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Gale Greenlee
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Post Number: 2198
 Posted on Monday, 01 August, 2011 - 06:14 pm:

Suppose we consider all the triangles whose sides are limited to integers only.

We could find two different triangles whose area is the same, for example (7,7,2) and (4,4,4) both have an area of 6.92820323 which is sqrt of 48 I believe.

Could there be a third?

Gale
Simba (Chris Bryant)
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Post Number: 2452
 Posted on Monday, 01 August, 2011 - 06:52 pm:

Have you come across Heron's formula ? This might be of use to you in your question.
Daniel Fretwell
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Post Number: 1514
 Posted on Monday, 01 August, 2011 - 06:55 pm:

Well we can use Heron's formula to show that we must have:

s(s-a)(s-b)(s-c) = 48

Where the sides of the triangle are a,b,c and s is the semi-perimeter (a+b+c)/2.

This can be turned into an integer equation which has finitely many solutions. So if there are other triangles then there can only be finitely many.

I haventlooked myself but i'll let you investigate possible solutions to this equation.
Daniel Fretwell
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Post Number: 1515
 Posted on Monday, 01 August, 2011 - 06:56 pm:

Ooops :p
Gale Greenlee
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Post Number: 2199
 Posted on Monday, 01 August, 2011 - 08:05 pm:

I don't know how to investigate that when Im only interested in integer sides a,b,and c.

Anyway, I wasn't particularily interested in 48 just the general case of more than two sets of integers.
Gale Greenlee
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Post Number: 2200
 Posted on Monday, 01 August, 2011 - 08:08 pm:

I used Heron's formaula to solve all 203 cases where the largest integer in the set of 3 is less than 13. I found several instances where there were two sets but never 3.

Could there be 3 or more in the general case where the largest integer in the set of 3 might be greater than 12?
Prolific poster

Post Number: 281
 Posted on Monday, 01 August, 2011 - 08:13 pm:

The number of pairwise non-congruent triangles with area A should be the number of unordered positive integer solutions (a,b,c) to the equation

s(s-a)(s-b)(s-c) = A^2, where s = (a+b+c)/2.

It might be interesting to figure out the maximum number of non-congruent triangles with equal area.
Gale Greenlee
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Post Number: 2201
 Posted on Monday, 01 August, 2011 - 09:17 pm:

Sorry, but I'm still lost.
Daniel Fretwell
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Post Number: 1516
 Posted on Monday, 01 August, 2011 - 10:29 pm:

The area of a triangle with side lengths $a,b,c$ is:

$\sqrt{s(s-a)(s-b)(s-c)}$

by Heron's formula.

Now say we have an integer sided triangle. It has an area $A$ so other triangles that share its area satisfy:

$\sqrt{s(s-a)(s-b)(s-c)} = A$

i.e.

$s(s-a)(s-b)(s-c) = A^2$

Writing $s = \frac{n}{2}$, we may clear the denominators. We are then left with an equation which we are seeking integer solutions to:

$n(n-2a)(n-2b)(n-2c) = 16A^2$

I'll let you take it from there.

Just to note though, this equation is not visibly solvable!

It isnt like $x+2 = 4$ where you can visibly "take the 2 over" then it is done, this equation is solved by using properties of integers. I can tell you how "I could solve it", given a value of A but I doubt there is an illuminating general solution to this.
azerbajdzan
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Post Number: 970
 Posted on Monday, 01 August, 2011 - 11:58 pm:

There are also cases with only one possibility. For example A2=20 only {3,3,4}.

three triangles with same area:
{{2, 11, 12}, {3, 8, 10}, {4, 5, 6}}
{{3, 19, 20}, {4, 15, 17}, {5, 11, 12}}

four triangles with same area:
{{2, 19, 19}, {3, 13, 14}, {6, 7, 7}, {6, 7, 11}}

five triangles with same area:
{{9, 25, 26}, {10, 23, 23}, {10, 23, 27}, {13, 18, 19}, {13, 18, 25}}
{{10, 22, 24}, {11, 20, 23}, {11, 23, 30}, {12, 19, 25}, {15, 16, 17}}

six triangles with same area:
{{9, 16, 17}, {9, 17, 22}, {9, 22, 29}, {11, 13, 16}, {11, 13, 18}, {13, 16, 27}}
azerbajdzan
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Post Number: 971
 Posted on Tuesday, 02 August, 2011 - 12:08 am:

the above cases contain at least such number of triangles...

