Inscribing Circles inside Integer Triangles
mre...@bangor.ca.boeing.com
The formula for a CIRcumscribing circle for a triangle is
R = a*b*c/(4*DELTA)
A similar formula for the largest circle that can be inscribed
inside a triangle is
r = 2*DELTA/(a + b + c)
where a,b,c are the sides of a triangle, and DELTA is the area
of the triangle.
DELTA = SQUARE ROOT{s(s-a)(s-b)(s-c)} where s =(a+b+c)/2
This brings to mind two questions:
1) Are there any pairs/groups of integer sided triangles that have
inscribing circles of the same radius? Unlike my entry on
circumscribing circles I dont have any "dumb luck" example to give.
2) If we consider pairs of integer sided triangles that have the
same CIRcumscribing circle, (and do a little algebra), a test
for such pairs having the same radius inscribing circle would
be:
a1*b1*c1 a2*b2*c2
---------- = -----------
a1 + b1 + c1 a2 + b2 + c2
I am going to go out on a limb here and predict....there are no pairs
of integer sided triangles that have BOTH the same circumscribing
circle, and the same inscribing circle. (Whats that quote??? Fools
rush in where angels fear to tread??? ) :-)
mre...@bangor.ca.boeing.com
Forgetting about integers for a moment.... I've been looking at
this relationship that holds for two triangles with the same
circumscribing circle and inscribed circles of equal radius:
a1*b1*c1 a2*b2*c2
----------- = ----------
a1 + b1 + c1 a2 + b2 + c2
The units are length cubed over length giving length squared or area.
These triangles fit inside of one circle of radius say "R" and area
pi*R^2 Each has an inscribed circle of radius say "r" and area
pi*r^2 One thing that will stay the same when considering each
triangle is the area in the annulus inside the circumscribing
circle and outside the inscribed circle. ie
Annulus area = pi*{R^2 - r^2 }
Could it be that for ANY triangle
a*b*c
---------- = K * pi*{R^2 - r^2}
a + b + c
where R = radius of the circumscribing circle and
r = radius of the largest inscribed circle?
If so, does anyone know the value of K ?
I have been trying the algebra for this and am either too tired to
get it right, or overlooking something. I did grossly construct
two triangles inside one circle, each triangle having the same
radius inscribing circle. a1,b1,c1 = 128mm,80.5mm,167.5mm
and a2,b2,c2 = 173mm,66mm,161.5mm
and R = 86.5mm and r = 27mm
a1*b1*c1/(a1+b1+c1) = 4590.2 a2*b2*c2/(a2+b2+c2) = 4604.3
4604.3/4590.2 = 1.0031 or less than a percent error.
Annulus area = pi*{7482.25 - 729} = 21216
21216/4590.2 = 4.622 for an approximate constant of proportionality
Now, (3/2)*pi = 4.712388 4.712388/4.622 = 1.0196 ie about 2% off
So maybe (3/2)*pi is the constant of proportionality, maybe not.
1+ root(13) = 4.60555 4.60555/4.622 = .9964 less than 1/2% off
Any suggestions?
Dick Tahta
Monte:
I think I can confirm your hunch that there are no integer-sided triangles
that have the same circum- and in-circle radii. To do this calls on some
algebra. You start with Pythagorean integers (ie sides of right-angled
triangles), say 2mn, m^2-n^2, m^2+n^2. You then shrink these by a factor
2mn leaving a right-angled triangle with a unit side. Now paste another of
these (say with parameters p,q) along the unit side. Enlarge by a factor
8mnpq and you now have an integer-sided triangle with sides 4mn(p^2+q^2),
4pq(m^2+n^2), 4(mq+np)(mp-nq). The circumradius of this triangle is
(p^2+q^2)(m^2+n^2) and the inradius is 4nq(mp-nq). The difference of these
will be (mp-nq)^2+(mq-np)^2+(2nq)^2 which cannot be zero, hence ....
The historically minded may wish to note that the pasting technique was
used by a number of famous mathematicians - Brahmagupta, Euler, Gauss. The
results enable us to calculate the various integer-sided triangles that
share a common circumradius. The circumradius is a sum of two squares
(mq+np)^2+(mp-nq)^2. It is known that not all numbers can be expressed as
a sum of two squares so there will be some circles which cannot
circumscribe any integer-sided triangle. Moreover, we could calculate the
precise number of triangles which share the same circumradius (for
example, the 12 possibilities you once listed for a circumradius 65 - which
you may note is a sum of two squares, in fact in two ways),
Mathematicians often pick up the problems they are working on from others.
