TRIANGLES WITH EQUIVALENT RELATIONS

BETWEEN THE ANGLES AND BETWEEN THE SIDES

Mogens Esrom Larsen Matematisk Institut, Copenhagen, Denmark

David Singmaster South Bank Polytechnic, London, UK

PART I GEOMETRY

0. Abstract and Introduction.

The simplest examples of equivalent relations between angles and sides for a triangle △ABC with sides a, b, and c are well known, e.g.,

(1)  ^{}A̸ = ^{}B̸ ⇔ a = b

and

(2)  ^{}A̸ = ^{}B̸ +^{}C̸ ⇔ a^{2} = b^{2} + c^{2} ,

because the angle–relation is equivalent to ^{}A̸ = \frac{π}{2} .

K. Schwering [7, 8, 9] and J. Heinrichs [3] (see also Dickson [2]) have studied relations of the form ^{}A̸ = n^{}B̸ and ^{}A̸ = n^{}B̸ + ^{}C̸ by the help of trigonometric functions and roots of unity. W. W. Willson [12] and R. S. Luthar [4] have considered the case n = 2, and recently J. E. Carroll and K. Yanosko [1] have generalized to the case of n rational. E. A. Maxwell [5, 6] has considered triangles with 2^{}A̸ = ^{}B̸ + ^{}C̸. In this paper we present elementary geometric proofs of the following equivalences:

(3)  ^{}A̸ = 2^{}B̸ ⇔ a^{2} = b^{2} + bc
(4)  ^{}A̸ = 2^{}B̸ +^{}C̸ ⇔ a^{2} = b^{2} + ac
(5)  2^{}A̸ = ^{}B̸ +^{}C̸ ⇔ a^{2} = b^{2} + c^{2} − bc
(6)  ^{}A̸ = 2(^{}B̸ +^{}C̸) ⇔ a^{2} = b^{2} + c^{2} + bc
(7)  ^{}A̸ = 2(^{}B̸ −^{}C̸) ⇔ ba^{2} = (b − c)(b + c)^{2}

Furthermore, we present the formulas for the complete set of integral solutions for each of the types of triangles.

1. All Triangles with ^{}A̸ = 2^{}B̸ .

In △ABC with ^{}A̸ > ^{}B̸ we draw a line from A to D on a such that ^{}C̸AD = ^{}B̸ as in Fig. 1.

Then △ABC ∼ △DAC, so that

(8)  \frac{b}{a} = \frac{d}{c} = \frac{a − x}{b}

A̸ = 2^{}B̸ ⇔ d = x

Figure 1.

or the two equalities

(9)  ad = bc
(10)  ax = a^{2} − b^{2}

from which we get the valid formula:

(11)  \frac{x}{d} = \frac{a^{2} − b^{2}}{bc} .

Hence, we conclude that ^{}A̸ = 2^{}B̸ iff △ADC is isosceles or x = d, i.e.,

(12)  ^{}A̸ = 2^{}B̸ ⇔ ^{}B̸AD = ^{}B̸ ⇔ x = d ⇔ a^{2} − b^{2} = bc .

This proves (3).

2. All Triangles with ^{}A̸ = 2^{}B̸ +^{}C̸ . In a triangle with ^{}A̸ > ^{}B̸ +^{}C̸ we draw a line from A to D on a such that ^{}C̸AD =

B̸ as in Fig. 2.

Again △ABC ∼ △DAC so that we have (10). Now,

(13)  ^{}A̸ = 2^{}B̸ +^{}C̸ ⇔ ^{}A̸ −^{}B̸ = ^{}B̸ +^{}C̸ .

A̸ = 2^{}B̸ +^{}C̸ ⇔ x = c

Figure 2.

But we have that

(14)  ^{}B̸AD = ^{}A̸ −^{}B̸

and

(15)  ^{}B̸DA = ^{}B̸ +^{}C̸ .

