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# Let ti be the length of the common external tangent between the circles C and Oi

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Problem 10483

(American Mathematical Monthly, Vol.102, November 1995) Proposed by S. Rabinowitz (USA).

Given an odd positive integer n, let A1, A2, . . . , An be a regular n-gon with circumcircle Γ. A circle Oi with radius r is drawn externally tangent to Γ at Ai for i = 1, 2, . . . , n. Let P be any point on Γ between An and A1. A circle C (with any radius) is drawn externally tangent to Γ at P . Let ti be the length of the common external tangent between the circles C and Oi. Prove that

Xn i=1

(−1)iti= 0.

Solution proposed by Roberto Tauraso, Scuola Normale Superiore, piazza dei Cavalieri, 56100 Pisa, Italy.

First, we prove the following Lemma:

Lemma. Let Γ be a circle of radius s and center O. Draw two circles CP and CQ externally tangent to Γ respectively at P and at Q. Then the common external tangent between the circles CP and CQ is

t= 2p

(s + r)(s + R)| sin(1 2P bOQ)|

where r and R are the radii respectively of CP and CQ.

Proof. We can assume without loss of generality that R ≥ r. Let A, B and D, C be the centers and the tangential points of the circles CP and CQ. In the trapezoid ABCD the angles B bCDand C bDA are right angles and therefore

t2= |AB|2− (|BC| − |DA|)2= |AB|2− (R − r)2. Moreover, if we apply the Carnot’s Theorem to the triangle AOB, we obtain

|AB|2 = |AO|2+ |OB|2− 2|AO||OB| cos(P bOQ)

= (s + r)2+ (s + R)2− 2(s + r)(s + R) cos(P bOQ).

Hence, it is easy to verify that

t2= 2(s + r)(s + R)(1 − cos(P bOQ)) = 4(s + r)(s + R) sin2(1 2P bOQ).

 Going back to our problem, assume that the circle Γ is centered in O with radius s and the circle C has radius R. Then, applying the Lemma to each couple of circles Oi and C for i = 1, 2, . . . , n,

Xn i=1

(−1)iti= 2p

(s + r)(s + R) Xn i=1

(−1)i| sin(1

2P bOAi)|.

This means that it suffices to solve the proposed problem when r = R = 0 and s = 1, i. e. we have to verify that

Xn i=1

(−1)i| sin(1

2P bOAi)| = 0.

(2)

Assume that A1, A2, . . . , An are the nth-complex roots of 1, hence A1OAb i = n(i − 1) for i = 1, 2, . . . , n. Let α = P bOA1 and n = 2m + 1 then 0 < α < n and

Xn i=1

(−1)i| sin(1

2P bOAi)| = Xm k=1

sin(1

2P bOA2k)−

Xm k=0

sin(1

2P bOA2k+1)

= Xm k=1

sin(α 2+π

n(2k − 1))+

Xm k=0

sin(α 2+π

n2k + π)

= Xm k=1

sin(α 2−π

n+2π n k)+

Xm k=0

sin(α 2−π

n+2π

n (k+m+1))

= Xn k=1

sin(α 2−π

n+2π n k) = 0.

The last sum corresponds to the imaginary part of the sum of the complex numbers A1, A2, . . . , An

rotated by α2πn which is zero. 

Riferimenti

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