Mathematical Excalibur, Volume 5, Number 3

Abstract

[evdate 2000-09-15] Mathematical Excalibur Volume 5, Number 3 May 2000-September 2000 Problem 1. Compute the sum computations will be similar for the right half.) Solution. (Due to Mok Tze Tao, I995 Hong Kong Team Member) Set the origin at Z and the x-axis on line AD. Let the coordinates of the circumcenters of trianglesAA4C and BND be (x1, 0) and (x2, Problem 2. Given the following triangular arrangement of circles: Each of the numbers 1,2,...,9 is to be P be a point on the line X{????}different from

written into one of these circles, so that ililols: 3R 6 & (CHEUNG Pak-Hong), Munsang College, HK #$ + & (KOTsz-Mei) % 3 6% (LEUNG Tat-Wing), Appt. Math Cept, HKPU $ & R (U Kin-Yin), Math Dept. HKUST $: #& & (NG Keng-Po Roger), ITC, HKPU H % X (YEUNG Sau-Ying Camille). MFA, CU he editors welcome contributions from all teachers and :uden& With your submission. please include your name, ddress, school, email, telephone and fax numbers (ii vailable). Electronic submissions, especially in MS Worcl IC encouraged. The deadline for receiving material for the ext issue is October 10,200O. Volume 5, Number 3 May 2000 - Septl2000 Mathematical Excalibur Volume 5, Number 4 May 2000-September 2000 Coordinate Geometry Olympiad Corner Kin Y. Li .YII Asia PaciJc Math Olympiad, March 2000: work on the left half of the figure: the Time allowed: 4 Hours some examples. It is important to stay away from messy computations ! Example 1. (1995 IMO) LetA, B, C A =r2{????}-x:,soQ =Q{????} and D be four distinct points on a line, in that order. The circles with diameters AC and BD intersect at the points X and I;. The line XI{????}meets BC at the point 2. Let Example 2. (1998 APMO) LetABC be a triangle and D the foot of the altitude Z. The line CP intersects the circle with each circle contains exactly one of these diameter AC at the points C and M and

numbers and the line BP intersects the circle with (i) the sums of the four numbers on each diameter BD at the points B and N. Prove

side of the triangle are equal; from A. Let E and F be on a line passing through D such that AE is perpendicular that the lines AM, DN, and XI; are to BE, AF is perpendicular to CF: and E concurrent. and F are different from D. Let h4 and N be the midpoints of the line segments BC and EF: respectively. Prove that AN is perpendicular to NM. rtirt: h .cknowledgment- Thanks IO Janet Wang M;\TH Dept, WJST for general assistance. us1 hkImalhematic~l~excalibur/ kn-line: http://w.math or individual subscriplion for the next five issues for the 00 1 academic year, send us five stamped self-addressed nvelopcs. Send all correspondence to (Remarks. Quite obvious we should DT Kin-Yin Li Depurtmenl of Mathematw oC Science and Technology Hong Kong D, E.F horizontal.) Hong Kong University Clear Water Bay, Kowlwn, Fax: 2358-1643 Em.4 ma!qli@ust hk set the origin at Z. Although the figure is not symmetric with respect to line XI{????} there are pairs such as A4, N and A, D and B, C that are symmetric in roles! So we (continued on page 4) When we do a geometry problem, we should first look at the given facts and the conclusion. If all these involve intersection points, midpoints, feet of perpendiculars, parallel lines, then there is a good chance we can solve the problem by coordinate geometry. However, if they involve two or more circles, angle bisectors and areas of triangles, then sometimes it is still possible to solve the problem by choosing a good place to put the origin and the x-axis. Below we will give O), and the circumradii be rj and rz$ respectively. Then the coordinates of A and C are (x1- rJ, 0) and (.T,+ rI, 0), respectively. Let the coordinates of P be (0, yd. Since AA4 I CP and the slope of CP is -y&x)+ r,), the equation of An4 works out to be (x, + r,)x-yav= x,: - q2 Let Q be the intersection ofAA4 with.riI;, then Q has coordinates (0, (rl{????}-XT) /y,) Similarly, let Q{????}e the intersection of DN with Xr;, then Q{????}has coordinates (O,(rz: -x;>/y,, . Since r,: - X,z = .Zx z (Remarks. We can set the origin at D and the x-axis on line UC. Then computing the coordinates of E and F will be a bit messy. A better choice is to set the line through Mathematical Excalibur, J&l. 5, No.3, A4ay 00 - Sept 00 Since 02A4=rJ and OJ4=rI, we get (b-s){????}({????} s{????}+ (r, -t){????}= r,{????}. c Solution. (Due to Cheung PokA4an, 1998 Hong Kong Team Member) Set the origin at A and the x-axis parallel to line 02 I MN and the slope of 0, 0, is (r2-

