Problem 1 A convex quadrilateral ABCD has perpendicular diagonals. The perpendicular bisectors of AB and CD meet at a unique point P inside ABCD. Prove that ABCD is cyclic if and only if triangles ABP and CDP have equal areas. Solution: Let AC and BD intersect at E. Suppose by symmetry that P is in ABE. Denote by M and N the respective feet of perpendiculars from P to AC and BD. Without assuming that PA  PB and PC  PD, we express the areas ABP and CDP as follows: 2ABP  2ABE  2PAE  2PBE  AM  PNBN  PM  AM  PNPM  BN  PMPN  AM  BN  PM  PN, 2CDP  2CDE  2PCE  2PDE  CM  PNDN  PM  CM  PNPM  DN  PMPN  CM  DN  PM  PN. Therefore 2ABP  CDP  AM  BN  CM  DN. We now assume that PA  PB and PC  PD. Suppose that ABCD is cyclic. Then the uniqueness of P implies that it must be the circumcenter. So M and N are the midpoints of AC and BD, respectively. Hence AM  CM and BN  DN,so(*) implies ABP  CDP. Conversely, suppose that ABP  CDP. Then, by (*), we have AM  BN  CM  DN.IfPA  PC, assume by symmetry that PA  PC. Then AM  CM and also BN  DN, because PB  PD. Thus AM  BN  CM  DN, a contradiction. Hence PA  PC, which implies that P is equidistant from A, B, C and D. We conclude then that ABCD is cyclic. Alternative Solution: Let AC and BD meet at E again. Assume by symmetry that P lies in BEC and denote ABE   and ACD  . The triangles ABP and CDP are isosceles. If M and N are the respective midpoints of their bases AB and CD, then PM  AB and PN  CD. Note that M, N and P are not collinear due to the uniqueness of P. Consider the median EM to the hypotenuse of the right triangle ABE.Wehave BEM  , AME  2 and EMP  90   2. Likewise, CEN  , DNE  2 and ENP  90   2. Hence MEN  90     , and a direct computation yields NPM  360   EMP  MEN  ENP  90       MEN. It turns out that, whenever AC  BD, the quadrilateral EMPN has a pair of equal opposite angles, the ones at E and P. We now prove our claim. Since AB  2EM and CD  2EN,wehaveABP   CDP  if and only if EM  PM  EN  PN,or,EM/EN  PN/PM. On account of MEN  NPM, the latter is equivalent to EMN  PNM. This holds if and only if EMN  PNM and ENM  PMN, and these in turn mean that EMPN is a parallelogram. But the opposite angles of EMPN at E and P are always equal, as noted above. So it is a parallelogram if and only if EMP  ENP;thatis,if 90   2  90   2. We thus obtain a condition equivalent to   ,ortoABCD being cyclic. Another Solution: Choose a coordinate system so that the axes lie on the perpendicular lines AC and BD, and so that the coordinates of A, B, C and D are 0,a, b,0, 0,c and d,0, respectively. By assumption, the perpendicular bisectors of AB and CD have a unique common point. Hence the linear system formed by their equations 2bx  2ay  b 2  a 2 and 2dx  2cy  d 2  c 2 has a unique solution, and it is given by x 0  cb 2  a 2   ad 2  c 2  2ad  bc , y 0  db 2  a 2   bd 2  c 2  2ad  bc . Naturally, x 0 ,y 0  are the coordinates of P. Since it is interior to ABCD, ABP and CDP have the same orientation. Then ABP  CDP if and only if ||  || This is equivalent to ax 0  by 0  ab  cx 0  dy 0  cd. Inserting the expressions for x 0 and y 0 , after the inevitable algebra work we obtain the equivalent condition ac  bda  c 2  b  d 2   0. Now, the choice of the coordinate system implies that a and c have different signs, as well as b and d. Hence the second factor is nonzero, so ABP  CDP if and only if ac  bd. The latter is equivalent to AE  CE  BE  DE, where E is the intersection point of the diagonals. However, it is a necessary and sufficient condition for