, fourth round, 1987) Let ∆ be the set of all triangles inscribed in a given circle k , whose degree angle measures are integers different from 45 , 90 and 135 . For each triangle

∆∈T , define )(Tf as the triangle with vertices the intersections of k with the altitudes of T (or with their extensions).

a) Prove that there exists a positive integer n such that for each triangle ∆∈T at least two among the triangles T , )(Tf ,

))(()()),...,(()( 12 TffTfTffTf nn −== are congruent. b) Determine the least value of n such that a) holds true. Problem 11 (National Olympiad, third round, 1988) Construct a circle Γ through the vertices A and B of the triangle ABC such that the common chord of Γ and the circumcircle of ABC has a given length .

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Problem 12 (National Olympiad, fourth round, 1988) Each point of the (three-dimensional) space is colored either red or white. Prove that there exists a tetrahedron whose vertices and centroid are of the same color. Problem 13 (National Olympiad, fourth round, 1988) Let A, B and C be noncollinear points. For each point D on the ray →AC , denote by E and F the points of tangency of the incirle of the triangle ABD with AD and BD . Prove that the line EF passes through a fixed point as D ranges over →AC . Problem 14 (National Olympiad, third round, 1989) Find all convex n -gons with the following property: There exist 2−n diagonals of the n -gon each of which bisects its area. Problem 15 (Winter Competition, 1983) The diagonals AC and BD of the cyclic quadrilateral ABCD intersect at E . Denote by P , Q the feet of the perpendiculars from E to AD , BC , and by M , N the midpoints of AB,CD . Prove that the line MN is the perpendicular bisector of the line segment PQ . This list provides ample evidence about an impressive genre diversity. Together with traditional classical settings (problems 6,13,15), we encounter instances of geometric loci (problem 1), geometric extrema in the plane and in space (problems 2, 9), Euclidian constructions (problem 11), classical solid geometry (problems 5,7), combinatorial geometry in the plane (problems 4,8,14) and in space (problems 3,12), even a combination of plane geometry and divisibility (problem 10). Since our attention is focused mostly on classical geometry, we present solutions to problems 6 and 15.

In Problem 6, the bisector property

gives .

QCQP

BCBP

ACAP

==

Hence A and B belong to the locus of points X such that the ratio XCXP : is equal to .QCQP It is

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well-known that this locus is a circle k with diameter on the line PC (the Apollonius circle of the line segment PC ), and Q is an endpoint of this diameter. Denote the other endpoint by R . Since QR divides the chord AB of k into two equal parts and is not perpendicular to AB (because

BCAC ≠ , AB is a diameter of k , too. Therefore 90=∠AQB . Now a standard computation yields the value of APB∠ . We have

,1802)(2

)(2−∠+γ=∠−∠+γ=

∠+∠+γ=∠+∠+∠=∠

APBAQBAPBPBQPAQPBCPACACBAPB

so .180 γ−=∠ APB

In Problem 15, it suffices to prove that P and Q are equidistant from M (by symmetry, the same will hold for N , implying the desired conclusion). To this end, take the midpoints U and V of AE and BE , respectively. We claim that MPU∆ and QMV∆ are congruent. Indeed, PU and QV are medians to the hypotenuses of the right triangles AEP and BEQ , hence

UEPU = and VEQV = . Next, CBDCAD ∠=∠ by inscribed angles, so the right triangles just mentioned are similar, which yields QVEPUE ∠=∠ . Finally, EUMV is a parallelogram, so EVMEUM ∠=∠ and MVUE = , MUVE = . It follows now that MVQPUMVQUMMVPU ∠=∠== ,, ,and the proof is complete.

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The proofs above are representative. Many classical geometry problems from the 1980ies do admit of analytic solutions. However, the very intention of the proposers and the jury was to promote synthetic methods in the spirit of the ancient Greek tradition. It was unlikely for a problem proposal to make a contest paper unless a synthetic proof was available in advance. Computational solutions were of course developed by some contestants. But such solutions were mostly regarded as an expression of poor geometrical thinking and not encouraged too much. As a rule, the proofs known to the jury were short and not particularly hard. Once a contestant reached a key idea, he or she was able to prepare a neat writeup in a limited amount of time. Summing up, some major features of the geometry problems from the 1980ies were the following:

• synthetic approaches prevailing; • great genre diversity; • concise and relatively easy solutions.

