jagomart
digital resources
picture1_Geometry Pdf 168127 | V9 N1 Item Download 2023-01-25 10-47-12


 134x       Filetype PDF       File size 0.40 MB       Source: www.math.hkust.edu.hk


File: Geometry Pdf 168127 | V9 N1 Item Download 2023-01-25 10-47-12
volume 9 number 1 january 2004 april 2004 olympiad corner geometry via complex numbers the sixth hong kong china kin y li mathematical olympiad took place on december 20 2003 ...

icon picture PDF Filetype PDF | Posted on 25 Jan 2023 | 2 years ago
Partial capture of text on file.
                                                                                                                                   
            Volume 9, Number 1                                                                                                                                                                         January 2004 – April 2004 
                
               Olympiad Corner                                                                            Geometry via Complex Numbers 
                
                The Sixth Hong Kong (China)                                                                                                                       Kin Y. Li 
                Mathematical Olympiad took place on 
                December 20, 2003.  Here are the                                                 Complex numbers are wonderful.  In this 
                problems.  Time allowed: 3 hours                                                                                                                                             Z +WW Z =W +W  
                                                                                                 article we will look at some applications                                                              1 2               1         2
                                                                                                                                                                               and       Z −W W Z = C −W W C  
               Problem 1.  Find the greatest real K such                                         of complex numbers to solving geometry                                        respectively.  1              2                     1 2
               that for every positive u, v and w with u2 >                                      problems.  If a problem involves points 
               4vw, the inequality                                                               and chords on a circle, often we can                                          By moving W  toward W  along the unit 
                                                                                                                                                                                                       2                  1
                         2            2              2                 2                         without loss of generality assume it is the 
                     (u  - 4vw)  > K(2v  - uw)(2w  - uv)                                                                                                                       circle, in the limit, we will get the 
               holds.  Justify your claim.                                                       unit circle.  In the following discussion,                                    equation of the tangent line at W  to the 
                                                                                                 we will use the same letter for a point to                                                                                             1
                                                                                                 denote the same complex number in the                                         unit circle.  It is Z +W 2Z = 2W . 
               Problem 2.  Let ABCDEF be a regular                                                                                                                                                                 1             1
               hexagon of side length 1, and O be the                                            complex plane.  To begin, we will study                                       Similarly, the equation of the line 
               center of the hexagon. In addition to the                                         the equation of lines through points.                                         through C perpendicular to this tangent  
               sides of the hexagon, line segments are                                           Suppose Z is an arbitrary point on the 
               drawn from O to each vertex, making a                                             line through W  and W .  Since the vector                                     line is Z −W 2Z = C −W 2C . 
                                                                                                                          1            2                                                           1                  1
               total of twelve unit line segments. Find                                          from W  to Z is a multiple of the vector 
                                                                                                              1                                                                For a given triangle A A A  with the unit 
               the number of paths of length 2003 along                                          from W  to W , so in terms of complex                                                                              1 2 3
                                                                                                              1           2                                                    circle as its circumcircle, in terms of 
               these line segments that start at O and                                           numbers, we get Z − W  = t(W  −W ) for 
                                                                                                                                         1            2       1                complex numbers, its circumcenter is the 
               terminate at O.                                                                   some real t. Now t  =  t  and so                                              origin O, its centroid is G = (A  + A  + 
                                                                                                                      Z −W             Z −W                                                                                           1         2
                                                                                                                               1                1                              A )/3, its orthocenter is H = A  + A  + A  
               Problem 3.  Let ABCD be a cyclic                                                                     W −W =W −W                                                   3                                                1        2        3
               quadrilateral.  K,  L,  M,  N are the                                                                    2       1        2       1                             (because OH = 3OG) and the center of its 
                                                                                                                                                                               nine point circle is N = (A  + A  + A )/2 
               midpoints of sides AB, BC, CD and DA                                              Reversing the steps, we can see that                                                                                        1        2        3
               respectively.  Prove that the orthocentres                                        every          Z satisfying the equation  (because N is the midpoint of OH). 
               of triangles AKN, BKL, CLM, DMN are                                               corresponds to a point on the line  Let us proceed to some examples. 
