Download A PARALLELOGRAM PROBLEM A diagonal angle of a

A PARALLELOGRAM PROBLEM
B.D.S. “DON” MCCONNELL
A diagonal angle of a parallelogram is an angle
bounded by a side and a diagonal. A parallelogram has eight diagonal angles, which break into
four pairs of congruent angles. In our
A'
B'
A'
B' discussion,
cc0 0
B'
A'
B'
A'
we take ABA B to be a parallelogram with diagαα α, β, α0 , and β 0 , and various
α'
α'
onal angle ββmeasures
α'
β'
β'β' aa bas
β'
b α'
segment lengths
shown
in Figure 1.
c
B'
α
β
β'
c'
A
Figure 1. Labeled Parallelogram
α'
α'
1. Fourβ'β' Easy
αα ββ
situations:AA ββ
Cases
B'
α0
β0
sin(
Case 3: α,
given; β,
sought.
c'
Case 4: β, β 0 given; α, αs0insought.
α
β'
B'
A'
B'
A'
We see easily that Case 1 and
in one of the quadrants of the
B' Case 2 are analogous:
A'
B'
A'
B'given two angles
B'
A'
β apply
A'
B'
parallelogram, α'α'
we seek the angles in another quadrant.
(Symbolically, αwe simply
the change ofα
c
B'
A'
0
0
α'
B making up Aone
β' analogous:
α'A given
sin( α+β
)
β'
variables α ↔ α and β ↔ β .) Case 3 andα'Case 4 are also
two angles
s in
α
β
α'
c'c'
(β
inin
corner of the ssparallelogram,
we
seek
the
angles
making
up
another
corner.
(Symbolically,
α ↔ β and
+β
a b α'
β'
β'
αα
0
0
'
)
α ↔ β .) In pointβ'β' of fact, Case 3 and Caseaa 4 are also analogous
α β angles appear in
b to Case 1: the given
A
c'
β' c'when four copies
β'
β'
quadrantsααof a secondary
parallelogram
that
arises
of
the
original
parallelogram
are joined
B'
A'β
a
β
b
β
α'
β
α Case β1.
β'α
α
α'
B
together.
we
may
focus
on
solving
β
AA Thus,
B
sin(
α+β
)
sin( α+β )
β'
α'
a
β'
B'
A
cA
BB
B
Case 2
A'
A'
Equivalence
1
A
β
B
β'
sin
CaseA' 4
B'
β'
EasyB'Cases.
β'
sin β
β'B
c
α'
α'
sin
A'
(β
+β
')
bα'
sin α
α
β'
α'
β'
of theα Four
β
A c'
sin
β
α
β'
α'
β
α
A
sin β
α'
Figure 2.β' The
α β
A
β
A' 3
Case
B'
B'
α'
β'
β'
α β
'
A
β
sin
α
')
A
sin
α α'
α
γ
α'
β'
αβ' β 1
β
A β
α
β
β
β'
sin α
α
γγ
sin
β'
α +α
Case 1
ααβ
α
A
β'
β'
β
β
β'
α'
BB
c'
a
b
A'
B'
sin(
AA
c'
sin
αα
β'
β'
ββ
a b
α'
)
β'
A'
α'
α'
α'
B
A'
B'
α
β
sin β'
B
c
sin
sin ββ
α'
α'
A'
A'
A'
Aβ
A
α
A
B'
'+β'
c
BB
β
ssiin
n ββ
B'
B'
B'
sin α'
α
sin(
A'
A'
β
ssiinn ββ
sin α'
α'
sin
AA
B'
B'
ssiinn αα
ssiinn(
( αα++
αα'' )
)
sin
β
Case 1: α , β given;
sought.
β'
β'
Case 2: α0 , β 0 given; α , β sought.
A'
α'
β'
β'
β0
B'
A'
sin α
α0 ,
B
c
')
following
β
α
β'
β'
α +α
Here,
β'
α'
There
the pairs of given
angles, leading to four cases that are easy to solve
A'
A'
B'
B'
and two that
A'
A'
B' are not-so-easy.
B'
α'
α'
β'
β'
αα ββthe
we’llAcover
A
ββ
c'
a
b
αα ββ
waysββto choose
BB
cc
α'
a b
AA
0
c'c'
c'c' Given two
Exercise.
ββ α ,
aa of the angle measures α, β,
bb
0
and β , determine
the remaining
two.
