## 1. Glide-reflexions

Glide-reflexions Y=G(X) are transformations of the plane (isometries or congruences) which are compositions of a reflexion Re, with respect to some line e (called the axis of the transformation) and a translation Tv, with respect to a vector v parallel to line e. It is easily seen that the order by which we apply these two transformations plays no role i.e. G = Tv*Re = Re*Tv.

## 2. Glide-reflexion representation through reflexions

Since every translation Tv by the vector v is a composition Tv=R2*R1 of two reflexions on two parallel lines {e1,e2} in mutual distance |v|/2 and direction orthogonal to v, the glide-reflexion has a canonical representation as a composition of three reflexions G = R2*R1*Re. The two lines e1, e2 being orthogonal to e and at distance |v|/2.

## 3. Glide-reflexion representation through reflexions II

Even in the case of a transformation G = R2*R1*Re, which is the composition of three reflexions on three lines {e,e1,e2} the last two being parallel, one can show that it reduces to a transformation of the previous case.
This means that one can find another tripple of lines {g,g1,g2} such that {g,g2} are both orthogonal to line g1 and the corresponding reflexions {Sg,S1,S2} on these lines satisfy Sg*S2*S1 = G.

To show this, start with the three reflexions on lines {e,e1,e2} and add a fourth reflexion Sg on a line g, which is orthogonal to line e. The composition is a Symmetry on a point D easily constructible from the data. In fact the product of the two first reflexions R1*Re is a rotation by an angle w and the product of the two other reflexions Sg*R2 is again a rotation by an angle of measure (pi/2)-w. Thus the whole product of the four reflexions is a rotation by an angle of (pi) i.e. a point-symmetry. The center of symmetry is easily determined by joining through a line f the two centers of rotation {A,C} and taking at these points respectively angles equal to -w and w-(pi/2), thus forming the right-angled triangle ACD. D is the symmetry center. Then write the symmetry as a product of two reflexions S2*S1 on two orthgonal lines {g1,g2} of which the second is parallel to g. Then
Sg*G = Sg*R2*R1*Re = S2*S1 ==> G = Sg*S2*S1.

## 4. Glide-reflexion representation through reflexions III

Even in the case of a transformation G = R3*R2*R1, which is the composition of three reflexions on three lines {e1,e2,e3} in general position, one can show that it reduces to a transformation of the first handled case.
This means that one can find another tripple of lines {g,g1,g2} such that {g1,g2} are both orthogonal to line g and the corresponding reflexions {Sg,S1,S2} on these lines satisfy Sg*S1*S2 = S2*S1*Sg = G.

In fact, this case reduces immediately to the previous one by turning the system of the two first lines {e1,e2} about their intersection point O, rigidly, i.e. without to alter their mutual angle at O, so that e2 obtains a position parallel to e3. The result follows then by applying the previous arguments.

Next section handles the very same case handled here but relates it to the triangle ABC formed by the three lines {e1,e2,e3} in general position and brings into the play the orthic triangle DEF of ABC.

## 5. Glide-reflexion representation through reflexions IV

Let ABC be a triangle and {Fc,Fb,Fa} be the reflexions respectively on its side-lines {BC, CA, AB}. Their composition S = Fc*Fb*Fa is a Glide-Reflexion, with axis the side FE of the orthic triangle DEF and translation-vector FF'' equal to the perimeter of the orthic triangle. The cyclic permutations of the three reflexions give glide-reflexions with respect to the other sides of the orthic triangle. The remaining permutations give inverse transformations to those three.

Taking into account the fact that the altitudes of ABC are bisectors of the angles of the orthic triangle DEF we find first the image F'' of F under the successive application of reflexions on the sides {Fc, Fb, Fa}. Segment FF'' has length equal to the perimeter of the orthic triangle DEF.
Join an arbitrary point X with F and the projection G on line FE to form triangle FGX. Then reflect again successively by the above three reflexions to obtain correspondingly the equal triangles {FG'X', F'G''X'', F''HS(X)}. The reflected F''HX* of the last triangle with respect to line FE is a parallel translate of the orginal triangle FGX and this proves the claim.

## 6. Shapiro's problem

The previous discussion solves the problem proposed by H. S. Shapiro (E929) in the American Mathematical Monthly:
Given three non-concurrent straight lines L1, L2, L3 in the plane. Let Ti denote reflection in Li and set T= T1*T2*T3 (* denoting composition). Show that T2 is a translation.

By the previous discussion T is a glide reflexion and it is a trivial property of every glide reflexion that T2 is a translation.

### References

The otto Dunkel Memorial Problem Book Four handred Best Problems (1918-1950) Amer. Math. Monthly, Vol. 64, No. 7, Part 2, pp. 9-64.

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