Suppose that ∆ABC = ∆A∗B∗C∗, and that you can’t get one triangle from the other by fewer than two
reflections. How can you tell whether two or three reflections are required?
The answer actually doesn’t have to do with whether the triangles are themselves symmetric (e.g., equilateral
or isosceles). Rather, what matters is the clockwise order of the vertices: is it A,B,C (which is the same
thing as B,C,A or C,B,A) or A,C,B (which is the same thing as B,A,C or C,A,B)? Any reflection will reverse
this clockwise order. Therefore, a composition of an even number of reflections will keep this order the same,
while a composition of an odd number of reflections will reverse it.
Now, it is true that if Tand T∗are equilateral or isosceles triangles and T∼
=T∗, then there exists a line ℓ
such that rℓ[T] = T∗. (That is, you can get from Tto T∗by a single reflection.) However — this is a subtle
point — you can’t control which vertex goes to which one. For example, if T= ∆ABC and T∗= ∆A∗B∗C∗
are the following pair of congruent triangles, then by the “clockwiseness” agument above, there can be no
reflection rℓsuch that rℓ(A) = A∗,rℓ(B) = B∗, and rℓ(C) = C∗. There does, however, exist a reflection rm
(shown) such that rℓ(A) = A∗,rℓ(B) = C∗, and rℓ(C) = B∗.
AC
B
B*
C*
A*
m
On the other hand, two congruent triangles don’t need to be isosceles (or equilateral) in order for one to be
transformable to the other by a single reflection. After all, you can take any old scalene triangle Tand any
old line ℓ, and define T∗=rℓ[T]; then Tand T∗are congruent scalene triangles related by one reflection.
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