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99.09 A family of Napoleon triangles associated with the Kiepert
conguration
Dao Thanh Oai
The Mathematical Gazette / Volume 99 / Issue 544 / March 2015, pp 151 - 153
DOI: 10.1017/mag.2014.22, Published online: 13 March 2015
Link to this article: http://journals.cambridge.org/abstract_S0025557215007822
How to cite this article:
Dao Thanh Oai (2015). 99.09 A family of Napoleon triangles associated with the Kiepert conguration. The Mathematical
Gazette, 99, pp 151-153 doi:10.1017/mag.2014.22
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NOTES 151
C
F
FIGURE 2: A physical device to find the maximal outscribed equilateral triangle.
At A, B and C, frictionless sleeves are free to pivot.
Through these sleeves pass telescoping spring-loaded bars.
The bars are welded rigidly to 60° formers at D, E and F.
References
1. Mark Levi, The mathematical mechanic, Princeton University Press
(2009).
2. Fengming Dong, Dongsheng Zhao and Weng Kin Ho, On the largest
outscribed equilateral triangle, Math. Gaz. 98 (March 2014) pp. 79-84.
doi: 10.1017/mag.2014.21 PHILIP TODD
Saltire Software, 12700 SW Hall Blvd, Tigard, OR 97223, USA
e-mail: philt@saltire.com
99.09 A family of Napoleon triangles associated with the
Kiepert configuration
A classic theorem in plane Euclidean geometry, often doubtfully
attributed to Napoleon, states that the centres of equilateral triangles erected
outwardly on the sides of a triangle ABC form an equilateral triangle. We
shall denote these centres by N A, N B and N C. This is easily proved from the
fact that the line of centres of two intersecting circles is the perpendicular
bisector of their common chord. It is also true that the new vertices of these
triangles form a triangle which is in perspective with ABC. These results
also hold when the initial triangles are erected facing inwards. The common
centre of these triangles is the centroid G of ABC.
152 THE MATHEMATICAL GAZETTE
A generalisation of this result is due to Ludwig Kiepert. This uses
similar isosceles triangles rather than equilateral triangles, and the new
vertices again form a triangle in perspective with ABC. As the base angles
vary (being allowed to become negative to give the inward-facing case), the
locus of the perspector is the Kiepert hyperbola. Details of all these results
can be found in [1].
In what follows, we produce some more equilateral triangles from this
configuration. Let the base angles of the isosceles triangles be α and let
t = tan α. Denote the new vertices formed by the isosceles triangles be
A0, B0 and C0 and define the point A∗ on the ray AA0 by setting
2
AA∗ = AA0, where a negative sign indicates that A∗ is on the
3 − 3t
opposite side of A to A0. Points B∗ and C∗ are defined analogously. In the
diagram which follows, α is taken as smaller than 30°, but this does not
affect the argument. Denote the midpoint of BC by A1.
C
NA
A0
A∗
A1
A B
FIGURE 1
Then a series of simple calculations shows that
A0A∗ 1 − 3t A1N A 1 AG
= , = and, of course, = 2
AA∗ 2 N AA0 3t − 1 GA1
so by Menelaus's theorem on AA1A0 it follows that G, A∗, N A are collinear.
Exactly the same argument works for G, B∗, N B and G, C∗, N C and
another application of Menelaus shows that the ratios GA∗ : GN A are
independent of which vertex is being used and therefore equal. It therefore
follows that, since G is the circumcentre of the Napoleon triangle, A∗B∗C∗
is homothetic to N AN BN C and is therefore equilateral with centre G. In the
special case when α = 30° the triangles A∗B∗C∗ and N AN BN C coincide.
If α = 60°, AA0 is parallel to A1N A and A∗ is at infinity.
2
If the same method is used with AA∗ = AA0, we obtain a
3 + 3t
∗ ∗ ∗
similar configuration with A B C homothetic to N aN bN c, the inner
Napoleon triangle, formed by the centres of inwardly-facing equilateral
triangles.
NOTES 153
Reference
1. G. Leversha, The geometry of the triangle, UKMT (2013).
doi: 10.1017/mag.2014.22 DAO THANH OAI
Cao Mai Doai, Quang Trung, Kien Xaong, Thai Binh Nam
e-mail: oaidt.slhpc@gmail.com
99.10 Proof without words: T1 + T2 + … + Tn = ( )
n+ 2
3
T1 T2 T3 … Tm − 1 Tm … Tn − 1 Tn
… …
↑ ↑ ↑
Underneath the symbols T1, T2, … , Tn place (n + 2) balls, with two
balls to the left of T1. Three of these balls can be selected in( )
n + 2
3
In the example shown, the second, fifth and (m + 2) th balls have been
ways.
selected.
We interpret the third of these as the triangular number Tm and then the
first two balls represent two numbers in the range from 1 to m + 1. We
shall interpret these two numbers as an instruction for defining a unique dot
amongst the Tm dots in the triangular array. The diagram below shows how
this is done in the case when m = 10, but it is clearly completely general.
FIGURE 1
The diagram shows the array representing Tm + 1, which can be thought
of as a top row of m + 1 dots below which is the array for Tm. The second
and fifth dots in the top row have been highlighted since the second and fifth
balls were selected. Now these are used to pinpoint the second dot in the
fourth row of T11 as the intersection of two diagonals in different directions.
This is also the second dot in the third row of T10. We now have a one-to-
one correspondence between the ( )
n + 2
3
choices of three objects from
n + 2 objects and the totality of points in all the triangular arrays from T1 to
Tn, and this is enough to prove the result.
