When , , α β γ are positive integers such that 2. 2. 2 α β γ. +. = , then the triangle BA. Πof. Figure 1 is a Pythagorean triangle and ( , , ) α β γ is a Pythagorean triple.
The Equation αsinx+βcosx =γ and a family of Heron Cyclic Quadrilaterals
Konstantine Zelator Department Of Mathematics College Of Arts And Sciences Mail Stop 942 University Of Toledo Toledo,OH 43606-3390 U.S.A.
1
Introduction
On the average, introductory texts in trigonometry may contain five to ten printed pages of material on trigonometric equations.
Only in passing one may come across an
equation such as sin x + cos x = 1 ; or, listed as an exercise in the exercise set. To study trigonometric equations and trigonometry in general, in more depth, one must resort to math books on advanced trigonometry. There are only a few of those around but they tent to be very good and thorough sources of information on the subject. Such are the two books listed in the references [1] and [2]. In [3] and [4], among other material, trigonometric equations (and systems of equations) are studied in some depth, and a variety of families of equations are analyzed. In [3], many types of problems of varying difficulty (both solved and unsolved) are presented. Let α , β , γ be fixed real numbers and consider the equation
α sin x + β cos x = γ
(1)
In Section 3, our analysis leads to the determination of the solution set S of Equation (1). In Section 3, via a simple construction process, we exhibit the angle θ that generates the solution set S of (1) in the case α 2 + β 2 = γ 2 (see Figure 1) and with α , β , γ > 0 ; the angle θ is part of an interesting quadrilateral Γ BΓ 2 Γ 1 . In Section 2 of this paper, we generate a family of quadrilaterals Γ BΓ 2 Γ 1 which have three integer side lengths, the fourth side length being rational, one integral diagonal length, on rational diagonal length, and with the four angles have rational tangent values and the four vertices lying on a circle; hence the term ‘cyclic’ in this paper’s title. In Section 8, we will see that a certain subfamily of the above family; consists solely of
1
Heron Quadrilaterals; quadrilaterals with integer side lengths and diagonals, as well as integral area.
2
The
Solution
Set
α sin x + β cos x = γ ; α , β ,γ ∈R
of
the
Equation
To find the solution set S (a subset of R ), we use the well known half-angle formulas, which are valid for any angle or x (typically measured in radians or degrees) not of the form 2kπ + π ; x ≠ 2kπ + π , k ∈ Z . Note that as a simple calculation shows, if the real numbers of form 2kπ + π are members of the solution set to (1), then it is necessary that β + γ = 0 . So, under the restriction x ≠ 2kπ + π , by using the above formulas and after a few algebra steps we obtain the equivalent equation ⎛x⎞ ⎛ x⎞ ( β + γ ) tan 2 ⎜ ⎟ − 2α tan ⎜ ⎟ + (γ − β ) = 0 . ⎝2⎠ ⎝2⎠
(2)
If β + γ = 0 , then straight from (1) one sees that the reals of the form 2kπ + π , k ∈ Z are solutions to (1). The other solutions are those reals which are
solutions to (2). We have,
Suppose β +γ =0; that is ,γ = − β . Then the solution set S of (1) is given by: (i) S = R, if α =β =γ =0. (ii) S = S1 , if α =0 and β ≠ 0. (iii) S = S1 U S 2 , if α ≠ 0, where
S1 = { x | x ∈ R, x = 2kπ + π , k ∈ Z } and S 2 = { x | x ∈ R, x = 2kπ + 2ϕ , k ∈ Z } ; where ϕ is the unique angle with -
π 2
γ 2 ) , the trinomial f (t )
distinct real roots, namely
has two
We have (the reader is urged to verify),
Suppose β +γ ≠ 0; then the solution set S of (1) is given by: (i) S = ∅, (empty set), if α 2 + β 2 < γ 2 . (ii) S = { x | x ∈ R, x = 2kπ + 2θ , k ∈ Z } , if α 2 + β 2 = γ 2 ; where θ is a unique angle such that -
π 2
n . Note: If one pursues Equation (5b) (Possibility 2) in combination with (6), one is led to
the equation δ 2 .(m + n) 2 .2.(m 2 + n 2 ) = k 2
From there, using a little bit of elementary
number theory on arrives at
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m 2 + n 2 = 2 L2 k = 2δ (m + n) L This leads to a second family of quadrilaterals ΓBΓ 2 Γ1 , which we will not consider here. We only point out that in this case, the triple (m, n, L) will be a positive integer solution to the three variable Diophantine equation X 2 + Y 2 = 2Z 2 , whose general solution has been well known in the literature. The interested reader should refer to [5].
