RESULTS
I.
VIVIANI’S THEOREM AND ITS EXTENSION
1. Vectors approach
Theorem 1.1 All regular polygons possess CVS property
Lemma 1:
Let N be a point on AB such that MN // AC. (1.1)
By Thales’ Theorem:
=
;
=
(1.2)
From (1.1) and (1.2):
⃗ =
⃗ +
⃗ (Q.E.D)
M is a point on side BC of triangle ABC. We have the equation:
⃗ =
⃗ +
⃗
2020 Singapore Mathematics Project Festival Viviani’s Theorem and its Related Problems
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Thales’ Theorem: Given triangle ABC, D and E are on AB and AC respectively such that DE is parallel
to BC. The following result holds:
Lemma 2:
Let A’ be the point of intersection of AI and BC. By characteristic of the bisector:
′
′
=
=
=
=
(2.1)
And
=
=
=
(2.2)
In triangle IBC, from Lemma 1:
′⃗ =
⃗ +
⃗ (2.3)
From (2.1) and (2.3):
′⃗ =
⃗ +
⃗
(2.4)
Also,
′⃗ = -
⃗ (2.5)
From (2.2) and (2.5):
′⃗ = -
⃗ (2.6)
From (2.4) and (2.6):
a ⃗
+ b ⃗ + c ⃗ = 0 (*)
Let AE = AF = x; BF = BD = y; CD = CE = z
y + z = a, z + x = b, x + y = c
The inscribed circle of triangle ABC touches each side BC, CA, AB at D, E, F, respectively. Denote
length of BC, AC, AB as a, b, c. We have the equation:
a ⃗ + a ⃗ +
⃗ =
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In triangle IBC, ICA, IAB, from Lemma 1:
⃗ = ⃗ + ⃗
⃗ = ⃗ + ⃗
⃗ =
⃗ +
⃗
a ⃗
+ a ⃗ +
⃗ = ( + ) ⃗ + ( + ) ⃗ + ( + ) ⃗
a ⃗
+ a ⃗ +
⃗ =
⃗ +
⃗ + ⃗ (**)
From (*) and (**):
a ⃗
+ a ⃗ +
⃗ = 0 (Q.E.D)
Lemma 3.
Given a convex polygon
….
with unit vectors
⃗
(1
≤ ≤ ) directing outwards of the polygon and are
perpendicular to
⃗
(considering
≡
) respectively.
We are going to prove that:
⃗ +
⃗ + ⋯ +
⃗ =
0⃗
For n=3, the polygon is a triangle. From lemma 2, the result holds
true for n=3 (3.1)
Supposing the theorem is true for all (
− 1)-sided polygon (n >
4) (3.2)
Construct a triangle A
1
A
n-1
A
n
outside the (
− 1)-sided polygon
such that the new n-sided polygon is a convex polygon.
Then, construct unit vector
⃗ perpendicular to
, directing
outwards of triangle
.
Apply (3.1) and (3.2) correspondingly to triangle
and (
− 1)-sided polygon:
⃗
+
(− ⃗) +
⃗ = 0⃗ (3.3)
⃗ +
⃗ + ⋯ +
⃗ = 0⃗ (3.4)
Take (3.3) + (3.4), it holds:
⃗ +
⃗ + ⋯ +
⃗ = 0⃗ (3.5)
Therefore, using induction, the theorem applies for n-sided convex polygons.
When the polygon is regular:
A
1
A
2
= A
2
A
3
= A
3
A
4
=…= A
n
A
1
= k
( k is the side length of the polygon)
The sum of unit vectors perpendicular to the sides of a regular polygon is a zero vector.
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From (3.5):
⃗ +
⃗ + ⋯ +
⃗ = 0⃗
. ⃗ + . ⃗ + ⋯ + . ⃗ = 0⃗
. ∑
⃗ = 0
∑
⃗ = 0
Proof of Theorem 1.
Construct unit vectors perpendicular and directed to the sides of given polygon from interior point P.
Let V: R → R be the function of the sum of distance from an interior point to sides of the polygon. Any
polygon that has constant sum of distance V is said to have CVS property. It holds:
∠
=
=
.
=
.
1 .
