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Home , Concyclic points, Median (geometry)

Volume 6, Number 1 January 2001 – March 2001

Olympiad Corner Concyclic Problems Kin Y. Li 17th Balkan Mathematical Olympiad, 3-9 May 2000: Near Christmas last year, I came across Now we look at ways of getting solutions Time allowed: 4 hours 30 minutes two beautiful problems. I was to these problems. Both are concyclic

Problem 1. Find all the functions f : informed of the first problem by a reporter, problems with more than 4 points. R → R with the property: who was covering President Jiang Generally, to do this, we show the points Zemin’s visit to Macau. While talking to are concyclic four at a time. For example, f (xf (x) + f ( y)) = ( f (x))2 + y, students and teachers, the President posed in the first problem, if we can show K, L, for any real numbers x and y. the following problem. M, N are concyclic, then by similar

Problem 2. Let ABC be a nonisosceles For any pentagram ABCDE obtained by reasons, L, M, N, O will also be concyclic acute and E be an interior extending the sides of a pentagon FGHIJ, so that all five points lie on the of the AD, D ∈ (BC). The point prove that neighboring pairs of the passing through L, M, N. F is the orthogonal projection of the circumcircles of ∆ AJF , BFG, CGH, DHI, There are two common ways of showing 4 point E on the straight BC. Let M be EIJ intersect at 5 K, L, M, points are concyclic. One way is to show an interior point of the segment EF, N N, O as in the figure. the sum of two opposite of the and P be the orthogonal projections of with the 4 points as vertices $ the point M on the straight lines AC and is 180 . Another way is to use the AB, respectively. Prove that the two converse of the intersecting chord straight lines containing the bisectrices theorem, which asserts that if lines WX ⋅ of the angles PMN and PEN have no and YZ intersect at P and PW PX = ⋅ common point. PY PZ , then W, X, Y, Z are concyclic. (The equation implies ∆PWY , PZX are Problem 3. Find the maximum number similar. Then ∠PWY = ∠PZX and the of rectangles of the dimensions 1×10 2 , conclusion follows.) which is possible to cut off from a For the first problem, as the points K, L, M, rectangle of the dimensions 50 × 90 , by N, O are on the circumcirles, checking the sum of opposite angles equal 180$ is using cuts parallel to the edges of the initial rectangle. The second problem came a week later. I likely to be easier as we can use the (continued on page 2) read it in the Problems Section of the theorem about angles on the same segment Editors: 張 百 康 (CHEUNG Pak-Hong), Munsang College, HK November issue of the American to move the angles. To show K, L, M, N 高 子 眉 (KO Tsz-Mei) are concyclic, we consider showing 梁 達 榮 (LEUNG Tat-Wing), Appl. Math Dept, HKPU Mathematical Monthly. It was proposed ∠ ∠ = $ 李 健 賢 (LI Kin-Yin), Math Dept, HKUST by Floor van Lamoen, Goes, The LMN + LKN 180 . Since the 吳 鏡 波 (NG Keng-Po Roger), ITC, HKPU ∠ Netherlands. Here is the problem. sides of LMN are in two circumcircles, Artist: 楊 秀 英 (YEUNG Sau-Ying Camille), MFA, CU it may be wise to break it into two angles A triangle is divided by its three medians Acknowledgment: Thanks to Elina Chiu, MATH Dept, LMG and GMN. Then the strategy is to HKUST for general assistance. into 6 smaller . Show that the change these to other angles closer to circumcenters of these smaller triangles On-line: http://www.math.ust.hk/mathematical_excalibur/ ∠LKN . lie on a circle. The editors welcome contributions from all teachers and Now ∠LMG = 180$ − ∠LFG = ∠LFA = students. With your submission, please include your name, ∠ address, school, email, telephone and fax numbers (if To get the readers appreciating these LKA . (So far, we are on track. We available). Electronic submissions, especially in MS Word, problems, here I will say, stop reading, try bounced ∠LMG to ∠LKA , which shares are encouraged. The deadline for receiving material for the ∠ ∠ next issue is April 15, 2001. to work out these problems and come a side with LKN .) Next, GMN = For individual subscription for the next five issues for the back to compare your solutions with those ∠GCN = ∠ACN . Putting these 01-02 academic year, send us five stamped self-addressed given below! together, we have envelopes. Send all correspondence to: ∠LMN + ∠LKN Dr. Kin-Yin Li Here is a guided tour of the solutions. The Department of Mathematics = ∠LKA + ∠ACN + ∠LKN first step in enjoying geometry problems Hong Kong University of Science and Technology = ∠ + ∠ Clear Water Bay, Kowloon, Hong Kong is to draw accurate pictures with compass AKN ACN .