The second case for five triangles in previous post can be extended by another 3 triangles so here are eight triangles with same area:
{{6, 38, 40}, {8, 30, 34}, {10, 22, 24}, {10, 33, 41}, {11, 20, 23}, {11, 23, 30}, {12, 19, 25}, {15, 16, 17}}

...again at least 8... maybe there is another one with same area, haven't check that
azerbajdzan
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Post Number: 972
 Posted on Tuesday, 02 August, 2011 - 12:32 am:

and here four triangles with same integer area A=60:
{{6, 25, 29}, {8, 15, 17}, {10, 13, 13}, {13, 13, 24}}
Gale Greenlee
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Post Number: 2202
 Posted on Tuesday, 02 August, 2011 - 12:38 am:

Thanks Daniel. Just as an example show me "how you would do it" given A=sqrt 200
Daniel Fretwell
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Post Number: 1518
 Posted on Tuesday, 02 August, 2011 - 11:49 am:

There is certainly a more efficient method but we are solving:

$n(n-2a)(n-2b)(n-2c) = 3200$

for integers $a,b,c$ where $n = a+b+c$.

We know that $n$ is a positive integer that is 3 or more because $a,b,c$ are positive integers. Also the equation above shows that $n$ is a factor of 3200. We may list the positive factors of 3200 that are 3 or more so we may find all possible values of $n$.

For each possible value of $n$ we may find all partitions into three positive integers $a,b,c$. For each partition you may check whether or not we have that:

$n(n-2a)(n-2b)(n-2c) = 3200$.

The ones that work give you all of the possible solutions to the area problem (obviously there is symmetry in $a,b,c$ so some of the solutions will represent the same triangle).

As I say, this is probably not the best way to solve this equation but it was just the method I saw immediately. It is obvious from this method that the number of solutions is finite (there are only finitely factors of the RHS and for each of these there are only finitely many partitions into three positive integers).

I would avoid doing it by hand, it is more the kind of thing a computer can do for you.
Jewbinson
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Post Number: 736
 Posted on Tuesday, 02 August, 2011 - 01:01 pm:

it seems that (arguably) the most interesting question is: how many cases are there with only one possibility?

Azerbajdzan gave the example A2=20 only {3,3,4}.

Are there a finite number of triangles with this property,or infinite?

How to prove...
azerbajdzan
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Post Number: 974
 Posted on Tuesday, 02 August, 2011 - 01:20 pm:

I think cases A=prime or A2=prime have only one possibility. And some other also, as you can see 20 is not prime.
Daniel Fretwell
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Post Number: 1521
 Posted on Tuesday, 02 August, 2011 - 01:36 pm:

There is trivially only one solution when we take $A^2 = 3$. I think other prime cases might be of use since things simplify nicely.

After a quick look at the prime case, I can easily narrow it down to 2 possible solutions taking into account symmetry (no matter which prime).
Daniel Fretwell
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Post Number: 1522
 Posted on Tuesday, 02 August, 2011 - 01:50 pm:

We have that $n=p$ in the prime case meaning that:

$(p-2a)(p-2b)(p-2c) = 1$

Hence we have that either:

1) All three of the factors on the right are 1 (giving $a=b=c=\frac{p-1}{2}$). But then $a+b+c = \frac{3(p-1)}{2}$ and this only equals $p$ when $p=3$.

or

2) Exactly two of them are -1.

Since we are dealing with symmetry here we may assume that:

$p-2a = -1$
$p-2b = -1$
$p-2c = 1$

Giving $(a,b,c) = (\frac{p+1}{2}, \frac{p+1}{2}, \frac{p-1}{2})$. Then $a+b+c = \frac{3p + 1}{2}$ which is strictly bigger than p.

So actually only case 1 can offer solutions and this only gives a solution when $p=3$.