Hence we have

(16)  ^{}A̸ = 2^{}B̸ +^{}C̸ ⇔ ^{}B̸AD = ^{}B̸DA ⇔ x = c ⇔ ac = ax ⇔ ac = a^{2} − b^{2} .

This proves (4).

3. All Triangles with ^{}A̸ = \frac{π}{3} or ^{}A̸ = \frac{2π}{3} . The angle relations in (5) and (6) are equivalent to the equations ^{}A̸ = \frac{π}{3} and ^{}A̸ = \frac{2π}{3}

respectively. The relations of the sides are, of course, just the cosine relations for these angles. But a closer look proves worthwhile.

Suppose ^{}A̸ = \frac{π}{3} and ^{}C̸ < ^{}A̸ < ^{}B̸. (Signs of equality gives the trivial case of equilateral triangles, a = b = c.) Then we draw two equilateral triangles with ^{}A̸ of sidelength’s respectively c and b as in Fig. 3.

A̸ = ^{}A̸′ = \frac{π}{3} and ^{}B̸DC = \frac{2π}{3}

Figure 3. Then we notice the pair of solutions △ABC and △A′BC with sides a, b, c and a, b,

b − c respectively. Furthermore, △DBC has ^{}B̸DC = \frac{2π}{3} and sides a, c, b − c. So, the solutions appear three at a time. An elementary solution of the cosine relation comes from Pythagoras applied to the

triangles △ABE and △BCE; i.e.,

(17)  c^{2} − (\frac{c}{2})^{2} = h^{2} = a^{2} − (b − \frac{c}{2})^{2}

⇒

(18)  c^{2} = a^{2} − b^{2} + bc

proving (5) from which

(19)  a^{2} = c^{2} + (b − c)^{2} + c(b − c)

proving (6).

4. All Triangles with ^{}A̸ = 2(^{}B̸ −^{}C̸) . This equation is similar to the previous ones so it is surprising that it is equivalent to

a third degree equation in the sides; and no longer surprising it is much harder to prove. We draw in a triangle with ^{}C̸ < ^{}B̸ the bisector from A to D on a as in Fig. 4.

A̸ = 2(^{}B̸ −^{}C̸) ⇔ d = c

Figure 4.

Then ^{}A̸DB = ^{}C̸ + \frac{1}{2}^{}A̸. Hence the relation is equivalent to △DAB being isosceles.

(20)  ^{}A̸ = 2(^{}B̸ −^{}C̸) ⇔ ^{}C̸ + \frac{1}{2}^{}A̸ = ^{}B̸ ⇔ c = d .

But this time we need some further lines for support. We extend AB over A to the point E such that AE = b, and we extend AD over D to the point F, chosen such that the angle ^{}D̸CF = \frac{1}{2}^{}A̸, as is shown in Fig. 5.

The first extension gives us the isosceles △CAE, and hence that ^{}A̸CE = ^{}C̸EA = \frac{1}{2}^{}C̸AB. Therefore

(21)  △DAB ∼ △CEB .

The second extension gives us

(22)  △DAB ∼ △CAF ∼ △DCF .

The angle relation is thus equivalent to the four similar triangles being simultaneously isosceles.

\underline{Proof of the Formula (7)}: The four similar triangles and that AD ∥ EC give two equal relations, namely

(23)  \frac{a − x}{b} = \frac{x}{c} = \frac{a}{b + c} = \frac{z}{y} = \frac{y}{d + z}

and

(24)  \frac{x}{d} = \frac{z}{a − x} .

Figure 5.

From (23) we get

(25)  y^{2} = z(d + z) = zd + z^{2} .

Using (24) we get

(26)  y^{2} = x(a − x) + z^{2} .

Now using (23) again, this becomes

(27)  y^{2} = \frac{c}{b}(a − x)^{2} + y^{2}(\frac{a}{b + c})^{2}

which can be rewritten as

(28)  y^{2}(1 − (\frac{a}{b + c})^{2}) = \frac{c}{b}(a − x)^{2} .

