E XAMPLES ON T RANSLATIONS AND ROTATIONS
(Lecture Notes for Math 532, taught by Michael Filaseta)
1. Let ∆ABC be given, and let MB denote the midpoint of side AC and let MC
←−−−→
←→
denote the midpoint of side AB. Show that MB MC is parallel to BC and that the
length of MB MC is one-half of the length of BC.
Solution. A picture would help here. From the theorem, f = Rπ,MB Rπ,MC is a
translation. One checks that f (B) = C. Hence, f is a translation which moves
B to C. Also, f (MC ) = MC0 where MC0 is the result of rotating MC about MB
by π. This means that MC MB and MB MC0 have the same length and the three
−−−−−→ −−→
points MC , MB , and MC0 are collinear. We get that MC MC0 = BC, and the result
follows. 2. Let A, B, C, and D be the vertices of an arbitrary quadrilateral. Show that the
midpoints of the sides of the quadrilateral form a parallelogram.
Solution. Let M1 be the midpoint of AB, M2 the midpoint of BC, M3 the mid−−−−→
point of CD, and M4 the midpoint of DA. Then from example (1), M1 M4 and
−−→
−−−−→
M2 M3 each have the same direction as BD and half its length. The desired conclusion follows.
Comment: Instead, one can let f = Rπ,M4 Rπ,M1 and g = Rπ,M2 Rπ,M3 . Then
f is a translation taking B to D and g is a translation taking D to B. One then
essentially repeats the argument in example (1). 3. In (2), consider instead midpoints of a 2n−gon.
Solution. The problem is a bit vague, but one can conclude the following using the argument in (2). Let M1 , M2 , . . . , M2n denote the midpoints along the
edges moving counterclockwise beginning with some edge. Then the segments
M1 M2 , M3 M4 , . . . , M2n−1 M2n can be translated (without rotating them) to form
an n−gon. Similarly, the segments M2n M1 , M2 M3 , . . . , M2n−2 M2n−1 can be
translated to form an n−gon. 4. Let ∆ABC be given. Draw an equilateral triangle exterior to ∆ABC with one
edge AB, an equilateral triangle exterior to ∆ABC with one edge BC, and an
equilateral triangle exterior to ∆ABC with one edge AC. Show that the centers of
these 3 equilateral triangles form the vertices of an equilateral triangle.
Solution. The argument is essentially the same as in the next problem. This is
worth going over separately, but we do not do so here. 1
5. Generalize (4) as follows. Let ∆ABC and real numbers α, β, and γ be given. Let
A0 , B 0 , and C 0 be points exterior to ∆ABC such that ∠BA0 C = α, ∠AB 0 C = β,
and ∠AC 0 B = γ. Also, suppose that the lengths of the sides BA0 and A0 C are the
same, the lengths of the sides AB 0 and B 0 C are the same, and the lengths of the
sides AC 0 and C 0 B are the same. Show that if α + β + γ = 2π, then the interior
angles of ∆A0 B 0 C 0 are 12 α, 12 β, and 12 γ.
Solution. Let f = Rα,A0 Rβ,B 0 Rγ,C 0 . Then f is a translation since α+β +γ = 2π.
Since f (B) = B, we get that f is the identity translation. Hence, f (C 0 ) = C 0 . Let
C 00 = Rβ,B 0 (C 0 ). Then ∠C 0 B 0 C 00 = β and the lengths of B 0 C 0 and B 0 C 00 are the
same. Also,
C 0 = f (C 0 ) = Rα,A0 Rβ,B 0 Rγ,C 0 (C 0 ) = Rα,A0 Rβ,B 0 (C 0 ) = Rα,A0 (C 00 ).
This means that ∠C 0 A0 C 00 = α and the lengths of A0 C 0 and A0 C 00 are the same.
One easily gets that the triangles ∆A0 B 0 C 0 and ∆A0 B 0 C 00 are congruent from
which it follows that ∠C 0 A0 B 0 = α/2 and ∠C 0 B 0 A0 = β/2. It follows that
∠A0 C 0 B 0 = π − (α/2) − (β/2) = γ/2, giving the desired result. Observe that the following is a consequence of the problem. Suppose that ∆ABC
is given and D, E, and F are points exterior to ∆ABC such that ∆DBC, ∆AEC,
and ∆ABF are similar so that ∠D, ∠E, and ∠F are the three angles associated
with these similar triangles. Let A0 , B 0 , and C 0 be the centers of the circumscribed
circles for ∆DBC, ∆AEC, and ∆ABF , respectively. Then ∆A0 B 0 C 0 is similar
to ∆DBC (and, hence, the other two exterior triangles as well).