                                                                                                 through  W  and W .  So this is the                                            
               the vertices of a parallelogram.                                                                     1                2
                                                    (continued on page 4)                        equation of a line through two points in                                      Example 1.  (2000 St. Petersburg City 
                                                                                                 the complex variable Z.                                                       Math Olympiad, Problem Corner 188) 
                                                                                                                                                                               The line S is tangent to the circumcircle 
                                                                                                 Next consider the line passing through a 
             Editors:  ஻ Ի ஶ (CHEUNG Pak-Hong), Munsang College, HK                              point  C and perpendicular to the line                                        of acute triangle ABC at B.  Let K be the 
                        ଽ υ ࣻ (KO Tsz-Mei)                                                       through W  and W .  Let Z be on this line.                                    projection of the orthocenter of triangle 
                        గ ႀ ᄸ (LEUNG Tat-Wing)                                                                     1            2
                        ؃ ୊ ፱ (LI Kin-Yin), Dept. of Math., HKUST                                Then the vector from C to Z is  ABC onto line S (i.e. K is the foot of 
                                                                                                 perpendicular to the vector from W  to                                        perpendicular from the orthocenter of 
                        ֔ ᜢ ݰ (NG Keng-Po Roger), ITC, HKPU                                                                                                      1
                                                                                                 W.  In terms of complex numbers, we get 
             Artist:    ྆ ؾ ़ (YEUNG Sau-Ying Camille), MFA, CU                                     2                                                                          triangle ABC to S).  Let L be the midpoint 
                                                                                                 Z − C = it(W  − W ) for some real t.  So  
             Acknowledgment:  Thanks to Elina Chiu, Math. Dept., HKUST                                                2         1                                              of side AC.  Show that triangle BKL is 
             for general assistance and to Lee Man Fui and Poon Ming Fung                                         Z −C            =       Z −C             .                   isosceles. 
              for typesetting.                                                                               i(W −W )                 i(W −W )                                  
             On-line:  http://www.math.ust.hk/mathematical_excalibur/                                              2         1              2         1                        Solution.  (Due to POON Ming Fung, 
             
             The editors welcome contributions from all teachers and                             Again reversing steps, we can conclude this                                   STFA Leung Kau Kui College, Form 6) 
             students.  With your submission, please include your name,                          is the equation of the line through C                                         Without loss of generality, let the 
             address, school, email, telephone and fax numbers (if available).  
             Electronic submissions, especially in MS Word, are encouraged.                      perpendicular to the line through W  and W . 
             The deadline for receiving material for the next issue is May 25,                                                                          1           2          circumcircle of triangle ABC be the unit 
             2004.                                                                                                                                                             circle on the plane.  Let A = a + bi, B = 
                                                                                                 In case the points W  and W  are on the 
             For individual subscription for the next five issues for the 03-04                                                      1             2                           −i, C = c + di.  Then the orthocenter is H 
             academic year, send us five stamped self-addressed envelopes.                       unit circle, we have                                                 .  
                                                                                                                                         WW =1=WW
             Send all correspondence to:                                                                                                    1 1               2    2           = A + B + C and K = (a + c) − i, L = (a + 
                                           Dr. Kin-Yin LI                                        Multiplying the numerators and c)/2 + (b + d)i/2.  Since 
                                   Department of Mathematics                                     denominators of the right sides of the                                                    1
                   The Hong Kong University of Science and Technology                            two displayed equations above by W W ,                                          LB=             (a + c)2 + (b + d + 2)2 = KL,  
                            Clear Water Bay, Kowloon, Hong Kong                                                                                                 1    2                     2
                                       Fax: (852) 2358 1643                                      we can simplify them to                                                       triangle BKL is isosceles. 