From this equation we get the equivalence

(29)  y = a − x ⇔ \frac{b − c}{b} = (\frac{a}{b + c})^{2} ,

which completes the proof.

Although this is considerably simpler than our original proof, we suspect a simpler proof can be found. †

5. Geometric Summary.

It is striking that the identical figure arises in cases 1 and 2 and a very similar figure arises in case 5. In cases 1 and 2, we considered two cases of equality between edges of △ABD. The trivial case, c = d, gives ^{}C̸ = 0.

† As a matter of fact, we have received a simpler proof from Professor Andy Liu in Edmonton on July 16, 1991. His proof refers to Figure 4 with the altitude from A added.

Since △ABC ∼ △DAC, an equality between sides of △DAC makes △ABC isosceles. We can make other simple constructions, e.g. find D so that ^{}A̸DB = ^{}B̸; but they do not lead to any other relations.

In case 5, we have the same basic construction, but with AD bisecting ^{}A̸. There are six possible equalities of sides in △ABD and △ACD. The five other cases do not lead to any new relations.

We have examined all linear relationships α^{}A̸ + β^{}B̸ + γ^{}C̸ = 0 with α, β, γ, ∈ {−2, −1, 0, 1, 2} and all of these reduce to the five cases we have considered. Initially we thought all these cases would lead to second degree relations among the sides, so case 5 and its difficulty were quite unexpected.

PART II DIOPHANTINE ANALYSIS

6. Integral Solutions.

One of the inspirations for this study is the fact that the 4, 5, 6 triangle has one angle double another [8, 10, 12]. It is very remarkable that the smallest integral solutions of most of our cases have sides which are consecutive integers.

(2, 3, 4) = (b, c, a) is a solution of (4).
(3, 4, 5) = (b, c, a) is a right triangle.
(4, 5, 6) = (b, c, a) is a solution of (3).
(6, 7, 8) = (c, a, b) is a solution of (7).
(3, 5, 7) = (b, c, a) is a solution of (6).
(3, 7, 8) = (c, a, b) is a solution of (5).
(5, 7, 8) = (c, a, b) is a solution of (5).

We might add (1, 2, 3) = (a, b, c) is a solution of (3) and it is the smallest integral–sided scalene triangle. An obvious question is: what problem has (5, 6, 7) as an integral solution?

We also know (13, 14, 15) is the smallest Heronian triangle with consecutive sides [2, 11], so a general question is: what can be said about the integral–sided triangles (b − 1, b, b + 1)? We have not been able to make any progress on these questions.

7. All Solutions to the Diophantine Equations. (1). The complete solution to a^{2} = b^{2} + bc is

(30)  (a, b, c) = r(pq, p^{2}, q^{2} − p^{2})

for (q, p) coprime, 2p > q > p, r arbitrary.

\underline{Proof}: If (a, b, c) are pairwise coprime, (3) is written

(31)  a^{2} = b(b + c) .

Hence b = p^{2} and b + c = q^{2}, q > p coprime. The triangular inequality a + b > c gives the condition q^{2} − pq − 2p^{2} < 0 or q < 2p.

q = 3, p = 2 gives a = 6, b = 4 and c = 5. (2). The complete solution to a^{2} = b^{2} + ac is

(32)  (a, b, c) = r(q^{2}, pq, q^{2} − p^{2})

for q > p coprime, r arbitrary.

\underline{Proof}: If (a, b, c) are pairwise coprime, (4) is written

(33)  b^{2} = a(a − c) .

Hence a = p^{2} and a − c = q^{2}, p > q coprime. p = 2, q = 1 gives a = 4, b = 2 and c = 3.

(3) AND (4). The complete solutions to (6) and (5) are respectively

(34)  (a, b, c) = \frac{r}{4}(3p^{2} + q^{2}, 4pq, |3p^{2} − q^{2}| − 2pq)

(a, b, b + c) and (a, b + c, c), (35)

when (p, q) are coprime odd numbers satisfying q ∉ [p, 3p], and r an arbitrary factor.