Have I made an error somewhere?
Daniel Fretwell
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Post Number: 1523
 Posted on Tuesday, 02 August, 2011 - 01:54 pm:

Assuming I havent then there is no triangle at all for the case $A^2 = p$ with $p>3$ prime.
Daniel Fretwell
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Post Number: 1524
 Posted on Tuesday, 02 August, 2011 - 02:01 pm:

Actually, it is easy to see by the same method that there are no solutions for $A^2$ being the square of a prime too.
Gale Greenlee
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Post Number: 2203
 Posted on Tuesday, 02 August, 2011 - 03:07 pm:

Thanks everyone. Take a look at this months "Ponder This" puzzle on the IBM site. It mixes the area of integer triangles with the circumference of integer triangles in a neat logic puzzle.
azerbajdzan
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Post Number: 975
 Posted on Tuesday, 02 August, 2011 - 04:08 pm:

Always the same... No creativity in this challenge.
I saw many times similar puzzle in which first person replay "I do not know the answer". Second person replay "I do not know too" and then the first person replay "Now, after knowing that you also do not know I know the answer".
Each of this type of puzzle must satisfy the same conditions to be solvable uniquely. If you once solved it you know what condition it is...
azerbajdzan
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Post Number: 976
 Posted on Tuesday, 02 August, 2011 - 04:29 pm:

...and I am suspicious that they made mistake because this challenge cannot be solved uniquely so therefore it cannot be solved at all...
I found two different examples of triangle for which the dialogue can occur exactly same.
Gale Greenlee
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Post Number: 2204
 Posted on Tuesday, 02 August, 2011 - 04:50 pm:

Really?

It goes like this.

c) I don't know.

a) Me either.

c) Now I know.

a) Me too.
azerbajdzan
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Post Number: 977
 Posted on Tuesday, 02 August, 2011 - 05:01 pm:

maybe I made mistake...I should check it
Daniel Fretwell
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Post Number: 1525
 Posted on Wednesday, 03 August, 2011 - 10:28 am:

Ha ha, we should make our own dialogue.

azerbajdzan : I cannot solve the problem uniquely
Me : Neither can I
azerbajdzan : Now I can
Me : Me too.
azerbajdzan
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Post Number: 978
 Posted on Wednesday, 03 August, 2011 - 11:59 am:

:-DDD
Daniel Fretwell
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Post Number: 1526
 Posted on Wednesday, 03 August, 2011 - 12:58 pm:

Has that helped you?
azerbajdzan
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Post Number: 979
 Posted on Wednesday, 03 August, 2011 - 02:02 pm:

In what way it should help me? :-)
I found one triangle that corresponds to the dialogue (original one, not yours ;-)) but I cannot prove that there isn't another one.

...or it wasn't reaction on me?
Gale Greenlee
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Post Number: 2205
 Posted on Wednesday, 03 August, 2011 - 05:32 pm:

Are you guys talking about "area and perimeter"

Or are you talking about "area and circumference of the circumcircle"?

Gale
Daniel Fretwell
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Post Number: 1527
 Posted on Wednesday, 03 August, 2011 - 05:36 pm:

It isnt clear which is meant in the problem.
Gale Greenlee
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Post Number: 2206
 Posted on Wednesday, 03 August, 2011 - 05:54 pm:

They changed it awhile ago. Now it says "perimeter".

Could it be solved under either interpretation?
azerbajdzan
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Post Number: 980
 Posted on Wednesday, 03 August, 2011 - 08:28 pm:

I understood it as one given value was sum of sides of triangle however they called it - circumference or perimeter.

So you do not agree with me that there can be more than one solution? ...and if you think there is only one then a proof of nonexistence of other is necessary to solve the puzzle completely?

Or is there something (which I might have overlooked) that ensures that if one triangle is found then there cannot be another?
Gale Greenlee
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Post Number: 2207
 Posted on Saturday, 06 August, 2011 - 07:46 pm:

I do not disagree that there may be only one solution. In fact I suspect there is only one. And if that is the case I imagine the solution is a rather small triangle.

If you think you have a good solution why don't you post it on the IBM site and see if your name shows up in a few days as a solver?
azerbajdzan
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Post Number: 982
 Posted on Thursday, 11 August, 2011 - 12:35 am:

Gale, I am already there... :-) See the list of people who answered correctly.
But I am not listed with my nick name but with real name.
One person on this forum surely can confirm that I am there.
... and if you wonder if I had to prove that only one triangle exist for this problem then no
my answer was that triangle with sides (x,y,z) comply with the dialogue but that I can not prove that there isn't another...
so probably they cannot prove it too when they published my name as good solver
we have to wait for the solution to be published to see how they explain it and if they proved something
Daniel Fretwell
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Post Number: 1543
 Posted on Friday, 12 August, 2011 - 08:20 am:

And there he is... :p
Gale Greenlee
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Post Number: 2208
 Posted on Saturday, 13 August, 2011 - 04:10 am:

Well great. I think I might have an answer then. I'll send it in and see, thanks, Gale