                                      Email: makyli@ust.hk 
       Mathematical Excalibur, Vol. 9, No. 1, Jan 04- Apr 04                                                                                Page 2
        
       Example 2.  Consider triangle ABC and           Example 4.  Let A ,  A ,  A  be the              Now the right side is 
                                                                               1   2    3
                                                       midpoints of W W ,  W W ,  W W  
       its circumcircle S.  Reflect the circle with                        2  3      3  1     1  2            Z  - Z  - Z Z W  +  Z Z W
                                                       respectively.  From A drop a                             1    2   1 3         2 3   . 
       respect to AB, AC and BC to get three new                                     i                        Z  - Z  - Z Z W  +  Z Z W
       circles S  ,  S  and S  (with the same          perpendicular to the tangent line to the                 1    3   1 2         2 3
                AB   AC       BC                       circumcircle of triangle W W W  at W. 
       radius as S).  Show that these three new                                     1  2  3      i      Multiplying the numerator and 
       circles intersect at a common point.            Prove that these perpendicular lines are         denominator by Z Z Z W  and  using 
                                                       concurrent.  Identify this point of                                  1 2 3
       Identify this point.                            concurrency.                                     ZiZi =  1=  WW , we get 
                                                        
       Solution.  Without loss of generality, we                                                              Z Z W −Z Z W −Z +Z
                                                       Solution.  Without loss of generality, let               2 3       1 3       2    1 . 
       may assume S is the unit circle.  Let the       the circumcircle of triangle W W W  be                 Z Z W −Z Z W −Z +Z
       center of S   be O´, then O´ is the mirror                                       1  2  3                 2 3       1 2       3    1
                  AB                                   the unit circle.  The line perpendicular to      This equals the left side (P − R)/(Q − R) 
       image of O with respect to the segment          the tangent at W  through A  = (W  + W )/2 
                                                                       1           1     2    3         and we complete the checking. 
       AB. So O´ = A + B (because segments             has equation                                      
       OO´ and AB bisect each other).  Similarly,                                                       Example 6. (2003 IMO, Problem 4) Let 
                                                                  2    W +W        2 W +W
       the centers of S   and S    are A + C and B          Z −W Z = 2       3 −W     2    3 .          ABCD be a cyclic quadrilateral.  Let P, Q 
                      AC       BC                                1        2       1     2               and R be the feet of the perpendiculars 
       + C respectively.  We need to show there                                                         from  D to the lines BC,  CA and AB 
       is a point Z such that Z is on all three new    Using WW =1, we may see that the right 
                                                                1 1                                     respectively.  Show that PQ = QR if and 
       circles, i.e.                                   side is the same as                              only if the bisectors of ∠ABC  and 
              |Z − (A + B)| = |Z − (A + C)|                  W +W +W           W +W +W                  ∠ADC meet on AC. 
                                                               1   2    3 −W2 1      2    3 .            
                  = |Z − (B + C)| = 1.                             2         1       2                  Solution.  (Due to SIU Tsz Hang, 2003 
       We easily see that the orthocenter of           From this we see that N = (W  + W  +             Hong Kong IMO team member)  Without 
       triangle ABC, namely Z = H = A + B + C,                                          1      2
                                                       W)/2 satisfies the equation of the line and      loss of generality, assume A, B, C, D lies 
       satisfies these equations.  Therefore, the        3
                                                       so N is on the line.  Since the expression       on the unit circle and the perpendicular 
       three new circles intersect at the              for N is symmetric with respect to W , W , 
                                                                                            1    2      bisector of AC is the real axis.  Let A = 
       orthocenter of triangle ABC.                    W, we can conclude that N will also lie on 
                                                         3                                              cosθ + isin θ, then C = A = cosθ −isinθ  
       Example 3.  A point A is taken inside a         the other two lines.  Therefore, the lines       so that AC = 1 and A + C = 2cosθ.  Since 
       circle.  For every chord of the circle          concur at N, the center of the nine point        the bisectors of ∠ABC  and ∠ADC  pass 
                                                       circle of triangle W W W . 
       passing through A, consider the                                     1  2 3                       through the midpoints of the major and 
       intersection point of the two tangents at       Example 5.  (Simson Line Theorem)  Let           minor arc AC, we may assume the 
                                                       W be on the circumcircle of triangle                           ∠ABC and ∠ADC pass 
       the endpoints of the chord.  Find the locus     Z Z Z  and P,  Q,  R be the feet of the          bisectors of 
       of these intersection points.                    1 2 3                                           through 1 and −1 respectively.  Let AC 
                                                       perpendiculars from W to Z Z , Z Z , Z Z  
                                                                                   3 1   1 2   2 3      intersect the bisector of ∠ABC at Z, then 
       Solution.  Without loss of generality we        respectively.  Prove that P, Q, R are            Z satisfies Z + ACZ = A+C , (which is 
       may assume the circle is the unit circle        collinear.  (This line is called the Simson      Z + Z = 2cosθ ), and                     . 