2 The 1990ies: Fall and Rise Bulgaria was hit by a severe overall crisis in the early 1990ies. The orderly system of extracurricular mathematical activities, which seemed to be established once and for all, suffered many blows and finally collapsed. In particular, the Ministry of Education gradually resigned from its obligations to mathematical contests at all levels. Financing was extremely scarce, if available at all. A multitude of dubious educational reforms decreased the intensity and the level of teaching high-school mathematics to a critical degree. The priviledges of the olympiad winners to enter the universities without entrance exams were abolished. The Union of Bulgarian Mathematicians was the institution to go on with the mathematical olympiads despite the extremal conditions. Without state funding and by many other reasons, it was already an achievement to save the major contests. However, preserving the entire movement with the same goals and structure as before was a fiction. Mass participation was unthinkable not only by trivial financial considerations. Most regional centers with established traditions had ceased to exist. Neither regular students nor their teachers had incentives and motivation to continue what had been started some two decades

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before. The base for selecting the national team was thus drasctically reduced to the two major mathematical schools in the capital Sofia. The limited task of preserving Bulgaria's positions at the international level was probably the only reasonable choice in that environment. Once the Union took over all the responsibility with respect to the olympiad, certain changes in the spirit of the competitions occurred. The main goal was already selecting a strong national team. This circumstance influenced in particular the geometry on the national contests. Here are examples of geometry problems from this period. Problem 16 (National Olympiad, third round, 1996) Let ABCD be a convex quadrilateral satisfying 180<∠+∠ BCDABC . The lines AB and CD intersect at E . Prove that ADCABC ∠=∠ if and only if

AEABCECDAC ..2 −= . Problem 17 (National Olympiad, third round, 1998) The convex quadrilateral ABCD satisfies CDAD = and 90<∠=∠ ABCDAB . The line through D and the midpoint of BC meets the line AB at E . Prove that DACBEC ∠=∠ . Problem 18 (Spring Tournament,2001) Let 1AA and 1BB be altitudes of the obtuse nonisosceles triangle ABC . Denote by O and 1O the circumcenters of the triangles ABC and CBA 11 , respectively. A line through C intersects the line segments AB and 11BA at D and 1D , respectively, and E is a point on the line 1OO such that 90=∠ECD . Prove that

CDCDEOEO :: 11 = . Problem 19 (National Olympiad, third round, 2001) The diagonals of the cyclic quadrilateral ABCD intersect at E . Prove that if

60=∠BAD and CEAE 3= then the sum of two of the sides of the quadrilateral is equal to the sum of the other two. Problem 20 (National Olympiad, fourth round, 2001) In a nonisosceles triangle ABC , the points of tangency of the incircle Γ with the sides AB , BC and CA are 1C , 1A and 1B , respectively. Let

139

2121 , BBBAAA =Γ=Γ , and let 3131 , BBAA be bisectors in the triangle ),( 113113111 CABCBACBA ∈∈ . Prove that:

a) 32 AA is the bisector of 121 CAB∠ ;

b) If P and Q are the common points of the circumcircles of the triangles 321 AAA and 321 BBB then the incenter of the triangle ABC lies on the line PQ . Problem 21 (Spring Tournament,1998) Let I and r the incenter and the inradius of the triangle ABC , respectively, and let N be the midpoint of its median through C . Prove that if INCNr −= then BCAC = or 90=∠ACB . Problem 22 (National Olympiad,third round,1997) The quadrilateral ABCD is cyclic. Let F be the common point of its diagonals AC and BD , and E the common point of the lines AD and BC . If M and N are the midpoints of AB and CD , prove that

.21

ABCD

CDAB

EFMN

−=

Problem 23 (Winter competition,2001) The points 11, BA and 1C are taken on the sides ABCABC, of the triangle ABC , respectively. The point G is the centroid of ABC∆ . Let cba GGG ,, be the centroids of 111111 ,, BCACBACAB ∆∆∆ , respectively, and 21, GG the centroids of

cba GGGCBA ∆∆ ,111 . Prove that: a) the points G , 1G and 2G are collinear; b) the lines aAG , bBG and cCG are concurrent if and only if the lines 1AA ,