EF. Let the coordinates of D, E, F be (d rJ/b, the equation of line MN is bn+(r2-

b), (e, b), fi bj, respectively. The case b=O (This line should pass leads to D=E, which is not allowed. So we through the midpoint of segment AB.) may assume b#O. Since BE IAE and the slope of AE is b/e, so the equation of line BE works out to be ex+by=e2+b2. Similarly, the equations of lines CF and and dx+by=&+b{????} , BC are fi+by=/+b{????}respectively. Solving the equations for BF and BC, we find B has coordinates (d+e, b-(de/b}). Similarly, C has coordinates (d+J b-(d$/b)). Then A4 has coordinates (d+(e+B/2, b-(de+df)/(2b)) and N has coordinates ((e+B/2, b). So the slope of AN is 2b/(e+J and the slope of h4N is -(e+j/(2b). Therefore, AX I MN. Example 3. (2000 IMO) Two circles r, and r, intersect at M and N. Let e be the common tangent to r, and r2 so that Mis closer to C than N is. Let C touch r, at A BD are y=p and x=4, respectively. Let

and r, at B. Let the line through A4 which the perpendicular at P to BA meets .4P=BP=r and CP=DP=s. Then the

parallel to e meet the circle r, again at C .4N produced. Prove that QO is coordinates of A, B, C, D are

perpendicular to BC. and the circle r2 again at D. Lines C4 and c-J7.7, P) : (q, Jq, 3 bp7, P) )

DB meet at E; lines AN and CD meet at P: (q,-$m~),respectively. lines BN and CD meet at (2. Show that Using the EP=EQ. Y (Remarks. Here if we set the x-axis on the line through the centers of the circles, then the equation of the line AB will be complicated. So it is better to have line AB on the x-axis.) Solution. Set the origin at A and the x- respectively. axis on line AB. Let B, A4 have Let the equation of BC be y=cx. Then B has coordinates @/(c-a), bc/(c-a)), C has coordinates (-b/&-a), -bc(c+a)), Ad has coordinates (ab/(c{????}a2), abc/(c -2))) A coordinates (b, 0), (s, t), respectively. Let the centers O,, 0, of r,, G be at (0, t-3, (b, rd, respectively. Then C, D have coordinates (-s, t). (2b -s, t), respectively. Since AB, CD are parallel, CD=2b=2AB implies A, B are midpoints of CE, DE, respectively. So E is at (s, -0. We see EM I CD. Example 5. (1998 IMO) In the convex quadrilateral ABCD: the diagonals AC and BD are perpendicular and the is the midpoint of segment PQ. Since OI rJ.v=b,+[+rJt. To get EP=EQ, it is now left to show A4 opposite sides AB and DC are not parallel.

