Proof 2 (Michel Cabart) Let A be a point inside the polygon, n i unit vectors
perpendicular to the ith side of the polygon, and H i the feet of
the perpendicular from A to the ith side. Since the polygon is
equiangular, the angles between successive vectors n i are equal,
so that ∑ n i =0.
The scalar product (AX, n i ), with X on the i th side, does not
depend on the position of X. The sum of distances from A to the sides of the polygon is:
S A =∑AH i =∑(AH i , n i ). For another point B with G i being the feet of the perpendicular
from B to the ith side of the polygon, the distance is:
S B =∑BG i =∑(BG i , n i ) =∑(BH i , n i ). So that:
S A − S B = (AB, ∑n i ) = 0.
5. Converse of Viviani’s theorem The converse of Viviani’s Theorem for equilateral triangles was proved to hold true (Zhibo Chen, Tian
Liang, 2012). The converse theorem states: The proof of Zhibo Chen and Tian Liang was greatly different from others since they approached the
theorem in an unique way other than the usual approach which is using formula for area sum. Besides, this
was also one of the earliest vector approach solution to Viviani’s theorem and its related problems. Proof Let P be a point in
ℛ, and let ⃗, ⃗, ⃗ be the unit vectors
from P perpendicular to the sides of the triangle (see
figure). Our goal is to show that the angle between any
two of these vectors is 120◦, from which it will follow that
each angle of the triangle is 60◦.
To this end, we first show that the sum of these
vectors,
⃗ =⃗ + ⃗ + ⃗, is 0. Suppose not. From our
hypothesis, it follows that there is a point P’ in R so that
,
′⃗ is parallel to ⃗. Let ⃗ denote the vector
′⃗, and let
θ be the angle between
⃗ and ⃗. Further, let d 1 , d 2 , and
d 3 be the distances from P to the sides of ABC, and let d’ 1 , d’ 2 , and d’ 3 be the corresponding distances from P’. Note that by hypothesis d 1 + d 2 + d 3 = d’ 1 + d’ 2 + d’ 3 On the one hand,
cos θ = | ⃗|
, while on the other hand,
cos θ = ⃗ . ⃗
| ⃗|
(since
⃗is a unit vector).
Theorem 6. If, inside ABC, there is a circular region for which the sum of the distances from a point P in R to the three sides of the triangle is independent of the position of P, then ABC is equilateral.
2020 Singapore Mathematics Project Festival Viviani’s Theorem and its Related Problems
9
Hence,
⃗ . ⃗ =
−
and by symmetry
⃗ . ⃗ =
−
⃗ . ⃗ =
−
It follows that
⃗ · ⃗ = 0, and since these two vectors are parallel, it must be that | ⃗| = 0, a contradiction.
From this it follows that, for i = 1, 2, and 3,
⃗ · ( ⃗ + ⃗ + ⃗) = 0. It is now straightforward to show that:
⃗ . ⃗= ⃗ . ⃗ = ⃗ . ⃗=− .
Consequently, the angle between any pair of these vectors is 2π3, and so ABC must be equilateral.