Volume 11, Number 4
November 2006 – December 2006
Olympiad Corner
Pole and Polar
The 9th China Hong Kong Math Olympiad was held on Dec. 2, 2006. The following were the problems. Problem 1. Let M be a subset of {1, 2, …, 2006} with the following property: For any three elements x, y and z (x < y < z) of M, x + y does not divide z. Determine the largest possible size of M. Justify your claim. Problem 2. For a positive integer k, let f1(k) be the square of the sum of the digits of k. (For example f1(123) = (1+2+3)2 = 36.) Let fn+1(k) = f1(fn(k)). Determine the value of f2007(22006). Justify your claim. Problem 3. A convex quadrilateral ABCD with AC ≠ BD is inscribed in a circle with center O. Let E be the intersection of diagonals AC and BD. If P is a point inside ABCD such that
∠PAB+ ∠PCB = ∠PBC + ∠PDC = 90o , prove that O, P and E are collinear. Problem 4. Let a1, a2, a3,… be a sequence of positive numbers. If there exists a positive number M such that for every n = 1, 2, 3, …, a 12 + a 22 + ... + a n2 < M a n2 + 1 ,
then prove that there exists a positive number M’ such that for every n = 1, 2, 3, …, a1 + a 2 + ... + a n < M ' a n +1 . Editors: Ի ஶ (CHEUNG Pak-Hong), Munsang College, HK ଽ υ ࣻ (KO Tsz-Mei) గ ႀ ᄸ (LEUNG Tat-Wing) ፱ (LI Kin-Yin), Dept. of Math., HKUST ֔ ᜢ ( ݰNG Keng-Po Roger), ITC, HKPU Artist:
྆ ( ़ ؾYEUNG Sau-Ying Camille), MFA, CU
Acknowledgment: Thanks to Elina Chiu, Math. Dept., HKUST for general assistance. On-line: http://www.math.ust.hk/mathematical_excalibur/ The editors welcome contributions from all teachers and students. With your submission, please include your name, address, school, email, telephone and fax numbers (if available). Electronic submissions, especially in MS Word, are encouraged. The deadline for receiving material for the next issue is January 25, 2007. For individual subscription for the next five issues for the 05-06 academic year, send us five stamped self-addressed envelopes. Send all correspondence to: Dr. Kin-Yin LI Department of Mathematics The Hong Kong University of Science and Technology Clear Water Bay, Kowloon, Hong Kong Fax: (852) 2358 1643 Email: makyli@ust.hk
Kin Y. Li Let C be a circle with center O and radius r. Recall the inversion with respect to C (see Mathematical Excalibur, vol. 9, no. 2, p.1) sends every point P≠O in the same plane as C to the image point P’ on the ray OP such that OP·OP’=r2. The polar of P is the line p perpendicular to the line OP at P’. Conversely, for any line p not passing through O, the pole of p is the point P whose polar is p. The function sending P to p is called the pole-polar transformation (or reciprocation) with respect to O and r (or with respect to C).
p r
P'
P
Y Following are some useful facts: (1) If P is outside C, then recall P’ is found by drawing tangents from P to C, say tangent at X and Y. Then P’ = OP ∩XY, where ∩ denotes intersection. By symmetry, OP⊥XY. So the polar p of P is the line XY. Conversely, for distinct points X, Y on C, the pole of the line XY is the intersection of the tangents at X and Y. Also, it is the point P on the perpendicular bisector of XY such that O, X, P, Y are concyclic since ∠ OXP = 90°= ∠ OYP. (2) (La Hire’s Theorem) Let x and y be the polars of X and Y, respectively. Then X is on line y ⇔ Y is on line x. Proof. Let X’, Y’ be the images of X, Y for the inversion with respect to C. Then OX·OX’ = r2 = OY·OY’ implies X, X’, Y, Y’ are concyclic. Now X is on y ⇔ ∠ XY’Y = 90° ⇔ ∠ XX’Y = 90° ⇔ Y is on x.
x∩y ⇔ z = XY. Proof. By La Hire’s theorem, Z on x∩y ⇔ X on z and Y on z ⇔ z = XY. (4) Let W, X, Y, Z be on C. The polar p of P = XY∩WZ is the line through Q = WX∩ZY and R = XZ∩YW. Proof. Let S, T be the poles of s = XY, t = WZ respectively. Then P = s∩t. By fact (3), S = x∩y, T = w∩z and p = ST. For hexagon WXXZYY, we have Q=WX∩ZY, S=XX∩YY, R=XZ∩YW,
X O
(3) Let x,y,z be the polars of distinct points X,Y,Z respectively. Then Z =
where XX denotes the tangent line at X. By Pascal’s theorem (see Mathematical Excalibur, vol. 10, no. 3, p.1), Q,S,R are collinear. Similarly, considering the hexagon XWWYZZ, we see Q,T,R are collinear. Therefore, p = ST = QR. Next we will present some examples using the pole-polar transformation. Example 1. Let UV be a diameter of a semicircle. P,Q are two points on the semicircle with UP<UQ. The tangents to the semicircle at P and Q meet at R. If S=UP∩VQ, then prove that RS⊥UV.
S R
Q
P K
U
V
Solution (due to CHENG Kei Tsi). Let K=PQ∩UV. With respect to the circle, by fact (4), the polar of K passes through UP∩VQ=S. Since the tangents to the semicircle at P and Q meet at R, by fact (1), the polar of R is PQ. Since K is on line PQ, which is the polar of R, by La Hire’s theorem, R is on the polar of K. So the polar of K is the line RS. As K is on the diameter UV extended, by the definition of polar, we get RS⊥UV.