Suppose that the point P, where the perpendicular bisectors of AB and DC meet, is inside ABCD. Prove that ABCD is a cyclic quadrilateral if and only if the triangles ABP and CDP have equal areas. -t){????}=r2{????}and Subtracting these equations, we get b2/2=bs+(rTrJt, which implies (b/2, 0) is on line MN. Since PQ, AB are parallel and linen&intersects AB at its midpoint, then Ad must be the midpoint of segment PQ. Together with EA4 I PQ, we get EP=EQ. Example 4. (2000 APMO) Let ABC be parallel to the axes will be helpll.) a triangle. Let Ad and N be the points in which the median and the angle bisector, respectively. at A meet the side BC. Let Q and P be the points in which the origin at P and the x-axis parallel to line perpendicular at N to NA meets A& and AC. Then the equations of lines AC and BA, respectively, and 0 the point in (Remarks. Here the equation of the angle bisector is a bit tricky to obtain unless it is the x-axis. In that case, the two sides of the angle is symmetric with respect to the x-axis.) Solution. (Due to Wong Chun Wai, 2000 Hong Kong Team Member) Set the origin at N and the x-axis on line NO. Let the equation of line AB be y=ax+b, then the equation of lines AC and PO are y=-ax-b and y=(-l/a)x+b, 0), 0 has has coordinates (-b/a, coordinates cab, 0) and Q has coordinates (0, ah/c). Then BC has slope c and QO has slope -l/c. Therefore, QO I BC. Page 2 P a P A D (Remarks. The area of a triangle can be computed by taking the half length of the cross product. A natural candidate for the origin is P and having the diagonals Solution. (Due to Leung Wing Chung, 1998 Hong Kong Team A4ember) Set the determinant formula for finding the area of a triangle, we see that the areas of triangles ABP and CDP are equal if and only if Jrz-,2-,, -Jn Since f(x)= is strictly decreasing when x _> lpi and (ql, equality of areas hold if and only if r=s, which is equivalent to A, B, C, D concyclic. After seeing these examples, we would like to remind the readers that there are pure geometric proofs to each of the problems. For examples (1) and (3), there are proofs that only take a few lines. We encourage the readers to discover these simple proofs. Although in the opinions of many people, a pure geometric proof is better and more beautiful than a coo&ante geometric proof, we should point out that sometimes the coordinate geometric proofs may be prefered when there are many cases. For examples (2), the different possible orderings of the points D, E, F on the line can all happen as some pictures will show. The coordinate geometric proofs above cover all cases. We welcome readers to submit solutions to the problems posed below for publication consideration. Solutions should be preceeded by the solver{????} name, home address and school affiliation. Please send submissions to Dr. Kin Y. Li, Problem 108. Circles C, and C, with centers 0, and O2 (respectively) meet at points A, B. The radii OIB and OIB intersect CI and C, at F and E. The line parallel to EF through B meets C, and C, at M and N, respectively. Prove that A4N=AE+AF. Problem 109. Show that there exists an increasing sequence a 1. a2, . . . of positive integers such that for every nonnegative integer k, the sequence k+al, k+az, k+aj . . . contains only finitely many prime numbers. (Source: Problem 110. In a park, 10000 trees have been placed in a square lattice. Determine the maximum number of trees that can be cut down so that from any stump, you cannot see any other stump. (Assume the trees have negligible radius compared to the distance between adjacent trees.) (Source: 1997 German Mathematical Olympiad) Mathematical Excalibur, Vol. 5, No. 3, A@ 00 - Sept 00 Page 3 n > I+ log3 (1999!/a). Problem Corner ***************** Solutions ******+********** of A4athematic.s, Hong of Science Water Department Kong University Technology, and Bay, Clear Kowloon. The deadline for submitting solutions is October 10, 2000. 1 1I4KN=904 (Source: 1999 Selection Fourth 10) ? (Source: 1999 National Math Competition in Croatia) Solution. FAN Wai Tong (St. Mark{????} School: Form 7), KO Man Ho (Wah Yan Problem 106. Find all positive integer ordered pairs (a, b) such that gcd(a,b)+lcm(a,b)=a+b+6, College, Kowloon, Form 6) and LAW Hiu Fai (Wah Yan College, Kowloon, Form 6). where gcd stands for greastest common divisor (or highest common factor) and lcm stands for least common multiple. Since (0.6a,-O.(3a,y + (0.8a,+0.6a,){????}= a, + 03 , lhe sum of the squares of the Problem 107. For a, b, c > 0: if abc=l,

