Integer-Sided Triangles with integral medians.
Konstantine Zelator
Department of Mathematics and Computer Science
Rhode Island College
600 Mount Pleasant Avenue
Providence, RI 02908
U.S.A
e-mails: konstantine_zelator@yahoo.com
kzelator@ric.edu
Abstract
In this paper we study triangles whose sides have integer lengths; and with at least one median also having
integer length. We produce three such families of triangles. In the first two, each triangle contains only one
median of integer or integral length. In the third family, each triangle contains two medians of integral
length. Further, in Proposition 2, we prove that each integer-sided triangle can have at most two medians of
integer length.
Key words:
1. Integer-sided triangles: these are triangles with each side having integer length.
2. Integer medians: these are medians of integer or integral length.
2000 Mathematics Subject Classification: 11A, 11D.
Page 1 of 10
1
Introduction
Virtually almost every science, math, or engineering college student must know, one would think, that in a
right triangle, the median which tends to the hypotenuse has length half of that of the hypotenuse. If a right
triangle is Pythagorean, that is a right triangle with integer sidelengths, then that median will either equal
half and odd integer in length, or its length will be an integer. For example, in the famous triangle with
integer sidelengths 3, 4, and 5, that median has length 52 = 2.5 . But in the triangle or triple (10, 24, 26), that
median has integral length, namely 262 = 13 .
The purpose of this paper on one hand is to establish that a triangle with integral sidelengths can have at
most two medians which are of integer length. (This is done in Proposition 2.) On the other hand, to present
examples of families of triangles having one median being of integer length; and families of triangles having
two medians of integral length (of course, all these triangles having integer sidelengths). The proofs in both
propositions are totally accessible to an undergraduate student of math or a math related subject. The proof
of Proposition 2 is very short; Proposition 2 being a direct consequence of Proposition 1. The proof of
Proposition 1 makes use of two simple facts from number theory.
Fact 1: Every positive integer has a unique representation (up to the order of two factors) as a product of a
power of 2 (with the exponent being a nonnegative integer) and an odd positive integer.
Fact 2: The square of any odd integer is congruent to 1 modulo 4 (in fact, 1 modulo 8; but we do not need
that).
In Section 3, we also use the following fact.
Fact 3: The n th root of a positive integer is a rational number if, and only if, that integer is an n th integer
power itself. The reader may refer to either of the two references: [2] and [3].
Page 2 of 10
2
A Formula from Geometry and Trigonometry
The formula we derive below is well known in
the literature of geometry and trigonometry, but
not widely known or used. Let α , β , γ be the
A
β
sidelengths of a triangle and µα , µ β , µγ the
sidelengths of the corresponding medians. Our
derivation is simple and it is based on the Law
of Cosines. Indeed, in Figure 1, M is a midpoint
Γ
ω
γ
M
µα
γ
µα
α
B
A′
of BΓ ; extending the median AM to the point
A′ ;
Figure 1
A′Γ = γ
= AB ,
BΓ = α
AΓ = β
so that AM = MA = µα ; so that AA′ = 2 µα . The quadrilateral BA ΓA′ is a parallelogram and A′Γ is parallel
to AB . Angle AΓˆ A′ = ω = (in degrees )180 0 − A ; thus, cosω = − cos A . By applying the Laws of Cosines to
∆
∆
triangles A B Γ and A Γ A′ , we find that
α 2 = β 2 + γ 2 − 2βγ cos A; and 


