Page 1

Forum Geometricorum

Volume 3 (2003) 145–159.

???

?

FORUM GEOM

ISSN 1534-1178

Rectangles Attached to Sides of a Triangle

Nikolaos Dergiades and Floor van Lamoen

Abstract. We study the figure of a triangle with a rectangle attached to each

side. In line with recent publications on special cases we find concurrencies

and study homothetic triangles. Special attention is given to the cases in which

the attached rectangles are similar, have equal areas and have equal perimeters,

respectively.

1. Introduction

In recent publications [3, 4, 10, 11, 12] the configurations have been studied

in which rectangles or squares are attached to the sides of a triangle. In these

publications the rectangles are all similar. In this paper we study the more general

case in which the attached rectangles are not necessarily similar. We consider a

triangle ABC with attached rectangles BCAcAb, CABaBcand ABCbCa. Let u

be the length of CAc, positive if Acand A are on opposite sides of BC, otherwise

negative. Similarly let v and w be the lengths of ABaand BCb. We describe the

shapes of these rectangles by the ratios

U =a

u,

V =b

v,

W =c

w.

(1)

The vertices of these rectangles are1

Ab= (−a2: SC+ SU : SB),

Ba= (SC+ SV : −b2: SA),

Ca= (SB+ SW : SA: −c2),

Consider the flank triangles ABaCa, AbBCband AcBcC. With the same rea-

soning as in [10], or by a simple application of Ceva’s theorem, we can see that the

triangle HaHbHcof orthocenters of the flank triangles is perspective to ABC with

perspector

?a

Ac= (−a2: SC: SB+ SU),

Bc= (SC: −b2: SA+ SV ),

Cb= (SB: SA+ SW : −c2).

P1=

u:b

v:c

w

?

= (U : V : W).

(2)

Publication Date: August 25, 2003. Communicating Editor: Paul Yiu.

1All coordinates in this note are homogeneous barycentric coordinates. We adopt J. H. Conway’s

notation by letting S = 2∆ denote twice the area of ABC, while SA =

SB = S cotB, SC = S cotC, and generally SXY = SXSY.

−a2+b2+c2

2

= S cotA,

Page 2

146N. Dergiades and F. M. van Lamoen

See Figure 1. On the other hand, the triangle OaObOcof circumcenters of the flank

triangles is clearly homothetic to ABC, the homothetic center being the point

P2= (au : bv : cw) =

?a2

U:b2

V

:c2

W

?

.

(3)

Clearly, P1and P2are isogonal conjugates.

A

BC

Ab

Ac

Bc

Ba

Ca

Cb

Ha

Hb

Hc

P1

Oa

Ob

Oc

P2

Figure 1

Now the perpendicular bisectors of BaCa, AbCband AcBcpass through Oa,

Oband Ocrespectively and are parallel to AP1, BP1and CP1respectively. This

shows that these perpendicular bisectors concur in a point P3on P1P2satisfying

P2P1: P1P3= 2S : au + bv + cw,

where S is twice the area of ABC. See Figure 2. More explicitly,

P3=(−a2V W(V + W) + U2(b2W + c2V ) + 2SU2V W

: − b2WU(W + U) + V2(c2U + a2W) + 2SUV2W)

: − c2UV (U + V ) + W2(a2V + b2U) + 2SUV W2)

This concurrency generalizes a similar result by Hoehn in [4], and was men-

tioned by L. Lagrangia [9]. It was also a question in the Bundeswettbewerb Math-

ematik Deutschland (German National Mathematics Competition) 1996, Second

Round.

From the perspectivity of ABC and the orthocenters of the flank triangles, we

see that ABC and the triangle A?B?C?enclosed by the lines BaCa, AbCband

AcBcare orthologic. This means that the lines from the vertices of A?B?C?to the

corresponding sides of ABC are concurrent as well. The point of concurrency is

the reflection of P1in O, i.e.,

(4)

Page 3

Rectangles attached to the sides of a triangle 147

P4= (−SBCU + a2SA(V + W) : ··· : ···).

