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Feb 25, 1998, 3:00:00 AM2/25/98

to

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??? ) :-)

Feb 26, 1998, 3:00:00 AM2/26/98

to

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?

Feb 26, 1998, 3:00:00 AM2/26/98

to

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

Feb 26, 1998, 3:00:00 AM2/26/98

to

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

Feb 26, 1998, 3:00:00 AM2/26/98

to

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

Feb 26, 1998, 3:00:00 AM2/26/98

to

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

Feb 26, 1998, 3:00:00 AM2/26/98

to

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

Feb 26, 1998, 3:00:00 AM2/26/98

to

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

Feb 27, 1998, 3:00:00 AM2/27/98

to

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

Feb 27, 1998, 3:00:00 AM2/27/98

to

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

Feb 27, 1998, 3:00:00 AM2/27/98

to

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".

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