\underline{Proof}: (5) follows from (19) and (6). To solve (6) we assume (a, b, c) to be pairwise coprime and hence b or c odd. By symmetry we may choose b as odd. Now we write (6) as

(36)  3b^{2} = (2a + 2c + b)(2a − 2c − b) .

A common prime factor p must be a factor in b and therefore odd. It is also a factor in the sum 4a and hence in a, and a factor in the difference 4c + 2b and hence in c, a contradiction.

So, the two factors are coprime, and we must have b = p · q, such that the following sets are equal:

(37)  {2a + 2c + b, 2a − 2c − b} = {3p^{2}, q^{2}} .

By addition we get 4a = 3p^{2} + q^{2} and by subtraction we get

(38)  4c + 2b = |3p^{2} − q^{2}|

from which (34) follows.

p = 3 and q = 1 gives (a, b, c) = (7, 3, 5) etc.

In order to require c > 0, we must have either q < p or 3p < q. Then p = 1 and q = 5 gives (a, b, c) = (7, 5, 3) etc. (5). The complete solution to (7) is

(39)  (a, b, c) = r(p(2q^{2} − p^{2}), q^{3}, q(q^{2} − p^{2}))

for p < q coprime integers and r arbitrary.

\underline{Proof}: A common factor in a and b will be a factor in c, so we may assume a and b to be coprime.

Let p be a prime factor in b to the power n, i.e.

(40)  b = p^{n} · d ,

such that p and d are coprime. Now let

(41)  c = p^{m} · f

m ≥ 0 is chosen such that p and f are coprime.

Then we substitute (40) and (41) in (7)

(42)  p^{n} · d · a^{2} = p^{3m}(p^{n−m}d − f)(p^{n−m}d + f)^{2} .

Hence n = 3m. We conclude that there is a q such that b = q^{3} and a g such that c = q·g. Hence we have

(43)  q^{3}a^{2} = q^{3}(q^{2} − g)(q^{2} + g)^{2} .

Let h = q^{2} − g, then from

(44)  a^{2} = h(2q^{2} − h)^{2}

we conclude that h is a square, say h = p^{2}.

Then a = p(2q^{2} − p^{2}), b = q^{3} and c = q(q^{2} − p^{2}). Now c > 0 requires p < q. For p = 1, q = 2 we get (a, b, c) = (7, 8, 6).

8. Right Angled and Isosceles Triangles. In the cases (5) and (6) we have obviously isosceles solutions and an even equilateral

in case (5). In these cases a right angle excludes integral solutions.

If the triangles in the cases (3) or (4) shall be isosceles, the only possibility will be

(45)  p^{2} − q^{2} = pq ,

but this equation has no rational solutions because

(46)  \frac{p}{q} = \frac{1 ± \sqrt{1 + 4}}{2} .

If a triangle in case (7) shall be isosceles, then ^{}A̸ = ^{}B̸ or ^{}A̸ = ^{}C̸. If ^{}A̸ = ^{}B̸ then 2^{}C̸ = ^{}A̸ = ^{}B̸ and we are in case (3) with ^{}B̸ = ^{}C̸, proved

impossible above.

If ^{}A̸ = ^{}C̸, then a = c and hence

(47)  2pq^{2} − p^{3} = q^{3} − qp^{2}

without integral solutions.

If a triangle in case (3) shall be right angled, then either ^{}B̸ or ^{}C̸ is right. In the first case the triangle becomes isosceles, in the second a 30^{∘}–60^{∘}–triangle, neither of these can be integral.

The case (4) can be rewritten as

(48)  ^{}A̸ = \frac{π}{2} + \frac{1}{2}^{}B̸

so these triangles are always obtuse–angled.

In the case (7) ^{}C̸ < ^{}B̸, so only ^{}B̸ or ^{}A̸ may be right. If ^{}B̸ is right, then ^{}A̸ = 0, so this is not the case. If ^{}A̸ is right, then ^{}B̸ = \frac{3π}{8} and ^{}C̸ = \frac{π}{8} , so one triangle exists.

