Global Journal of Advanced Research
on Classical and Modern Geometries
ISSN: 2284-5569, Vol.3, (2014), Issue 2, pp.62-65
ANOTHER PROOF OF FLOOR VAN LAMOEN’S GENERALIZED THEOREM
NGUYEN MINH HA
A BSTRACT. Another proof of Floor van Lamoen’s generalized theorem will be introduced by using the algebraic form of Law of sine.
MSC 2010: 51M04, 51M25.
Keywords: signed distances, centroid, circumcenter.
1. I NTRODUCTION
In 2000, Dutch mathematician Floor van Lamoen gave a nice theorem on American Mathematical Monthly [1].
Theorem 1. If G is the centroid of triangle ABC and AG, BG, CG intersects with BC, CA, AB
at A1 , B1 , C1 respectively, then the circumcenters of six triangles GBC1 , GB1 C, GCA1 , GC1 A,
GAB1 , GA1 B are concyclic.
Theorem 1 is called Floor van Lamoen’s theorem, whose proof can be found in [1], [2],
[3] and [4].
In 2002, theorem 1 was expanded by A. Myakishev and then re-stated by Barry Wolk [5].
Theorem 2. If two triangles ABC and A1 B1 C1 have the same centroid and AA1 , BB1 , CC1 are
concurrent at O, then the circumcenters of six triangles OBC1 , OB1 C, OCA1 , OC1 A, OAB1 ,
OA1 B are concyclic.
Theorem 2 is called Floor van Lamoen’s generalized theorem, which was proven for the
first time in 2003 by Darij Grinberg [6].
In this article, I will introduce a different proof of theorem 2 by using the algebraic form
of Law of sine.
As in [7], the signed distances from the point A to the point B denoted by AB.
2. P ROOF OF THEOREM 2
Three lemmas are required.
−
→ −
−
→
→
→
→
Lemma 1. If −
α , β 6= 0 and k 6= 0 then sin(k−
α, β)=
k
|k|
−
→
→
sin(−
α , β ).
The author would like to thank Dao Thanh Oai for providing useful information for completing this
article.
62
Another proof of Floor van Lamoen’s generalized theorem
Proof.
In this proof, the following signs ↑↑ and ↑↓ refer to the co-directionality and contradirectionality of two vectors.
There are two cases to consider.
→
→
Case 1. k > 0. Noting that k−
α ↑↑ −
α , we have
−
→
−
→
−
→
−
→
−
→
k
→
→
→
→
→
→
→
sin(−
α , β ).
sin(k−
α , β ) = sin((k−
α,−
α ) + (−
α , β )) = sin(0 + (−
α , β )) = 1. sin(−
α, β)=
|k|
−
→
−
→
Case 2. k < 0. Noting that k α ↑↓ α , we have
−
→
−
→
−
→
−
→
−
→
k
→
→
→
→
→
→
→
sin(−
α , β ).
sin(k−
α , β ) = sin((k−
α,−
α ) + (−
α , β )) = sin(π + (−
α , β )) = (−1) . sin(−
α, β)=
|k|
−
→ →
→
Lemma 2. Given triangle ABC. −
α, β,−
γ are the direction vectors of lines BC, CA, AB respectively. Then
1)
2)
−→ −→
−
→ −→
−→ −
→
sin( AB, AC )
BA)
CB)
= sin( BC,
= sin(CA,
.
BC
CA
AB
−
→ −
−
→
→
−
→
−
→
−
→
sin( β , γ )
α, β)
γ, α)
= sin(CA
= sin(AB
.
BC
Proof.
−
→ →
→
Without the loss of generality, assume that ∆ABC has a positive direction and −
α, β,−
γ
are unit vectors.
According to Law of sine and lemma 1, we have
−→
[
sin( AB,
sin BAC
BC
=
=
−
→
CA
[
sin CBA
sin( BC,
Therefore,
Likewise,
In short,
1)
−
→
−→
→
AC )
γ , AC. β )
sin( AB.−
=
−→ =
→
→
sin( BC.−
α , BA.−
γ)
BA)
−
→ →
−→ −→
−
→ −→
−
→
→
sin( β , −
γ)
sin( AB, AC )
sin( BC, BA)
γ ,−
α)
=
and
= sin(CA
.
BC
CA
BC
−
→
−→ −
→
−
→ −→
−
→
−
→
−
→
α, β)
CB)
γ, α)
sin( BC, BA)
= sin(CA,
and sin(CA
= sin(AB
.
CA
AB
AB AC
AB . AC
BC BA
BC . BA
−
→
→
sin(−
γ, β)
.
→
→
sin(−
α,−
γ)
−→ −→
−
→ −→
−→ −
→
sin( AB, AC )
sin( BC, BA)
sin(CA,CB)
=
=
.
BC
CA
AB
−
→
−
→ →
→
→
→
sin(−
α, β)
sin( β , −
γ)
sin(−
γ ,−
α)
=
=
.
BC
CA
AB
2)
Lemma 2 is actually the algebraic form of Law of sine.
Lemma 3. Given triangle ABC and the following pairs of points (A1 , A2 ), (B1 , B2 ), (C1 , C2 )
belonging to lines BC, CA, AB respectively. If the following groups of four points (B1 , B2 , C1 ,
C2 ), (C1 , C2 , A1 , A2 ), (A1 , A2 , B1 , B2 ) are concyclic each, then six points A1 , A2 , B1 , B2 , C1 ,
C2 lie on the same circle.