11 CCBB are concurrent. Problem 24 (Spring Tournament,2000) The point 1A is chosen on the side BC of the triangle ABC so that the triangles 1ABA and 1ACA have congruent incircles. Denote the diameters of these circles by ad . The diameters bd and cd are defined analogously. Let

140

pcbacABbCAaBC 2,,, =++=== . If cba hhh ,, are the altitudes through CBA ,, , respectively, and d is the diameter of the incircle of ABC , prove

that:

a) aa hda

appd =−

+)( ;

b) cbacba hhhpddd ++≥+++ . Problem 25 (Winter competition,1997) A triangle ABC is given whose angles with vertices A and B are greater than or equal to 60 .Let BL be the bisector of ABC∠ and H the foot of the altitude through A. Find AHL∠ if AHLBLC ∠=∠ 3 . Problem 26 (Winter competition,2000) Given is a convex quadrilateral ABCD whose diagonals meet at M . It is known that DMDB 3= and MCAM = . a) Express BC and CD in terms of the sides of the triangle ABD . b) Prove that if 1802 =∠−∠ ABDADB then BDCDBC ∠=∠ 2 . Problem 27 (Winter competition,1999) Let O and Rbe the circumcenter and the circumradius of the triangle ABC . The incircle of this triangle is tangent to BC , CA and AB at 11, BA and 1C , respectively. The lines determined by the midpoints of the line segments 1AB and 11, BCAC and 11, CABA and 1CB intersect at 22 , BA and 2C . Prove that the triangle 222 CBA has circumcenter O and circumradius

2rR + .

Problem 28 (Spring Tournament,1996) The point D lies on the arc

^BC of the circumcircle of the triangle ABC which

does not contain A, and CDBD ≠≠ , . Points E and F are taken on the rays →BD and →CD so that ACBE = and ABCF = . Let M be the midpoint of the

line segment EF . a) Prove that BMC∠ is a right angle.

b) Find the locus of the points M as D ranges over the arc ^

BC .

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We present two solutions of Problem 17 now, a synthetic and a computational one.

Naturally, a synthetic approach would not rely on the initial configuration alone. Additional constructions are needed for a purely geometric solution. To begin with, observe that if AD and BC meet at N then BNAN = . Also, extend EC to meet AN at P . Denote α=∠=∠ DACACD and

β=BAC .Then β+α=∠ABC , and the desired equality α=∠BEC is equivalent to β=∠BCE , or to

β=∠NCP . On the other hand, an easy angle computation shows that β=∠ 2DCN . Hence the problem reduces to proving that CP is the bisector of DCN . Note that even after these key steps one may very well get stuck here. To move on, we have to realize that ratios of line segments are likely to help rather than angle computations. More exactly, a repeated application of Menelaus's theorem yields the conclusion. Consider DMN∆ and the line AB first, then DMN∆ again and the line CE . By Menelaus's theorem,

1.... ==PDNP

CNMC

EMDE

ADNA

BNMB

EMDE .

Now, taking the equalities CMBMBNAN == , and CDAD = into account, we infer that PNPDCNCD = . Hence CP is the bisector of DCN∠ , as needed. The problem is by far not among the hardest ones posed on Bulgarian competitions. Still, a solution like above demands ingenuity and may be a matter of chance. On the other hand, a straightforward computational proof takes no inventiveness at all.