Fax: 2358-1643 and ruler! Email: [email protected] Mathematical Excalibur, Vol. 6, No. 1, Jan 01 – Mar 01 Page 2

Now if we can only show A, K, N, C are theorem. For a triangle ABC, let G, D, E, this point, you can see the angles of $ ∆ concyclic, then we will get 180 for the F be the , the of sides KO2O3 equal the three angles with displayed equations above and we will BC, CA, AB, respectively. Let O1 , O2 , vertices at G on the left side of segment finish. However, life is not that easy. This O , O , O , O be the circumcenters of AD. 3 4 5 6 turned out to be the hard part. If you draw triangles DBG, BFG, FAG, AEG, ECG, Now we try to put these three angles the circle through A, C, N, then you see it CDG, respectively. together in another way to form another goes through K as expected and triangle. Let M be the point on line AG surprisingly, it also goes through another such that MC is parallel to BG. Since point, I. With this discovery, there is new ∠MCG = ∠BGF , ∠MGC = ∠FGA hope. Consider the arc through B, I, O. (and ∠GMC = ∠BGD , ) we see On the two sides of this arc, you can see ∆KO O , MCG are similar. there are corresponding point pairs (A, C), 2 3 ∆ (K, N), (J, H), (F, G). So to show A, K, N, The sides of MCG are easy to compute C are concyclic, we can first try to show N in term of AD, BE, CF. As AD and BE is on the circle through A, C, I, then in that occurred in the ratio of KO1 and KO4 , Well, should we draw the 6 circumcircles? argument, if we interchange A with C, K, this is just what we need! Observe that It would make the picture complicated. ∆ with N and so on, we should also get K is MCD , GBD are congruent since The do not seem to be helpful at ∠ = ∠ on the circle through C, A, I. Then A, K, N, MCD GBD (by MC parallel to GB), this early stage. We give up on drawing ∠ = ∠ C (and I) will be concyclic and we will CD = BD and MDC GDB . So the circles, but the circumcenters are 2 finish. MG = 2GD = AD, important. So at least we should locate 3 Wishful thinking like this sometimes them. To locate the circumcenter of 2 works! Here are the details: ∆ MC = GB = BE FAG , for example, which two sides do 3 ∠ACN = ∠GCN = 180$ − ∠GHN we draw bisectors? Sides 2 (and CG = 3 CF. Incidentally, this means = ∠ = ∠ = $ − ∠ AG and FG are the choices because they NHD NID 180AIN . the three medians of a triangle can be put are also the sides of the other small So N is on the circle A, C, I. together to form a triangle! Actually, this triangles, so we can save some work later. Interchanging letters, we get similarly K is well-known and was the reason we Trying this out, we discover these is on circle C, A, I. So A, K, N, C (and I) considered ∆MCG .) We have perpendicular bisectors produce many are concyclic. Therefore, K, L, M, N, O parallel lines and ! KO3 / KO2 = MG/MC = AD/BE = are indeed concyclic. KO1 / KO4 . (History. My friend C.J. Lam did a search Since circumcenters are on perpendicular ⋅ ⋅ on the electronic database JSTOR and bisectors of chords, lines O3 O4 , O6 O1 So KO1 KO2 = KO3 KO4 , which came across an article titled A Chain of are perpendicular bisectors of AG, GD, implies O1 , O2 , O3 , O4 are concyclic. Circles Associated with the 5-Line by J.W. respectively. So they are perpendicular to Similarly, we see that O , O , O , O Clawson published in the American 1 2 3 4 5 line AD and are 2 AD units apart. Mathematical Monthly, volume 61, concyclic (using the formed Similarly, the two lines O O , O O number 3 (March 1954), pages 161-166. 1 2 4 5 by the lines O1O2 , O4O5 , O2O3 , O5O6 are perpendicular to line BE and are 1 BE There the problem was attributed to the 2 instead) and O3 , O4 , O5 , O6 are nineteenth century geometer Miquel, who units apart. Aiming in showing O1 , O2 , concyclic. published the result in Liouville’s Journal O , O are concyclic by the converse of de Mathematiques, volume 3 (1838), 3 4 the intersecting chord theorem, let K be pages 485-487. In that paper, Miquel Olympiad Corner proved his famous theorem that for four the intersection of lines O1 O2 , O3 O4 pairwise intersecting lines, taking three of and L be the intersection of the lines (continued from page 1) the lines at a time and forming the circles O4 O5 , O6 O1 . Since the of the Problem 4. We say that a positive through the three intersecting points, the integer r is a power, if it has the form r = parallelogram KOL4 O1 is four circles will always meet at a common s ≥ 1 1 t where t and s are integers, t 2 , point, which nowadays are referred to as AD ⋅ KO = BE ⋅ KO , ≥ 2 4 2 1 s 2 . Show that for any positive integer the Miquel point. The first problem was n there exists a A of positive integers, we get KO1 /KO4 = AD/BE. then deduced as a corollary of this Miquel which satisfies the conditions: theorem.) Now that we get ratio of KO and KO , 1 4 1. A has n elements; For the second problem, as the 6 we should examine KO2 and KO3 . 2. any element of A is a power; ∆ Trying to understand KO2O3 , we first ≤ ≤ circumcenters of the smaller triangles are 3. for any r1 , r2 , …, rk (2 k n ) find its angles. Since KO⊥ BG, not on any circles that we can see 2 r + r + + r ⊥ ⊥ from A the number 1 2  k immediately, so we may try to use the OFG2O3 and KO3 AG, we see that ∠ ∠ ∠ k converse of the intersecting chord KO2O3 = BGF and KO3O2 = ∠ ∠ ∠ is a power. FGA . Then O2 KO3 = DGB . At Mathematical Excalibur, Vol. 6, No. 1, Jan 01 – Mar 01 Page 3

