2. Equilateral Triangles
Recall the well-known theorem of van Schooten.
Theorem 1
If ABC is an equilateral triangle and M is a
point on the arc BC of C(ABC) then
|M A| = |M B| + |M C|.
Proof
(Figure 1)
Use Ptolemy on the cyclic quadrilateral ABM C.
In fact, using the Ptolemy inequality for quadrilaterals, we get
the following van Schooten inequality.
Figure 1:
Theorem 2
Let ABC be an equilateral triangle. Then if M is any
point in the plane of ABC we have
|M A| ≤ |M B| + |M C|.
1
Pompeiu Triangle
We get the following well-known theorem of D. Pompeiu as an immediate
consequence of the previous inequality.
Theorem 3
Let M be any point in the plane of an equilateral triangle
ABC. Then the distances |M A|, |M B| and |M C| can be the sidelengths of a
triangle.
1
Proof
It follows immediately since
|M A| ≤ |M B| + |M C|
The triangle is degenerate if M lies on the arc BC of the circumcircle
C(ABC).
A triangle with side lengths |M A|, |M B| and |M C| is called a Pompeiu triangle. When M is in the interior of ABC, then the pompeiu triangle can be
explicitly constructed.
Locate N so that the triangle BN M is equilateral. Now consider
the triangles
We have
and
AM B and BN C,
AB = BC,
BM = BN
b
b = C BN
b .
M BA = 60◦ − M BC
Thus the triangles are similar, in fact CBN is got by rotating
triangle ABM through 60◦ anti-clockwise in the diagram.
Thus AM = CN,
Figure 2:
and so the triangle N M C has side lengths equal to
|M A|, |M B| and |M C|. Thus N M C is the Pompeiu triangle.
The measure of the angles of the Pompeiu triangle in terms of the angles
subtended at M by the vertices of ABC and the area of the Pompeiu triangle
are given by the following result of the distinguished Romanian born Sabin
Tabirca
Theorem 4 (Tabirca)
If ABC is an equilateral triangle, and M is an
interior point of ABC, then the angles of the Pompeiu triangle and its area
are as follows:
cC − 60◦ , C M
cA − 60◦ and AM
cB − 60◦ ;
(a) the 3 angles are the angles B M
2
(b) the area S (Pompeiu
√ triangle) is
1
3
(area of ABC)−
|M O|2 ,
3
4
where O is circumcentre of the triangle ABC.
Proof
(a) In the triangle N M C,
cN
CM
cB − N M
cB
= CM
cB − 60◦ ,
= CM
bM
CN
bB − MN
bB
= CN
b M − 60◦
= CN
cB − 60◦ ,
= AM
Figure 3:
b
and, finally, M CN
Notation:
S(ABC).
=
=
=
=
cN + C N
bM}
180 − {C M
◦
◦
cB − 60 + AM
cB − 60◦ }
180 − {C M
cC}
300◦ − {360◦ − AM
cC − 60◦ }.
{AM
◦
The area of a triangle ABC is denoted by
(b) In the diagram, N M C is the Pompeiu triangle which we
now denote by Tp . Then:
S(Tp )
=
=
=
=
=
where a = |BC| = |CA| = |AB|.
3
1
cN )
(|CM |.|M N |) sin(C M
2
1
cB − 60◦ )
|CM |.|BM | sin(C M
2
√
1
1
3
c
c
|CM |.|BM |{sin(C M B). − cos(C M B).
}
2
2√
2
1
cB) − 3 |CM ||M B| cos(C M
cB)
|CM |.|BM | sin(C M
4
4
√
1
3
S(CM B) −
{|CM |2 + |BM |2 − a2 },
2
8
√
Thus S(Tp ) = S(CM B) −
3
{|CM |2 + |BM |2 − a2 }.
8
Similarly we can show that
√
1
3
S(Tp ) = S(CM A) −
{|CM |2 + |M A|2 − a2 },
2
8 √
1
3
and S(Tp ) = S(BM A) −
{|BM |2 + |M A|2 − a2 }.
2
8
Adding, we get
√
3
1
3S(Tp ) = S(ABC) −
{2(|M A|2 + |M B|2 + |M C|2 ) − 3a2 }.