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Another numerical example
If in (9b) we put t1 = 2, t2 = 1 , then we obtain L = 5 and m=4 > n=3 as required. Since the condition in (6) is satisfied if we take δ = 5, then we have, by (5a),
α = 120, β = 35, γ = 125 ; and from (8) we obtain the quadrilateral ΓBΓ 2 Γ1 with the following specifications: (i) Sidelengths: ΓB = BΓ 2 = 120, Γ 2 Γ1 = 200, x = ΓΓ1 = 56. (ii) Diagonal lengths: BΓ1 = 160, y = Γ Γ 2 = 92.
(
)
(
(
tan ΓΓ1Γ 2
)
)
24 8 , tan BΓΓ1 = − , 7 3 24 8 = , tan Γ1Γ 2 B = . 7 3
(iii) tan ΓBΓ 2 = −
Also, from tan θ =
(
α β +γ
)
=
24 3 = , and with the aid of a scientific calculator, we find 32 4
that ϕ = ω = θ = 36.86989765o ; ϕ = ω = θ = 36.87o Remark: Of the six lengths in (8), namely ΓB , BΓ 2 , Γ 2 Γ1 , x, BΓ1 , and y , four are
always integers as (8) clearly shows. These are the lengths ΓB , BΓ 2 , Γ 2 Γ1 , and BΓ1 .
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On the other hand, the lengths x and y are rational but not integral unless δ is a multiple of L . This follows from the conditions m + n ≡ 1(mod 2)and (m,n)=1 . These two conditions and m 2 + n 2 = L2 (see (7) or (8)), imply that the integer L must be relatively prime or coprime to the product 2m(m 2 − n 2 ) as well as to the product 4nm 2 ; the proof of this is a standard exercise in an elementary number theory course. Therefore, according to (8), the rational number x will be an integer precisely with L is a divisor of δ .
7
Two results from number theory
Let a, b, n be positive integers. Fact 1: If a n is a divisor of b n , then a is a divisor of b . Fact 2: The integer a is the nth power of a rational number if and only if it is the nth
power of an integer. These two results can be typically found in number theory books. We cite two sources. First, W. Sierpinski’s voluminous book Elementary Theory of Numbers (see reference [7] for details); and Kenneth H. Rosen’s number theory text Elementary Number Theory and Its Applications, (see [6] for details).
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A family of Heron Cyclic Quadrilaterals
If we compute the areas of the triangles B Γ1 Γ 2 , B Γ A and Γ A Γ1 by using formulas (8) we find,
12
2δ mnδ (m2 − n2 ) 2 = δ .mn(m2 − n2 ) 2 2 β 2.sin(180o − 2ω) β 2 sin 2ω β 2 δ 2 (m2 − n2 )2 mn Area of (isosceles) triangle Γ AΓ1 = = = αγ= 2 2 2 m2 + n2 2 2 2 2 α (β + γ ) 2δ mn ⎡⎣δ (m + n ) + δ (m − n )⎤⎦ Area of (right) triangle B Γ1 Γ2 = = = 2δ 2 .n.m3 2 2 Area of (right) triangle B Γ A =
αβ
=
(10)
The sum of the three areas in (10) is equal to the area A of the quadrilateral BΓΓ1Γ 2
⎡ 2 (m2 − n 2 )2 2 2 2⎤ . . 2 = − + + A δ m n ⎢m n m ⎥ m2 + n2 ⎣ ⎦
(11)
In the winter 2005 issue of Mathematics and Computer Education (see [8]), K.R.S. Sastry presented a family of Heron quadrilaterals. These are quadrilaterals with integer sides, integer diagonals, and integer area.
