Triangle Bordered With Squares

Triangle Bordered With SquaresC˘at˘alin Barbu Abstract The triangle determined by the centers of the squares built outside on the sides of an acute triangle is called the exterior Vecten

Triangle Bordered With Squares

C˘at˘alin Barbu

Abstract The triangle determined by the centers of the squares built outside on the sides of an acute triangle is called the exterior Vecten triangle Similarly we define the interior Vecten triangle if the squares are built to the interior of the triangle In this article we give properties for squares with sides equal to half of triangles’s sides that are built with respect to the midpoints of the sides of the triangle.

Denote by Oa0, O0b, O0c the centers of the squares AbA0bA0cAc, BcBc0Ba0Ba, CaCa0Cb0Cb, Ab, Ac ∈ [BC], Bc, Ba ∈ [CA], Ca, Cb ∈ [AB] constructed externally on the sides BC, CA, AB; by O”

a, Ob”, Oc” the centers of the squares AbA”bA”cAc, BcBc”Ba”Ba, CaCa”Cb”Cb constructed internally; by Ma, Mb,

Mc, A0a, B0b, Cc0 the midpoints of the segments BC, CA, AB, B0aCa0, Cb0A0b, A0cBc0; by G0a, G0b, G0c,

G”a, G”b, G”c the centroids of triangles Ob0MaOc0, O0cMbO0a, Oa0McOb0, O”bMaOc”, O”cMbO”a, O”aMcO”b;

by G01, G02, G03, G”1, G”2, G”3 the centroids of triangles Oa0Ob”O”c, O0bO”cO”a, Oc0O”aO”b, Oa”Ob0O0c, Ob”Oc0O0a,

Oc”O0aOb0, and {A0} = Bc0Ba0 ∩ Ca0Cb0, {B0} = Ca0Cb0∩ A0bA0c, {C0} = A0bA0c∩ Bc0Ba0

Let O be the circumcenter of triangle ABC Consider a complex plane with origin at O Denote by the corresponding lowercase letter the coordinate of a point denoted by an uppercase letter Then

ma= b + c

2 , ab=

3b + c

4 , ac=

3c + b

Point Oa0 is obtained from point Ab by a rotation of center Ma and angle 90◦ Hence

o0a= ma+ i(ab− ma) = b + c

2 + i ·

b − c

4 . Likewise,

o0b = c + a

2 + i ·

c − a

4 , o

0

c= a + b

2 i ·

a − b

o”a = b + c

2 − i ·

b − c

4 , o

”

b = c + a

2 − i ·

c − a

4 , o

”

c = a + b

2 − i ·

a − b

a0b = 3b + c

4 + i ·

b − c

2 , a

0

c= 3c + b

4 + i ·

b − c

2 b

0

c= 3c + a

4 + i ·

c − a

2 ,

b0a= 3a + c

4 + i ·

c − a

2 , c

0

a = 3a + b

4 + i ·

a − b

2 , c

0

b = 3b + a

4 + i ·

a − b

a”b = 3b + c

4 − i ·

b − c

2 , a

”

c = 3c + b

4 − i ·

b − c

2 , b

”

c= 3c + a

4 − i ·

c − a

2 ,

b”a= 3a + c

4 − i ·

c − a

2 , c

”

a= 3a + b

4 − i ·

a − b

2 , c

”

b = 3b + a

4 − i ·

a − b

The coordinates of the midpoints Aa, Bb, Cc of the segments BaCa, CbAb, AcBc are

a0a= 6a + b + c

c − b

4 , b

0

b = 6b + c + a

a − c

4 , c

0

c= 6c + a + b

b − a

Theorem 1 Lines AOa0, BO0b and CO0c are concurrent

Proof Because triangles BO0aC, CO0bA and AOc0B are similar isosceles triangles, the concurrence follows from Kiepert’s Theorem [1]

Theorem 2 Lines AOa”, BO”b and COc” are concurrent

Proof The proof is similar to the previous one

Theorem 3 Lines AA0, BB0 and CC0 are concurrent

Proof The proof follows from the fact that triangles ABC and A0B0C0 have parallel sides