Let β=∠α=∠=∠ BACDACACD , , as in the previous proof, and let

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vCDEuAED =∠=∠ , .ThenvuDCEADC −−β−α+π=∠α−π=∠ ,2

Apply the law of sines for ADE∆ and CDE∆ first. In view of CDAD = , we

obtain

.)sin(

sin)sin(

sinvu

vDECD

DEADu

++α−β===

β+α

Next, we have vuBCE −−β+α=∠ . Since CMBM = , the law of sines for BME∆ and CME∆ gives

.)sin(

sin)sin(

sinvu

vMECM

MEBMu

−−β+α===

β+α

Hence

,)sin(

)sin(sinsin

)sin()sin(

vuvu

vu −−β+αβ+α

==++α−β

β+α

and it follows that )sin()sin( vuvu −−β+α=++α−β . This can be rewritten as 0)sin(cos =α−+β vu . Since β is acute, we easily conclude that

α=+ vu . Problem 22 is an appropriate representative of the computational aspect of the period. The triangles ABE and CDE are similar, because ABCD is cyclic. Hence λ===

EDEB

ECEA

CDAB , a positive number which is not 1 (or else AD and

BC would not intersect). This initial observation already provides the base of a vector solution for which no figure is necessary.

Let dEDcEC == , . Define i

as the unit vector collinear with →

EC and

having the same direction, that is, ECci

→=

1 .Likewise, let EDdj

→=

1 . It is now

a standard (but somewhat tedious) task to express EF→ and MN

→ as linear

combinations of i

and j

.Note first that jcEA

λ=→

and idEB

λ=→

. Since F lies on both AC and BD , there are numbers x and y such that

.)1()1(

,)1()1(

jdidyyy

jcxicxxEAx

EDEBEF

ECEF

λ−+λ=−+=

λ+−=−+=→→→

→→→

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Equating the coefficients of i

and j

and eliminating y yields c

cdx)1( 2 −λ

−λ= .

Therefore

[ ].)()(12

jcdidcEF

−λ+−λ−λ

λ=

→

On the other hand,

[ ] .)()(21

21

21

jcdidc

EBECEAEDBCADMN

λ−+λ−=

−+−=

+=

→→→→→→→

It is clear now that the ratio of 2MN and 2EF equals the ratio of 41 and

12

2

−λ

λ ,

that is,

.411

41 22

2

2

−=

λ−λ=

ABCD

CDAB

EFMN

The claim follows. Standard as they appear to be, these two third-round problems were solved by very few contestants. They are examples of seemingly standard configurations, rich in connections and well-known properties but objectively hard to be handled synthetically. We presented a synthetic proof of Problem 17 above, and should add that Problem 22 admits of such a proof as well (far and away harder to devise).However, it is exactly searching for a synthetic approach and estimating the difficulty arising that confirms a general claim.The effort to argue synthetically is neither proportional to the theoretical significance of the question nor reasonable in the conditions of a time constraint. The majority of the problems 15-28 are similar. Not only is their average difficulty higher than the one of Problems 1-15. Their spirit is different. It was a conscious, continuous effort to impose such a spirit. There is no pursuit of diversity any more, although no topic is ever permanently closed. A brief look at the problem statements says enough: these are exclusively questions in classical geometry, and questions intentionally difficult. Writing down a solution alone may take quite a while, the exposition is prone to computational errors of any kinds. Consequently, it happens that no contestant solves a certain geometry problem sometimes-and the hardest problem on a contest in Bulgaria is almost never a geometry one.

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The resemblance with most IMO style geometry proposals is fairly evident, which was the intention of the proposers and the leaders of the national team. Purely geometric proofs are neither prohibited nor impossible to find in principle. Only the time is long gone when the geometry IMO problems were the hardest challenge for the Bulgarian participants, on which they could afford spending a long time during the contest. Arguing synthetically is welcome, if the insight comes quickly. On the other hand, being on the safe side becomes increasingly more important. By safe we mean being able to carry out reasonable amounts of standard computations to get the solution to a geometry problem as fast as possible. Because the real challenges to decide the IMO golds are still to be handled, and these are rarely geometrical questions. The key characteristics of the Bulgarian geometry problems in the 1990ies are therefore the following:

• analytic approaches prevailing;

• settings almost exclusively classical;

