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].