(2 µα )2 = β 2 + γ 2 − 2 βγ cos ω 
An algebra step in combination with the fact that cos ω = − cos A leads to 4 µα2 = 2(β 2 + γ 2 ) − α 2 ; and by
cyclicity of letters we have
(
)
(
)
(
)
4µα2 = 2 β 2 + γ 2 − α 2 ,4µ β2 = 2 α 2 + γ 2 − β 2 ,4µγ2 = 2 β 2 + α 2 − γ 2 .
Page 3 of 10
3
Three Special Families
It can be shown that in any Pythagorean triangle
with hypotenuse length α , the medians µ β and
B
µ γ are irrational; the proof is difficult and
involved though. The interested reader should
refer to [4].In Figure 2, the median AM has
length µ α = α2 . Thus, when α is an even integer,
and only then, will µα be an integer.
γ
M
Now, as it is well known, the entire family of
Pythagorean triangles or triples (β , γ , α ) can be
A
Γ
β
described (without loss of generality) by the
parametric formulas:
Figure 2
β = 2δ mn , γ = δ (m − n ) , α = δ (m + n ) ;
2
2
2
2
Where m, n, δ are positive integers such that m > n , (m, n ) = 1 (i.e., m and n are relatively prime) and
m + n ≡ 1(mod 2) (i.e., one of m, n is odd; the other even). The reader may want to refer to [2] and [3] .This
leads to the first family:
The Pythagorean triangles with µα being an integer can be described by,
Family F1 :
β = 2δmn , γ = δ (m 2 − n 2 ) , α = δ (m 2 + n 2 ) ; with m, n, α being
integers such that m > n ≥ 1 , δ ≥ 2 , δ ≡ 0(mod 2) , m + n ≡ 1(mod 2)
and with µα =
δ (m 2 + n 2 )
2
.
A
Next, let us look at Figure 3, depicting an
isosceles triangle with integer sidelengths
α , β , γ ; γ = β ; and the median length µα being
∆
β
γ
an integer. If we look at either of the two
∆
µα
congruent right triangles A B M and A M Γ , we
see that
α
2
B
= γ 2 − µα2 .
α
2
M
α
Γ
2
Figure 3
Clearly, if µα is an integer, the square root γ 2 − µα2 , according to Fact 3 in the introduction, can be an
integer but not a proper rational number (i.e., one which when reduced into lowest terms has denominator
greater than 1). Hence, the rational number α2 must be an integer, and therefore α must be even; and so,
Page 4 of 10
(α2 , µα , γ ) will be a Pythagorean triple with
γ being the hypotenuse. Moreover, neither of the other median
lengths µ β , µγ can be an integer; because if one is integral, so must be the other, by virtue of µ β = µ γ . But
that would mean that all three median would be integers, thereby violation Proposition 2. We have,
The set of all isosceles triangles with integral sidelenghts α , β , γ ; with β = γ and
median length µα = µ being an integer, can be parametrically expressed as follows:
Either by,
(
)
(
)
that γ = β = δ ⋅ m 2 + n 2 , α = 4δmn , µ α = δ m 2 − n 2 , for positive integers
Family F2 :
m, n, δ , such
(m, n) = 1, m > n , and m + n ≡ 1(mod 2) .
Or alternatively, by (second subfamily),
γ = β = δ (m 2 + n 2 ), α = 2δ (m 2 − n 2 ), µα = 2δmn , again for positive integer
m, n, δ , with m > n , m + n ≡ 1(mod 2) , and (m, n ) = 1 .
Now, let us take a look at Figure 4. Here the situation is
different. Again we have an isosceles triangle with the
integer sidelengths satisfying β = γ . But, unlike the
A
previous case, here we have µ β = µ γ = integer; while
µα is not an integer. According to the formulas for the
lengths of the medians we derived in the previous
section, we see that if we set µ β = µ γ = µ and β = γ
we obtain 4 µ 2 = 2α 2 + β 2 ⇔ (2 µ ) = β 2 + 2α 2 which
2
shows that the triple
(β , α ,2µ ) is
M
N
a positive integer
solution to the diophantine equation x 2 + 2 y 2 = z 2 (in
three variables). In reference [1] one can find the general
µγ
µγ
B
solution in positive integers to the latter diophantine
equation given by
α
Γ
Figure 4
(
x = δ ⋅ k 2 − 2 L2 , y = 2δ ⋅ k ⋅ L , z = δ . k 2 + 2 L2
)
For positive integer-valued parameters k , L , δ such that (k , L ) = 1 . We have the following
Page 5 of 10
The family of all isosceles triangles with integral sidelenghts α , β , γ with
β = γ and median length µ β = µ γ = µ being an integer, can be parametrically
expressed as follows:
Family F3
β = γ = δ ⋅ k 2 − 2 L2 , α = 2δkL , µ =
δ (k 2 + 2 L2 )
, for positive
2
integers δ , k , L, such that (k , L ) = 1 and δ k 2 + 2 L2 ≡ 0(mod 2 ) and
(
)
with k 2 − 2 L2 > kL so that the triangle inequality β + γ = 2 β > α is
satisfied
4
Propositions 1 and 2, and their proofs
Proposition 1 : If four positive integers k , n1 , n 2 , n3 , satisfy the relation or condition,
4k 2 = 2(n 22 + n32 ) − n12 ;
then, and only then
k = 2 m ⋅ M , n1 = 2 e1 ⋅ a , n 2 = 2 e2 .b , n3 = 2 e3 ⋅ c ,
Where M , a , b , c are odd positive integers; e1 , e2 , e3 are positive integers, and m is a nonnegative integer
such that either
e1 = e2 < e3 , m = e1 − 1, and 
 2