Proof of lemma 3 is very simple and hence, not presented here.
Return to the proof of theorem 2.
Proof.[Proof of theorem 2]Let Ab , Ac , Bc , Ba , Ca , Cb be the circumcenters of triangles
OBC1 , OB1 C, OCA1 , OC1 A, OAB1 , OA1 B, respectively and X, Y, Z be the intersections
of the following pairs of lines (Ab Cb , Bc Ac ), (Bc Ac , Ca Ba ), (Ca Ba , Ab Cb ) respectively. Let
H, K be the projections of X on Ac Ca , Ab Ba respectively and let M, M1 , N, N1 be the
midpoints of OB, OB1 , OC, OC1 respectively.
63
Nguyen Minh Ha
−
→ → −
→
→
→
Direct the lines AA1 , BB1 , CC1 , YZ, ZX, XY respectively by unit vectors −
a, b,−
c, →
x,−
y,−
z.
Evidently, Bc Cb k YZ; XH k BB1 ; XK k CC1 . Therefore, without the loss of generality,
−
→ →
→
direct the lines Bc Cb , XH, XK respectively by vectors −
x, b,−
c.
Clearly, YZ, ZX, XY are perpendicular to AA1 , BB1 , CC1 respectively. Therefore, without
−
→
→
→
→
→
→
the loss of generality, assume that (−
x,−
a ) ≡ (−
y , b ) ≡ (−
z,−
c ) ≡ π2 ( mod2π ).
Since Bc Cb k YZ, according to lemma 2,
XBc
XCb
=
→
→
sin(−
x ,−
y)
−
→
−
sin( x , →
z)
(1).
Noting that XK = NN1 ; XH = MM1 ; Ab K k XY; XK k CC1 ; Ac H k XZ; XH k BB1 ,
−
→
→
→
→
according to lemma 2, noting that sin(−
z ,−
c ) = sin(−
y , b ) = 1, we have
→
sin(−
z,
XAb
XAb NN1 XH
=
=
.
.
−
→
sin( z ,
XK MM1 XAc
XAc
O
M
K
Ab
M1
Cb
H Ac
B
−
→
CC1
z)
−
→ = − BB .
1
b)
C1
N C N1
X
−
→
→
c ) 12 CC1 sin(−
y,
.
.
−
→
1
−
→
y ) 2 BB1 sin( y ,
Z
Ca
Bc
Ba
Y
B1
A1
A
Figure 1.
Because triangles ABC and A1 B1 C1 have the same centroid,
−
→
−−→
−−→
−→
−
→ −−→
→
c .
0 = AA1 + BB1 + CC1 = AA1 + BB1 b + CC1 −
−−→
→
From this, noting that −
x ⊥ AA1 , we can deduce that
64
Another proof of Floor van Lamoen’s generalized theorem
−
→
−
→
−
→ → −−→
→
→
→
→
→
→
→
0 =−
x.0 =−
x . AA1 + BB1 cos(−
x , b ) + CC1 cos(−
x ,−
c ) = BB1 cos(−
x , b ) + CC1 cos(−
x ,−
c ).
Therefore,
Hence,
−
→
→
(−
x, b)
= cos
−
→
→
cos( x , −
c)
→
→
sin(−
x ,−
y)
(2).
→
→
sin(−
x ,−
z)
1
− CC
BB1
XAb
XAc
=
=
−
→ →
→
→
cos (−
x ,−
y ) + (−
y, b)
→
→
→
→
cos (−
x ,−
z ) + (−
z ,−
c)
(
From (1) and (2), we can deduce that
)
XBc
XCb
=
=
→
→
cos((−
x ,−
y ) + π2 )
−
→
−
→
cos ( x , z ) + π
(
2
)
=
→
→
sin(−
x ,−
y)
.
→
→
sin(−
x ,−
z)
XAb
.
XAc
In other words, XAb .XCb = XAc .XBc .
This means that four points Ab , Ac , Bc , Cb are concyclic.
Similarly, the following groups of four points (Bc , Ba , Ca , Ac ) and (Ca , Cb , Ab , Ac ) are
concyclic as well.
In short, by lemma 3, Ab , Ac , Bc , Ba , Ca , Cb belong to the same circle.
R EFERENCES
[1] F. M. van Lamoen, Problem 10830, Amer. Math. Monthly, 2000 (107) 863; solution by the Mon thly editor,
2002 (109) 396-397.
[2] K. L. Li, Concyclic problem, Mathematical Excalibur, 6 (2001) Number 1, 1-2; available at
http://www.math.ust.hk/excalibur.
[3] A. Myakishev and Pter Y. Woo, On the Circumcenters of Cevasix Configuration, Forum Geom, 3 (2003)
57-63.
[4] Nguyen Minh Ha, Another Proof of van Lamoen’s Theorem and Its Converse, Forum Geom, 5 (2005)
127-132.
[5] A. Myakishev and Barry Wolk, Hyacinthos message 5612, May 31, 2002.
[6] Darij Grinberg, Hyacinthos message 7351, Jul 13, 2003.
[7] Roger A. Johnson, Advanced Euclidean Geomertry, p.2, Dover Publications, Inc, N.Y. (1960).
H ANOI U NIVERSITY OF E DUCATION ,
H ANOI , V IETNAM
E-mail address: [email protected]
65