Mathematical Reflections 5 (2010) On distances in regular polygons Abstract

On distances in regular polygonsAbstract This paper shows a method for solving exercises at the math olympics level involving distances to the vertices in a regular polygon.. Furthermore

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| =p(a − c)2+ (b − d)2, and let the vertices in a regular n−sided polygon be given by

Ak = R · ei(2kπn +φ) = Rcos 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 A0 the first vertex counting in the counter clockwise direction, by A1 the second vertex counting in the counter clockwise direction, and so on, until we reach the nth vertex denoted by An−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

M Ak =

s



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

M Ak =

s

R2+ p2− 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

M Ak =

s

r2+ p2− 2rp cos 2kπ

n − θ

 , for k = 0, 1, , n − 1 (2)

If the point M is lies on the circumcircle, it is easy to show that (2) can be written as

M Ak= 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 A1A2A3· · · An inscribed in a circle of radius R is given If S is a point

on the circle, calculate

T =

n

X

k=1

SA2k

(IMO longlist 1989)

Solution: From (2) we have

n

X

k=1

SA2k =

n−1

X

k=0



R2+ l2− 2rl cos 2kπ

n − θ



= n(R2+ l2) − 2Rl

n−1

X

k=0

cos 2kπ

n − θ



= n(R2+ l2) − 2Rl cos θ

n−1

X

k=0

cos 2kπ

n

 + sin θ

n−1

X

k=0

sin 2kπ n

!

Since

n−1

X

k=0

cos 2kπ

n



=

n−1

X

k=0

sin 2kπ n



the sum has value n(R2+ l2)

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

M A · M C = M B2− AB2 (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

M A = 2R sin θ

2

 , M B = 2R sin θ

2 − π n

 , and M C = 2R sin θ

2 − 2π n

 ,

because it is clear that 2π − 4πn ≥ θ ≥ 4π

n Now taking k = 0 and θ = 2πn in (3) we see that

AB = 2R sin πn
, i.e., the size of each side of the polygon Combining the above results

(and recalling the identities cos(α − β) − cos(α + β) = 2 sin α sin β, cos(2α) = 1 − 2 sin2α)

we have

M B2− AB2 = 4R2sin2 θ

2 − π n



− 4R2sin2π

n



= 2R2



1 − 2 sin2π

n



− 1 + 2 sin2 θ

2 − π n



= 2R2

 cos 2π n



− cos



θ − 2π n



= 4R2sin θ

2

 sin θ

2− 2π n



= M A · M C

3 Let A1, A2, , An 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

Y

k=1

M Ak≤ (OM2+ R2)n2

(Mathematical Reflections, problem S128 Proposed by Dorin Andrica)

Solution: Let d = OM Applying in (2) the AM–GM inequality to the numbers M A2k we have

n

X

k=1

M A2 k

n

!n

≥

n

Y

k=1

M A2k



d2+ R2− 2dR

n (A cos θ + B sin θ)

n

≥

n

Y

k=1

M A2k,

where A =

n−1

P

k=0

cos2kπn and B =

n−1

P

k=0

sin2kπn Finally, we again apply 4) and see

d2+ R2
n

= OM2+ R2
n

≥

n

Y

k=1

M A2k,

and the conclusion follows

4 Let d1, d2, , dndenote the distances of the vertices A1, A2, , Anof the regular n−gon

A1A2 An from an arbitrary point P on the minor arc A1An of the circumcircle Prove that

1

d1d2 +

1

d2d3 + · · · +

1

dn−1dn =

1

d1dn. (The IMO Compendium Group)

Solution: Since P is on the minor arc A1An, it’s clear that −2πn < θ < 0 So from (3) we find

n−1

X

k=1

1

dkdk+1 =

1 4R2

n−2

X

k=0

1 sin kπ

n − θ

2
sin(k+1)π

n − θ 2



4R2

n−2

X

k=0

csc kπ

n −θ 2

 csc (k + 1)π

2



Using the identity

csc(α) csc(β) = 1

sin(α − β)(cot(α)−cot(β)), ∀α 6= β and α 6=

nπ

2 , β 6=

nπ

2 , n = 0, ±1, ±2, ,

where α = (k+1)πn − θ

2 and β = kπn −θ

2, (5) can be written

n−1

X

k=1

1

dkdk+1 =

1 4R2

n−2

X

k=0

1 sin πn

 cot (k + 1)π

2



− cot kπ

n −θ 2



The above sum is telescopic, therefore

n−1

X

k=1

1

dkdk+1

4R2

1 sin πn

 cot (n − 1)π

2



− cot



−θ 2



Using the identity again,

n−1

X

k=1

1

dkdk+1 =

1 4R2 csc (n − 1)π

2

 csc



−θ 2



d1dn,

since from (3) we see that

d1 = 2r sin



−θ 2

 and dn= 2r sin (n − 1)π

2



Now we shall prove the following lemma:

