Volume 3 (2003) 145–159.
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,
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
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 , 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
Ac= (−a2: SC: SB+ SU),
Bc= (SC: −b2: SA+ SV ),
Cb= (SB: SA+ SW : −c2).
= (U : V : W).
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.
= S cotA,
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) =
Clearly, P1and P2are isogonal conjugates.
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 , and was men-
tioned by L. Lagrangia . It was also a question in the Bundeswettbewerb Math-
ematik Deutschland (German National Mathematics Competition) 1996, Second
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.,
Rectangles attached to the sides of a triangle 147
P4= (−SBCU + a2SA(V + W) : ··· : ···).
Remark. We record the coordinates of A?. Those of B?and C?can be written down
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
. 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)),
148N. Dergiades and F. M. van Lamoen
which is the complement of the Kiepert perspector
It follows that as t varies, this perspector traverses the Kiepert hyperbola of the
medial triangle. See .
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 .
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).
, it is easy to see that P1= K, P2= G, and
2This point is not in the current edition of .
Rectangles attached to the sides of a triangle149
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 .