                                                       line of triangle Z Z Z  from W.)                                           Z + BZ = B+1
       and A is on the real axis.  Let WX be a                          1 2 3                           Solving for Z, we get 
       chord passing through A with W and X on         Solution.  Without loss of generality, we                       2Bcosθ −B−1
       the circle.  The intersection point Z of the    may assume the circumcircle of triangle                    Z  =                 . 
       tangents at W and X satisfies  Z Z Z  is the unit circle.                                                            B − 1
              2                       2                 1 2 3
        Z +W Z =2W  and  Z + X Z =2X .                 Then  Z = Z = Z = W =1.  Now P is                Similarly, the intersection point Z′ of AC 
       Solving these equations together for Z, we              1     2     3                            with the bisector of ∠ADC is 
                                                       on the line Z Z  and the line through W 
       find               .                                          3 1
            Z = 2/(W + X)                              perpendicular to Z Z .  So P satisfies the                     2Dcosθ +D−1
                                                                          3 1                                     Z'=                   . 
       Since  A is on the chord WX, the real           equations  P+ Z Z P = Z +Z  and  P –                                 D+1
                                                                        1 3      1     3
       number A satisfies the equation for line        Z Z P =W −Z Z W .  Solving these 
                                                         1 3           1 3                              Next, R is on the line AB and the line 
       WX, i.e. A + WXA = W + X.  Using                together for P, we get                           through  D perpendicular to AB.  So 
       WW = 1 = XX , we see that                                   Z  + Z  +W - Z Z W                   R+ABR=A+B andR−ABR=D−ABD .  
                                                              P  =  1      3        1 3    .            Solving for R, we find 
           ReZ =    1   +   1    =WX+1= 1 .                                   2
                 W +X W+X W+X A                        Similarly,                                                R  = A+ B+ D− ABD . 
       So the locus lies on the vertical line                      Z  + Z  +W - Z Z W                                         2
       through 1/A.                                           Q  =   1     2        1 2                 Similarly, 
       Conversely, for any point Z on this line,                              2
       draw the two tangents from Z to the unit        and                                                       P  = B +C + D−BCD  
       circle and let them touch the unit circle at                Z  + Z  +W - Z Z W                                          2
       the point W and X.  Then the above                     R  =  2      3        2 3    .            and 
       equations are satisfied by reversing the                               2
       argument.  In particular, A + WXA = W +         To show P, Q, R are collinear, it suffices to             Q  = C + A+ D−CAD . 
       X and so A is on the chord WX.  Therefore,      check that                                                              2
       the locus is the line perpendicular to OA                      P-R = P-R .                                         (continued on page 4) 
       at a distance 1/OA from O.                                     Q-R    Q-R
       Mathematical Excalibur, Vol. 9, No. 1, Jan 04- Apr 04                                                                                 Page 3
        
       Problem Corner                                   Prove that there was a moment when the           Problem 192.  Inside a triangle ABC, 
                                                        difference between the distances he had          there is a point P satisfies ∠PAB = ∠
       We welcome readers to submit their               covered moving to the east and moving to         PBC = ∠PCA = φ.  If the angles of the 
       solutions to the problems posed below            the west was at least half of the length of      triangle are denoted by α,  β and γ, 
       for publication consideration.  The              the equator.                                     prove that  
       solutions should be preceded by the                                                                    1        1         1         1
       solver’s name, home (or email) address                     *****************                        sin2ϕ = sin2α + sin2 β + sin2γ . 
       and school affiliation.  Please send                            Solutions 
       submissions to Dr. Kin Y. Li,                               ****************                      Solution.  LEE Tsun Man Clement (St. 