• solutions relatively hard and difficult to expose. 3 The Future: New Trends and a Cautious Hope The period of the late 1990ies was a relatively stable one for the olympiad movement. With a lot of effort, the Union of Bulgarian Mathematicians managed to establish a strict organization of all major initiatives, for proper selection and coaching the national team. New and encouraging signs can be observed. Exchange of valuable experience started with Russian and Chinese colleagues as well as with co-workers from the Balkan countries. This includes inviting teams from abroad to Bulgarian contests, exchanging visits, and so on. Forms for early selection of talented young mathematical minds play an increasingly important role. Such are the Junior Balkan Olympiad and the annual junior contest in Hong Kong. The latter is regarded by many as something of a junior IMO despite the differences in the format. Bulgaria is an active participant in both of these since their beginning. The Union of Bulgarian Mathematicians resumes publishing appropriate books and materials for extracurricular mathematical work. Slowly and with great difficulty, several regional mathematical centers outside Sofia gradually revive their previous activity. Some universities agreed to accept IMO participants without entrance exams again, as before. Finding sources of funding is a constant major concern. Though randomly, foundations would

145

provide the means for certain indispensable needs like travel expenses. Private and corporate sponsorship is modest but present. Unfortunately, the Ministry of Education remains impassive to the mathematical olympiad. This is a brief outline of the atmosphere in which yet another tendency was conceived. To justify our observations, we continue with several more geometry problems. Problem 29 (National Olympiad,fourth round,1997) Let G be the centroid of the triangle ABC . Prove the inequality

32sinsin ≤∠+∠ CBGCAG

a) if the circumcircle of the triangle ACG is tangent to the line AB; b) for an arbitrary triangle ABC .

In case the question were naturally stated, it would have read: Determine the maximum value of the sum

CBGCAG ∠+∠ sinsin , where G is the centroid of the triangle ABC .

However, this problem was found too hard by the jury, so a hint for the key step was explicitly provided in the problem statement. But why is the special case in a) crucial? Suppose that ABC∆ is arbitrary, and let M be the midpoint of AB. There are two circles passing through C and G which are tangent to the line AB. Let the corresponding points of tangency be 11 , BA ,on the rays →→ MBMA , , respectively. Since 2

121 . MBMCMGMA == by the power-of-a-point theorem

and, in addition, GGMCG ,1:2: = is the centroid of CBA 11 as well. Moreover, A and B are exterior to the two circles unless the two triangles coincide. It is straightforward now that GCBCBGGCACAG 11 , ∠≤∠∠≤∠ . Thus, assuming GCA1∠ and GCB1∠ acute and the special case a) already settled, we have

.3

2sinsinsinsin 11 ≤+≤∠+∠ GCBGCACBGCAG

Thus we are left with the proof of a) and with the subcase of b) where one of GCA1∠ and GCB1∠ is right or obtuse.

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So, let the circumcircle of ACG∆ touch AB. We use the standard notation for the elements of ABC∆ . By the power-of-a-point theorem and the well-known median formula,

),22(121

31.

422222

2cbamMCMGMAc

c −+====

yielding 222 2cba =+ . The median formula again gives ambm ba 23,

23

== .

Then

[ ] [ ] [ ] [ ]

.3

sin)(

.2

.2sinsin

22

abba

amABC

bmABC

BGBCBCG

AGACACGCBGCAG

ba

γ+=

+=+=∠+∠

The law of cosines, combined with 222 2cba =+ , implies γ=+ cos422 abba . Therefore ,

322sin

32sinsin ≤γ=∠+∠ CBGCAG and a) follows.

Finally, let, for instance, 901 ≥∠ GCA ; then GCB1∠ is acute. Denote by

111 ,, cba and 1γ the side lengths of CBA 11∆ and 11CBA∠ , respectively. We obtain from GCA1∆ that 2

121

2 GACACG +> , that is,

).22(91)22(

91 2

121

21

21

21

21

21 acbbcba −++>−+

By a), we have 21

21

21 2cba =+ , and the above inequality takes the form

21

21 7ba > . Now set 2

121 abx = . The argument in the proof of a) also gives

[ ]