Lemma: If zk, for k = 0, 1, , n − 1, are the complex roots of unity of order n, where n

is an integer, then

n−1

Y

k=0

(A − Bzk) = An− Bn

for all complez numbers A and B

Proof: If B = 0, the result is trivial If B 6= 0, taking using the identity

n−1

Y

k=0

(z − zk) = zn− 1,

with z = BA we find

n−1

Y

k=0

 A

B − zk



= A B

n

− 1 ⇒

n−1

Y

k=0

(A − Bzk) = An− Bn,

the desired identity

Taking the norm on both sides and letting M = A = peiθ and B = R, we see from (2) that

n

Y

k=1

M Ak =

n

Y

k=1

|M − Bzk| =

n−1

Y

k=0

s

R2+ p2− 2Rp cos 2kπ

n − θ



On the other hand,

|Mn− Bn| = |pneinθ− Rn| =pp2n+ R2n− 2Rnpncos(nθ)

Equating both expressions we obtain

n

Y

k=1

M Ak =

n−1

Y

k=0

s

R2+ p2 − 2Rp cos 2kπ

n − θ



=pp2n+ R2n− 2Rnpncos(nθ) (6)

If R = p, the result is reduced to

n

Y

k=1

M Ak=

n−1

Y

k=0

2R

sin kπ

n − θ 2



= 2Rn

sin nθ 2



5 A1A2 An is a regular polygon inscribed in the circle of radius R and center O P is

a point on line OA1 extended beyond A1 Show that

n

Y

i=1

P Ai = P On− Rn

(Putnam 1955)

Solution: It is enough to take θ = 0 and p = P O ≥ R in (6) The conclusion follows

6 Let A1A2 An be a regular polygon with circumradius 1 Find the maximum value of

n

Q

k=1

P Ak 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

sin2(x) sin2(nx)

n2sin2(x) − sin2(nx).

(Mathematical Reflections, problem U143)

Solution: Taking natural logarithm in (7), differentiating twice with respect to θ and omitting all θ for which kπn − θ

2 = 0 we find

n−1

X

k=0

csc2 kπ

n − θ 2



= n2csc2 nθ

2



Evaluating at k = 0, and taking the limit as θ → 0 the previous expression is equivalent to

n−1

X

k=1

csc2 kπ

n



= lim

θ→0

n−1

X

k=1

csc2 kπ

n −θ 2



= lim

θ→0



n2csc2 nθ

2



− csc2 θ

2



By [1],

n−1

X

k=1

csc2 kπ

n



= n

2− 1

3 ,

it follows that

lim

θ→0



n2csc2 nθ

2



− csc2 θ

2



= n

2− 1

3 .

Therefore, taking x = θ

2, the desired limit has a value of

3

n2− 1.

8 A regular n−gon inscribed in a circle of radius 1 is given Let a2, , an−1be the distances from one vertex of the polygon to all other vertices Show that

(5 − a22)(5 − a23) · · · (5 − a2n) = Fn2, where Fn denotes the nth Fibonacci number

(Iberoamerican Mathematical Olympiad for University Students, 2006)

Solution: Without loss of generality we can take the vertex A0 as the reference vertex and multiply both sides by 5 to get

n

Y

k=1

(5 − a2k) = 5Fn2

Taking in (2) θ = 0, R = p = 1 we have

n

Y

k=1

(5 − a2k) =

n−1

Y

k=0



3 + 2 cos 2kπ

n



We need to find the values of A and B satisfying A2+ B2 = 3 and AB = −1 We can take

A > B, and simultaneously solving the equations above we obtain:

A = 1 +

√ 5

1 −√ 5

Squaring (6), we obtain for the given values

n

Y

k=1

(5 − a2k) =

n−1

Y

k=0



3 + 2 cos 2kπ

n



=

1 +√ 5 2

!n

√ 5 2

!n!2

= 5Fn2

because

Fn= √1

5

1 +√ 5 2

!n

√ 5 2

!n! ,

and the conclusion follows

1 Two regular n−gons A1A2 An and B1B2 Bn are in the same plane P and have the same center

a) Show that

n

Q

j=1

BiAj =

n

Q

i=1

AjBi, ∀i, j ∈ {1, 2, , n}

b) Find min

M ∈P{M A1· M A2· · M An+ M B1· M B2· · M Bn}

(Romanian mathematical competition, shortlist 2008)

2 Let A0, A1, , A2n be a regular polygon with circumradius equal to 1 and consider

a point P on the circumcircle Prove that

n−1

X

k=0

P A2k+1P A2n+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 y2 = 4px, with focus F A regular n−gon A1A2· · · An has center at F and no one of its vertices lies on the x axis The rays F A1, F A2, , F An cut the parabola at points B1, B2, , Bn

Prove that F B1+ F B2 + · · · + F Bn> 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

sin2(x) sin2(nx)

n2sin2(x) − sin2(nx).

(Mathematical Reflections, problem U143)

Solution: Taking...

(Putnam 1 955 )

Solution: It is enough to take θ = and p = P O ≥ R in (6) The conclusion follows

6...

(Romanian Mathematical Regional Contest “Grigore Moisil”, 1992)

Solution: Taking R = in (7), we see that the maximum value is

7 For a positive integer n > 1, determine

lim