ABCD to be cyclic. Problem 2 Let ABCD be a cyclic quadrilateral. Let E and F be variable points on the sides AB and CD, respectively, such that AE : EB  CF : FD.LetP be the point on the segment EF such that PE : PF  AB : CD. Prove that the ratio between the areas of triangles APD and BPC does not depend on the choice of E and F. Solution: We first assume that the lines AD and BC are not parallel and meet at S.SinceABCD is cyclic, ASB and CSD are similar. Then, since AE : EB  CF : FD, ASE and CSF are also similar, so that DSE  CSF. Moreover, we have SE SF  SA SC  AB CD  PE PF , which implies that ESP  FSP. Thus, ASP  BSP,andsoP is equidistant from the lines AD and BC. Therefore, APD : BPC  AD : BC, which is a constant. Next, assume that the lines AD and BC are parallel. Then ABCD is an isosceles trapezoid with AB  CD, and we have BE  DF.LetM and N be the midpoints of AB and CD, respectively. Then ME  NF and clearly, E and F are equidistant from the line MN. Thus P, the midpoint of EF, lies on MN. It follows that P is equidistant from AD and BC, and hence APD : BPC  AD : BC. Alternative Solution: Let AE : EB  CF : FD  a : b,wherea  b  1. Since PE : PF  AB : CD,we have d  P,AD   CD AB  CD  d  E,AD   AB AB  CD  d  F,AD  , d  P,BC   CD AB  CD  d  E,BC   AB AB  CD  d  F,BC  , where dX,YZ stands for the distance from the point X to the line YZ. Thus, we obtain APD  CD AB  CD AED  AB AB  CD AFD  a  CD AB  CD ABD  b  AB AB  CD ACD, BPC  CD AB  CD BEC  AB AB  CD BFC  b  CD AB  CD BAC  a  AB AB  CD BDC. Next, since A, B, C and D areconcyclic,wehavesinBAD  sinBCD and sinABC  sinADC. Thus, APD BPC  a  CD ABD  b  AB ACD b  CD BAC  a  AB BDC  a  CD  AB  AD  sinBAD  b  AB  CD  AD  sinADC b  CD  AB  BC  sinABC  a  AB  CD  BC  sinBCD  AD BC  a  sinBAD  b  sinADC b  sinABC  a  sinBCD  AD BC . Problem 3 Let I be the incenter of triangle ABC.LetK,L and M be the points of tangency of the incircle of ABC with AB,BC and CA, respectively. The line t passes through B and is parallel to KL. The lines MK and ML intersect t at the points R and S. Prove that RIS is acute. Solution: Since the lines KL and RS are parallel, we have, in BKR , BKR  90° A/2, KBR  90° B/2, BRK  90° C/2. Hence, by the law of sine, BR  cosA/2 cosC/2  BK. # Similarly, we have, in BLS, BLS  90° C/2, BSL  90° A/2, LBS  90° B/2. so that BS  cos  C/2  cos  A/2   BL  cos  C/2  cos  A/2   BK. # Notice now that BI  RS and IK  AB. Then, on account of (1) and (2), we obtain IR 2  IS 2  RS 2   BI 2  BR 2    BI 2  BS 2    BR  BS  2  2  BI 2  BR  BS   2  BI 2  BK 2   2IK 2  0. So,bythelawofcosine,RIS is acute. Alternative Solution: Let W be the midpoint of KL and Q the midpoint of KM. Then Q  AI, W  BI, AIKM and BIKL. We first prove that AWRI and CWSI. Since RBI  RQI  90  , the points R, B, I, Q are concyclic, which implies BRI  BQI. Also, in the right triangles AIK and BIK,wehave IQ  IA  KI 2  IW  IB, so that IQ/IW  IB/IA. Hence AIW and BIQ are similar. It follows that BRI  BQI  AWI.SinceBR  IW, this implies that RIAW.Byasimilar argument, we can prove that SICW. Thus, RIS  180  AWC. It remains to prove that AWC is obtuse. Let T be the midpoint of AC.Then2 WT  WC  WA  LC  KA.SinceLC and KA are not collinear, we have WT  LC  KA 2  CM  AM 2  AC 2 . This implies that W is inside the circle with diameter AC,andsoAWC  90°. Therefore, RIS is acute. Comment: Another proof for the fact AW  RI is as follows. Since RBI  RQI  90°, RBIQ is cyclic and its circumcircle is orthogonal to the diameter RI. Consider the inversion  with respect to