2 (e3 − e1 )+1
2
2
⋅ c2 
M = 2b − a + 2
or alternatively
e1 = e3 < e2 , m = e1 − 1, and 
 2

2 (e2 − e1 )+1
2
2
⋅ b2 
M = 2c − a + 2
Proof: (1) Necessity. Suppose that k , n1 , n 2 , n3 satisfy the said condition.
4k 2 = 2(n22 + n32 ) − n12
(1)
Page 6 of 10
According to Fact 1 (stated in the introduction), each of the four numbers k , n1 , n2 , n3 must have a unique
representation of the form 2 t ⋅ ω , where t is a nonnegative integer and ω is an odd integer. Accordingly, we
have,
k = M ⋅ 2 m , n1 = a.2 e1 , n 2 = b ⋅ 2 e2 , n3 = c ⋅ 2 e3 ,
(2)
for odd integers M , a , b , c ; and nonnegative integers m , e1 , e2 , e3 .
Note that obviously, by equation (1), the integer n1 must be even; thus e1 must be, in fact, a positive integer:
e1 ≥ 1 . Substituting for k , n1 , n2 , and n3 in (1) we obtain,
[
]
2 2(m +1) ⋅ M 2 = 2 ⋅ b 2 ⋅ 2 2e2 + c 2 ⋅ 2 2e3 − 2 2e1 ⋅ a 2
(3)
We claim that the two exponents e2 and e3 must be distinct; for if they are equal, we arrive at a
contradiction.
Indeed, if e2 = e3 , (3) implies
(
)
M 2 ⋅ 2 2( m +1) = 2 2e2 +1 ⋅ b 2 + c 2 − 2 2e1 . a 2 .
(4)
Clearly, 2e2 + 1 ≠ 2e1 , since obviously an odd number cannot equal an even one. Moreover, the highest
powers of 2 dividing the left-hand side of (4) must be equal to the highest power of 2 dividing the right-hand
side of (4) which implies, 2(m + 1) = 2e1 and (since 1 ≤ e1 ) 2 ≤ 2e1 < 2e2 + 1 . (Obviously, 2(m + 1) cannot
equal 2e2 + 1 .) Hence
(
)
(4) ⇒ M 2 = 2 2(e2 −e1 )+1 ⋅ b 2 + c 2 − a 2 .
(5)
Since 2e1 < 2e2 + 1 , and e1 , e2 are integers, we deduce that 2e1 + 1 ≤ 2e2 + 1 ⇒ 2(e2 − e1 ) + 1 ≥ 1. Moreover,
(
)
since b 2 + c 2 is an even integer (because b ≡ c ≡ 1(mod 2) , it follows that the integer 2 2(e2 − e1 )+1 ⋅ b 2 + c 2 is a
multiple of 4. Therefore, (5) ⇒ M 2 + a 2 ≡ 0(mod 4 ) , which is a contradiction, since by Fact 2 (of the
introduction) we must have M 2 ≡ a 2 ≡ 1(mod 4 ) and consequently M 2 + a 2 ≡ 1 + 1 ≡ 2(mod 4) . We have
shown e2 ≠ e3 .
Let us go back to equation (3) with e2 ≠ e3 . There are two possibilities: either e2 < e3 or e3 < e2 . We only
need to consider the first possibility, e2 < e3 . The second possibility simply leads to the “ or alternative” part
of the conclusion of Proposition 1. After all, equation (1) is symmetric with respect to n2 and n3 .
Page 7 of 10
Under e2 < e3 , Equation (3) implies
[
]
2 2(m +1) ⋅ M 2 = 2 2e2 +1 ⋅ b 2 + 2 2(e3 −e2 ) ⋅ c 2 − 2 2 e1 ⋅ a 2 .
(6)
Again, a similar argument to that given before, demonstrates that 2 2 e1 must be the highest power of 2
dividing the right-hand side of (6); and hence,
(2(m + 1) = 2e1
and 2 ≤ 2e1 < 2e2 + 1) ⇔ (m − e1 = 1, and 1 ≤ e1 ≤ e2 )
Consequently, equation (6) implies,
[
]
M 2 = 2 2(e2 −e1 )+1 ⋅ b 2 + 2 2(e3 − e2 ) ⋅ c 2 − a 2 .
(7)
Since M 2 + a 2 ≡ 1 + 1 ≡ 2(mod 4 ) , equation (7) is possible or noncontradictory only if e2 − e1 = 0 (otherwise,
if e2 − e1 ≥ 1 , we arrive at the impossibility M 2 + a 2 ≡ 0(mod 4 ).
Hence e1 = e2 and (7) ⇒ M 2 = 2b 2 − a 2 + 2 2 (e3 − e2 )+1 ⋅ c 2 ; and since we have already established m = e1 − 1
and e1 = e2 , we are done.
(2) Sufficiency This is the easy part: for if the odd positive integers a, b, c, M , and the positive integer
exponents e1 = e2 and e3 > e1 , satisfy the equation, M 2 = 2b 2 − a 2 + 2 2 (e3 − e1 )+1 ⋅ c 2 ; by multiplying both sides
by 2 2 e1 and setting k = 2 m ⋅ M , where m = e1 − 1; and n1 = 2 e1 ⋅ a , n 2 = 2 e1 ⋅ b , n3 = 2 e3 ⋅ c , the result readily
follows. We omit the few algebraic steps.
Proposition 2: There exists no integer-sided triangle with all three median lengths being integral.
Equivalently every integer-sided triangle can have at most two medians which are integers.
Proof: By contradiction, if such a triangle exists, then all three relations or conditions
(
(
(
)
)
)
 4µ α2 = 2 β 2 + γ 2 − α 2