(5)

O

A

BC

A?

B?

C?

Ab

Ac

Bc

Ba

Ca

Cb

P1

Oa

Ob

Oc

P2

Ma

Mb

Mc

P3

P4

Figure 2

Remark. We record the coordinates of A?. Those of B?and C?can be written down

accordingly.

A?=(−(a2S(U + V + W) + (a2V + SCU)(a2W + SBU))

:SCS(U + V + W) + (b2U + SCV )(a2W + SBU)

:SBS(U + V + W) + (a2V + SCU)(c2U + SBW)).

2. Special cases

We are mainly interested in three special cases.

2.1. The similarity case. This is the case when the rectangles are similar, i.e., U =

V = W = t for some t. In this case, P1 = G, the centroid, and P2 = K, the

symmedian point. As t varies,

P3= (b2+ c2− 2a2+ 2St : c2+ a2− 2b2+ 2St : a2+ b2− 2c2+ 2St)

traverses the line GK. The point P4, being the reflection of G in O, is X376in

[7]. The triangle MaMbMcis clearly perspective with ABC at the orthocenter H.

More interestingly, it is also perspective with the medial triangle at

((SA+ St)(a2+ 2St) : (SB+ St)(b2+ 2St) : (SC+ St)(c2+ 2St)),

Page 4

148N. Dergiades and F. M. van Lamoen

which is the complement of the Kiepert perspector

?

1

SA+ St:

1

SB+ St

:

1

SC+ St

?

.

It follows that as t varies, this perspector traverses the Kiepert hyperbola of the

medial triangle. See [8].

The case t = 1 is the Pythagorean case, when the rectangles are squares erected

externally. The perspector of MaMbMcand the medial triangle is the point

O1= (2a4− 3a2(b2+ c2) + (b2− c2)2− 2(b2+ c2)S : ··· : ···),

which is the center of the circle through the centers of the squares. See Figure 3.

This point appears as X641in [7].

A

BC

Ma

Mb

Mc

Ab

Ac

Bc

Ba

Ca

Cb

A1

B1

C1

O1

Figure 3

2.2. Theequiareal case. Whenthe rectangles haveequal areasT

?

P4=(a2(−SBC+ SA(b2+ c2)) : ··· : ···)

=(a2(a4+ 2a2(b2+ c2) − (3b4+ 2b2c2+ 3c4)) : ··· : ···)

is the reflection of K in O.2The special equiareal case is when T = S, the

rectangles having the same area as triangle ABC. See Figure 4. In this case,

P3= (6a2− b2− c2: 6b2− c2− a2: 6c2− a2− b2).

2,i.e.,(U,V,W) =

2a2

T,2b2

T,2c2

T

?

, it is easy to see that P1= K, P2= G, and

2This point is not in the current edition of [7].

Page 5

Rectangles attached to the sides of a triangle149

O

A

BC

A?

B?

C?

Ab

Ac

Bc

Ba

Ca

Cb

K

G

Ma

Mb

Mc

P3

P4

Figure 4

2.3. Theisoperimetric case. Thisisthecase whentherectangles have equal perime-

ters 2p, i.e., (u,v,w) = (p − a,p − b,p − c). The special isoperimetric case is

when p = s, the semiperimeter, the rectangles having the same perimeter as trian-

gle ABC. In this case, P1= X57, P2= X9, the Mittenpunkt, and

P3=(a(bc(2a2− a(b + c) − (b − c)2) + 4(s − b)(s − c)S) : ··· : ···),

P4=(a(a6− 2a5(b + c) − a4(b2− 10bc + c2) + 4a3(b + c)(b2− bc + c2)

− a2(b4+ 8b3c − 2b2c2+ 8c3b + c4) − 2a(b + c)(b − c)2(b2+ c2)

+ (b + c)2(b − c)4) : ··· : ···).

These points can be described in terms of division ratios as follows.3

P3X57: X57X9=4R + r : 2s,

P4I : IX57=4R : r.

