the fermat point configuration

The Fermat point configurationAuthor(s): C. J. BRADLEYSource: The Mathematical Gazette, Vol. 92, No. 524 (July 2008), pp. 214-222Published by: The Mathematical AssociationStable URL: http://www.jstor.org/stable/27821769 .

Accessed: 31/05/2014 05:15

Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at .http://www.jstor.org/page/info/about/policies/terms.jsp

.JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range ofcontent in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new formsof scholarship. For more information about JSTOR, please contact [email protected].

.

The Mathematical Association is collaborating with JSTOR to digitize, preserve and extend access to TheMathematical Gazette.

http://www.jstor.org

This content downloaded from 217.150.211.10 on Sat, 31 May 2014 05:15:23 AMAll use subject to JSTOR Terms and Conditions

http://www.jstor.org/action/showPublisher?publisherCode=mathas

http://www.jstor.org/stable/27821769?origin=JSTOR-pdf

http://www.jstor.org/page/info/about/policies/terms.jsp


214 THE MATHEMATICAL GAZETTE

The Ferm?t point configuration C. J. BRADLEY

In this article a theorem similar to Napoleon's theorem is established for the Ferm?t point configuration of a triangle. Areal coordinates are used

throughout, with ABC as triangle of reference. For a full account of how to define and use these coordinates see Bradley [1]. Alternative synthetic proofs, generously supplied by the referee, are also included. The

presentation of the paper with its emphasis on coordinates was designed in

part to show that an algebraic treatment of the Ferm?t point configuration is

possible, as well as the more familiar synthetic or complex number treatments. The theorem is as follows:

Theorem

Let ABC be a non-equilateral triangle, with no angle equal to 120?, with centroid G and Ferm?t point F, which, in a triangle all of whose angles are less than 120?, we define to be the point F such that ZBFC = ZCFA =

ZAFB = 120?. When, say ZBAC > 120?, F is defined so that ZBFC = 120?, ZCFA = ZAFB = 60?, with similar changes when the other angles of ABC exceed 120?. Let 7\, T2, T3 be the nine-point centres of

triangles BFC, CFA, AFB respectively, then Tx, T2, T3 are the vertices of an

equilateral triangle and circle T{T2T3 has FG as a diameter.

The Ferm?t point may also be defined as the common point of intersection of the circumcircles of equilateral triangles BCA, CAB!, ABC drawn external to triangle ABC. Then AA', BB, CC are concurrent at F. The associated Ferm?t point F is defined in a similar way by drawing equilateral triangles folded over triangle ABC. We state later what happens to the above theorem when F replaces F. It is assumed that these alternative definitions are known.

In the course of establishing the above theorem a number of intermediate results are proved. The first of these is concerned with Figure 1, in which K is a. general point (that is, not on the sides or their extensions) in the plane of a triangle ABC, with centroid G.

Result 1 :

Let Gi, G2, G3 be the centroids of triangles BKC, CKA, AKB

respectively then lines through G\, G2, G3 parallel to AK, BK, CK

respectively are concurrent at G.

Proof: Let K have normalised areals (/, m, n), then G\ has unnormalised areals

(/, m + 1, n + 1), AK has equation mz = ny and the equation of any line

parallel to AK has equation mz - ny + k (x + y + z) = 0. If such a line

passes through G\, then m(n + 1) -

n(m + 1) + 3k = 0, and k = (n -

m)/3.



THE FERMAT POINT CONFIGURATION 215

Thus the line through G\ parallel to AK has equation

(n -

m)x -

(m + 2n)y + (2m + n)z = 0. This line passes through G(l, 1, l). Similar working shows that the other two lines also pass through G.

FIGURE 1

Result 2:

Triangles ABC and GiG2G3 are homothetic with scale factor 3 and centre of (inverse) enlargement the point J on KG with KJ = 3JG.

Proof: AGi has equation (ra+l)z =

(?+l)y and this passes through the

point J with normalised areals J(/+l,m+l,w+l) and, using an obvious notation, J =

\K + |G, showing KJ = 3/G. By symmetry / also lies on BG2 and CG3.

Alternative proofs of Results 1 and 2

If we take L to be the midpoint of BC and consider triangle LAK, we have LG : LA = 1 : 3 = LGi : LK. Thus GGL is parallel to AK, and

3GGi = AK. This, with the corresponding cyclic results, establishes the

homothety and shows that its centre is on GK dividing it in the ratio 1:3.




We next prove a property of a triangle that has an angle of 120?. This is

undoubtedly a known result, but as it is possibly not well known, we give a

proof, which is illustrated in Figure 2a. PQR is a triangle with ZQPR = 120?, incentre /, circumcentre O, orthocentre H and nine-point centre T.

Pt

FIGURE 2a FIGURE 2b

Result 3:

PI is parallel to the Euler line OTH and triangle OPH is isosceles, so that PT is perpendicular to OTH.

Proof: We have PO = /?>, the circumradius and PH =

\2I% cos ZQPR\ =

since ZQPR = 120?. Hence PO = PH and triangle OPH is isosceles.