2
2
2
2
 4µ β = 2 α + γ − β

2
2
2
2
 4µ γ = 2 α + β − γ





(8)
Page 8 of 10
will be simultaneously satisfied with all six α , β , γ , µ α , µ β , µ γ , being positive integers. Let
h(α ), h(β ) , h(γ ), be the nonnegative integer exponents of the highest powers of 2 dividing α , β and γ ,
respectively. By Proposition 1, the three equations in (8) imply, respectively,
((h(α ) = h(β ) < h(γ )) or (h(α ) = h(γ ) < h(β )))
(9i)
((h(β ) = h(α ) < h(γ )) or (h(β ) = h(γ ) < h(α )))
(9ii)
((h(γ ) = h(α ) < h(β )) or (h(γ ) = h(β ) < h(α )))
(9iii)
We see that any two of (9i) , (9ii) , (9iii) are logically consistent with each other; but all three are not. In
fact, there are 2 3 = 8 possibilities (since in each of them, the conjunctive “or” allows for two choices or
possibilities), all contradictory. The proof is concluded.
5
The Family F4 of all integer-sided non-isosceles triangles with two integral
median lengths.
Proposition 1 allows us to describe the set of non-isosceles triangles which have integral sidelenghts and two
medians of integral length.
All non-isosceles triangles with two median integral lengths µ α , µ β , and integer
sidelenghts α , β , α , can be described by the conditions,
α ≠ β , α = 2 e ⋅ a , β = 2 e ⋅ b , γ = 2 e ⋅ c , µα = M ⋅ 2 e −1 , µ β = N ⋅ 2 e −1 ; where
1
2
3
1
1
a, b, c, M , N , are odd positive integers and the exponents e1 , e2 , e3 are positive
Family F4
integers such that e1 =e 2 < e3 ; and the two conditions,
 M 2 = 2b 2 − a 2 + 2 2 (e3 − e1 )+1 ⋅ c 2 ; and 
 2

 N = 2a 2 − b 2 + 2 2 (e3 −e1 )+1 ⋅ c 2 ,

and with α + β > γ , β + γ > α , γ + α > β (triangle inequalities)
Page 9 of 10
References
[1]
L.E . Dickson, History of the Theory of Numbers, Vol. II, Second Edition, Chelsea Publishing Co.,
New York, NY (1952). Bottom of page 420 and top of page 421.
[2]
Kenneth H.Rosen, Elementary Number Theory and Its Applications, third edition, 1993, AddisonWesley Publishing Co., ISBN:0-201-57889-1; (there is now a fourth edition as well). Pages 436-440.
Fact 3 is a consequence of Th. 2.11 on page 96.
[3]
W.Sierpinsi, Elementary Theory of Numbers, Warsaw, 1994. Also, there is a newer edition (1988),
by Elsevier Publishing, and distributed by North-Holland. North-Holland Mathematical Library, 32,
Amsterdam (1988). This book is now only printed upon demand, but it is available in various
libraries. ISBN: 0-598-52758-3. Pages 36-40. For fact 3, see Th. 7 on page 16.
[4]
Konstantine Zelator, The Seventeen Elements of Pythagorean Triangles, 26 pp.
arxiv.org; arxiv: 0809.0902, September 2008
Page 10 of 10