then show that triple of numbers before and after an operation stays the same. Since so (2,8, IO) L-- b+c+c+o+a+b>J;;+J-b+&+3. J;; fi Solution. FAN Wai Tong (St. Mark{????} School, Form 7) 3{????}42+17{????}~t82+1#i cannot be obtained. Let M{????}and N{????}e the midpoints of chords BC and BD respectively. From the midpoint theorem we see that B&XV{????}is a parallelogram. Now Problem 102. Let a be a positive real number and (x,$,,~ be a sequence of real numbers such that xJ =a and mN=m蔃+90{????} n-1 Y >(n+2)x~-~Cku,;foralln51. n+l - k-1 =LKMB+90{????}(Source: 17th Iranian Mathematical Olympiad) Then = LKMM. Let a=LNBD=LNAB. fBC a3. KN{????}FAN Wai Tong (St. Mark{????} MB N{????}= N{????}tana = 1 :BD tana Now LMCB = .&fCB = 3 LCAB 1997 Math Olympiad of Czech and Slovak Republics) 2 9xn.2 -LIMB) +o{????}j=I , . . . . n-1 are true. Then x,, >3x,,.J 2... and =90{????}-.&xl? = 90{????}-a so . 2 (17 + 2) - p$ k=l 1 BCcota MM{????}CM{????}ota. 2 BN{????}-=-= MX Z(n+Z)-()I-l)(++$+...) = 2 3.L So the case j-n is also true. Comments. You may think of the trees being placed at (x. y), where x, Since a > 0, we can take y=O,1,2 I..., 99. Problem 101. A triple of numbers (aJ, a2, a,)=(3, 4. 12) is given. We now perform the following operation: choose two numbers a, and aj, (i f j), and exchange them by 0.6ai-O.&j and 0.8a,+0.6aJ. Is it possible to obtain after several steps the (unordered) triple (2,8, Problem 103. Two circles intersect in points A and B. A line 1 that contains the point A intersects again the circles in the points C, D: respectively. Let M, Nbe the midpoints of the arcs BC and BD, which do not contain the point A, and let k{????}be the midpoint of the segment CD. Show that Romanian Examination) Show that there exists a positive integer n such that X, > 1999! (Source: 1999 Romanian Third Selection Examination) Solution. School, Form 7). We wih prove by induction that q+J23xj for every positive integerj. The case j=l is true by the given inequality. Assume the cases regular if after a number of legal transpositions it becomes (1,2, . . . . no). For which numbers n is the permutation (1, n, n-l, . . . . 3, 2, 0) regular ? Then KN{????} N{????} = MM{????}M K SO triangles A&fX~ KN{????} are similar. Then m RX4 = .LN{????}K and = LtLV{????} - (LV{????}K i- LN{????}Ar) =90{????}Other commended solvers: WONG Chun Wai (Choi Hung Estate Catholic Secondary School, Form 7). Tong (St. Mark{????} School, Form 7). Problem 104. Find all positive integers n such that 2{????}1 is a multiple of 3 and (.?{????}I)~3 is a divisor of 4n8+I for some integer nz. (Source: 1999 Korean Mathematical Olympiad) Solution. (Official Solution) (Some checkings should suggest n is a power of 2.) Now 2{????}1 is a multiple of 3 if and only if (-1){????}27{????}nod 3): that is n is even. Suppose for some even n, (2{????} I)/3 is a divisor of Im{????}l for some nz. Assume n has an odd prime divisor d. Now zd-1=3 (mod 4) implies one of its prime divisor p is of the form 4k+3. Then p divides 241, which divides 2{????} I, which divides 41$+ I. Then p and 2m are relatively prime and so 1=(2m)P-{????}= (4nt2j2{????}{????}=--I (mod p), a contradiction. So n cannot have any odd prime divisor. Hence n=2{????}for some positive integerj. Conversely, suppose n=Y. Let F, = 2?{????}+ 1. Using the factorization 22b-1=(2b-1)(2b+l) repeatedly on the numerator, we get = F,F2 ... F,_, . 