triangle with integer sides and angle bisectors

[quasi]

Conjecture:

There does not exist a triangle such that the sides
and angle bisectors all have integer lengths.

quasi

[quasi]

Here's a stronger claim:

Conjecture:

If two sides of a triangle have unequal, rational lengths,
then the length of the angle bisector of the included angle
is irrational.

quasi

[Timothy Murphy]

quasi wrote:

> Conjecture:
>
> If two sides of a triangle have unequal, rational lengths,
> then the length of the angle bisector of the included angle
> is irrational.

This sounds very improbable to me.
Suppose you fix the length of the two sides, say 1 and 2,
and vary the angle between them.
The length of the bisector will vary continuously
as you vary the angle continuously.
So the length of the bisector must pass through rational values.


--
Timothy Murphy
e-mail: gayleard /at/ eircom.net
tel: +353-86-2336090, +353-1-2842366
s-mail: School of Mathematics, Trinity College Dublin

[1treePetrifiedForestLane]

retrodict to all sides rational.

thus:
no-one, as far as I know, since Ahrrenius' 1896 coinage
of "glass housing," has ever modeled a glass house
at *a* lattitude ... other than glass house makers;
there is tons of that "how to do it" crappola
in the passive solar biz, as well as the datum
for PVs, "tracking Sun."

[quasi]

Timothy Murphy wrote:
>quasi wrote:
>>
>> Conjecture:
>>
>> If two sides of a triangle have unequal, rational lengths,
>> then the length of the angle bisector of the included
>> angle is irrational.
>
>This sounds very improbable to me. Suppose you fix the
>length of the two sides, say 1 and 2, and vary the angle
>between them. The length of the bisector will vary
>continuously as you vary the angle continuously. So the
>length of the bisector must pass through rational values.

Oops.

Your argument above clearly disproves the conjecture, but
I realize now that I inadvertently misstated it.

I intended all three sides to have rational lengths in
_addition_ to hypothesis about two unequal sides and the
included angle.

Even the corrected version may break, but maybe not as
easily.

In any case, the original conjecture (the impossibility of
having all sides and angle bisectors rational) is very much
alive.

quasi

[quasi]

I'll restate the original conjecture and also give a
corrected statement of the stronger claim.

Conjecture (1):

There does not exist a triangle such that the sides
and angle bisectors all have integer lengths.

Conjecture (2):

If the sides of a triangle have rational lengths, then
the length of the bisector of an angle between two unequal
sides is irrational.

quasi

[quasi]

As stated above, conjecture (2) fails.

As a simple counterexample, the angle bisector of the larger
acute angle of a 7,24,25 right triangle has rational length.

By adapting that example, conjecture (1) also fails. All
angle bisectors of the isosceles triangle with side lengths
14,25,25 have rational lengths.

There are scalene triangles as well with rational side lengths
and rational angle bisector lengths -- for example, the lengths
of all angle bisectors of a triangle with side lengths
84,125,169 are rational.

The two conjectures were based on some empirical testing, but
my program had a bug. Once the bug was corrected, both
conjectures were easily defeated.

However, I can at least prove the following ...

Theorem:

In a right triangle with rational side lengths, at most one
angle bisector has rational length.

quasi

[quasi]

On Wed, 23 May 2012 03:58:10 -0500, quasi <qu...@null.set> wrote:

>However, I can at least prove the following ...
>
>Theorem:
>
>In a right triangle with rational side lengths, at most one
>angle bisector has rational length.

Based on light empirical testing, I'll conjecture the
following extension of the above theorem,

Conjecture:

In a right triangle with relatively prime integer side
lengths, some angle bisector has rational length if and
only if the length of the hypotenuse is a perfect square.

quasi

[Dr J R Stockton]

In sci.math message <0rnnr79ugq2788vl5...@4ax.com>, Tue,
22 May 2012 13:50:38, quasi <qu...@null.set> posted:

>There does not exist a triangle such that the sides
>and angle bisectors all have integer lengths.

The limiting case of a triangle with zero altitude can, I think, have
that property.


--
(c) John Stockton, Surrey, UK. ?@merlyn.demon.co.uk Turnpike v6.05 MIME.
Web <http://www.merlyn.demon.co.uk/> - FAQish topics, acronyms, & links.
Proper <= 4-line sig. separator as above, a line exactly "-- " (SonOfRFC1036)
Do not Mail News to me. Before a reply, quote with ">" or "> " (SonOfRFC1036)

[Michael Press]

In article <qjbpr7dgd09fh21o8...@4ax.com>,

Let a,b,c be the sides of a triangle.
Here is a relation among a,b,c,d
where the internal bisector at C has length d.

There is a constant k such that

k.a + k.b = c

Then
k = c/(a+b)

Law of cosines.
kk.aa = aa + dd - 2.a.d.cos C/2
kk.bb = bb + dd - 2.b.d.cos C/2

Multiply by b, a respectively and subtract.
(kk - 1)(aa.b - bb.a) = (b-a)dd
(kk - 1)a.b(a - b) = (b-a)dd
(1 - kk)a.b = dd
((a+b)^2 - cc)a.b = dd(a+b)^2

So we need for ((a+b)^2 - cc)a.b to be a square.