But the sides must satisfy (7) and Pythagoras. Eliminating a^{2} from the equation, we obtain

(49)  b(b^{2} + c^{2}) = (b − c)(b + c)^{2} ,

with the only solution

(50)  b = (\sqrt{2} + 1)c .

9. Heronian Triangles.

Some of the triangles turn out to have integral areas. Of course, none of the triangles of (5) or (6) can avoid a factor \sqrt{3} so these are less interesting.

It proves useful to rewrite the parameterizations (30), (32) and (39) as follows.

Now (p, q) are coprime, p < q.

a  b  c
A̸ = 2^{}B̸  pq  p^{2}  q^{2} − p^{2}
A̸ = 2^{}B̸ +^{}C̸  q^{2}  pq  q^{2} − p^{2}
A̸ = 2(^{}B̸ −^{}C̸)  2pq^{2} − q^{3}  q^{3}  q^{3} − qp^{2}

These forms make it easy to give a useful table of possible sides:

\begin{matrix}p & q & p^{2} & pq & q^{2} − p^{2} & q^{2} & 2pq^{2} − p^{3} & q^{3} & q^{3} − qp^{2} \\ 1 & 2 & ∗1 & 2 & 3 & 4 & 7 & 8 & 6 \\ 1 & 3 & ∗1 & 3 & 8 & 9 & 17 & 27 & 24 \\ 2 & 3 & 4 & 6 & 5 & 9 & 28 & 27 & 15 \\ 1 & 4 & ∗1 & 4 & 15 & 16 & 31 & 64 & 60 \\ 3 & 4 & 9 & 12 & 7 & 16 & 69 & 64 & 28 \\ 1 & 5 & ∗1 & 5 & 24 & 25 & 49 & 125 & 120 \\ 2 & 5 & ∗4 & 10 & 21 & 25 & 92 & 125 & 105 \\ 3 & 5 & 9 & 15 & 16 & 25 & 123 & 125 & 80 \\ 4 & 5 & 16 & 20 & 9 & 25 & 136 & 125 & 45 \\ 1 & 6 & ∗1 & 6 & 35 & 36 & 71 & 216 & 210 \\ 5 & 6 & 25 & 30 & 11 & 36 & 235 & 216 & 66 \\ 1 & 7 & ∗1 & 7 & 48 & 49 & 97 & 343 & 336 \\ 2 & 7 & ∗4 & 14 & 45 & 49 & 188 & 343 & 315 \\ 3 & 7 & ∗9 & 21 & 40 & 49 & 267 & 343 & 252 \\ 4 & 7 & 16 & 28 & 33 & 49 & 328 & 343 & 231 \\ 5 & 7 & 25 & 35 & 24 & 49 & 365 & 343 & 168 \\ 6 & 7 & 36 & 42 & 13 & 49 & 372 & 343 & 91 \\ 1 & 8 & ∗1 & 8 & 63 & 64 & 127 & 512 & 504 \\ 3 & 8 & ∗9 & 24 & 55 & 64 & 357 & 512 & 440 \\ 5 & 8 & 25 & 40 & 39 & 64 & 515 & 512 & 312 \\ 7 & 8 & 49 & 56 & 15 & 64 & 553 & 512 & 120 \end{matrix}

The ∗ means that a case (3) triangle does not exist because q ≥ 2p.

Of course, it is not obvious whether any of these have integral area. It is convenient to make use of the formula of Heron,

(51)  △ = Area = \sqrt{s(s − a)(s − b)(s − c)} ,

where s = \frac{1}{2}(a + b + c).