the incircle of ABC.Since takes B and Q into W and A, respectively, it takes the circumcircle of RBIQ into the line AW.Since leaves the line RI invariant, we have AW  RI. Problem 4 Let M and N be points inside triangle ABC such that MAB  NAC MBA  NBC. Prove that AM  AN AB  AC  BM  BN BA  BC  CM  CN CA  CB  1. Solution: Let K be the point on the ray BN such that BCK  BMA. Note that K is outside ABC, because BMA  ACB.SinceMBA  CBK,wehaveABM  KBC;so, AB BK  BM BC  AM CK . # Then, since ABK  MBC,andAB/KB  BM/BC, we see that ABK  MBC. Hence AB BM  BK BC  AK CM . # Now we have CKN  MAB  NAC. Consequently, the points A, N, C and K are concyclic. By Ptolemy’s theorem, AC  NK  AN  CK  CN  AK,or AC  BK  BN   AN  CK  CN  AK. # From (1) and (2), we find CK  AM  BC/BM, AK  AB  CM/BM and BK  AB  BC/BM. Inserting these expressions in (3), we obtain AC  AB  BC BM  BN  AN  AM  BC BM  CN  AB  CM BM , or AM  AN AB  AC  BM  BN BA  BC  CM  CN CA  CB  1. Alternative Solution: Let the complex coordinates of A, B, C, M and N be a, b, c, m and n, respectively. Since the lines AM, BM and CM are concurrent, as well as the lines AN, BN and CN, it follows from Ceva’s theorem that sinBAM sinMAC  sinCBM sinMBA  sinACM sinMCB  1. # sinBAN sinNAC  sinCBN sinNBA  sinACN sinNCB  1. # By hypotheses, BAM  NAC and MBA  CBN. Hence BAN  MAC and NBA  CBM. Combined with (1) and (2), these equalities imply sinACM  sinACN  sinMCB  sinNCB. Thus, cos  NCM  2ACM   cosNCM  cos  NCM  2NCB   cosNCM. and hence ACM  NCB. Since BAM  NAC, MBA  CBN and ACN  MCB, the following complex ratios are all positive real numbers: m  a b  a : c  a n  a , m  b a  b : c  b n  b and m  c b  c : a  c n  c . Hence each of these equals its absolute value, and so AM  AN AB  AC  BM  BN BA  BC  CM  CN CA  CB   m  a  n  a   b  a  c  a    m  b  n  b   a  b  c  b    m  c  n  c   b  c  a  c   1. Comment: Contestants who are familiar with the notion of isogonal conjugate points may skip over the early part of the proof. Problem 5 Let ABC be a triangle, H its orthocenter, O its circumcenter, and R its circumradius. Let D be the reflection of A across BC, E be that of B across CA,andF that of C across AB. Prove that D,E and F are collinear if and only if OH  2R. Solution: Let G be the centroid of ABC,andA  , B  and C  be the midpoints of BC, CA and AB, respectively. Let A  B  C  be the triangle for which A, B and C are the midpoints of B  C  , C  A  and A  B  , respectively. Then G is the centroid and H the circumcenter of A  B  C  . Let D  , E  and F  denote the projections of O on the lines B  C  , C  A  and A  B  , respectively. Consider the homothety h with center G and ratio 1/2. It maps A, B, C, A  , B  and C  into A  , B  , C  , A , B and C, respectively. Note that A  D   BC, which implies AD : A  D   2:1 GA : GA  and DAG  D  A  G. We conclude that hD  D  . Similarly, h  E   E  and h  F   F  . Thus, D, E and F are collinear if and only if D  , E  and F  are collinear. Now D  , E  and F  are the projections of O on the sides B  C  , C  A  and A  B  , respectively. By Simson’s theorem, they are collinear if and only if O lies on the circumcircle of A  B  C  . Since the circumradius of A  B  C  is 2R, O lies on its circumcircle if and only if OH  2R. Alternative Solution: Let the complex coordinates of A, B, C, H and O be a, b, c, h and 0, respectively. Consequently, aa   bb   cc   R 2 and h  a  b  c.SinceD is symmetric to A with respect to line BC, the complex coordinates d and a satisfy d  b c  b  a  b c  b ,or  b   c   d   b  c  