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Compute the sum of the squares of distances from the center of O to the incenters of the triangles of∆ and show that this sum is independent of

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Problem 11116

(American Mathematical Monthly, Vol.111, November 2004) Proposed by D. Veljan (Croatia).

Let P be a convex n-gon inscribed in a circle O, and let∆ be a triangulation of P without new vertices. Compute the sum of the squares of distances from the center of O to the incenters of the triangles of∆ and show that this sum is independent of ∆.

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

Let Ik, rk, ak, bk and ck be respectively the incenter, the radius and the three sides of the kth triangle of ∆ for k = 1, · · · , n − 2. Then, by Euler’s Theorem, |OIk|2 = R(R − 2rk) and we have

that nX−2

k=1

|OIk|2=

nX−2 k=1

R(R − 2rk) = (n − 2)R2− 2R

n−2

X

k=1

rk.

Moreover, by Carnot’s Theorem, the signed sum of perpendicular distances from the circumcenter O to the sides of any triangle of ∆ is

Oak+ Obk+ Ock= R + rk.

Note that the sign of the distance is chosen to be positive if and only if the segment intersects the interior of the triangle and therefore, since the n-gon is convex and cyclic, we have that

n−2

X

k=1

(Oak+ Obk+ Ock) = (n − 2)R +

nX−2 k=1

rk = d(O, P )

where d(O, P ) is the signed sum of the perpendicular distances from the circumcenter O to the sides of the n-gon P , which does not depend on the triangulation. Thus also the sum of the squares of distances from the center of O to the incenters of the triangles of ∆ is independent on ∆ and it is equal to

nX−2 k=1

|OIk|2= (n − 2)R2− 2R(d(O, P ) − (n − 2)R) = R (3(n − 2)R − 2d(O, P )) .

This statement riminds a well known Sangaku problem (also called Japanese Theorem) which says that for this kind of triangulations the sum of the inradii is constant. 

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