3. A pair of homothetic triangles

Let A1, B1and C1be the centers of the rectangles BCAcAb, CABaBcand

ABCbCarespectively, and A2B2C2the triangle bounded by the lines BcCb, CaAc

and AbBa. Since, for instance, segments B1C1and BcCbare homothetic through

3These points are not in the current edition of [7].

Page 6

150N. Dergiades and F. M. van Lamoen

A, the triangles A1B1C1and A2B2C2are homothetic. See Figure 5. Their homo-

thetic center is the point

P5=?−a2SA(V + W) + U(SB+ SW)(SC+ SV ) : ··· : ···?.

Ba

Ca

A

BC

A2

B2

C2

Ab

Ac

Bc

Cb

A1

B1

C1

P6

P5

Figure 5

Forthe Pythagorean case with squares attached to triangles, i.e.,U = V = W =

1, Toshio Seimiyaand Peter Woo[12]have proved the beautiful result that the areas

∆1and ∆2of A1B1C1and A2B2C2have geometric mean ∆. See Figure 5. We

prove a more general result by computation using two fundamental area formulae.

Proposition 1. For i = 1,2,3, let Pibe finite points withhomogeneous barycentric

coordinates (xi: yi: zi) with respect to triangle ABC. The oriented area of the

triangle P1P2P3is

??????

x1

x2

x3

y1

y2

y3

z1

z2

z3

??????

(x1+ y1+ z1)(x2+ y2+ z2)(x3+ y3+ z3)· ∆.

A proof of this proposition can be found in [1, 2].

Proposition 2. For i = 1,2,3, let ?ibe a finite line with equation pix+qiy+riz =

0. The oriented area of the triangle bounded by the three lines ?1, ?2, ?3is

??????

p1

p2

p3

D1· D2· D3

q1

q2

q3

r1

r2

r3

??????

2

· ∆,

Page 7

Rectangles attached to the sides of a triangle 151

where

D1=

??????

∆1∆2

∆2

1

p2

p3

1

q2

q3

1

r2

r3

??????

,D2=

??????

p1

1

p3

q1

1

q3

r1

1

r3

??????

,D3=

??????

p1

p2

1

q1

q2

1

r1

r2

1

??????

.

A proof of this proposition can be found in [5].

Theorem 3.

=(U+V +W−UV W)2

4(UV W)2

.

Proof. The coordinates of A1, B1, C1are

A1=(−a2: SC+ SU : SB+ SU),

B1=(SC+ SV : −b2: SA+ SV ),

C1=(SB+ SW : SA+ SW : −c2).

By Proposition 1, the area of triangle A1B1C1is

∆1=S(U + V + W + UV W) + (a2V W + b2WU + c2UV )

4SUV W

The lines BcCb, CaAc, AbBahave equations

· ∆.

(6)

(S(1 − V W) − SA(V + W))x + (S + SBV )y + (S + SCW)z =0,

(S + SAU)x + (S(1 − WU) − SB(W + U))y + (S + SCW)z =0,

(S + SAU)x + (S + SBV )y + (S(1 − UV ) − SC(U + V ))z =0.

By Proposition 2, the area of the triangle bounded by these lines is

∆2=

S(U + V + W − UV W)2

UV W(S(U + V + W + UV W) + (a2V W + b2WU + c2UV ))·∆. (7)

From (6, 7), the result follows.

?

Remarks. (1) The ratio of homothety is

−S(U + V + W − UV W)

2(S(U + V + W + UV W) + (a2V W + b2WU + c2UV )).

(2) We record the coordinates of A2below. Those of B2and C2can be written

down accordingly.

A2=(−a2((S + SAU)(V + W) + SU(1 − V W)) + (SB+ SW)(SC+ SV )U2

: (S + SAU)(SUV + SC(U + V + W))

: (S + SAU)(SUW + SB(U + V + W))).

From the coordinates of A2B2C2we see that this triangle is perspective to ABC

at the point

?

P6=

1

SA(U + V + W) + SV W: ··· : ···

?

.