Since T is the midpoint of OH, it follows that PT is perpendicular to the Euler line OTH. Now it is known that the line PI bisects the angle between the lines PO and PH (one of the facts used in showing that O and H are

isogonal points). Since PT is the internal bisector of ZOPH and is at right angles to OTH, it follows that PI is parallel to the Euler line and ZTPI = 90?.

Result 3 also applies if ZQPR = 60? (although the role of / and the

proof are very slightly different). This is illustrated in Figure 2b.

We now identify the point K in Results 1 and 2 with the Ferm?t point F of triangle ABC, which, in all that follows, we take to be a non-equilateral triangle with no angle equal to 120?. And since ZBFC = ZCFA = ZAFB = 120? or one of these angles is 120? and the other two 60?, we may identify the triangle PQR in Result 3 or its extension with each of the three triangles BFC, CFA, AFB in turn, with F taking the part of P in each case. See Figures 3 and 4, in which G\, G2, G3 are the centroids of the three triangles, 0\, 02, 03 are the circumcentres of the three triangles, Tu T2, T3 are the nine

point centres of the three triangles and e{, e2, e3 are the Euler lines of the




three triangles. Triangle BC? is the equilateral triangle erected on BC.

Figures 3 and 4 illustrate Napoleon's theorem, that O1O2O3 is equilateral and

they also illustrate the main theorem of this article, stated at the outset.

Figure 3 shows F internal to ABC and Figure 4 shows F external to ABC. A

We first show how to obtain the areal coordinates of the various points in the figures. Let AF = u, BF = v, CF = w. Then, when F is internal to the triangle ABC, because of the 120? angles we have

v2 + vw + w = a

w + wu + u - b

+ V = c

(1) u + uv

When F is not internal, a sign convention on u, v, w, about to be explained, means that these equations remain true.

Also [BFC] = \vw sin 120?, [CFA] = \wu sin 120?, [AFB] = \uv sin 120?, where [BFC] denotes the area of triangle BCF etc. Hence the normalised areal coordinates of F are (vw, wu, uv)/(vw + wu + uv), so that in relation to

Results 1 and 2 when K is identified with F, we have / = vwl(vw + wu + uv), m = wu/(vw + WM + uv), n = uvl(vw + wu + uv). Note that the only




difference in the analysis between Figures 3 and 4 is that in Figure 3 w, v, w are positive, but in Figure 4 m is negative and v, w are positive. In all cases,

provided this type of sign convention is used, it is a known theorem that AA' = Bff = CC = u + v + w, where A', ??, C are the third vertices of the equilateral triangles with bases BC, CA, AB respectively. In terms of

w, v, w the normalised areals of G\ are

(vw, vw + 2wu + wv, vw + wu + luv)

3( VW + wu + uv ) (2)

with similar expressions for the areals of G2 and G3 by cyclic change of x, y, z and u, v, w.




Result 1 tells us that the lines through Gu G2, G3 parallel to AF, BF, CF

respectively are concurrent at G. Now the angles at F subtended by the vertices of triangle ABC are such that, as a consequence of Result 3 or its extension, the Euler line of triangle BFC is parallel to AF. But this Euler line contains G\ and so by Result 1 passes through G. When the same result is applied to triangles CFA and AFB it follows that the three Euler lines eu e2, e3 all pass through G and are equally inclined to each other. We have

proved

Result 4:

The Euler lines of triangles BFC, CFA, AFB are concurrent at G.

As a corollary to this we have

Result 5:

If T\, T2, T3 are the nine-point centres of triangles BFC, CFA, AFB

respectively, then 7\, T2, T3 are the feet of the perpendiculars from F on to ex, e2, e3 respectively. Since the feet of the perpendiculars from a point on to three equally inclined lines form an equilateral triangle, it follows that

triangle T\T2T3 is equilateral. From the proof of Result 1, we also have the equations of e\, e2, e3,

which are:

ex : (v -

w)x -

(w + 2v)y + (2w + v)z = 0;

e2 : (2? + w)x + (w -

u)y -

(u + 2w)z = 0; (3)

e3 : -(v + 2u)x + (2v + u)y + (u -

v)z = 0. In order to work out the coordinates of A', we use the fact that AF = u,

FA' = v + w and the known coordinates of A and F to obtain the normalised areals of A' as

(-(v2 + vw + w2), (u + v + w)w, (u + v + w)v) vw + wu + uv

Now 0\ is the centroid of triangle BCK and therefore has normalised areals

(-(v2 + vw + w2), 2vw + 2wu + uv + w2, 2vw + wu + 2uv + v2) 3 ( vw + ww + wv)

The areals of 02 and O3 may be obtained by cyclic change of x, y, z and u, v, w and it is easy to check that 0\, 02, 03 do actually lie on the Euler lines e\, e2, e3.