The investigation of integer-sided triangles stems from an observation made
by the Alexandrian mathematician Heron who derived the area formula. He
noted that a 13,14,15 triangle had an integer area, namely 84. A number
of famous mathematicians (Brahmagupta, Euler, Gauss) took up the general
problem. Others extended the problem in various ways - for example W
Whitworth, C19, noted that Heron's example is the only triangle with
sides and an altitude in consecutive integers. You added the fertile
observation that a number of such triangles could share the same
circumradius. And there is still your question about triangles sharing the
same incircle ....
many thanks, and best wishes, Dick
John Conway
On Thu, 26 Feb 1998, Dick Tahta wrote:
> Monte:
>
> I think I can confirm your hunch that there are no integer-sided triangles
> that have the same circum- and in-circle radii. To do this calls on some
> algebra. You start with Pythagorean integers (ie sides of right-angled
> triangles), say 2mn, m^2-n^2, m^2+n^2. You then shrink these by a factor
> 2mn leaving a right-angled triangle with a unit side. Now paste another of
> these (say with parameters p,q) along the unit side. Enlarge by a factor
> 8mnpq and you now have an integer-sided triangle with sides 4mn(p^2+q^2),
> 4pq(m^2+n^2), 4(mq+np)(mp-nq). The circumradius of this triangle is
> (p^2+q^2)(m^2+n^2) and the inradius is 4nq(mp-nq). The difference of these
> will be (mp-nq)^2+(mq-np)^2+(2nq)^2 which cannot be zero, hence ....
Of course you should start by remarking that since a,b,c and
R = abc/4DELTA are supposed to be rational, DELTA must also be rational
(so that in the traditional terminology, the triangle is "Heronian"),
which entails that its altitudes are also rational, so that it MUST indeed
be obtained by gluing two Pythagorean triangles together in the manner you
described.
Forgive me if this point had already been made in some discussion
I didn't follow. I make it because it might help some other people
to understand your argument.
John Conway
John Conway
On Thu, 26 Feb 1998, Dick Tahta wrote:
> Monte:
>
> I think I can confirm your hunch that there are no integer-sided triangles
> that have the same circum- and in-circle radii.
/\
I've just noticed that the word "rational" was inadvertently
omitted from this statement, although it was of course used in
Dick Tahta's argument.
Let me study the more general problem as stated above. We
have R = abc/4DELTA, r = DELTA/s = abc/4Rs, in the usual
notation
[R = circumradius, r = inradius, s = semiperimeter (a+b+c)/2].
Hmm. I see no easy way to begin. So let me propose this as a
problem for others! I expect that the lists of triangles with
equal circumradii that several people have drawn up will probably
contain some examples (all one has to do is see whether any two
such triangles have the same perimeter).
John Conway
John Conway
On 26 Feb 1998 mre...@bangor.ca.boeing.com wrote:
> Forgetting about integers for a moment.... I've been looking at
> this relationship that holds for two triangles with the same
> circumscribing circle and inscribed circles of equal radius:
>
> a1*b1*c1 a2*b2*c2
> ----------- = ----------
> a1 + b1 + c1 a2 + b2 + c2
>
> The units are length cubed over length giving length squared or area.
>
> These triangles fit inside of one circle of radius say "R" and area
> pi*R^2 Each has an inscribed circle of radius say "r" and area
> pi*r^2 One thing that will stay the same when considering each
> triangle is the area in the annulus inside the circumscribing
> circle and outside the inscribed circle. ie
>
> Annulus area = pi*{R^2 - r^2 }
>
>
> Could it be that for ANY triangle
>
> a*b*c
> ---------- = K * pi*{R^2 - r^2}
> a + b + c
No. R^2 = (abc/4DELTA)^2
r^2 = (DELTA/s)^2
and so what you're asking is whether
(abc/4DELTA)^2 - (DELTA/s)^2 (*)
bears a fixed proportion to abc/s.
But (by Heron's formula) (*) is a rational function of a,b,c,
which, unlike abc/s, becomes infinite when a+b=c.
I note that in my recent reply to Dick Tahta I mistakenly
said that one could pick out examples with equal inradii from
those with equal circumradii by testing for equal perimeter
(or equal s). Of course one has to check equality of the
values of abc/s. Sorry!
John Conway
mre...@bangor.ca.boeing.com
Dick,
Thanks very much for the information, and for the historical
perspective. I am going to study your formula and equations and
see what I can absorb. In my opinion the historical information
makes geometry even more interesting than when it is considered alone.
So Heron was in Alexandria...! About what year would you say? Somehow
I had the impression that algebraic notation didnt develop until
800 AD and on. Heron's area formula root(s(s-a)(s-b)(s-c)) contains
a "root", so evidently he knew about roots. Anyway, I wanted to
say Thanks, and in my opinion, the more history, the better!
Best wishes,
Monte Evans
Everett, Washington
mre...@bangor.ca.boeing.com
John Conway,
Thanks for straightening me out on that erroneous idea that
a*b*c
--------- = k *{R^2 - r^2} R = circumradius r = in-radius
a + b + c
I dont know how I got stuck on that thing, except it was late
at night..... :-)
Sometimes the best course is to go home and go to sleep.
A little different topic here.... I showed your post to a co-worker
who noted the Princton e-mail address. He found the Princton web site
and located you in the staff area. When he showed me your name, it
included the middle initial H. and suddenly it hit me that this
person is John H. Conway ! ....inventor of the LIFE simulation!