).(11434

14

13

3sin

.2

sin

22

11

21

21

1

1

1

11

11

11

xfxxbaba

ab

ab

GBCBCGBGCB

=−−=

+−=

γ==∠

Since 71<x , it follows that 71)71()( =< fxf . Therefore

147

.3

2711sin1sinsin 1 <+<∠+<∠+∠ GCBCBGCAG

Problem 30 (National Olympiad,fourth round,1999) The vertices of a triangle have integer coordinates and one of its sides has length n , where n is a square-free positive integer. Prove that the ratio of the circumradius and the inradius of the triangle is irrational. The setting suggests that the question can probably be quickly reduced to divisibility considerations with a minimum use of geometry, which is indeed the case at first glance. However, this impression is deceitful. It proves hard (or impossible) to reach the conclusion without switching back to geometry at a certain point. Let R, r and S be the circumradius, the inradius and the area of the triangle, respectively. It is not hard to observe that S is rational. Assume that the ratio rR is rational. Without loss of genereality, let one endpoint of the side with length n be at the origin, and let the remaining two vertices have coordinates ),( yx and ),( tz , with tzyx ,,, integers. Then the side lengths of the triangle are of the form CcBbAa === ,, , where

222222 )()(,, tyzxCtzByxnA −+−=+=+== .By hypothesis, the number

28)(

2.

4 Scbaabc

Scbaq

Rabc

rR ++

=++

==

is rational. Then so is also 28)()( qScbaabcCABCABABC =++=++ , because S is rational. Upon squaring ABCqSCABBCA −=+ 28 , we readily observe that AB is rational; by symmetry, the same holds for BC and CA. Hence CABCAB ,, are perfect squares. Let

2212

212

21 ,, ccCbbBaaA === , where 222 ,, cba are square-free. Since

( ) 222

11 babaAB = , it follows that 22 ba = . Likewise, 22 cb = , so 21

21

21 ,, mcCmbBmaA === for some square-free positive integer m . Now,

21manA == and n is square-free, which gives 1, 1 == anm . Thus

.)()(,, 2

1222

12222 nctyzxnbtznyx =−+−=+=+

It appears that a little more work based on the above equalities will yield the desired contradiction. However, one more geometric consideration is needed. Namely, we proved so far that the three side lengths are

148

11 ,, cncbnbna === . Then 111 cb >+ and 111 bc >+ by the triangle inequality, implying 11 cb = . This last observation proves critical indeed, since one can write 22222222 )()(,, ndtyzxndtznyx =−+−=+=+ for some d . These yield )(222 ytxzyxn +=+= . Now the well-known Lagrange's identity

222222 )()())(( yzxtytxztzyx −++=++ suggests considering the number )(2 yzxtk −= . We obtain

( )[ ] 2222222222 4))((4)(4 dntzyxyzxtytxzkn =++=−++=+ , or )14( 222 −= dnk . But then 14 2 −d turns out to be a perfect square, a contradiction. Problem 31 (Spring Tournament,2000) There are given 4≥n points in the plane such that their pairwise distances are all integers. Prove that at least

61 of these distances are divisible by 3.

We first prove the statement for :4=n if DCBA ,,, are four points such that their pairwise distances are integers then at least one of these 6 distances is a multiple of 3. Assume on the contrary that this is false. The square of an integer not divisible by 3 is congruent to 1 modulo 3. Thus, )3(mod1222222 ≡≡≡≡≡≡ CDBDBCADACAB .

Suppose that C lies in BAD∠ and denote β=∠α=∠ DACBAC , . The law of cosines for ADCABC ∆∆ , and ABD∠ yields

.)cos(.2

,cos.2

,cos.2

222

222

222

BDADABADAB

CDADACADACBCACABACAB

−+=β+α

−+=β

−+=α

The left-hand sides are integers congruent to 1 modulo 3. Therefore

)3(mod1coscos..4 2 ≡βαADABAC . In addition, since )3(mod22 2 ≡AC , we obtain )3(mod2)cos(..4 2 ≡β+αADABAC . Subtracting the last two congruences implies that

149

[ ] βα=β+α−βα sinsin..4)cos(coscos..4 22 ADABACADABAC is an integer congruent to 2 modulo 3. On the other hand, αcos and βcos are clearly rational. Moreover, if

dc

ba

=β=α cos,cos , with dcba ,,, integers and 1),(),( == dcba , then none

of dcba ,,, is a multiple of 3. Then, once the number

bdcdab

ADABACADABAC))((

..4sinsin..42222

22 −−=βα

is an integer not divisible by 3 , the same holds for the number ))(( 2222 cdab −− . But )3(mod12222 ≡≡≡≡ dcba , implying that

))(( 2222 cdab −− is a multiple of 3. The contradiction proves the claim for 4=n .