∠
=
⃗. ⃗
=
⃗. ⃗
Similarly:
=
⃗. ⃗
=
⃗. ⃗
…
=
⃗. ⃗
The sum of the unit vectors from interior point P that are perpendicularly directed to all the sides of the
polygon is
( )
= ∑
⃗ ∙ ⃗
Choosing another interior point Q, similarly, the sum of distances from interior point Q to sides of the
polygon is
( ) = ∑
⃗ ∙ ⃗
The polygon possesses CVS property if the sum of distances of any given interior point of the polygon is a
constant. Therefore, in order to prove that the regular polygon possesses CVS property, we have to prove
V(P)= V(Q).
V(P) – V(Q) = ∑
⃗ ∙ ⃗ − ∑
⃗ ∙ ⃗
= ∑
⃗ (
⃗ −
⃗ ) = ⃗ ∙ ( ∑
⃗ ) = 0⃗ (*)
From Lemma 3, the sum of unit vectors perpendicular to the sides of a regular polygon is a zero vector.
Hence, the above result is a zero vector.
From here we can conclude that for any two random points P and Q chosen in the given polygon, ( )
=
( )
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Thus, the regular polygon possesses CVS property.
Similarly, to the proof of Theorem 1.1, let P and Q be two arbitrary points in the given polygon.
It holds:
( ) = ∑
⃗ ∙ ⃗
( ) = ∑
⃗ ∙ ⃗
V(P) – V(Q) = ∑
⃗ ∙ ⃗ − ∑
⃗ ∙ ⃗ = ∑
⃗ (
⃗ −
⃗) = ⃗ ∙ ( ∑
⃗ ) = 0⃗ (*)
Since
∑
⃗ ) = 0⃗, (*) holds true: V(P) = V(Q). Thus, the polygon possesses CVS property.
Similarly, P and Q are interior points of the given polygon. From the assumption:
V(P) = V(Q) = k (a constant)
V(P) = ∑
⃗ ∙ ⃗
V(Q)
= ∑
⃗ ∙ ⃗
V(P) – V(Q) =
0⃗
( ) − ( ) = ∑
⃗ ∙ ⃗ − ∑
⃗ ∙ ⃗ =
⃗ ∙ ( ∑
⃗ ) = 0⃗ (*)
Since P and Q are two distinct points,
⃗ is not a zero vector.
However, from here, we cannot conclude that the sum of unit vectors that are perpendicular to the sides of
the polygon is a zero vector even though vector PQ is a non zero vector. Noted that this conclusion is the
key condition for a convex polygon to have CVS property.
In fact, from (*), the deduction should be:
∑
⃗ = 0⃗ (4.1)
Or ∑
⃗ = ⃗ and ⃗ ⊥
⃗ (4.2)
When
⃗is perpendicular to the sum-vector. This results in ⃗ ∙ ∑
⃗ = 0, regardless of whether or not
the polygon has CVS property.
Choosing another arbitrary point Z inside the given polygon. Since the polygon possesses CVS property:
V(P)=V(Q)=V(Z)
V(P) – V(Q) = 0
∑
⃗ ∙ ⃗ − ∑
⃗ ∙ ⃗ = ∑
⃗ ∙ ⃗ = ⃗ ∙ (∑
⃗) = 0⃗ (4.3)
Theorem 1.3 If a polygon possesses CVS property, the sum of unit vectors
perpendicular to the sides of the polygon is a zero vector.
Theorem 1.2 If the sum of unit vectors perpendicular to the sides of the polygon is a
zero vector, the polygon possesses CVS property.
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V(Q) – V(Z)=0
∑
⃗ ∙ ⃗ − ∑
⃗ ∙ ⃗ = ∑
⃗ ∙ ⃗ = ⃗ ∙ (∑
⃗) = 0⃗ (4.4)
If P, Q and Z are collinear, the result (4.3) obviously leads to (4.4) without the polygon possessing CVS
property. In this case, choosing another point Z simply does not make any more significant deduction than
using two points P and Q.
If P, Q and Z are non-collinear, we can conclude that the sum of unit vectors is a zero vector. If the sum of
unit vector is a non-zero vector,
⃗ and ⃗ must be both perpendicular to this sum-vector. Therefore, P,
Q and Z are on the same line (which is contradictory to our assumption).
Thus, using 3 non-collinear points, our hypothesis on the converse of Theorem 1.2 has been proved.
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