I would like you to know that in the early 70's when you published
it in Scientific American, my college roommate and I had a great time
with a huge paper grid, and scads of pennies, looking for blinkers
and such. We programmed it up, and at that time were using IBM
punch cards! How time moves along. A few years ago I read a book
that dealt with assembling a computer, if I recall correctly, based
on gliders carefully positioned etc. It was interesting, I loaned
it away and dont have it anymore. Anyway, I wanted to take this
opportunity to say "Thanks for the enjoyment of LIFE!"
Monte Evans
Everett, Washington
Pat Ballew
MOnte,
Seems like if K was a constant for ALL triangles, I would work with
an equilateral triangle. And will immediately after I get this class
of kids to work.
Pat Ballew
Misawa, Japan
______________________________ Reply Separator _________________________________
Subject: Inscribing Circles inside Integer Triangle
Author: mre...@bangor.ca.boeing.com at EDU-INTERNET
Date: 2/26/98 1:34 AM
Forgetting about integers for a moment.... I've been looking at
this relationship that holds for two triangles with the same
circumscribing circle and inscribed circles of equal radius:
a1*b1*c1 a2*b2*c2
----------- = ----------
a1 + b1 + c1 a2 + b2 + c2
The units are length cubed over length giving length squared or area.
These triangles fit inside of one circle of radius say "R" and area
pi*R^2 Each has an inscribed circle of radius say "r" and area
pi*r^2 One thing that will stay the same when considering each
triangle is the area in the annulus inside the circumscribing
circle and outside the inscribed circle. ie
Annulus area = pi*{R^2 - r^2 }
Could it be that for ANY triangle
a*b*c
---------- = K * pi*{R^2 - r^2}
a + b + c
Pat Ballew
Monte, and other interested parties;
I have convinced (or deluded) myself with a Geometer's Sketchpad
sketch that there would be an infinite number of triangles that had
the same radius inscribed circle and circumscribed circle... I can not
show that any of them have integer sides... but I think the first
statement can be illustrated by a thought experiment (or reproduced on
an interactive geometry program)...
Imagine a Circle (radius r) inside a circle (Radius R) Pick a
point on the outer circle and draw two tangents to the inner circle.
By drawing a third tangent to the circle at some point a triangle
is formed making r the radius of the inscribed circle.. I have found
that I am able to move this inner circle around and get the two other
vertices to lie on the outer circle...
Now if I move the free tangent point we have created, we form a
triangle with different angles, and thus different side relations..
and yet we can still move the circle around and get a inscribed
triangle with this different shape...
comments
Pat Ballew
Misawa, Jp
______________________________ Reply Separator _________________________________
Subject: Inscribing Circles inside Integer Triangles
Author: mre...@bangor.ca.boeing.com at EDU-INTERNET
Date: 2/25/98 3:57 PM
The formula for a CIRcumscribing circle for a triangle is
R = a*b*c/(4*DELTA)
A similar formula for the largest circle that can be inscribed
inside a triangle is
r = 2*DELTA/(a + b + c)
where a,b,c are the sides of a triangle, and DELTA is the area
of the triangle.
DELTA = SQUARE ROOT{s(s-a)(s-b)(s-c)} where s =(a+b+c)/2
This brings to mind two questions:
1) Are there any pairs/groups of integer sided triangles that have
inscribing circles of the same radius? Unlike my entry on
circumscribing circles I dont have any "dumb luck" example to give.
2) If we consider pairs of integer sided triangles that have the
same CIRcumscribing circle, (and do a little algebra), a test
for such pairs having the same radius inscribing circle would
be:
a1*b1*c1 a2*b2*c2
---------- = -----------
a1 + b1 + c1 a2 + b2 + c2
wo...@ccm.umanitoba.ca
Pat Ballew <Pat_B...@ccmail.odedodea.edu> wrote
>>>
I have convinced (or deluded) myself with a Geometer's Sketchpad
sketch that there would be an infinite number of triangles that had
the same radius inscribed circle and circumscribed circle... I can not
show that any of them have integer sides... but I think the first
statement can be illustrated by a thought experiment (or reproduced on
an interactive geometry program)...
Imagine a Circle (radius r) inside a circle (Radius R) Pick a
point on the outer circle and draw two tangents to the inner circle.
By drawing a third tangent to the circle at some point a triangle
is formed making r the radius of the inscribed circle.. I have found
that I am able to move this inner circle around and get the two other
vertices to lie on the outer circle...
Now if I move the free tangent point we have created, we form a
triangle with different angles, and thus different side relations..
and yet we can still move the circle around and get a inscribed
triangle with this different shape...
comments
<<<
If a triangle has R and r as above, then it is an old result (probably
Euler) that the distance between the centers of the two circles is
d = sqrt(R^2-2Rr). The converse is a result of Steiner: if you start
with two circles of radii R and r, with distance d between their
centers, where d = sqrt(R^2-2Rr), then there are infinitely many
triangles that have the two given circles as their circumcircle and
incircle. You can start with any point A on the large circle, and drop
tangents as described above, to get such a triangle ABC.
--
Barry Wolk | Did you know that the word "pizza"
Dept of Mathematics | is a mathematical formula? It is
University of Manitoba | the volume of a circular pizza of
Winnipeg Manitoba Canada | radius "z" and thickness "a".