In the case (3) we obtain

(52)  \begin{aligned}s & = \frac{1}{2}(pq + p^{2} + q^{2} − p^{2}) = \frac{1}{2}q(p + q) \\ s − a & = \frac{1}{2}qp + \frac{1}{2}q^{2} − pq = \frac{1}{2}q(q − p) \\ s − b & = \frac{1}{2}qp + \frac{1}{2}q^{2} − p^{2} = \frac{1}{2}(q − p)(q + 2p) \\ s − c & = \frac{1}{2}qp + \frac{1}{2}q^{2} − q^{2} + p^{2} = \frac{1}{2}(q + p)(2p − q) . \end{aligned}

So, we get

(53)  △^{2} = (\frac{1}{4})^{2} · q^{2} · (p + q)^{2} · (q − p)^{2} · (2p + q)(2p − q) .

For △ to be an integer, (2p + q)(2p − q) = 4p^{2} − q^{2} must be a square. Considerations (mod 4) show that q must be even, which makes △^{2} an integer and also makes p odd.

Any common factor of 2p + q and 2p − q must divide their sum 4p and their difference 2q, but gcd(4p, 2q) can only be 2 or 4. So either

(54)  2p + q = 4s^{2} ∧ 2p − q = 4t^{2}

or

(55)  2p + q = 2s^{2} ∧ 2p − q = 2t^{2} .

So, we have two possibilities,

(56)  p = s^{2} + t^{2} ∧ q = 2(s^{2} − t^{2})

and

(57)  p = \frac{s^{2} + t^{2}}{2} ∧ q = s^{2} − t^{2}

with (56) to apply for s, t coprime of different parity, and (57) for s, t coprime, both odd. The area becomes then either

(58)  △ = q · (q^{2} − p^{2}) · s · t

or

(59)  △ = \frac{1}{2} · q(q^{2} − p^{2}) · s · t ,

where s, t have different parity in (58) and s, t are both odd in (59).

So, we can make the following table of Heronian triangles:

a  b  c  △
s  t  p  q  pq  p^{2}  q^{2} − p^{2} (\frac{1}{2})q · c · s · t
2  1  5  6  30  25  11  132
3  1  5  8  40  25  39  468
4  1  17  30  510  289  611  73320
5  1  13  24  312  169  407  24420
5  2  29  42  1218  841  923  387660
6  1  37  70  2590  1369  3531  1483020

In the case of (4) we obtain:

(60)  \begin{aligned}s & = \frac{1}{2}(q^{2} + pq + q^{2} − p^{2}) = \frac{1}{2}(q + p)(2q − p) \\ s − a & = q^{2} + \frac{1}{2}(pq − p^{2}) − q^{2} = \frac{1}{2}p(q − p) \\ s − b & = q^{2} + \frac{1}{2}(pq − p^{2}) − pq = \frac{1}{2}(q − p)(2q + p) \\ s − c & = q^{2} + \frac{1}{2}(pq − p^{2}) − q^{2} + p^{2} = \frac{1}{2}p(q + p) . \end{aligned}

So, we get

(61)  △^{2} = (\frac{1}{4})^{2} · p^{2} · (q^{2} − p^{2})^{2} · (2q − p) · (2q + p) .

We can use the solutions (56) and (57) with p and q interchanged. We get for the area

(62)  △ = p · (q^{2} − p^{2}) · s · t (s, t even)

or

(63)  △ = \frac{1}{2} · p · (q^{2} − p^{2}) · s · t (s, t odd)

and the following table:

a  b  c  △
s  t  p  q  q^{2}  pq  q^{2} − p^{2} (\frac{1}{2})q · c · s · t
3  2  10  13  169  130  69  4140
4  3  14  25  625  350  429  72072
5  3  16  17  289  272  33  3960
5  4  18  41  1681  738  1357  488520
6  5  22  61  3721  1342  3237  2136420
7  5  24  37  1369  888  793  333060
7  6  26  85  7225  2210  6549  7151508

In the case of (7) we obtain;