2
8
Recall the Leibniz formula which states that for any triangle ABC with
centroid G and point M
1
|M A|2 + |M B|2 + |M C|2 = 3|M G|2 + {|AB|2 + |BC|2 + |CA|2 }
3
In the case of an equilateral triangle, G = 0, the centre of the circumcircle
and a2 = |AB|2 = |BC|2 = |CA|2 so |M A|2 +|M B|2 +|M C|2 = 3|M O|2 +a2 .
Thus
√
1
3
3S(Tp ) = S(ABC) −
{6|M O|2 + 2a2 − 3a2 }
2
8
√
√
3
3 2
1
6|M O|2 +
a.
= S(ABC) −
2
8
8
√
3 2
Since ABC is equilateral with side length a, S(ABC) =
a , so
4
√
1
3 3
1
3S(Tp ) = S(ABC) −
|M O|2 + S(ABC)
2
4
√ 2
3 3
= S(ABC) −
|M O|2 .
4
√
1
3 3
Thus S(Tp ) = S(ABC) −
|M O|2 , as required.
3
4
4
2
Fermat-Toricelli Point
Let ABC be any triangle and on each side construct externally three equilateral triangles ABC1 , BCA1 and CAB1 .
Then we have the following theorem.
Theorem 5
(a) The three circumcircles of the equilateral triangles intersect in a point T , i.e.
C(ABC1 ) ∩ C(BCA1 ) ∩ C(CAB1 ) = {T }.
(b) The lines AA1 , BB1 and CC1 are concurrent at T , i.e.
AA1 ∩ BB1 ∩ CC1 = {T }.
(c) |AA1 | = |BB1 | = |CC1 | = |T A| + |T B| + |T C|.
Figure 4:
(d) For all points M in the plane of ABC,
|M A| + |M B| + |M C| ≥ |AA1 | = |T A| + |T B| +
|T C|.
i.e. the point T minimises the expression |M A| + |M B| + |M C|. The point
T is called the Toricelli-Fermat point.
Proof
(a) Let {A, T } be the intersection points of the circles
C(ABC1 ) and C(ACB1 ).
Then since
C1 BT A is cyclic,
b1 B = 120◦ .
ATbB = 180◦ − AC
Because
B1 AT C is cyclic,
ATbC = 180◦ − AB1 C = 120◦ .
Thus
Thus
BT C = 120◦ also.
b1 C = 180◦ .
B TbC + B A
5
Figure 5:
and so BA1 CT is cyclic, i.e. T ∈ C(A1 BC)
(b) We claim that C1 , T, C are collinear points.
ATbC = 120◦ ,
b 1 = 60◦
ATbC1 = ABC
giving AT C + AT C1 = 180◦ , i.e.
C, T and C1
are collinear.
Similarly A, T, A1 and B, T, B1 are
collinear.
(c) We claim that |CC1 | = |T A| + |T B| + |T C|.
Since T ∈ C(AC1 B) and AC1 B is equilateral, then by
van Schooten’s theorem
|T C1 | = |T A| + |T B|.
Thus |CC1 | = |CT | + |T C1 | = |T C| + |T A| + |T B|, as
required. Similarly for |AA1 | and |BB1 |.
(d) Now let M be any point in the plane of ABC. Then,
since ABC1 is equilateral:
|M C1 | ≤ |M A| + |M B|
Thus
|M A| + |M B| + |M C|
≥ |M C| + |M C1 |
≥ |CC1 |
= |T A| + |T B| + |T C|
So the point of a triangle which minimises the sum of the distances to the
three vertices is the Toricelli-Fermat point. One could ask the question of
weighted distances to the vertices and ask which point(s) minimise weighted
sums. This is the question we now investigate.
6
Generalised Fermat-Toricelli Theorem
Let x, y and z be the side length of a triangle
αβγ with x the length of the side opposite vertex
α, y the length of the side opposite β and z the
length of the side opposite γ.
On an arbitrary triangle ABC construct externally 3 triangles similar to αβγ with vertices positioned as indicated in Figure 6.
(a) Then their circumcircles intersect at a point
T1 , i.e.
C(ABC1 ) ∩ C(BCA1 ) ∩ C(CAB1 ) = {T1 }
(b) The lines AA1 , BB1 and CC1 are concurrent,
i.e.