Now the general case follows as an immediate consequence. Indeed, take n arbitrary points in the plane, 4≥n , whose pairwise distances are all integers. One can choose ( )n

4 quadruples of points among these, each quadruple containing a pair of points at distance divisible by 3. Each such pair is counted ( )2

2−n times. Therefore the total number of distances determined by the given

points and divisible by 3 is at least ( )( ) ( )n

n

n

222

4

61

=−

, that is: at least 61 of the total

number of distances in question. Problem 32 (Winter Competition,2001) A 906030 −− triangle T with hypotenuse 1 is given. A point is chosen on each side of T so that the traingle with vertices at these points is a right triangle. Determine the least value of its hypotenuse. Problem 33 (National Olympiad,fourth round,1998) A square, a regular n -gon and a regular m-gon are constructed externally on the sides of the nonobtuse triangle ABC . Their centers are the vertices of an equilateral triangle. Prove that 6== mn and find the angles of ABC . One may ask here: Why are the last five problems considered separately. Do the observations in Section 2 not apply to them? True, their difficulty is

150

definitely high, probably too high from a certain point of view. But being simply ``hard" is not the point. With most questions in Section 2, computations are quite long and yet reasonable. The major difficulty is translating the information into a certain analytic language (trigonometry, vector algebra, complex numbers, standard coordinate geometry, etc.) Once the translation is done, the rest is a matter of time and patience. A reasonable amount of effort suffices. The story is different about Problem 29, for instance. Not having the crucial hint stated in a), one is probably destined to face more technical trouble than he or she could possibly take during a contest. It is not the mechanical translation that matters, nor the subsequent manipulations. Essential geometrical thinking precedes the computations, thinking hard to be replaced by other means. In this case, the key idea is a purely geometric observation: The angle subtended by a fixed line segment AB on one side of a given angle at a variable point C on the other side of this angle is a maximum if and only if the circumcircle of

ABC∆ is tangent to the second side (at the point C ). Much technical work is still to come, but work manageable, guided by a proper understanding of the situation. A similar consideration is the core of Problem 32. If a right triangle T with minimal hypotenuse is inscribed in a given triangle 0T then the circumcircle of T is tangent to the side of 0T containg the vertex of the right angle. As above, the remaining work is tedious but within reach. In Problem 30, it proved necessary to amend the insufficient information in a divisibility argument by an unexpected use of the modest triangle inequality. A fact which may be regarded as a true fine point of this proof. In Problem 31, insight is needed that the case of a general n reduces to

4=n by a counting argument, which is the combinatorial part of the matter. As for 4=n itself, this is where analytic geometric means are essentially involved. This special case alone is a substantial problem in itself. The last Problem 33 is also remarkable. It takes little to express the sides of the triangle known to be equilateral by hypothesis, and to equate the expressions obtained. However, by doing this we only end up with equalities promising no more than hard work without any perspective. And indeed, the few solutions found by contestants rested on a preliminary geometric observation again. It is a well-known classical fact (Napoleon's theorem): If equilateral triangles are constructed externally on the sides of a given triangle then their centers are the vertices of an equilateral triangle.

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These five examples are not unique, and we believe that a certain new trend has come into being. This is posing geometry problems not simply hard but also natural and hence not isolated from the vast mathematical reality. With them, it is never clear a priori what additional consideration will be needed-be it a geometric observation, a combinatorial reasoning, a divisibility consideration, or a combination of all these. We are flattered to think that such a trend brings the olympiad movement closer to genuine mathematical work in the proper sense.

REFERENCES 1. Ganchev.I., (1971), For the mathematical problems, Sofia.(In Bulgarian) 2. Kenderov, (1990), Bulgarian Olympiads, Sofia.(In Bulgarian) 3. Journal "Mathematics plus", 1993-2001 issues, ISSN 0861-8321, Sofia.(In Bulgarian)