(64)  \begin{aligned}s & = \frac{1}{2}(2pq^{2} − p^{3} + q^{3} + q^{3} − qp^{2}) = \frac{1}{2}(q + p)(2q^{2} − p^{2}) \\ s − a & = pq^{2} + q^{3} − \frac{1}{2}p^{3} − \frac{1}{2}qp^{2} − 2pq^{2} + p^{3} = \frac{1}{2}(q − p)(2q^{2} − p^{2}) \\ s − b & = pq^{2} + q^{3} − \frac{1}{2}p^{3} − \frac{1}{2}qp^{2} − q^{3} = \frac{1}{2}p(q − p)(2q + p) \\ s − c & = pq^{2} + q^{3} − \frac{1}{2}p^{3} − \frac{1}{2}qp^{2} − q^{3} + qp^{2} = \frac{1}{2}p(q + p)(2q − p) . \end{aligned}

So, we get

(65)  △^{2} = (\frac{1}{4})^{2} · p^{2} · (q^{2} − p^{2})^{2} · (2q^{2} − p^{2})^{2} · (2q + p) · (2q − p) .

We can again use the solutions (56) and (57) with p and q interchanged. We get for the area the formulas

(66)  △ = p · (q^{2} − p^{2})(2q^{2} − p^{2}) · s · t (s, t even)

or

(67)  △ = \frac{1}{2}p(q^{2} − p^{2})(2q^{2} − p^{2}) · s · t (s, t odd)

and the table:

a  b  c  △
s  t  p  q  2pq^{2} − p^{3}  q^{3}  q^{3} − qp^{2} (\frac{1}{2}) · \frac{c}{q} · a · s · t
3  2  10  13  2380  2197  897  985320
4  3  14  25  14756  15625  10725 75963888
5  3  16  17  5152  4913  561  1275120
5  4  18  41  54684  68921  55637 1484123760
7  5  24  37  51888  50653  29341 720075720

10. Conclusion.

d = x or x = c or d = c

Figure 6. It is striking that the common feature of the problems investigated are the isosceles

triangles involved. Some line segment from A to D on A is drawn, and then either x = d, d = c, c = x or even d = c = x. Because of this similarity, the interpretations as angle relations are very similar, but in spite of this the side relations are different and in particular the case (7) with d = c is surprisingly complicated.

Nevertheless, the question of possible Heronian triangles is answered by the application of the very same Diophantine equation in the three solvable cases,

(68)  4p^{2} − q^{2} = r^{2}

proving a sort of similarity between the side relations too.

\underline{References}

1. J. E. Carroll and K. Yanosko, “The Determination of a Class of Primitive Integral Triangles,” Fibonacci Quarterly 29 (1991) 3–6.

2. L. E. Dickson, History of the Theory of Numbers, Volume II, Diophantine Analysis. Rational or Heron Triangles and Triangles with Rational Sides and a Linear Relation Between the Angles, G. E. Stechert & Co., New York, 1934, pp. 191–201, 213–214.

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6. E. A. Maxwell, “Triangles Whose Sides are in Geometric Progression,” Math. Gazette 42 (1958) 114.

7. K. Schwering, “Über Dreiecke, in denen ein Winkel das Vielfache eines andern ist,” Jahresbericht über das Königliche Gymnasium Nepomucenianum zu Coesfeld im Schuljahre 1885–86, 85 (1886) 3–7.

8. K. Schwering, 100 Aufgaben aus der niederen Geometrie nebst vollständigen Lösungen, Aufgabe 56: Von einem Dreieck ist gegeben: die Grundlinie BC, der zugehörige Höhenfußpunkt D und die Bestimmung α = 2β, Herdersche Verlagshandlung, Freiburg im Breisgau 1891, 88–89.

9. K. Schwering, “Ganzzahlige Dreiecke mit Winkelbeziehungen,” Archiv der Mathematik und Physik (3) 21 (1913) 129–136.

10. T. Sole, The Ticket to Heaven and Other Superior Puzzles, Penguin, London, 1988, 70, 79.

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12. W. W. Willson, “A Generalization of a Property of the 4, 5, 6 Triangle,” Math. Gazette 60, No. 412 (1976) 130–131.