2{????}-1 - Since Fi divl3des F-2 for i < j the F~鷔 are pairwise relaiively prime. By the Chinese remainder theorem, there is a positive integer x satisfying the simultaneous equations x= 0 (mod 2) and x s z2{????}{????}(mod Fi) for i=I, 2, . . , j-l. Then x=2m for some positive integer m and 4m2+ 1 =x2+ I = 0 (mod FJ for i=l, 2, ,., j-l. So 4m2+l is divisble by F,F2... &=(2{????}1)/3. n! Problem 4. Let n, k be given positive integers with n > k. Prove that 105. A Problem rectangular parallelopiped (box) is given, such that its intersection with a plane is a regular hexagon. Prove that the rectangular parallelopiped is a cube. (Source: 1999 Problem 5. Given a permutation (a~, National Math Olympiad in Slovenia) Solution. (Official Solution) Y X As in the figure; an equilateral triangle XFZ is formed by extending three alternate sides of the regular haxagon. The right triangles XBZ and YBZ are congruent as they have a common side BZ and the hypotenuses have equal length. So BX=BY and similarly BX-BZ. As the pyramids XBYZ and OB{????}Z are similar and 0,~ = ;kT i it follows B/Z =~BZ. Thus we have BB &~BZ similarly and m = ~BX and We get Bx=BY=Bz, since CB = 2By. AB=B;=BB{????} Other commended solvers: FAN Wai Olympiad Corner (continued from page 1) (ii) the sums of the squares of the four numbers on each side of the triangle are equal. Find all ways in which this can be done. Problem 3. LetABC be a triangle. Let A4 and N be the points in which themedian and the angle bisector, respectively, at.4 meet the side BC. Let Q and P be the points in which the perpendicular at N to NA meets m and BA, respectively, and 0 the point in which the perpendicular at P to BA meets AN produced. Proved that QO is perpendicular to BC. nn n{????}1 -. n+l <k!(n-k)!<kk(n-k){????}k J$@,-k){????} al, . . . . a,) of the sequence O:l,...: n. A transposition of ai with aj is called legal if i > 0: ai=O and a.i+l=aj. The permutation (a*, al, . . . . an) is called 2000 APMO and IMO In April this year, Hong Kong IMO trainees participated in the XII Asia Pacific Mathematical Olympiad, The winners were Fan Wai Tong (Form 7; St Mark{????} School) Wong Chun Wai (Form 7: Choi Hung Estate Catholic Secondary School) Chao Khek Lun (Form 5, St. Paul{????} College) Award Law Ka Ho (Form 7; Queen Elizabeth Sc11001) Ng Ka Chun (Form 5, Queen Elizabeth School) Yu Hok Pun (Form 4, SKI-I Bishop Baker Secondary School) Chan Kin Hang (Form 6, Bishop Hall Jubilee School) Ng Ka Wing (Form 7, STFA Leung Kau Kui College) Chau Suk Ling (Form 5, Queen Elizabeth School) Choy Ting Pong (Form 7: Ming Kei College) Based on the APMO and previous test results, the following trainees were selected to be the Hong Kong team members to the 2000 International Mathematical Olympiad, which was held in July in South Korea. Wong Chun Wai (Form 7, Choi Hung Estate Catholic Secondary School) Ng Ka Wing (Form 7, STFA Leung Kau Kui College) Law Ka Ho (Form 7, Queen Elizabeth School) Chan Kin Hang (Form 6, Bishop Hall Jubilee School) Yu Hok Pun (Form 4, SKH Bishop Baker Secondary School) Fan Wai Tong (Form 7, St. Mark{????} School) Editors:張 百 康 (CHEUNG Pak-Hong), Munsang College, HK 高 子 眉 (KO Tsz-Mei) 梁 達 榮 (LEUNG Tat-Wing), Appl. Math Dept, HKPU 李 健 賢 (LI Kin-Yin), Math Dept, HKUST 吳 鏡 波 (NG Keng-Po Roger), ITC, HKPU Artist: 楊 秀 英 (YEUNG Sau-Ying Camille), MFA, CU