6. Let n be an odd positive integer, and let P1 , . . . , Pn be n (not necessarily distinct)
points. Let A = A0 be an arbitrary point. For j ∈ {1, . . . , n}, define Aj as the
point you get by rotating Aj−1 about Pj by π. For j ∈ {n + 1, . . . , 2n}, define Aj
as the point you get by rotating Aj−1 about Pj−n by π. Prove that A2n = A.
Solution. Write n = 2k + 1 where k is some nonnegative integer. Let f denote
the composition of the first n rotations about P1 , . . . , Pn each by π. Then we want
to show that f (f (A)) = A. Note that every two rotations by π are equivalent to a
translation and the composition of translations is a translation. Hence, we can view
f as T(a,b) Rπ,P1 for some (a, b) (where a = b = 0 if k = 0). By a homework
problem, we can rewrite this as
Rπ,(a/2,b/2) Rπ,(0,0) Rπ/2,P1 Rπ/2,P1 .
Taking the product of the matrices as indicated by the parentheses above, we get
from the theorem that f is equivalent to a rotation about some point by π. It is clear
then that f (f (A)) = A, completing the argument. Comment: The situation when n is even is that A is translated since the compositions of the rotations is a translation. It is possible that the translation is the identity
translation in which case A2n = A.
2
7. Let n be an odd positive integer. Suppose that we are given the n midpoints of
the sides of an n−gon. Show how one can construct an n−gon with these given
midpoints along its edges.
Solution. Let M1 , . . . , Mn be the midpoints. Consider the composition f of the rotations about M1 , . . . , Mn each by π. By the solution to (6), we get that f is equivalent to a rotation about some point, say A, by π. We can construct A as follows.
Take any point B and apply f to it. Note that this is done by successively taking
the rotations about Mj by π with straightedge and compass for j = 1, 2, . . . , n. We
get some point C = f (B). Since B is obtained from C by a rotation about A by π,
we deduce that A must be the midpoint of BC. Since f is equivalent to a rotation
about A, we have that f (A) = A. Set A0 = A and rotate it about M1 to obtain
a new point. Call the new point A1 and rotate it about M2 to obtain another point
A2 . Continue rotating Aj about Mj+1 to obtain Aj+1 for j ∈ {1, 2, . . . , n − 1}.
Since f (A) = A, we get that An = A0 , and the points A1 , . . . , An are the vertices
of an n−gon as desired. Comments: (i) There are many such n−gons since the order of the Mj one chooses
to do the above construction will affect the outcome.
(ii) There is another approach to the problem which may be worth discussing. For
example, suppose n = 5 (though any odd n works here). Call the given midpoints
M1 , . . . , M5 . Let A0 , . . . , A4 be the points we are trying to construct with Mj
along edge Aj−1 Aj for j ∈ {1, . . . , 5} where A5 = A0 . Then the vertices A0 , A1 ,
A2 , and A3 are the vertices of a quadrilateral and three of its midpoints M1 , M2 ,
and M3 are known. Using the information from example (2), it is not difficult to
construct the midpoint M 0 of A0 A3 . Now, we know the midpoints of the sides of
triangle ∆A0 A3 A4 . Using the information from example (1), we can construct the
vertices A0 , A3 , and A4 . One can modify this argument to obtain the other vertices
or use the approach in the solution above.
8. Let P1 , P2 , P3 , and P4 be 4 (not necessarily distinct) points. Let A be an arbitrary
point. Beginning with A0 = A, for j ∈ {1, 2, 3, 4}, define Aj as the point you get
by rotating Aj−1 about Pj by π. Set Q1 = P3 , Q2 = P4 , Q3 = P1 , and Q4 = P2 .
Beginning with B0 = A, for j ∈ {1, 2, 3, 4}, define Bj as the point you get by
rotating Bj−1 about Qj by π. Prove that A4 = B4 . (See Figure 1.)
Solution. A rotation about P1 by π followed by a rotation about P2 by π is a
translation, say TR . Similarly, a rotation about P3 by π followed by a rotation
about P4 by π is a translation, say TS . The problem amounts to establishing that
TS TR = TR TS . This is easy to establish (but note that in general the product of
two matrices does not commute). 9. Let A, B, C, and D be the vertices of a convex quadrilateral labelled counterclockwise. Consider 4 squares exterior to the quadrilateral, one square with an edge AB,
3
A3
B1
B3
A1 P3
P4
P2
P1
A2
A
B2
A4 = B4
Figure 1
one square with an edge BC, one square with an edge CD, and one square with an
edge DA. Let M1 be the center of the square with edge AB, let M2 be the center
of the square with edge BC, let M3 be the center of the square with edge CD, and
let M4 be the center of the square with edge DA. Show that the length of M1 M3
←−−→
←−−→
is the same as the length of M2 M4 and that M1 M3 and M2 M4 are perpendicular.