       Department of Mathematics, The Hong                                                               Paul’s College), POON Ming Fung (STFA 
       Kong University of Science &                     Due to an editorial mistake in the last          Leung Kau Kui College, Form 6), SIU Ho 
       Technology, Clear Water Bay, Kowloon,            issue, solutions to problems 186, 187, 188       Chung (Queen’s College, Form 5) and 
                                                        by POON Ming Fung (STFA Leung Kau                Yufei ZHAO (Don Mills Collegiate 
       Hong Kong.  The deadline for                                                                      Institute, Tornoto, Canada, Grade 10). 
                               May 25, 2004.            Kui College, Form 6) were overlooked              
       submitting solutions is                          and his name was not listed among the            Let AP meet BC at X.  Since ∠XBP = ∠
                                                        solvers.  We express our apology to him          BAX and ∠BXP = ∠AXB, triangles 
       Problem 196.  (Due to John                       and point out that his clever solution to        XPB and XBA are similar.  Then XB/XP = 
       PANAGEAS, High School “Kaisari”,                 problem 188 is printed in example 1 of the       XA/XB.  Using the sine law and the last 
       Athens, Greece)  Let x ,x ,...,x  be 
                                 1  2      n                    Geometry via Complex Numbers” 
       positive real numbers with sum equal             article “                                        equation, we have 
       to 1.  Prove that for every positive             in this issue.                                          sin2ϕ     sin2 ∠XAB     XB2
                                                                                                                       =              =       
       integer m,                                       Problem 191.  Solve the equation                        sin2 β    sin2 ∠XBA     XA2
             n≤nm(xm +xm +...+ xm).                                 3                                                     XP⋅XA XP
                       1      2          n                         x −3x = x+2.                                          =        =     
                                                                                                                            XA2     XA
                                                        Solution.  Helder Oliveira de CASTRO 
       Problem 197.  In a rectangular box, the          (ITA-Aeronautic Institue of Technology,          Using [ ] to denote area, we have 
       length of the three edges starting at the        Sao Paulo, Brazil) and Yufei ZHAO (Don                  XP [XBP] [XCP] [BPC] 
       same vertex are prime numbers.  It is            Mills Collegeiate Institute, Toronoto,                  XA = [XBA] = [XCA] = [ABC]
                                                        Canada, Grade 10). 
       also given that the surface area of the           
       box is a power of a prime.  Prove that           If x < -2, then the right side of the equation   Combining the last two equations, we 
       exactly one of the edge lengths is a             is not defined. If x > 2, then                   have  sin2ϕ/sin2 β =[BPC]/[ABC].    By 
                                    k                                    3                               similar arguments, we have 
       prime number of the form 2  - 1.                        3        x +3x(x+2)(x−2)
                                                              x −3x=              4                               sin2ϕ     sin2ϕ    sin2ϕ
       Problem 198.  In a triangle ABC, AC =                                                                              +        +
       BC. Given is a point P on side AB such                           x3                                        sin2α     sin2φ    sin2γ
       that ∠ACP = 30∘.  In addition, point                           > 4 > x+2.                                  [APB]     [BPC]     [CPA]
       Q outside the triangle satisfies ∠CPQ                                                                    =[ABC]+[ABC]+[ABC] 
       =  ∠CPA + ∠APQ = 78∘.  Given that                So the solution(s), if any, must be in [-2, 2].           [ABC]
       all angles of triangles ABC and QPB,             Write x = 2 cos a, where 0 ≤ a ≤π .  The                =         =1
       measured in degrees, are integers,               equation becomes                                          [ABC]
       determine the angles of these two                      8cos3 a −6cosa =    2cosa+2.               The result follows. 
       triangles.                                       Using the triple angle formula on the left       Other commended solvers: CHENG Tsz 
                                                        side and the half angle formula on the right     Chung (La Salle College, Form 5), LEE 
                                 +
       Problem 199.  Let  R   denote the                side, we get                                     Man Fui (CUHK, Year 1) and Achilleas 
       positive real numbers.  Suppose                                            a                      P. PORFYRIADIS (American College 
        f : R+ → R+ is a strictly decreasing                      2cos3a = 2cos    (≥ 0).                of Thessaloniki “Anatolia”, Thessaloniki, 
       function such that for all  x, y∈R+ ,                                      2                      Greece). 