A/P. 3d 3d 3d Problem Corner 2d = a + b + c ≤ + + < 2d, Solution 1. CHAO Khek Lun (St. Paul’s 6 5 4 We welcome readers to submit their College, Form 6). a contradiction. Therefore, two of the solutions to the problems posed below sides are equal. Draw four rectangles on the sides of the for publication consideration. Solutions quadrilateral and each has height A/P Problem 118. Let R be the real numbers. should be preceded by the solver’s name, pointing inward. The sum of the of Find all functions f : R → R such that for home (or email) address and school the rectangles is A. Since at least one all real numbers x and y, affiliation. Please send submissions to interior of the quadrilateral is less f (xf (y) + x) = xy + f (x). Dr. Kin Y. Li, Department of than 180$ , at least two of the rectangles Mathematics, Hong Kong University of will overlap. So the union of the four Solution 1. LEUNG Wai Ying (Queen Elizabeth School, Form 6). Science & Technology, Clear Water Bay, rectangular regions does not cover the Kowloon. The deadline for submitting interior of the quadrilateral. For any point Putting x = 1, y = -1 – f (1) and letting a = solutions is April 15, 2001. in the interior of the quadrilateral not f (y) + 1, we get

covered by the rectangles, the distance Problem 121. Prove that any integer f (a) = f ( f (y) + 1) = y + f (1) = -1. between the point and any side of the greater than or equal to 7 can be written Putting y = a and letting b = f (0), we get quadrilateral is greater than A/P. So we as a sum of two relatively prime integers, can draw a desired circle with that point as b = f (xf (a) + x) = ax + f (x), both greater than 1. center. so f(x) = -ax + b. Putting this into the (Two integers are relative prime if they equation, we have share no common positive divisor other Solution 2. CHUNG Tat Chi (Queen Elizabeth School, Form 4) and LEUNG a 2 xy − abx − ax + b = xy − ax + b. than 1. For example, 22 and 15 are Wai Ying (Queen Elizabeth School, Form = ± relatively prime, and thus 37 = 22 + 15 6). Equating coefficients, we get a 1 and represents the number 37 in the desired b = 0, so f (x) = x or f (x) = -x. We can Let BCDE be a quadrilateral with area A way.) (Source: Second Bay Area easily check both are solutions. and perimeter P. One of the diagonal, say Mathematical Olympaid) Solution 2. LEE Kai Seng (HKUST). BD is inside the quadrilateral. Then either Problem 122. Prove that the product of ∆BCD or ∆BED will have an area Setting x = 1, we get the lengths of the three angle bisectors of greater than or equal to A/2. Suppose this f ( f ( y) +1) = y + f (1). ∆ a triangle is less than the product of the is BCD . Then BCDE contains the For every a, let y = a – f (1), ∆ lengths of the three sides. (Source: 1957 incircle of BCD , which has a of then f (f (y) + 1) = a and f is surjective. In Shanghai Junior High School Math 2[BCD] particular, there is b such that