Interestingly, a subfamily of the family of
quadrilaterals described in (8); consists exclusively of heron quadrilaterals. First note that, for any choice of the positive integer δ , the lengths ΓB , BΓ 2 , Γ 2 Γ1 , BΓ1 are always integral, while Γ Γ1 and ΓΓ 2 , just rational. Clearly, if we take δ such that
δ ≡ 0 (mod L), then Γ Γ1 , ΓΓ 2 will be integers as well; and as (11) easily shows, the area
A
will
also
be
an
integer,
since
by
(7)
L2 = m 2 + n 2 and so δ ≡ 0 (mod L) => δ 2 ≡ 0 (mod L2 ) => δ 2 ≡ 0 (mod(m 2 + n 2 )).
Conclusion: When δ is a positive integer multiple of L in (8); the quadrilaterals obtained in (8) are Heron ones. The smallest such choice for
δ is δ =L . From (8) we then obtain
Γ B = BΓ 2 = 2 Lmn, Γ 2Γ1 = 2m.L2 , x = Γ Γ1 = 2m(m 2 − n 2 ), BΓ1 = 2 Lm 2 , Γ Γ 2 = y = 4nm 2
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(12)
⎡ ⎤ (m2 − n 2 ) 2 And area A=L2 .mn ⎢ m 2 − n 2 + + 2m 2 ⎥ ; and since L2 = m 2 + n 2 we arive at, 2 2 m +n ⎣ ⎦ A = mn ⎡⎣(m 2 + n 2 ). ( m 2 − n 2 ) + (m 2 − n 2 ) 2 + 2m 2 .(m 2 + n 2 ) ⎤⎦
(13)
A = mn ⎡⎣ m 4 − n 4 + m 4 − 2m 2 n 2 + n 4 + 2m 4 + 2m 2 n 2 ⎤⎦ = 4nm5
Using (12),(13), and (9a,9b) we arrive at the following table(with t1t2 being the smallest possible; one choice with t1 even t2 odd; the other with t1 odd, t2 even). t1
t2
m
n
δ =L
2
1
4
3
5
3
2
12
5
13
Also note that from tan θ =
ΓΓ1
Γ1Γ 2
Γ2 B
120
56
200
120
1560
2856
4056
1560
BΓ
α β +γ
=
ΓΓ 2
Area A
160
192
12888
3744
2880
4.125.5
BΓ1
2δ mn n = ; 2 2 δ (m − n ) + δ (m + n ) m 2
2
3 we obtain (when n=3, m =4) tanθ = ;θ = ω = ϕ ≈ 36.86989765o ; 4 5 and (when n =5, m =12) tanθ = ;θ = ω = ϕ ≈ 22.61986495o. 12
9
References
[1] C.V. Durell and A. Robson, Advanced Trigonometry, 352 pp., Dover Publications, (2003), ISBN: 0486432297. [2] Kenneth S. Miller, Advanced Trigonometry, Krieger Publishing Co., (1977), ISBN: 0882753916. [3] Konstantine D. Zelator, A Trigonometric Primer: From Elementary to Advanced Trigonometry, published by Brainstorm Fantasian Inc., January 2005, ISBN: 0-9761810-1-0. P.O. Box 4280, Pittsburg, PA 15203, U.S.A. [4] Marinos Zevas, Trigonometry (transliterated from Greek) Gutenberg Press, Athens, Greece, (1973), no ISBN.
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[5] L.E. Dickson, History of the Theory of Numbers, Vol. II, pages 435-437 (Also pages 426 and 427 for the more general equation ***), AMS Chelsea Publishing, ISBN: 08218-1935-6; 1992. [6] Kenneth H. Rosen, Elementary Number Theory and Its Applications, third edition, 1993, Addison-Wesley Publishing Co., ISBN: 0-201-57889-1; (there is now a fourth edition, as well). [7] W. Sierpinski, Elementary Theory of Numbers, Warsaw, 1964. Fact 1 can be found on page 15, listed as Corollary 1 to Theorem 6a; Fact 2 can be found on page 16, listed as Theorem 7. For a better understanding the reader may also want to study the preceding material, Theorem 1 to Theorem 6 (pages 10-15). Also, there is a newer edition (1988), by Elsevier Publishing, and distributed by North-Holland, NorthHolland Mathematical Library, 32, Amsterdam (1988). This book is now only printed upon demand, but it is available in various libraries. [8] K.R.S. Sastry, A description of a family of Heron Quadrilaterals, Mathematics and Computer Education, Winter, 2005, pp. 72-77.
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