On distances in regular polygons Abstract This paper shows a method for solving exercises at the math olympics level involving distances to the vertices in a regular polygon. Using basic expressions, exercises and solutions of differents levels are presented. We also establish a lemma which simplifies the solutions in many cases. Let the distance between two numbers in the complex plane (z = a + bi and w = c + di) be defined by |z − w|, equivalent to the ordinary distance |z − w| = (a − c) 2 + (b −d) 2 , and let the vertices in a regular n−sided polygon be given by A k = R · e i ( 2kπ n +φ ) = R cos 2kπ n + φ + i sin 2kπ n + φ , k = 0, 1, . . . , n − 1 where R is the radius of the polygon’s circumcircle, and φ is the angle of rotation about the real plane. Furthermore we denote by A 0 the first vertex counting in the counter clockwise direction, by A 1 the second vertex counting in the counter clockwise direction, and so on, until we reach the n th vertex denoted by A n−1 . We can also find that the distance between the arbitrary point M , with coordinates x = p cos(θ) and y = p sin(θ), and the vertices in a regular polygon is MA k = R cos 2kπ n + φ − p cos(θ) 2 + R sin 2kπ n + φ − p sin(θ) 2 . This expression, using trigonometrical identities can be written as MA k = R 2 + p 2 − 2Rp cos 2kπ n + φ −θ , for k = 0, 1, . . . , n − 1. (1) In the exercises that we will present, without loss of generality we can let φ = 0 and therefore (1) becomes MA k = r 2 + p 2 − 2rp cos 2kπ n − θ , for k = 0, 1, . . . , n − 1. (2) Mathematical Reflections 5 (2010) 1 If the point M is lies on the circumcircle, it is easy to show that (2) can be written as MA k = 2R sin kπ n − θ 2 , for k = 0, 1. . . . n − 1, (3) With this we can solve the following exercices: 1. A regular n−gon A 1 A 2 A 3 ···A n inscribed in a circle of radius R is given. If S is a point on the circle, calculate T = n k=1 SA 2 k . (IMO longlist 1989) Solution: From (2) we have n k=1 SA 2 k = n−1 k=0 R 2 + l 2 − 2rl cos 2kπ n − θ = n(R 2 + l 2 ) − 2Rl n−1 k=0 cos 2kπ n − θ = n(R 2 + l 2 ) − 2Rl cos θ n−1 k=0 cos 2kπ n + sin θ n−1 k=0 sin 2kπ n . Since n−1 k=0 cos 2kπ n = n−1 k=0 sin 2kπ n = 0, (4) the sum has value n(R 2 + l 2 ). 2. Let A, B, C be three consecutive vertices of a regular polygon and let us consider a point M on the major arc AC of the circumcircle. Prove that MA ·M C = MB 2 − AB 2 . (Andreescu T. and Andrica D. Complex Numbers from A to . . . Z) Solution: Without loss of generality, we let k = 0, k = 1, and k = 2 correspond to the points A, B and C respectively. As M is on the major arc AC we plug k = 0, k = 1, and k = 2 into (3) to get MA = 2R sin θ 2 , MB = 2R sin θ 2 − π n , and MC = 2R sin θ 2 − 2π n , because it is clear that 2π − 4π n ≥ θ ≥ 4π n . Now taking k = 0 and θ = 2π n in (3) we see that AB = 2R sin π n , i.e., the size of each side of the polygon. Combining the above results Mathematical Reflections 5 (2010) 2 (and recalling the identities cos(α −β) −cos(α + β) = 2 sin α sin β, cos(2α) = 1 −2 sin 2 α) we have MB 2 − AB 2 = 4R 2 sin 2 θ 2 − π n − 4R 2 sin 2 π n = 2R 2 1 − 2 sin 2 π n − 1 + 2 sin 2 θ 2 − π n = 2R 2 cos 2π n − cos θ − 2π n = 4R 2 sin θ 2 sin θ 2 − 2π n = MA ·M C. 3. Let A 1 , A 2 , . . . , A n be a regular n−gon inscribed in a circle with center O and radius R. Prove that for each point M in the plane of the n−gon the following inequality holds: n k=1 MA k ≤ (OM 2 + R 2 ) n 2 . (Mathematical Reflections, problem S128. Proposed by Dorin Andrica) Solution: Let d = OM. Applying in (2) the AM–GM inequality to the numbers MA 2 k we have n k=1 MA 2 k n n ≥ n k=1 MA 2 k d 2 + R 2 − 2dR n (A cos θ + B sin θ) n ≥ n k=1 MA 2 k , where A = n−1 k=0 cos 2kπ n and B = n−1 k=0 sin 2kπ n . Finally, we again apply 4) and see d 2 + R 2 n = OM 2 + R 2 n ≥ n k=1 MA 2 k , and the conclusion follows. 