Page 8

152N. Dergiades and F. M. van Lamoen

4. Examples

4.1. The similarity case. If the rectangles are similar, U = V = W = t, then

1

3SA+ St:

traverses the Kiepert hyperbola. In the Pythagorean case, the homothetic center P5

is the point

((SB−S)(SC−S)−4SBC: (SC−S)(SA−S)−4SCA: (SA−S)(SB−S)−4SAB).

P6=

?

1

3SB+ St:

1

3SC+ St

?

A

BC

A2

B2

C2

Ab

Ac

Bc

Ba

Ca

Cb

A1

B1

C1

P6

P5

Figure 6

4.2. The equiareal case. For (U,V,W) = (2a2

T,2b2

T,2c2

T), we have

P6=

?

1

T(a2+ b2+ c2)SA+ 2Sb2c2: ··· : ···

?

.

This traverses the Jerabek hyperbola as T varies. When the rectangles have the

same area as the triangle, the homothetic center P5is the point

(a2((a2+ 3b2+ 3c2)2− 4(4b4− b2c2+ 4c4)) : ··· : ···).

5. More homothetic triangles

Let CA, CBand CCbe the circumcricles of rectangles BCAcAb, CABaBcand

ABCbCarespectively. See Figure 7. Since the circle CApasses through B and C,

its equation is of the form

a2yz + b2zx + c2xy − px(x + y + z) = 0.

Page 9

Rectangles attached to the sides of a triangle 153

Since the same circle passes through Ab, we have p =SAU+S

same method we derive the equations of the three circles:

U

= SA+S

U. By the

a2yz + b2zx + c2xy = (SA+S

U)x(x + y + z),

a2yz + b2zx + c2xy = (SB+S

V)y(x + y + z),

a2yz + b2zx + c2xy = (SC+S

W)z(x + y + z).

From these, the radical center of the three circles is the point

J =

?

1

SA+S

U

:

1

SB+S

V

:

1

SC+S

W

?

=

?

U

SAU + S:

V

SBV + S:

W

SCW + S

?

.

A

BC

Ab

Ac

Bc

Ba

Ca

Cb

A1

B1

C1

A3

B3

C3

J

Figure 7

Note that the isogonal conjugate of J is the point

J∗=

?

a2SA+ S ·a2

U: b2SB+ S ·b2

V

: c2SC+ S ·c2

W

?

.

It lies on the line joining O to P2. In fact,

P2J∗: J∗O = 2S : au + bv + cw = P2P1: P1P3.

Page 10

154N. Dergiades and F. M. van Lamoen

The circles CBand CCmeet at A and a second point A3, which is the reflection

of A in B1C1. See Figure 8. In homogeneous barycentric coordinates,

?

Similarly we have points B3and C3. Clearly, the radical center J is the perspector

of ABC and A3B3C3.

A3=

V + W

SA(V + W) − S(1 − V W):

V

SBV + S:

W

SCW + S

?

.

A

BC

A2

B2

C2

Ab

Ac

Bc

Ba

Ca

Cb

A1

B1

C1

A3

B3

C3

J

M1

Ma

N

O1

Figure 8

Proposition 4. The triangles ABC and A2B2C2are orthologic. The perpendicu-

lars from the vertices of one triangle to the corresponding lines of the other triangle

concur at the point J.

Proof. As C1B1bisects AA3, we see A3lies on BcCband AJ ⊥ BcCb. Similarly,

we have BJ ⊥ CaAcand CJ ⊥ AbBa. The perpendiculars from A, B, C to the

corresponding sides of A2B2C2concur at J.

On the other hand, the points B, C3, B3, C are concyclic and B3C3is antiparal-

lel to BC with respect to triangle JBC. The quadrilateral JB3A2C3is cyclic, with

JA2as a diameter. It is known that every perpendicular to JA2is antiparallel to

Page 11

Rectangles attached to the sides of a triangle 155

B3C3with respect to triangle JB3C3. Hence, A2J ⊥ BC. Similarly, B2J ⊥ CA

and C2J ⊥ AB.

It is clear that the perpendiculars from A3, B3, C3to the corresponding sides

of triangle A2B2C2intersect at J. Hence, the triangles A2B2C2and A3B3C3are

orthologic.