                  f (x + y) + f (f (x) + f (y))         Then 3a ± (a/2) = 2nπ for some integer n.        Comments: Professor Murray 
            = f (f (x + f (y)) + f (y + f (x))).        Since 3a ± (a/2) ∈ [-π/2, 7π /2], we get         KLAMKIN (University of Alberta, 
       Prove that f (f (x)) = x for every x > 0.                                                         Edmonton, Canada) informed us that the 
                                                        n = 0 or 1.  We easily checked that a = 0,                 2       2       2        2 
       (Source: 1997 Iranian Math                                                                        result csc φ = csc α + csc β + csc γ in the 
                                                        4π /5, 4π/7 yield the only solutions x = 2,      problem is a known relation for the 
       Olympiad)                                        2cos(4π/5), 2cos(4π/7).                          Brocard angle φ  of a triangle.  Also 
                                                                                                         known is cot φ = cot α + cot β + cot γ.  He 
       Problem 200.  Aladdin walked all over            Other commended solvers: CHUNG Ho                mentioned these relations and others are 
       the equator in such a way that each              Yin (STFA Leung Kau Kui College, Form 
       moment he either was moving to the               7),  LEE Man Fui (CUHK, Year 1),                 given in R.A. Johnson, Advanced 
                                                        LING Shu Dung, POON Ming Fung                    Euclidean Geometry, Dover, N.Y., 1960, 
       west or was moving to the east or                (STFA Leung Kau Kui College, Form 6),            pp. 266-267.  (For the convenience of 
       applied some magic trick to get to the           SINN Ming Chun (STFA Leung Kau Kui               interested readers, the Chinese translation 
       opposite point of the Earth.  We know            College, Form 4), SIU Ho Chung 
       that he travelled a total distance less          (Queen’s College, Form 5), TONG Yiu              of this book can be found in many 
                                                        Wai (Queen Elizabeth School), YAU Chi            bookstore.–Ed) LEE Man Fui and 
       than half of the length of the equator           Keung (CNC Memorial College, Form 7)             Achilleas PORFYRIADIS gave a proof 
       altogether during his westward moves.            and  YIM Wing Yin (South Tuen Mun                of the cotangent relation and use it to 
                                                        Government Secondary School, Form 4). 
        Mathematical Excalibur, Vol. 9, No. 1, Jan 04- Apr 04                                                                                                   Page 4
         
        derive the cosecant relation, which is the             (Kiangsu-Chekiang College Shatin, Teacher),            There are  n
                                                                                                                                    C  pigeonholes.  For each 
        equation in the problem, by trigonometric              Helder Oliveira de CASTRO                                              3
        manipulations.                                         (ITA-Aeronautic Institute of Technology, Sao           segment joining a pair of endpoints, that 
                                                               Paulo, Brazil), LEE Tsun Man Clement (St.              segment will be in n − 2 pigeonholes.  So 
        Problem 193.  Is there any perfect square,             Paul’s College), LING Shu Dung, POON                   if  x(n − 2) ≥ 2Cn +1, that is 
        which has the same number of positive                  Ming Fung (STFA Leung Kau Kui College,                                    3
                                                               Form 6), SIU Ho Chung (Queen’s College,                             n
        divisors of the form 3k + 1 as of the form             Form 5), YEUNG Wai Kit (STFA Leung Kau                           2C +1 n(n−1)(n−2)+3
                                                               Kui College), Yufei ZHAO (Don Mills                         x ≥     3      =                       , 
        3k + 2?  Give a proof of your answer.                  Collegiate Institute, Toronto, Canada, Grade 10)                  n−2              3(n−2)
        Solution 1.  K.C. CHOW (Kiangsu-Chekiang               and the proposer.                                      then by the pigeonhole principle, there is 
        College Shatin, Teacher), LEE Tsun Man                 If S, N, T are collinear, then triangles SO N          at least one triangle having these line 
        Clement (St. Paul’s College), SIU Ho Chung                                                           1
        (Queen’s College, Form 5) and Yufei ZHAO               and  SOT are isosceles and share the                   segments as edges. 