4. Let d 1 , d 2 , . . . , d n denote the distances of the vertices A 1 , A 2 , . . . , A n of the regular n−gon A 1 A 2 . . . A n from an arbitrary point P on the minor arc A 1 A n of the circumcircle. Prove that 1 d 1 d 2 + 1 d 2 d 3 + ··· + 1 d n−1 d n = 1 d 1 d n . (The IMO Compendium Group) Mathematical Reflections 5 (2010) 3 Solution: Since P is on the minor arc A 1 A n , it’s clear that − 2π n < θ < 0. So from (3) we find n−1 k=1 1 d k d k+1 = 1 4R 2 n−2 k=0 1 sin kπ n − θ 2 sin (k+1)π n − θ 2 = 1 4R 2 n−2 k=0 csc kπ n − θ 2 csc (k + 1)π n − θ 2 . (5) Using the identity csc(α) csc(β) = 1 sin(α − β) (cot(α)−cot(β)), ∀α = β and α = nπ 2 , β = nπ 2 , n = 0, ±1, ±2, . . . , where α = (k+1)π n − θ 2 and β = kπ n − θ 2 , (5) can be written n−1 k=1 1 d k d k+1 = 1 4R 2 n−2 k=0 1 sin π n cot (k + 1)π n − θ 2 − cot kπ n − θ 2 . The above sum is telescopic, therefore n−1 k=1 1 d k d k+1 = 1 4R 2 1 sin π n cot (n − 1)π n − θ 2 − cot − θ 2 . Using the identity again, n−1 k=1 1 d k d k+1 = 1 4R 2 csc (n − 1)π n − θ 2 csc − θ 2 = 1 d 1 d n , since from (3) we see that d 1 = 2r sin − θ 2 and d n = 2r sin (n − 1)π n − θ 2 . Now we shall prove the following lemma: Lemma: If z k , for k = 0, 1, . . . , n − 1, are the complex roots of unity of order n, where n is an integer, then n−1 k=0 (A − Bz k ) = A n − B n for all complez numbers A and B. Mathematical Reflections 5 (2010) 4 Proof: If B = 0, the result is trivial. If B = 0, taking using the identity n−1 k=0 (z − z k ) = z n − 1, with z = A B we find n−1 k=0 A B − z k = A B n − 1 ⇒ n−1 k=0 (A − Bz k ) = A n − B n , the desired identity. Taking the norm on both sides and letting M = A = pe iθ and B = R, we see from (2) that n k=1 MA k = n k=1 |M − Bz k | = n−1 k=0 R 2 + p 2 − 2Rp cos 2kπ n − θ . On the other hand, |M n − B n | = |p n e inθ − R n | = p 2n + R 2n − 2R n p n cos(nθ). Equating both expressions we obtain n k=1 MA k = n−1 k=0 R 2 + p 2 − 2Rp cos 2kπ n − θ = p 2n + R 2n − 2R n p n cos(nθ). (6) If R = p, the result is reduced to n k=1 MA k = n−1 k=0 2R sin kπ n − θ 2 = 2R n sin nθ 2 . (7) 5. A 1 A 2 . . . A n is a regular polygon inscribed in the circle of radius R and center O. P is a point on line OA 1 extended beyond A 1 . Show that n i=1 P A i = P O n − R n . (Putnam 1955) Solution: It is enough to take θ = 0 and p = P O ≥ R in (6). The conclusion follows. Mathematical Reflections 5 (2010) 5 6. Let A 1 A 2 . . . A n be a regular polygon with circumradius 1. Find the maximum value of n k=1 P A k as P ranges over the circumcircle. (Romanian Mathematical Regional Contest “Grigore Moisil”, 1992) Solution: Taking R = 1 in (7), we see that the maximum value is 2. 