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Concyclic Problems Kin Y 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 geometry 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 triangle and E be an interior point extending the sides of a pentagon FGHIJ, so that all five points lie on the circle of the median 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 line BC. Let M be EIJ intersect at 5 concyclic points 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 angles of the and P be the orthogonal projections of quadrilateral 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 triangles. 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 centroid, the midpoints 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 circles 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 perpendicular 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 parallelograms! 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 parallelogram 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 1 are perpendicular to line BE and are 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.
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