?

Proposition 5. The perpendiculars from A2, B2, C2to the corresponding sides of

A3B3C3meet at the reflection of J in the circumcenter O3of triangle A3B3C3.

Proof. Since triangle A3B3C3is the pedal triangle of J in A2B2C2, and A2J

passes through the circumcenter of triangle A2B3C3, the perpendicular from A2

to B3C3passes through the orthocenter of A2B3C3and is isogonal to A2J in

triangle A2B2C2. This line therefore passes through the isogonal conjugate of J

in A2B2C2. We denote this point by J!. Similarly, the perpendiculars from B2, C2

to the sides C3A3and A3B3pass through J!. The circumcircle of A3B3C3is the

pedal circle of J. Hence, its circumcenter O3is the midpoint of JJ!. It follows

that J!is the reflection of J in O3.

?

Remark. The point J and the circumcenters O and O3of triangles ABC and

A3B3C3are collinear. This is because |JA · JA3| = |JB · JB3| = |JC · JC3|,

say, = d2, and an inversion in the circle (J,d) transforms ABC into A3B3C3or

its reflection in J.

Theorem 6. The perpendicular bisectors of BcCb, CaAc, AbBaare concurrent at

a point which is the reflection of J in the circumcenter O1of triangle A1B1C1.

Proof. Let M1and Mabe the midpoints of B1C1and BcCbrespectively. Note that

M1is also the midpoint of AMa. Also, let O1be the circumcenter of A1B1C1,

and the perpendicular bisector of BcCbmeet JO1at N. See Figure 8. Consider

the trapezium AMaNJ. Since O1M1is parallel to AJ, we conclude that O1is

the midpoint of JN. Similarly the perpendicular bisectors of CaAc, AbBapass

through N, which is the reflection of J in O1.

?

We record the coordinates of O1:

((c2U2V − a2V W(V + W) + b2WU(W + U)

+ UV W((SA+ 3SB)UV + (SA+ 3SC)UW))S

+ c2SBU2V2+ b2SCU2W2− a4V2W2

+ (S2+ SBC)U2V2W2+ 4S2U2V W)

: ··· : ···)

In the Pythagorean case, the coordinates of O1are given in §2.1.

6. More triangles related to the attached rectangles

Write U = tanα, V = tanβ, and W = tanγ for angles α, β, γ in the range

(−π

2,π

2). The point A4for which the swing angles CBA4and BCA4are β and γ

Page 12

156N. Dergiades and F. M. van Lamoen

respectively has coordinates

(−a2: SC+ S · cotγ : SB+ S · cotβ) =

It is clear that this point lies on the line AJ. See Figure 9. If B4and C4are

analogously defined, the triangles A4B4C4and ABC are perspective at J.

?

−a2: SC+S

W: SB+S

V

?

.

A

BC

Ab

Ac

Bc

Ba

Ca

Cb

A1

B1

C1

A3

B3

C3

J

A4

B4

C4

α

β

β

γ

γ

α

α

Figure 9

Note that A3, B, A4, C are concyclic since ∠A4BC = β = ∠ABcV =

∠A4A3C.

Let d1= BcCb, d2= CaAc, d3= AbBa, d?

1= AA4, d?

2= BB4, d?

3= CC4.

Proposition 7. The ratiosdi

precisely,

d?

i, i = 1,2,3, are independent of triangle ABC. More

d1

d?

1

=1

V+

1

W,

d2

d?

2

=

1

W+1

U,

d3

d?

3

=1

U+1

V.

Proof. Since AA4⊥ CbBc, the circumcircle of the cyclic quadrilateral A3BA4C

meets CbBcbesides A3at the antipode A5of A4. See Figure 10. Let f, g, h

denote, for vectors, the compositions of a rotation byπ

2, and homotheties of ratios

Page 13

Rectangles attached to the sides of a triangle 157

A

BC

Ab

Ac

Bc

Ba

Ca

Cb

A1

B1

C1

A3

B3

C3

J

A4

B4

C4

A5

B5

C5

P7

Figure 10

1

U,1

V, and

1

Wrespectively. Then

g(− − →

V. Similarly, h(− − →

1

W.