β'
β'
α'
α'
αα
are six
AA
A'
β'
β
α
A a sin β'
β'
B
2
B.D.S. “DON” MCCONNELL
Two straightforward trigonometric relations will come in handy. The first relates the side of a triangle
to its adjacent elements; for instance,
(1)
c = a cos β + b cos α
and the second (which is provable by the Law of Cosines) relates the parallelogram’s sides:
2a2 + 2b2 = c2 + c02
(2)
Finally, the Law of Sines guarantees that
sin α
sin β
sin (π − α − β)
=
=
a
b
c
Convenient scaling of the diagram allows us to assume that the Law of Sines ratio is 1. That is,
(3)
(4)
a = sin α
c = sin (π − α − β) = sin (α + β)
b = sin β
At this point, we may proceed directly to our computations: Substituting back through Equations (2)
and (1), we arrive at the following results. (To conserve space, we represent “sin θ” by “[θ]”.)
cos α0 =
b − a cos (α + β)
[β] − [α] cos (α + β)
=p
0
c
[α]2 + [β]2 − 2[α][β] cos (α + β)
cos β 0 =
a − b cos (α + β)
[α] − [β] cos (α + β)
=p
0
c
[α]2 + [β]2 − 2[α][β] cos (α + β)
Note: Via Equation (2), we get the implicit relation “c02 = [α]2 + [β]2 − 2[α][β] cos (α + β)”. Because
our side lengths are non-negative we are allowed to invoke the inverse function expressing c0 as an unsigned
square root. Likewise, our angle measures being bounded by 0 and π allow us to apply the inverse cosine
function to the above results to obtain
!
[β]
−
[α]
cos
(α
+
β)
α0 = acos p
[α]2 + [β]2 − 2[α][β] cos (α + β)
β 0 = acos
!
[α] − [β] cos (α + β)
p
[α]2 + [β]2 − 2[α][β] cos (α + β)
For completeness, here are the results for the analogous Cases:
Case 2:
α = acos
[β 0 ] − [α0 ] cos (α0 + β 0 )
p
!
β = acos
[α0 ]2 + [β 0 ]2 − 2[α0 ][β 0 ] cos (α0 + β 0 )
!
[α0 ] − [β 0 ] cos (α0 + β 0 )
p
[α0 ]2 + [β 0 ]2 − 2[α0 ][β 0 ] cos (α0 + β 0 )
Case 3:
β = acos
[α0 ] − [α] cos (α + α0 )
!
β = acos
p
[α]2 + [α0 ]2 − 2[α][α0 ] cos (α + α0 )
!
[α] − [α0 ] cos (α + α0 )
0
p
[α]2 + [α0 ]2 − 2[α][α0 ] cos (α + α0 )
Case 4:
α = acos
[β 0 ] − [β] cos (β + β 0 )
p
[β]2 + [β 0 ]2 − 2[β][β 0 ] cos (β + β 0 )
!
0
α = acos
[β] − [β 0 ] cos (β + β 0 )
p
[β]2 + [β 0 ]2 − 2[β][β 0 ] cos (β + β 0 )
!
A PARALLELOGRAM PROBLEM
3
2. Two Not-So-Easy Cases
Here, we’ll cover the situations in which the given angles appear in a kind of “pinwheel” arrangement:
Case 5: α , β 0 given; α0 , β sought.
Case 6: α0 , β given; α , β 0 sought.
Figure 3 shows these Cases to be analogous —symbolically, we apply the change of variables α ↔ β and
α0 ↔ β 0 — so we need only consider Case 5.
B'B'
B'
B'B'
A'A'
A'A'
ββ
β'β'
αα
α'α'
ββ
β'β'
cc
B'B'
A'A'
αα
α'α'
A'A'
αα
α'α'ββ
β'β'
α'α'
β'β'
α'α'
α'
β'β'β'β' aa bb α'
ββ
αα
ββ
αα
α ββ
α
AA c'β
AA
BB
c'c'
AA
BB
c'β
a
b
a
b
Case 5
Case 6
β'
A'A'
B'B'
α'
β'
A'
α'
B'
A'
B'
A'
B'
A'
β'β' B'
B'
A'
A'
ββ
ααB'
Figure 3. The Equivalence of the Two Not-So-Easy
α'α' β'β' Cases.
α
Aα'
β
Aα'
BBα
cc β
α'α'
α'
β' α'observe that this β'
Before
we
Case has a built-in ambiguity,
unlike the
first four Cases which
β'β' weβ' begin, β'
ββ
αα
β
α
A
each
had
a
unique
solution.
The
“secondary
parallelogram”
appearing
in
Cases
3
and
4 of Figure 2 has the
β
α
β
α
A
c
αα differently positioned
A α βarrangement as the original figure, yet the angles
ββ α0 and β are
0 pinwheel
AAA' same
A
αβ
β
β
ββ
α
0
relative to thatα'arrangement.
β' Thus, for anyα'given angleβ' measures α and β , we can find (at least) two
β'
a b α' solutions to the parallelogram problem. We’ll see this fact echoed naturally
β' α'α'in our symbolic solutions,
β
α
β'β
β'β'
α
β'
where we’ll find that there are exactly two possible
parallelograms.