        (Don Mills Collegiate Institute, Toronto,              common angle OST, which imply they are                 If f (n) = (n(n − 1)(n − 2) + 3) / (3(n − 2)) 
        Canada, Grade 10). 
                                                                                 ∠SON = ∠SOT and so 
                                                               similar.  Thus           1                             is an integer, then 3(n − 2) f (n) = n(n − 
                                         2            a
        No.  For a perfect square m , let m = 3 b              lines O N and OT are parallel.  Similarly, 
                                               2    2a 2               1                                              1)(n − 2) + 3 implies 3 is divisible by n − 
        with b not divisible by 3.  Then m  = 3 b .            lines  O2N and OS are parallel.  Hence,                2.  Since n > 3, we must have n = 5.  Then 
        Observe that divisors of the form 3k + 1               OONO is a parallelogram and OO  = 
                          2            2                           1    2                                   2         f (5) = 7.  For the five vertices A, B, C, D, 
        or 3k + 2 for m  and for b  consist of the             ON = O S, OO  = O N = O T.  Therefore, 
        same numbers because they cannot have                    1       1       1     2       2                      E of a regular pentagon, if we connected 
                                                               SO /OO  = OO /TO .  Conversely, if 
                                        2                         1     1          2     2                            the six segments BC, CD, DE, EA, AC, 
        any factor of 3.  Since b  has an odd                  SO /OO  = OO /TO , then using OO  = OS 
        number of divisors and they can only be                   1     1       2     2                  1            BE, then there is no triangle.  So a 
                                                               − O S and OO  = OT − O T, we get 
        of the form 3k + 1 or 3k + 2, so the                       1            2            2                        minimum of f (5) = 7 segments is needed 
                                                                              OS        OT−OT                         to get a triangle. 
        number of divisors of the form 3k + 1                                   1     =         2  ,                   
                                                                            OS−OS          O T
        cannot be the same as the number of                                        1         2                                                        K. C. CHOW 
                                                                                                                      Other commended solvers: 
        divisors of the form 3k + 2.  Therefore,               which reduces to O S + O T = OS.  Then                 (Kiangsu-Chekiang College Shatin, 
                                                                                      1       2                       Teacher) and 
        the same is true for m2.                               OO = OS − OS = O T = O N and OO  =                                      POON Ming Fung (STFA 
                                                                   1            1       2       2            2        Leung Kau Kui College, Form 6). 
        Solution 2.  Helder Oliveira de CASTRO                 OT − O T = O S = O N.  Hence OO NO  is                  
        (ITA-Aeronautic Institute of Technology, Sao                    2      1       1                1    2                                                          
        Paulo, Brazil),                                        again a parallelogram. Then 
                        LEE Man Fui (CUHK, Year                                                                        
        1),  LING Shu Dung, POON Ming Fung                            ∠O NS +∠O NO +∠O NT
        (STFA Leung Kau Kui College, Form 6),                            1          1    2       2                    Olympiad Corner 
                                                                   =∠OSN+∠O NO +∠O TN
        Achilleas P. PORFYRIADIS (American                               1          1    2       2                                       (continued from page 1) 
        College of Thessaloniki “Anatolia”,                        = 1∠OO N +∠O NO +1∠OO N                             
        Thessaloniki, Greece), Alan T.W. WONG                        2       1         1    2    2       2
        (Markham, Ontario, Canada) and YIM Wing                    =180°.                                             Problem 4.  Find, with reasons, all 
        Yin (South Tuen Mun Government Secondary                                                                      integers a, b, and c such that 
        School, Form 4).                                       Therefore, S, N, T are collinear.                       1                                           3
                                                                                                                         (a + b) (b + c) (c + a) + (a + b + c)  
        No.  For a perfect square, its prime                   Other commended solver:  TONG Yiu                       2
        factorization is of the form  Wai (Queen Elizabeth School).                                                   = 1 – abc. 