Remark 1 If the squares have the sides equal with the sides of the triangle, then lines AOa0, BO0b and COc0 are concurrent in the external Vecten point, lines AOa”, BO”b and COc” are concurrent in the internal Vecten point, AA0, BB0 and CC0 are concurrent in the point of Lemoine, the A0B0C0 triangle beeing Grebe’s triangle of triangle ABC [1]

Theorem 4 Triangles ABC, Oa0Ob0O0c, Oa”Ob”Oc”, A0bBc0Ca0, A0cBa0Cb0, A”bBc”Ca”, A”cBa”Cb”have the same centroid G

Proof Triangles ABC, OaObOc, Oa”Ob”Oc”, AbBcCa, AcBaCb, A”bBc”Ca”, A”cBa”Cb” have the same cen-troid because

o0a+ o0b+ o0c

o”

a+ o”

b + o” c

a0b+ b0c+ c0a

a0c+ b0a+ c0b

a”b+ b”c+ c”a

a”c+ b”a+ c”b

a + b + c

Theorem 5 Lines A0aO0a, Bb0O0b and Cc0O0c are concurrent

Proof The equations of lines A0aOa0, B0bOb0 and Cc0Oc0 are:

zo0a− a0

a



− zo0a− a0a+ o0aa0a− a0ao0a= 0,

z



o0b− b0b− zo0b− b0b+ o0bb0b− b0bo0b = 0, and

zo0c− c0

c



− zo0c− c0c+ o0cc0c− c0co0c= 0

By adding up the previous equations we obtain

o0aa0a− a0ao0a+ o0bb0b− b0bo0b+ o0cc0c− c0co0c= 0

The last relation is equivalent to

[a(b + c) − a(b + c)] + [b(c + a) − b(c + a)] + [c(b + a) − c(b + a)] = 0,

which is clear

Theorem 6 Lines A0aO”a, Bb0O”b and Cc0Oc” are concurrent

Proof The proof is similar with Theorem’s 6 proof

Theorem 7 Lines A0aMa, Bb0Mb and Cc0Mc are concurrent

Proof The proof is similar with Theorem’s 6 proof

Theorem 8 Lines A0Oa0, B0Ob0 and C0Oc0 are concurrent

Proof The proof is similar with Theorem’s 6 proof

Theorem 9 Lines A0Oa”, B0Ob” and C0O”c are concurrent

Proof The proof is similar with Theorem’s 6 proof

Theorem 10 The following relations are true: AOa0⊥Ba0Ca0, BOb0⊥A0bCb0, COc0⊥A0cBc0 and AOa0 =

Ba0Ca0, BOb0 = A0bCb0, COc0 = A0cBc0

Proof Because b0a − c0a = i b+c−2a2 + ib−c4

= i(o0a − a), we have AOa0⊥Ba0Ca0 and

b0a− c0a

=

i(o0a− a)

=

(o0a− a)

, hence AOa0 = Ba0Ca0 In the same way we demonstrate the other relations

Theorem 11 If by A[XY Z] we denote the area of triangle XY Z, then

A[O0

a O0bO0c]= 7

16A[ABC]+

1 16

 3R2+BC

2+ CA2+ AB2 2



Proof Because

A[ABC]= i

4

= i

4[bc + ca + ab − ac − ba − cb]

and

BC2+ CA2+ AB2 = |c − b|2+ |a − c|2+ |b − a|2 = (c − b)(c − b) + (a − c)(a − c) + (b − a)(b − a) = 2[3R2− (bc + ca + ab + ac + ba + cb)],

we have

bc + ca + ab + ac + ba + cb = 3R2−BC

2+ CA2+ AB2 2

(where |a| = |b| = |c| = R is circumradius of triangle ABC) We obtain

A[O0

a O0bO0c]= i

4

o0a o0a 1

o0b o0b 1

o0c o0c 1

= i

4 ·

1

16 · [7(bc + ca + ab − ac − ba − cb) + 4i(bc + ca + ab + ac + ba + cb) − 24iR

2],

hence

A[O0

a O0bO0c]= 1

16

 7A[ABC]−

 3R2−BC

2+ CA2+ AB2

2

 + 6R2



= 1 16

 7A[ABC]+ 3R2+BC

2+ CA2+ AB2 2



Theorem 12 If by A[XY Z] we denote XY Z triangle’s area, then

A[O”