AA4) = g(− →

AC) + g(− − →

AA4) =− − − →

CA4) =− − →

CBc+− − →

A5C =− − − →

A5Bc,

andA5Bc

d1

d?

AA4

V+

=

1

CbA5, andCbA5

AA4

=

1

W. It follows that

1=1

The coordinates of A5can be seen immediately: Since A4A5is a diameter of

the circle (A4BC), we see that ∠BCA5= −π

A5= (−a2: SC− SW : SB− SV ).

Similarly, we have the coordinates of B5and C5. From these, it is clear that

A5B5C5and ABC are perspective at

?

2+ ∠BCA4, and

P7=

?

1

SA− SU:

For example, in the similarity case it is obvious from the above proof that

the points A5, B5, C5are the midpoints of BcCb, CaAc, AbBa. Clearly in the

Pythagorean case, the points A4, B4, C4coincide with A1, B1, C1respectively.

1

SB− SV

:

1

SC− SW

?

=

?

1

cotA − U:

1

cotB − V

:

1

cotC − W

?

.

Page 14

158N. Dergiades and F. M. van Lamoen

In this case, J is the Vecten point and from the above proof we have d1 = 2d?

d2= 2d?

1,

2, d3= 2d?

3and P7= X486.

7. Another interesting special case

If α + β + γ = π, then U + V + W = UV W. From Theorem 3 we conclude

that ?2= 0, and the points A2, B2, C2, A3, B3, C3coincide with J, which now

is the common point of the circumcircles of the three rectangles. Also, the points

A4, B4, C4lie on the circles CA, CB, CCrespectively.

A

BC

A4

B4

C4

Ab

Ac

Bc

Ba

Ca

Cb

A5

B5

C5

P7

A1

B1

C1

J

Figure 11

In Figure 11 we illustrate the case α = β = γ =π

point. The triangles BCA4, CAB4, ABC4are the Fermat equilateral triangles,

and the angles of the lines AA4, BB4, CC4, BcCb, CaAc, AbBaaround J areπ

The points A5, B5, C5are the mid points of BcCb, CaAc, AbBa. Also, d?

d?

point X18in [7].

3. In this case, J is the Fermat

6.

1= d?

2=

3, and d1= d2= d3=2√3

3d?

1. In this case, P7is the second Napoleon point, the

Page 15

Rectangles attached to the sides of a triangle 159

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[3] Z.ˇCerin, Loci related to variable flanks, Forum Geom., 2 (2002) 105–113.

[4] L. Hoehn, Extriangles and excevians, Math. Magazine, 67 (2001) 188–205.

[5] G. A. Kapetis, Geometry of the Triangle, vol. A (in Greek), Zitis, Thessaloniki, 1996.

[6] C. Kimberling, Triangle Centers and Central Triangles, Congressus Numerantium, 129 (1998)

1–285.

[7] C. Kimberling, Encyclopedia of Triangle, Centers, July 1, 2003 edition, available at

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[8] S. Kotani, H. Fukagawa, and P. Penning, Problem 1759, Crux Math., 16 (1990) 240; solution,

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[9] L. Lagrangia, Hyacinthos 6948, April 13, 2003.

[10] F. M. van Lamoen, Friendship among triangle centers, Forum Geom., 1 (2001) 1–6.

[11] C. R. Panesachar and B. J. Venkatachala, On a curious duality in triangles, Samasy¯ a, 7 (2001),

number 2, 13–19.

[12] T. Seimiya and P. Woo, Problem 2635, Crux Math., 27 (2001) 215; solution, 28 (2002) 264–

266.

Nikolaos Dergiades: I. Zanna 27, Thessaloniki 54643, Greece

E-mail address: ndergiades@yahoo.gr

Floor van Lamoen: St. Willibrordcollege, Fruitlaan 3, 4462 EP Goes, The Netherlands

E-mail address: f.v.lamoen@wxs.nl