How about (a,b,c,d) = (6,1,5, 12/7)?

--
Michael Press

[quasi]

On Wed, 23 May 2012 21:40:26 -0700, Michael Press <rub...@pacbell.net>
wrote:

There are two things wrong with that ...

Firstly, (a,b,c) = (6,1,5) is not a triangle.

Secondly, both the theorem and the conjecture
above refer to _right_ triangles.

auasi

[quasi]

Dr J R Stockton wrote:

>quasi wrote:
>>
>>There does not exist a triangle such that the sides
>>and angle bisectors all have integer lengths.
>
>The limiting case of a triangle with zero altitude can,
>I think, have that property.

If the altitude has zero length, it's a degenerate triangle
which, in the context of this problem, is not a valid
triangle.

quasi

[Michael Press]

In article <pnorr712ulv8vq7hg...@4ax.com>,

Okay. Here is a nontrivial, non-right triangle
counter-example to the original conjecture.
(6, 8, 7, 6).

--
Michael Press

[none Rouben Rostamian]

In article <0rnnr79ugq2788vl5...@4ax.com>,

I am not offering a solution, but the following may provide
a lead.

The Conjecture is equivalent to the one where "integer" is
replaced by "rational", so that's what I will consider from
now on.

Let a,b,c be the lengths of the triangle's sides, and let u,v,w
be the lengths of the bisectors. It is known, see, e.g.,
http://www.proofwiki.org/wiki/Length_of_Angle_Bisector/Proof_1
that
(b+c)^2 u^2 = bc [ (b+c)^2 - a^2].

The expression inside the square brackets factors as
(b+c)^2 - a^2 = (b+c+a)(b+c-a) = (b+c+a)(b+c+a-2a).

Let's write p = (a+b+c)/2 for the triangle's semiperimeter.
Then the above reduces to (b+c)^2 - a^2 = 4p(p-a) and the
equation for the length of the bisector changes to
(b+c)^2 u^2 = 4bcp(p-a).

Similarly,
(c+a)^2 v^2 = 4cap(p-b),
(a+b)^2 w^2 = 4abp(p-c).

Multiplying the previous three equations, we arrive at:

[(a+b)(b+c)(c+a) u v w]^2 = 64 (a b c)^2 p^2 p(p-a)(p-b)(p-c).

Heron's formula says A^2 = p(p-a)(p-b)(p-c), where A is the
triangle's area. We conclude that

(a+b)(b+c)(c+a) u v w = 8 a b c p A.

This leads to the following theorem.

Theorem: Consider a triangle with rational side lengths. Then
the product of the bisector lengths is rational if and only if
the triangle's area is rational.

A triangle with rational side lengths and area is called a
Heronian triangle; see, e.g..
http://en.wikipedia.org/wiki/Heronian_triangle
That web page gives formulas for the side lengths of a general
Heronian triangle.

We conclude that if a triangle with rational bisector lengths
exists, then it must be a Heronian triangle, and its side
lengths must be according to the side lengths of a Heronian
triangle.

--
Rouben Rostamian

[quasi]

Rouben Rostamian wrote:

>quasi wrote:
>>
>>Conjecture:
>>
>>There does not exist a triangle such that the sides
>>and angle bisectors all have integer lengths.

That was the _original_ conjecture, but in a previous
reply, I've already shown it to be false.

For example, the following triangles

14, 25, 25
84, 125, 169

are such that the lengths all angle bisectors are rational,
and (as you note), any such triangle can be scaled by an
integer factor to get a triangle such that the lengths of
all the sides and angle bisectors are integers.

The "only if" direction above is OK, but not the "if"
direction.

For example a 3-4-5 right triangle has rational side lengths
and rational area but irrational angle bisector lengths.

>A triangle with rational side lengths and area is called a
>Heronian triangle; see, e.g..
> http://en.wikipedia.org/wiki/Heronian_triangle
>That web page gives formulas for the side lengths of a
>general Heronian triangle.
>
>We conclude that if a triangle with rational bisector
>lengths exists, then it must be a Heronian triangle,
>and its side lengths must be according to the side lengths
>of a Heronian triangle.

Yes, nice result, but as I noted, not every Heronian
triangle has rational angle bisector lengths.

In any case, the original (failed) conjecture has been
revised.

In a separate reply, I'll post two (currently) live
conjectures.

quasi

[quasi]

Here are the current versions of the conjectures ...

Conjecture (1):

In a right triangle with relatively prime integer side

lengths, there is an angle bisector with rational length

if and only if the length of the hypotenuse is a
perfect square.

Conjecture (2):

If a triangle with relatively prime integer side lengths
is such that all angle bisectors have rational lengths,
then the length of one of the sides of the triangle is a
perfect square.

Conjecture (3):

If a triangle with relatively prime integer side lengths
is such that all angle bisectors have rational lengths,
then the length of one of the sides of the triangle is
the square of a prime congruent to 1 mod 4.

Remark:

I don't have much hope for the survival of Conjecture (3).

quasi

[quasi]

I'll add one more conjecture ...