a O ” O ” ]= 7

16A[ABC]−

1 16

 3R2+BC

2+ CA2+ AB2 2



Proof The proof is similar to the previous one

Theorem 13 If {P1} = O0aO0c∩ BC, {P10} = O0aO0b∩ BC, {Q1} = O0bOa0 ∩ CA, {Q01} = Ob0Oc0∩ CA, {R1} = Oc0O0b∩ AB, {R01} = O0cO0a∩ AB, then points P1, P10, Q1, Q01, R1 and R01 are on the same

Proof Because ABC and OaObOc are homological triangles (See Theorem 1), then by using Salmon’s theorem [2] the conclusion follows

Theorem 14 If P2} = O”

aO”c∩ BC, {P20} = O”

aO”b∩ BC, {Q2} = O”

bOa”∩ CA, {Q02} = O”

bO”c∩ CA, {R2} = O”

cOb”∩ AB, {R02} = O”

cO”a∩ AB, then points P1, P10, Q1, Q01, R1 and R02 are on the same conic

Proof The proof is similar to the previous one

Theorem 15 The following relations are true: G0aG⊥BC, G0bG⊥CA, G0cG⊥AB and BC = 12G0aG,

CA = 12G0bG, AB = 12G0cG

Proof We have ga0 = ma +o0b+o0c

3 = g + i · c−b12, hence g

0

a −g c−b = i ·121 Since G0aG⊥BC and

ga0−g c−b

=

i · 121= 121 , we obtain 12G0aG = BC

Theorem 16 The following relations are true: G”aG⊥BC, G”bG⊥CA, G”cG⊥AB and BC = 12G”aG,

CA = 12G”bG, AB = 12G”cG

Proof We have ga”= ma+o

” +o”

3 = g − i · c−b12 , then G”aG⊥BC and 12G”aG = BC

Corollary 1 The centroid of the triangle ABC is the midpoint of the segments G0aG”a, G0bG”b and

G0cG”c

Theorem 17 Triangles G0aG0bG0c, G”aG”bG”c and ABC have the same centroid G

Proof We have g

0

a +g0b+gc0

3 = g”a +g ” +g ”

3 = a+b+c3 , as desired

Theorem 18 The following relations are true: G0aG0b⊥AMa, G0cG0a⊥BMb, G0aG0b⊥CMcand AMa= 6G0bG0c, BMb = 6G0cG0a, CMc= 6G0aG0b

Proof We have gb0 − gc0 = 6i(a − ma), then G0bG0c⊥AMaand AMa= 6 · G0bG0c

Theorem 19 The following relations are true: G”bG”c⊥AMa, G”cG”a⊥BMb, G”aG”b⊥CMcand AMa= 6G”bG”c, BMb= 6G”cG”a, CMc= 6G”aG”b

Proof We have gb”− g”

c = −6i(a − ma), then G”bG”c⊥AMa and AMa= 6 · G”bG”c Remark 2 The sides of triangles G0aG0bG0c and G”aG”bG”c have lenghts equal to one-sixth of the lengths if the medians of triangle ABC.The existence of triangles G0aG0bG0c and G”aG”bG”c implies, also, the existence of a medians triangle(a triangle that has sides of equal lenght with medians the lengths of the triangle ABC) [1]

Corollary 2 Quadrilaterals G0bG0cG”bG”c, G0cG0aG”cG”a, G0aG0bG”aG”b are parallelograms

Proof By Theorems 20 and 21 we get

G0bG0ck G”bG”c, G0cG0ak G”cG”a, G0aG0b k G”aG”b and

G0bG0c= G”bG”c, G0cG0a= G”cG”a, G0aG0b = G”aG”b

Corollary 3 Triangles GaGbGc and G”aG”bG”c are congruent.