f (b) = -1. Competition) BC + CD + DB Also, if f (c) = f (d), then Problem 123. Show that every convex 2[BCD] c + f (1) = f ( f (c) + 1) > quadrilateral with area 1 can be covered BC + CD + DE + EB = f ( f (d) +1) by some triangle of area at most 2. A = + ≥ , d f (1). (Source: 1989 Wuhu City Math P So c = d and f is injective. Taking x = 1, y Competition) where the brackets denote area. Hence, it = 0, we get f (f (0) +1) = f (1). Since f is contains a circle of radius A/P. Problem 124. Find the least integer n injective, we get f (0) = 0. Comment: Both solutions do not need the such that among every n distinct numbers For x ≠ 0, let y = − f (x) / x , then convexity assumption. a1 , a2 , …, an , chosen from [1, 1000], f (xf ( y) + x) = 0 = f (0). Problem 117. The lengths of the sides of there always exist ai , a j such that By injectivity, we get xf (y) + x = 0. Then < − < + 3 a quadrilateral are positive integers. The 0 ai a j 1 3 ai a j . − = = − = length of each side divides the sum of the f ( f (x) / x) f ( y) 1 f (b) (Source: 1990 Chinese Team Training other three lengths. Prove that two of the and so –f (x)/x = b for every x ≠ 0 . That Test) sides have the same length. is, f (x) = -bx. Putting this into the given

Problem 125. Prove that Solution. CHAO Khek Lun (St. Paul’s equation, we find f (x) = x or f (x) = -x, 2 $ + 2 $ + 2 $ + + 2 $ College, Form 6) and LEUNG Wai Ying which are checked to be solutions. tan 1 tan 3 tan 5  tan 89 (Queen Elizabeth School, Form 6). is an integer. Other commended solvers: CHAO Khek Suppose the sides are a, b, c, d with a < b Lun (St. Paul’s College, Form 6) and NG ***************** Ka Chun Bartholomew (Queen < c < d. Since d < a + b + c < 3d and d Elizabeth School, Form 6). Solutions divides a + b + c, we have a + b + c = 2d. ***************** Now each of a, b, c divides a + b + c + d Problem 119. A circle with center O is internally tangent to two circles inside it Problem 116. Show that the interior of a = 3d. Let x = 3d/a, y = 3d/b and z = 3d/c. at points S and T. Suppose the two convex quadrilateral with area A and Then a < b < c < d implies x > y > z > 3. perimeter P contains a circle of radius So z ≥ 4, y ≥ 5, x ≥ 6. Then circles inside intersect at M and N with N Mathematical Excalibur, Vol. 6, No. 1, Jan 01 – Mar 01 Page 4

closer to ST. Show that OM ⊥ MN if and Solution 1. CHAO Khek Lun (St. Paul’s b − c c − a College, Form 6), CHAU Suk Ling = f (a) + f (b), only if S, N, T are collinear. (Source: − − (Queen Elizabeth School, Form 6) and b a b a 1997 Chinese Senior High Math CHUNG Tat Chi (Queen Elizabeth which is what we will get if we solve for Competition) School, Form 4). f (c) in the two inequalities in the

Solution. LEUNG Wai Ying (Queen Applying the inclusion-exclusion statement of the lemma. Elizabeth School, Form 6). principle, we see there are 82 integers on In brief the lemma asserts that the slopes