Solution. Let
g = Rπ/2,M1 Rπ/2,M2 ,
h = Rπ/2,M3 Rπ/2,M4 ,
and
f = gh.
Then f (A) = A, and we get that f is the identity translation. Also, there are points
P1 and P2 such that g is a rotation about P1 by π and h is a rotation about P2 by
π. It is easy to see (draw a picture) that since a rotation about P1 by π followed by
a rotation about P2 by π is the identity, we must have P1 = P2 . Next, we observe
that P1 is the only point such that g(P1 ) = P1 . It follows that ∆M2 P1 M1 is a
isosceles right triangle labelled counterclockwise (since P1 so located is mapped
to itself by g). Similarly, ∆M4 P2 M3 is a isosceles right triangle labelled counterclockwise. Since P1 = P2 , we get that ∆P1 M2 M4 is obtained from ∆P1 M1 M3
by a rotation about P1 by π/2. This implies that the length of M1 M3 is the same
←−−→
←−−→
as the length of M2 M4 . Let Q1 be the point of intersection of M1 M3 and M2 M4
←−−→
←−−→
and Q2 the point of intersection of M2 M4 and M1 P1 (convince yourself these exist). Then ∠M1 Q2 Q1 = ∠P1 Q2 M2 and ∠Q2 M1 Q1 = ∠P1 M2 Q2 , and it follows
←−−→
←−−→
that ∠M1 Q1 Q2 = ∠M2 P1 Q2 = π/2. Thus, the lines M1 M3 and M2 M4 are
perpendicular. Comment: It can be shown that if the original quadrilateral is a parallelogram, then
M1 , M2 , M3 , and M4 form the vertices of a square.
4
10. Let ∆ABC be a given triangle with the angles ∠ABC, ∠BCA, and ∠CAB all
acute. Let AP be an altitude drawn from A so that P is on BC. Similarly, let
BQ be the altitude drawn from B and CR the altitude drawn from C. Show that
∆P QR is a triangle with minimum perimeter that can be inscribed in ∆ABC (that
is show that if ∆U V W is a triangle with U on BC, V on AC, and W on AB, then
its perimeter is at least that of ∆P QR).
Solution. First, we will show that ∠P RB = ∠QRA, ∠RP B = ∠QP C, and
∠P QC = ∠RQA. We establish one of these and the other two can be done in the
same way. Let D be the intersection of the altitudes. Since ∠DRA = ∠DQA =
π/2, the points A, Q, D, and R all lie on a circle. Hence,
∠DRQ = ∠DAQ =
π
− ∠ACP.
2
Since ∠DRB = ∠DP B = π/2, the points B, P , D, and R all lie on a circle.
Hence,
∠DRP = ∠DBP =
π
π
− ∠BCQ = − ∠ACP = ∠DRQ.
2
2
Since ∠ARC = ∠BRC, we get that ∠P RB = ∠QRA. As mentioned, in a
similar fashion, one obtains ∠RP B = ∠QP C and ∠P QC = ∠RQA.
If you haven’t already started drawing pictures, get out those crayons. To match
some of the discussion here, you should label your points along the triangle as A,
B, and C in a counterclockwise direction. We begin with triangle ∆ABC and
reflect it about side BC to get a new tringle ∆A1 BC. The 2 triangles are distinct,
congruent, and share the edge BC. Next, we reflect ∆A1 BC about side A1 C to
get a new tringle ∆A1 B1 C. Then we reflect ∆A1 B1 C about side A1 B1 to get a
new tringle ∆A1 B1 C1 . Next, we reflect ∆A1 B1 C1 about side B1 C1 to get a new
tringle ∆A2 B1 C1 . Finally, we reflect ∆A2 B1 C1 about side A2 C1 to get a new
tringle ∆A2 B2 C1 . If your crayoning technique is mastered, this is what should
happen. Let α = ∠BAC and β = ∠ABC. Every point along segment AB is first
rotated about B by 2π − 2β, then about A1 by 2π − 2α, then about B1 by 2β, and
finally about A2 by 2α. We can conclude that the segment AB has been translated
←→
←−−→
to A2 B2 . This means that AB is parallel to A2 B2 . If we draw in ∆P QR and its
reflections, the information from the first paragraph implies that the points R, P ,
Q1 (the first reflection of Q), R2 (the next reflection of R - there was already an R1
from the first reflection), P2 (the next reflection of P ), Q3 (the next reflection of
Q), and R4 (the last reflection of R) are all collinear. Also, the segment RR4 has
length twice the perimeter of ∆P QR. If one similar reflects ∆U V W , one finds
that W is translated to some W4 . Here, W W4 and RR4 have the same length (since
R and W went through the same translation), and the length of W W4 is ≤ twice
the perimeter of ∆U V W . The desired conclusion follows. 5
A
C
B
Figure 2
6