A
B
a
b
c'
A
α'
A'β' B' B'
B' β' A'
β
α
Aα'
c
α'B
β'
β'
β
β
α
A
β
B
B'
A'A'
B'B' A'
A'A'
α'α'
sisninA'
A'
( (ββ+
β+ββ' ' )
α
c'
A
c'
b
s
i
bα'α' α )
sinn
β
α
α
β'β'
α'α'
β'β'
β' ααα'
ββ
β'
β'
β
α
a sin β'
β
α
a
A
A
BB
sin
β'
β'
A
B
sin(
α+β
)
β'
A sin( α+β )
B
B'
A'
α
A'
ββ Solution.
Figure 4. Theα Secondary
B'
A'
B'
A'
sinα'α'
BB
α' AA
sin
β'
s
i
α' Our analysis is adapted
from that nof
B'
A'
( β Matthew Fahrenbacher, who presented this problem in the first
+β
α'α'
place.
'
ssinin γ
c'
αα “quadrant”
bα' the) Law of Sines applied to the two
sin We begin with an observation based
γ
triangles of the
α
β'
parallelogram:
α'
β'
11
β
α
ββ
sisnin
β'
α
β
α
0
α' ' β'
A a sinsin
B
a
sin α
β' α
(5) )
= =
β'
A sin( α+β
B
β
α
sin
β
b
sin β 0
β
α
sinαα
β
α
γγ
sin
A
B
B
cc
'
A' A
A
B
sin α'
α'α'
α'
β'β'β'
B'B'
n β'
sisinnββ
')
α +α
sin(
ssin
inββ'
'
ssin
inββ
β
sin α
sin
ssininαα
ssin(
in(α+
α+αα'
'))
ssininββ
sin β
')
α'++ββ' )
ssinin(( α'
si
α'
4
B.D.S. “DON” MCCONNELL
Hence
sin α sin β 0 = sin α0 sin β
(6)
Noting that α + β + α0 + β 0 = π (as the angle sum of 4ABA0 , for instance), we can replace β in Equation
(6) with π − α − β 0 − α0 and manipulate as follows:
sin α sin β 0 = sin α0 sin β
= sin α0 sin (π − α − β 0 − α0 )
= sin α0 sin (α + β 0 + α0 )
1
cos ((α + β 0 + α0 ) − α0 ) − cos ((α + β 0 + α0 ) + α0 )
=
2
1
=
cos (α + β 0 ) − cos (α + β 0 + 2α0 )
2
so that we have one of our sought angles implicitly related to our given angles:
(7)
cos (α + β 0 + 2α0 ) = cos (α + β 0 ) − 2 sin α sin β 0 = 2 cos (α + β 0 ) − cos (α − β 0 )
Likewise, replacing α0 in Equation (6) with π − α − β 0 − β yields
cos (α + β 0 + 2β ) = 2 cos (α + β 0 ) − cos (α − β 0 )
(8)
Because Equations (7) and (8) are identical (except for the appearance of α0 or β, respectively), the
angle measures α0 and β are not inherently distinguishable. This matches the geometric ambiguity we
mentioned earlier.
Now, since either (or both) of the angle sums α + β 0 + 2α0 (call this “φ”) and α + β 0 + 2β (call this “ψ”)
threatens to be larger than π, we cannot simply apply the inverse cosine to each side of Equations (7) and
(8). To understand our situation better, we make a few observations about these angle sums.
Comparing them to partial angle sums from triangles in our parallelogram, we have
0 ≤ α + β 0 + 2α0 = (α + α0 ) + (β 0 + α0 ) ≤ π + π = 2π
0
0
0 ≤ α + β + 2β = (α + β ) + (β + β ) ≤ π + π = 2π
and also, we have
(α + β 0 + 2α0 ) + (α + β 0 + 2β) = 2(α + β + α0 + β 0 ) = 2π
=⇒
0 ≤ φ ≤ 2π
=⇒
0 ≤ ψ ≤ 2π
=⇒
φ + ψ = 2π
Therefore, φ and ψ are precisely the two solutions for θ in the equation
cos θ = 2 cos (α + β 0 ) − cos (α − β 0 )
in the range 0 ≤ θ ≤ 2π, although we don’t know which is which. This implies that
1
0 1
0
0
acos x − α − β ,
2π − acos x − α − β
, with x = 2 cos (α + β 0 ) − cos (α − β 0 )
α ,β =
2
2
where, again, we don’t know which angle measure is which.
For completeness, the solution to Case 6 is
1
1
0
0
0
0
0
α, β =
acos x − α − β ,
2π − acos x − α − β , with x0 = 2 cos (α0 + β) − cos (α0 − β)
2
2
which concludes our analysis.