a    bc   b  c   0. # Since b   c    R 2  b  c  bc and bc   b  c  R 2  b 2  c 2  bc , by inserting these expressions in (1), we obtain d  bc  ca  ab a  k  2bc a . d   R 2  a  b  c  bc  R 2  h  2a  bc . where k  bc  ca  ab. Similarly, we have e  k  2ca b , e   R 2  h  2b  ca , f  k  2ab c andf   R 2  h  2c  ab . Since  dd  1 ee  1 ff  1  e  de   d  f  df   d    ba  k2ab  ab R 2  ab  h2c  abc  ca  k2ca  ca R 2  ac  h2b  abc  R 2  c  a  a  b  a 2 b 2 c 2    ck  2abc  h  2c   bk  2abc    h  2b   R 2  b  c  c  a  a  b  hk  4abc  a 2 b 2 c 2 and h   R 2 k/abc, it follows that D,EandFarecollinear 0  hk  4abc  0  hh   4R 2  OH  2R. [...]... ggn  n), and so takes distinct primes to distinct primes Hence a lower bound for g 1998  g2  3 3  37  g2g3 3 g37 is obtained when g2, g3, g37 are the three smallest primes 2, 3, 5, with g3  2 Thus, g 1998  3  2 3  5  120 for each g  S There is, however, a function g  S with g 1998  120; this proves that the minimum in question is 120 Set g1  1, and define g...  10 and 6m  10  1mod3 is already in the list Finally, we have a 4m8  9m  16, becsase no y  2mod3 is in the list The formulae for a n has been established by induction From 1998  4  499  2, we find that a 1998  9  499  3  4494 Comment: It is easy to see that x  y  3z has no solution in the set of positive integers  1 Multiply the set by 3 still has the same property Thus one takes... arbitrary n  p  1 p  2 p  k  N by setting 1 2 k gn  gp  1 p  2 p  k   gp 1   1 gp 2   2 gp k   k The conditions in (2) are satisfied 1 2 k (with a  1), so g  S Clearly, g 1998  120, which completes the proof Comment: 1 Actually, each function f considered above is of the form fn  agn, where g is a multiplicative involution of N; that is, a function satisfying (2)... is defined as follows: a 1  1 and for n  1, a n1 is the smallest integer greater than a n such that a i  a j  3a k for any i, j and k in 1, 2, 3,  , n  1, not necessarily distinct Determine a 1998 Solution: We investigate the first few values for a n , writing ‘not m’ to mean that the next value for a n cannot be m We have a 1  1, but not 2, because 1  2  3  1 We next find that a 2  3,... generating polynomials f n x does not work in itself To prove the mentioned algebraic relation presents no less difficulty than the original problem Problem 13 Determine the least possible value of f 1998 , where f is a function from the set N of positive integers into itself such that for all m, n  N, fn 2 fm  mfn 2 Solution: Denote by S the set of functions considered Let f be any of them,... consisting of positive integers  1, 3, 4, 7 Ordering this set as an increasing sequence a n The subtle point is to show that this sequence indeed is what we want, then it is easy to get the value of a 1998 from the formula for a n Problem 18 Determine all positive integers n for which there exists an integer m such that 2 n  1 is a divisor of m 2  9 Solution: First we show that if 2 n  1 divides . 1. Comment: Contestants who are familiar with the notion of isogonal conjugate points may skip over the early part of the proof. Problem 5 Let ABC be a triangle, H its orthocenter, O its circumcenter,. expressions for x 0 and y 0 , after the inevitable algebra work we obtain the equivalent condition ac  bda  c 2  b  d 2   0. Now, the choice of the coordinate system implies that a and c. being cyclic. Another Solution: Choose a coordinate system so that the axes lie on the perpendicular lines AC and BD, and so that the coordinates of A, B, C and D are 0,a, b,0, 0,c and