[104, 208] that are divisible by 2, 3, 5 or of chords are increasing as the chords are 7. There remain 23 other integers on the moving to the right. Now we are ready to interval. If 28 integers are chosen from proof the majorization inequality. the interval, at least 28 – 23 = 5 are Suppose among the 82 integers that are divisible (x1, x2 , ..., xn ) ( y1, y2 , ..., yn ). by 2, 3, 5 or 7. So there will exist two ≥ ≥ Since xi xi+1 and yi yi+1 for i = 1, that are both divisible by 2, 3, 5 or 7. 2, …, n – 1, it follows from the lemma Solution 2. CHAN Yun Hung (Carmel that the slopes Divine Grace Foundation Secondary f (x ) − f ( y ) m = i i School, Form 4), KWOK Sze Ming i x − y (Queen Elizabeth School, Form 5), LAM i i ≥ ≤ ≤ Shek Ming (La Salle College, Form 5), satisfy mi mi+1 for 1 i n – 1. LEUNG Wai Ying (Queen Elizabeth (For example, if y ≤ y ≤ x ≤ x , Consider the tangent lines at S and at T. School, Form 6), WONG Tak Wai Alan i+1 i i+1 i (Suppose they are parallel, then S, O, T (University of Toronto) and WONG then applying the lemma twice, we get Wing Hong (La Salle College, Form 3). will be collinear so that M and N will be f (x ) − f ( y ) m = i+1 i+1 equidistant from ST, contradicting N is There are 19 prime numbers on the i+1 − xi+1 yi+1 closer to ST.) Let the tangent lines meet at interval. The remaining 86 integers on ∠ = $ = ∠ f (x + ) − f ( y ) K, then OSK 90OTK implies O, the interval are all divisible by at least ≤ i 1 i − S, K, T lie on a circle with OK. one of the prime numbers 2, 3, 5, 7, 11 xi+1 yi 2 = 2 Also, KS KT implies K is on the and 13 since 13 is the largest prime less f (x ) − f ( y ) ≤ i i = m radical axis MN of the two inside circles. than or equal to 208 . So every number − i xi yi So M, N, K are collinear. on the interval is a multiple of one of ∠ and similarly for the other ways yi+1 , yi , If S, N, T are collinear, then SMT = these 25 primes. Hence, among any 26 ∠ + ∠ ∠ + ∠ x , x are distributed.) SMN TMN = NSK KTN = integers on the interval at least two will i+1 i 180$ − ∠SKT . So M, S, K, T, O are have a common prime divisor. For k = 1, 2, …, n, let concyclic. Then ∠OMN = ∠OMK = X = x + x + + x ∠OSK = 90$ . k 1 2  k

⊥ ∠ and Conversely, if OM MN , then OMK A Proof of the Majorization Inequality = + + + Yk y1 y2  yk . = 90$ = ∠OSK implies M, S, K, T, O are Kin Y. Li ≥ Since X k Yk for k = 1, 2, …, n – 1 and concyclic. Then Quite a few readers would like to see a X = Y , we get ∠SKT = 180$ − ∠SMT proof of the majorization inequality, n n n $ − ∠ − ∠ which was discussed in the last issue of − − ≥ = 180SMN TMN ∑(X k Yk )(mk mk+1 ) 0, the Mathematical Excalibur. Below we k=1 = 180$ − ∠NSK − ∠KTN . will present a proof. We will first make where we set m + = 0 for convenience. Thus, ∠TNS = 360$ − ∠NSK − ∠SKT - n 1 one observation. ∠ = $ Expanding the sum, grouping the terms KTN 180 . Therefore, S, N, T are Lemma. Let a < c < b and f be convex involving the same m ’s and letting collinear. on an interval I with a, b, c on I. Then the k Comments: For the meaning of radical following are true: X 0 = 0 = Y0 , we get axis, we refer the readers to pages 2 and 4 − − n f (c) f (a) ≤ f (b) f (a) ∑ − − + ≥ of Math Excalibur, vol. 4, no. 3 and the (X k X k−1 Yk Yk −1 )mk 0, c − a b − a k=1 corrections on page 4 of Math Excalibur, and which is the same as vol. 4, no. 4. f (b) − f (c) f (b) − f (a) n ≤ . ∑(x − y )m ≥ 0. Other commended solvers: CHAO Khek − − k k k Lun (St. Paul’s College, Form 6). b c b a k=1 Proof. Since a < c < b, we have c = (1 – − = − Since (xk yk )mk f (xk ) f ( yk ), Problem 120. Twenty-eight integers are t)a + tb for some t ∈(0, 1). Solving for t, we get chosen from the interval [104, 208]. we get t = (c – a)/(b – a). Since f is n ∑ − ≥ convex on I, (f (xk ) f ( yk ))mk 0. Show that there exist two of them having k=1 a common prime divisor. f (c) ≤ (1 – t) f (a) + t f (b) Transferring the f ( y ) terms to the k right, we get the majorization inequality.