Theorem 20 The pairs of triangles G0aG0bG0c and ABC, G”aG”bG”c and ABC are bilogic

Proof The solutions results from Theorems 16-21, the common center of orthology being centroid

of the triangle ABC

Theorem 21 If by A[XY Z] we denote XY Z triangle’s area, then A[G0

a G0bG0c] = A[G”

a G ” G ” ] =

1

48A[ABC]

Proof Triangle G0aG0bG0cis similar with the medians triangle M1M2M3, so m( \G0aG0bG0c) = m(M\1M2M3) =

180◦− m(\BGC) We have

A[G0

a G0bG0c]= G

0

aG0b· G0aG0c· sin \G0aG0bG0c

CM c

6 ·BMb

6 · sin(\BGC)

1

16A[BGC]=

1

16 ·

1

3A[ABC]=

1

48A[ABC]. Triangles G0aG0bG0c and G”aG”bG”c are congruent, so A[G0

a G0bG0c]= A[G”

a G”G ” ]= 481A[ABC]

Theorem 22 The following relations are true: G1G⊥BC, G2G⊥CA, G3G⊥AB and BC = 6G1G,

CA = 6G02G, AB = 6G03G

Proof We have g10 = o

0

a +o ” +o ”

3 = g + i · b−c6 , so g

0

1 −g b−c = 6i Because G01G⊥BC and

g10−g b−c

=i 6

, we

obtain BC = 6G01G

Theorem 23 The following relations are true: G”1G⊥BC, G”2G⊥CA, G”3G⊥AB and BC = 6G”1G,

CA = 6G”2G, AB = 6G”3G

Proof The proof is similar to the one of the previous theorem

Theorem 24 Triangles G01G02G03, G”1G”2G”3 and ABC have the same centroid

Proof We have g

0

1 +g20+g30

3 = g”+g3”+g” = a+b+c3 = g, as desired Corollary 4 The centroid of the triangle ABC is the midpoint of the segments G0aG”

a, G0bG”

b,

G0cG”c, G01G”1, G02G”2 and G03G”3

Corollary 5 Points G”1, G0a, G, G”a, G01 are collinear and G”1G0a= G0aG = GG”a= G”aG01

Theorem 25 Quadrilaterals G01G”3G02G, G”1G03GG02 and G03G”2G01G are parallelograms

Proof Because g + g”3 = g01 + g20 = 2g0c, quadrilateral G01G”3G02G is a parallelogram Similarly, quadrilaterals G”1G03GG02 and G03G”2G01G are parallelograms

Corollary 6 Hexagons G”aG0cG”bG0aG”cG0b and G01G”3G02G”1G03G”2 are homotetic, the center of ho-motethy being the centroid of triangle ABC

Corollary 7 The points G0a, G0b, G0c, G”a, G”b and G”c are same conic

Proof We have G0aG”c k G”

aG0c, G0bG”a k G”

bG0a, G0cG”b k G”

bG0c As two parallel lines have the same infinity it follows that the lines are intersected in the same point of the infinity Then, the inter-section points between the pair of parallel lines enumerated above belong to the infinite line and from Pascal’s reverse theorem the conclusion follows

Corollary 8 Points G01, G02, G03, G”1, G”2 and G”3 are on the same conic

References

[1] C Barbu, Teoreme fundamentale din geometria triunghiului, Ed Unique, Bac˘au, 2008

[2] G Salmon, Trait´e de g´eom´etrie analytique, Ed Gauthier –Villars, Paris, 1903

[3] T Andreescu, D Andrica, Complex Numbers from A to Z, Birkh¨auser, Boston, 2006

[4] D Andrica, K Nguyen, A note on the Nagel and Gergonne points, Creative Math & Inf., 17 (2008), 127-136

[5] R Musselman, The triangle bordered with squares, American Mathematical Monthly, 43 (1936), 539-548

[6] J Neuberg, Bibliographie du Triangle et du T´etraˇcdre, Mathesis, 37 (1923), 289-293

C˘at˘alin Barbu

“Vasile Alecsandri” College,

Bac˘au, Romania

E-mail address: kafka mate@yahoo.com

Corollary Quadrilaterals G0bG0cG”bG”c,...

48A[ABC]

Proof Triangle G0aG0bG0cis similar with the medians triangle M1M2M3,... data-page="6">

Corollary Triangles GaGbGc and G”aG”bG”c are congruent.

Theorem 20 The pairs of triangles