7. For a positive integer n > 1, determine lim x→0 sin 2 (x) sin 2 (nx) n 2 sin 2 (x) − sin 2 (nx) . (Mathematical Reflections, problem U143) Solution: Taking natural logarithm in (7), differentiating twice with respect to θ and omitting all θ for which kπ n − θ 2 = 0 we find n−1 k=0 csc 2 kπ n − θ 2 = n 2 csc 2 nθ 2 . Evaluating at k = 0, and taking the limit as θ → 0 the previous expression is equivalent to n−1 k=1 csc 2 kπ n = lim θ→0 n−1 k=1 csc 2 kπ n − θ 2 = lim θ→0 n 2 csc 2 nθ 2 − csc 2 θ 2 . By [1], n−1 k=1 csc 2 kπ n = n 2 − 1 3 , it follows that lim θ→0 n 2 csc 2 nθ 2 − csc 2 θ 2 = n 2 − 1 3 . Therefore, taking x = θ 2 , the desired limit has a value of 3 n 2 − 1 . 8. A regular n−gon inscribed in a circle of radius 1 is given. Let a 2 , . . . , a n−1 be the distances from one vertex of the polygon to all other vertices. Show that (5 − a 2 2 )(5 − a 2 3 ) ···(5 − a 2 n ) = F 2 n , where F n denotes the n th Fibonacci number. (Iberoamerican Mathematical Olympiad for University Students, 2006) Mathematical Reflections 5 (2010) 6 Solution: Without loss of generality we can take the vertex A 0 as the reference vertex and multiply both sides by 5 to get n k=1 (5 − a 2 k ) = 5F 2 n . Taking in (2) θ = 0, R = p = 1 we have n k=1 (5 − a 2 k ) = n−1 k=0 3 + 2 cos 2kπ n . We need to find the values of A and B satisfying A 2 + B 2 = 3 and AB = −1. We can take A > B, and simultaneously solving the equations above we obtain: A = 1 + √ 5 2 and B = 1 − √ 5 2 . Squaring (6), we obtain for the given values n k=1 (5 − a 2 k ) = n−1 k=0 3 + 2 cos 2kπ n = 1 + √ 5 2 n − 1 − √ 5 2 n 2 = 5F 2 n because F n = 1 √ 5 1 + √ 5 2 n − 1 − √ 5 2 n , and the conclusion follows. Mathematical Reflections 5 (2010) 7 Exercises: 1. Two regular n−gons A 1 A 2 . . . A n and B 1 B 2 . . . B n are in the same plane P and have the same center. a) Show that n j=1 B i A j = n i=1 A j B i , ∀i, j ∈ {1, 2, . . . , n}. b) Find min M∈P {MA 1 · MA 2 · . . . · MA n + MB 1 · MB 2 · . . . · MB n }. (Romanian mathematical competition, shortlist 2008) 2. Let A 0 , A 1 , . . . , A 2n be a regular polygon with circumradius equal to 1 and consider a point P on the circumcircle. Prove that n−1 k=0 P A 2 k+1 P A 2 n+k+1 = 2n. (Andreescu T. and Andrica D. Complex Numbers from A to . . . Z) 3. Consider an integer n ≥ 3 and the parabola of equation y 2 = 4px, with focus F . A regular n−gon A 1 A 2 ···A n has center at F and no one of its vertices lies on the x axis. The rays F A 1 , F A 2 , . . . , F A n cut the parabola at points B 1 , B 2 , . . . , B n . Prove that F B 1 + F B 2 + ··· + F B n > np. (Romanian mathematical competition 2004) References [1] Some remarks on problem U23, Dorin Andrica and Mihai Piticari, Mathematical Re- flections 4(2008). N. Javier Buitrago A. Universidad Nacional de Colombia, Departamento de F´ısica Ciudad Universitaria, Bogot´a D.C., Colombia njavierbuitragoa@gmail.com, njbuitragoa@unal.edu.co Mathematical Reflections 5 (2010) 8 . On distances in regular polygons Abstract This paper shows a method for solving exercises at the math olympics level involving distances to the vertices in a regular polygon. Using basic. positive integer n > 1, determine lim x→0 sin 2 (x) sin 2 (nx) n 2 sin 2 (x) − sin 2 (nx) . (Mathematical Reflections, problem U143) Solution: Taking natural logarithm in (7), differentiating twice. n−gon the following inequality holds: n k=1 MA k ≤ (OM 2 + R 2 ) n 2 . (Mathematical Reflections, problem S128. Proposed by Dorin Andrica) Solution: Let d = OM. Applying in (2) the AM–GM inequality