AA1 ∩ BB1 ∩ CC1 = {T1 }
(c) x|AA1 | = y|BB1 | = z|CC1 |
= x|AT1 | = y|BT1 | = z|CT1 |
Figure 6:
(d) For any point M in the plane of ABC,
x|M A| + y|M B| + z|M C| ≥
x|AA1 | = x|AT1 | + y|BT1 | + z|CT1 |
Thus the point T1 minimises the weighted distances
of a point to the vertices.
Proof
The construction of the proof is similar to the proofs in the special case when x = y =
z.
(a) Let C(ABC1 ) ∩ C(ACB1 ) = {A, T1 }.
7
Figure 7:
Since
AT1 BC1 is cyclic,
ATb1 B = 180◦ − γ
b,
and
AT1 CB1 is cyclic, so
b
ATb1 C = 180◦ − β.
Thus
B Tb1 C
= 360◦ − {ATb1 B + ATb1 C}
= 360◦ − {180◦ − γ
b + 180◦ − βb
b =α
= 180◦ − {b
γ + β}
b.
Thus T1 BA1 C is cyclic, i.e. T1 ∈ C(BA1 C), as required.
(b) We claim that ATb1 C1 + ATb1 C = 180◦ and from
this it follows that T1 lies on CC1 . In a similar way, we get that T1 also
belongs to the line segments BB1 and AA1 .
To show that ATb1 C1 + ATb1 C = 180◦ , we have, since, AT1 CB1 is cyclic,
b
AT1 C = 180◦ − β.
b 1 = 180◦ , as required.
Also, ATb1 C1 + ABC
(c) Since the triangles A1 BC and αβγ are similar,
then
|A1 B|
z
=
|A1 C|
|BC|
=
.
y
x
Thus
z
|A1 B| = |BC| ,
x
and
y
|A1 C| = |BC| .
x
Since T1 ∈ C(BCA1 ), then, by Ptolemy,
Figure 8:
8
|T1 A1 |.|BC|
= |BT1 ||CA1 | + |CT1 ||BA1 |
y
z
=|BT1 |.|BC| + |CT1 ||BC|
x
x
Dividing across by |BC| and multiplying by x, we
get x|T1 A1 | = y|BT1 | + z|CT1 |.
Thus
x|T1 A1 | + y|BT1 | + z|CT1 |
= x|T1 A| + x|T1 A1 | = x|AA1 |.
Similarly, we can show that x|T1 A|+y|T1 B|+z|T1 C| =
y|BB1 | = z|CC1 |
(d) Now take a point M 6∈ C(BCA1 ). Then, by the Ptolemy inequality,
|BM ||CA1 | + |CM |.|BA1 | > |M A1 ||BC|
Proceeding as in (c) above, we get
x|M A| + y|M B| + z|M C| > x|AA1 | = x|T1 A| + y|T1 B| + z|T1 C|
as required.
Remarks
1. Now suppose that the positive weights x, y and z are not the sides of a
triangle, i.e. suppose
x≥y+z
What then is the point which minimises the quantity
x|M A| + y|M B| + z|M C|?
where M is any point in the plane of ABC.
To decide this, consider
9
x|M A| + y|M B| + z|M C| ≥ y(|M A| + |M B|) + z(|M A| + |M C|), since x ≥ y + z
≥ y|AB| + z|AC|
= x|AA| + y|AB| + z|AC|
Thus the point A(vertex) minimises the quantity x|M A| + y|M B| +
z|M C|
2. Suppose we take
b = sin A,
b
x = sin(B AC)
b = sin B,
b
y = sin(ABC)
b
b
z = sin(B CA) = sin C,
then the weighted expression
b
b
b
sin(A)|M
A| + sin(B)|M
B| + sin(C)|M
C|
is minimised when M = O the centre of the circumcircle of ABC
b y = sin(B)
b and z = sin(C),
b then
3. If we take x = sin(A),
b
b
b
sin(A)|M
A| + sin(B)|M
B| + sin(C)|M
C|
is minimised when M = H, the orhthocentre.
b
b
b
A
B
C
4. If x = sin( ), y = sin( ) and z = sin( )
2
2
2
then
b
b
b
C
A
sin B
sin( )|M A| + (
)|M B| + sin( )|M C|
2
2
2
is minimised when M = I, the incentre.
10