          2e  2e   2e                                                                                                  
        2 13 25 3 ···.  Let x, y, z be the number              Problem 195.  (Due to Fei Zhenpeng,                                                                      
        of divisors of the form 3k, 3k + 1, 3k + 2             Yongfeng High School, Yancheng City,                    
        for this perfect square respectively.  Then            Jiangsu Province, China)  Given n (n > 3)              Geometry via Complex Numbers 
        x + y + z = (2e  + 1) (2e  + 1) (2e  + 1) ··· is       points on a plane, no three of them are                                       (continued from page 2) 
                        1          2         3                 collinear,  x pairs of these points are 
        odd.  Now divisor of the form 3k has at                connected by line segments.  Prove that if                                       
        least one factor 3, so x = (2e  + 1) (2e ) 
                                           1           2                                                              By Simson’s theorem, P,  Q,  R are 
        (2e  + 1) ··· is even.  Then y + z is odd.                           n(n −1)(n − 2) + 3
            3                                                          x ≥                             ,              collinear. So PQ = QR if and only if Q = 
        Therefore y cannot equal z.                                                 3(n − 2)                          (P + R)/2.  In terms of A, B, C, D, this may 
                                                               then there is at least one triangle having             be simplified to 
        Other commended solvers:  CHENG Tsz 
        Chung (La Salle College, Form 5) and YEUNG             these line segments as edges.  Find all                     C+A−2B=(2CA−AB−BC)D. 
        Wai Kit (STFA Leung Kau Kui College).                  possible values of integers n > 3 such that  
                                                                n(n − 1)( n − 2) + 3                                  In terms of B, D, θ, this is equivalent to 
        Problem 194.  (Due to Achilleas Pavlos                                             is an integer and          (2cosθ − 2B)D = 2 − 2Bcos θ.  This is 
        PORFYRIADIS, American College of                              3(n − 2)                                        easily checked to be the same as 
        Thessaloniki “Anatolia”, Thessaloniki,                 the minimum number of line segments 
        Greece)  A circle with center O is                     guaranteeing a triangle in the above                         2cosθ − B−1 = 2Dcosθ + D−1, 
        internally tangent to two circles inside it,           situation is this integer.                                        B−1                 D+1
        with centers O  and O , at points S and T               
                         1        2                            Solution.  SIU Ho Chung (Queen’s                       i.e.  Z = Z'. 
        respectively.  Suppose the two circles                 College, Form 5), Yufei ZHAO (Don 
        inside intersect at points M,  N with N                Mills Collegiate Institute, Toronto,  Comments:  The official solution by 
        closer to ST.  Show that S,  N,  T are                 Canada, Grade 10) and the proposer. 
                                                                                                                      pure geometry is shorter, but it takes a 
        collinear if and only if SO /OO  =                     For every three distinct points A,  B,  C, 
                                              1     1          form a pigeonhole containing the three                 fair amount of time and cleverness to 
        OO/TO . 
            2     2                                            segments  AB,  BC,  CA.  (Each segment                 discover. Using complex numbers as 
        Solution.    CHENG Tsz Chung (La Salle                 may be in more than one pigeonholes.)                  above reduces the problem to straight 
        College, Form 5), K. C. CHOW                                                                                  computations. 
The words contained in this file might help you see if this file matches what you are looking for:

...Volume number january april olympiad corner geometry via complex numbers the sixth hong kong china kin y li mathematical took place on december here are wonderful in this problems time allowed hours z ww w article we will look at some applications and c problem find greatest real k such of to solving respectively that for every positive u v with if a involves points vw inequality chords circle often can by moving toward along unit without loss generality assume it is uw uv limit get holds justify your claim following discussion equation tangent line use same letter point denote let abcdef be regular hexagon side length o plane begin study similarly center addition lines through perpendicular sides segments suppose an arbitrary drawn from each vertex making since vector total twelve multiple given triangle paths so terms as its circumcircle these start t circumcenter terminate now origin centroid g orthocenter h abcd cyclic quadrilateral l m n because oh og nine midpoints ab bc cd da re...

no reviews yet
Please Login to review.