We now have the coordinates of 0\ and G\, so we may obtain the coordinates of Tx, the nine-point centre of triangle BFC. Since GiGi = 2GiTi we obtain the normalised areals of T\ as

(v2 + 4vw + w2, vw + 4ww + 2uv - w2, vw + 2wu + 4wv - v2) 6 (vw + ww + mv)




Result 6 (Napoleon):

0,0, - 0203 - 0,0, - - " +

Strictly speaking Napoleon's theorem is concerned with the equilateral triangles, rather than the triangles BFC, CFA, AFB, but this statement is

equivalent to the standard theorem.

Result 6 may be proved using the coordinates derived above and the areal metric together with equations (1). It shows that the circle 0\0203 has radius (AF + BF + CF)/3, where the sign convention is being used when F is external to ABC. I expect this is a known result, though it does not

appear in any of the books I possess. Clearly e\ passes through 0\ and the

midpoint of 0203 and similarly for the other two Euler lines.

For those unfamiliar with the areal metric, it is defined by

(dsf = -a(dy)(dz) -

b2(dz)(dx) -

c2(dx)(dy), (7) for the square of the length of the vector displacement dr =

(dx, dy, dz), (see [1]). With complicated expressions for the displacements, such calculations are tedious, and are best performed by using a computer algebra package, such as DERIVE.

Result 7:

T\, T2, T3 lie on a circle with FG as diameter.

Proof: This follows at once from Result 5 and the fact that the angle in a semi

circle is a right angle. Using the areal metric we find

\Ju2 + v2 + w2 - vw - wu - uv ~

-3

Result 8:

_ m m _ _ _ FG\l3 sju2 + v2 + w2 - vw - wu - uv 7^ = 7^ = 7^ = ?

=-W3-.

Again this may be proved using the coordinates derived above and the areal metric together with equations (1). Note that when ABC is isosceles one of

Tx, T2, T3 coincides with F.

Alternative proofs of Results 6-8

In Figure 5 we show the Ferm?t configuration when F is an internal point and v > u > w. Proofs of results need minor modifications in other cases. First note that ZOxCA' = Z02CA = 30? and, because of the equilateral triangles, CA' = \/3COx and CA = V3C02. It follows that triangles OlC02 and A'CA are similar. Hence 0\02 = -tA4r = 4=(w + v + w).




Now CF is perpendicular to 0\02 and e3 is the perpendicular bisector of

0\02. The corresponding lines with respect to triangle ACA\ namely the

perpendicular through C to AA' and the line parallel to it through the

midpoint of AA! (shown as pecked lines in Figure 5) are distance apart equal to \{u + V + w)

- (u + \w) = \(v

- w). Hence, by the similarity,

FT3 = (v -

w)/(2V5). Now in this configuration, in quadrilateral FT2GT3 we have ZT2FT3 = 120? with ZFT2G = ZFT3G = 90?, so FT2 =

(u- w)l(l\?) and hence by the cosine rule

_ _2 (u -

w)2 + (v -

uf + (u -

w)(v -

u) u2 + v2 + w2 -vw -wu- uv T2T3 =-=-.

12 12 The situation is not entirely symmetric, in that FTX = (v

- w)/(2\/3), but the

fact that ZT2FTi = 60?, rather than 120?, means that we finally get the same expressions for both T3T\ and T\T2 (but we know that anyway, since

triangle T{T2T3 has already been shown to be equilateral).

A'

FIGURE 5




Formulae in terms of u, v, w may be re-expressed in terms of a, b, c and the area [ABC] by means of equations (1), which gives

l{u + v2 + w2) + uv + vw + wu = a + b2 + c2, (8) and from the areal coordinates of F, which gives

The result is

and

4 [ABC] uv + vw + wu = ?~i=?.

(9)

2 = ? + * + S _ 2V3 [ABC] 18 9 K J

T{Tl = T27% = T3T2 = \FG\ (11) Finally, if the associated Ferm?t point F' is used instead of F and the

triangle has no angle equal to either 120? or 60?, the nine-point centres of the triangles BFC, CFA and AF'B also form an equilateral triangle, whose circumcircle has F G as a diameter. The working involved is very similar to that for F and is not repeated here.

I am greatly indebted to the referee for making some corrections to the initial manuscript, for suggestions for improvement and for providing the

synthetic proofs of results, where alternative proofs are given.

Reference 1. C. J. Bradley, Challenges in geometry for mathematical olympians,

Oxford University Press (2005). C. J. BRADLEY

6A Northcote Road, Bristol BS8 3HB

Thirteenth Annual Mathematical Gazette Writing Awards

Once again, I wish to thank all who voted and also all those who wrote for the Gazette. The results for articles and notes were as follows:

Winner

Barry Lewis

Runners up

Nick Lord

2007 Article of the Year

Trigonometry and Fibonacci numbers

Intriguing integrals: an Euler inspired Odyssey Graham Hoare Leonhard Euler

2007 Note of the Year Winner

Mark Blyth

Prize Draw Winner

Did Kepler know this?

M Bataille. GERRY LEVERSHA



the fermat point configuration

Documents