Conjecture (4):

If a scalene triangle with relatively prime integer side

lengths is such that all angle bisectors have rational

lengths, then none of the angle bisectors lengths is an
integer.

quasi

[quasi]

Michael Press wrote:

>Okay. Here is a nontrivial, non-right triangle
>counter-example to the original conjecture.
>(6, 8, 7, 6).

You may have been confused by the many modifications
of the original conjecture, but as far as I can see,
the example you show above is not a counterexample to
any conjecture that was not already previously disproved.

The original conjecture was this ...

Conjecture:

There does not exist a triangle such that the sides
and angle bisectors all have integer lengths.

Note that I had previously posted the example

(a,b,c) = (14,25,25)

which has integer sides lengths and rational angle
bisector lengths, so the above triangle, scaled up by an
appropriate integer factor, yields a counterexample to
the original conjecture, and hence the conjecture was
implicitly already withdrawn.

But also note, the original conjecture required that all
sides and _all_ angle bisectors have integer lengths, so
your example is not a counterexample to the original
conjecture.

There was the followup conjecture ...

Conjecture:

If the sides of a triangle have rational lengths, then
the length of the bisector of an angle between two
unequal sides is irrational.

and for that conjecture, your example

(a,b,c) = (6,8,7)

is a counterexample, however I had already noted that
the example

(a,b,c) = (14,25,25)

is a counterexample to that conjecture as well, hence that
conjecture was also already implicitly withdrawn.

But there are some "live" conjectures ready to proved or
disproved -- see my latest replies in this thread.

Remarks:

All the currently live conjectures are based only on light
empirical evidence -- I've checked all triangles with
perimeters less than 1000.

However, other than the empirical evidence, I have no
intuition about whether they should be true or false.
Thus, empirical evidence notwithstanding, I wouldn't
be surprised if they are all false.

quasi

[none Rouben Rostamian]

In article <mqitr71lb53rddcsf...@4ax.com>,

quasi <qu...@null.set> wrote:
>Rouben Rostamian wrote:
>>quasi wrote:
>>>
>>>Conjecture:
>>>
>>>There does not exist a triangle such that the sides
>>>and angle bisectors all have integer lengths.
>
>That was the _original_ conjecture, but in a previous
>reply, I've already shown it to be false.
>
>For example, the following triangles
>
> 14, 25, 25
> 84, 125, 169
>
>are such that the lengths all angle bisectors are rational,
>and (as you note), any such triangle can be scaled by an
>integer factor to get a triangle such that the lengths of
>all the sides and angle bisectors are integers.

Oh, OK, that's neat.

>>Theorem: Consider a triangle with rational side lengths.
>>Then the product of the bisector lengths is rational if
>>and only if the triangle's area is rational.
>
>The "only if" direction above is OK, but not the "if"
>direction. For example a 3-4-5 right triangle has rational
>side lengths and rational area but irrational angle bisector
>lengths.

The theorem is correct as stated. Note that it does not
say that bisector lengths are rational. It says that
their product is rational.

--
Rouben Rostamian

[quasi]

Ah yes -- I misread it.

quasi

[quasi]

On Fri, 25 May 2012 00:15:45 -0500, quasi <qu...@null.set> wrote:

>On Thu, 24 May 2012 21:36:32 -0500, quasi <qu...@null.set> wrote:
>
>>Here are the current versions of the conjectures ...
>>
>>Conjecture (1):
>>
>>In a right triangle with relatively prime integer side
>>lengths, there is an angle bisector with rational length
>>if and only if the length of the hypotenuse is a
>>perfect square.
>>
>>Conjecture (2):
>>
>>If a triangle with relatively prime integer side lengths
>>is such that all angle bisectors have rational lengths,
>>then the length of one of the sides of the triangle is a
>>perfect square.
>>
>>Conjecture (3):
>>
>>If a triangle with relatively prime integer side lengths
>>is such that all angle bisectors have rational lengths,
>>then the length of one of the sides of the triangle is
>>the square of a prime congruent to 1 mod 4.
>>
>>Remark:
>>
>>I don't have much hope for the survival of Conjecture (3).

And in fact, conjectures (2) and (3) both fail.

The triangle

(a,b,c) = (125,507,578)

yields a counterexample for both conjectures.

However conjectures (2) and (3) are still alive if
"perfect square" is replaced by "perfect power".

Even with that modification, I doubt they'll survive.

>I'll add one more conjecture ...
>
>Conjecture (4):
>
>If a scalene triangle with relatively prime integer side
>lengths is such that all angle bisectors have rational
>lengths, then none of the angle bisectors lengths is an
>integer.

Conjecture (4) fails as well using the counterexample

(a,b,c) = (350,429,625)

quasi

[quasi]

On Fri, 25 May 2012 05:07:37 -0500, quasi <qu...@null.set> wrote:

>>>Conjecture (1):
>>>
>>>In a right triangle with relatively prime integer side
>>>lengths, there is an angle bisector with rational length
>>>if and only if the length of the hypotenuse is a
>>>perfect square.

Conjecture (1) is true, but the proof is essentially trivial
so it's not very interesting.

quasi