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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,229

Hi;

Not necessary Agnishom found the link below.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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Hello Bobbym,

Would you mind if I go to bed now?

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'You have made another human being happy. There is no greater accomplishment.' -bobbym

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,229

Hi;

Of course you can get some sleep see you later.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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Well, are you seeing this link?

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'You have made another human being happy. There is no greater accomplishment.' -bobbym

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,229

Hi;

Oh you found it, thanks. I could not remember where I copied it from. I cleaned it up a little in my notes.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**noelevans****Member**- Registered: 2012-07-20
- Posts: 236

Hi Agnishom!

There are other triangles other than right triangles that work. I was just pointing out that there are

only two that are right triangles. The proof of that is probably much easier than the proof for triangles

in general. It will be interesting to see the general proof.

P.S. I tried your link and found the proof. Thanks!

*Last edited by noelevans (2012-12-11 09:26:34)*

Writing "pretty" math (two dimensional) is easier to read and grasp than LaTex (one dimensional).

LaTex is like painting on many strips of paper and then stacking them to see what picture they make.

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Hi noelevans!

Will you explain the proof?

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'You have made another human being happy. There is no greater accomplishment.' -bobbym

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**noelevans****Member**- Registered: 2012-07-20
- Posts: 236

Hi! This is Dan's proof of 4 years ago from your link.

Let a,b,c denote the sides of triangle ∆ABC and P and A its perimeter and area respectively.

Note that Herons formula states that the area of ∆ABC is: A = √[s(s-a)(s-b)(s-c)]

Where s denotes the semi-perimeter, s = (a+b+c)/2.

To find all such triangles such that P=A we must have

√(a+b+c)(b+c-a)(a+c-b)(a+b-c) = 4(a+b+c).

Put x=b+c-a, y=a+c-b and z=a+b-c, where x,y,z are positive integers (true by the triangle

inequality). We see that Heron reduces to the Diophantine equation 16(x+y+z)=xyz.

We see from above that x,y,z must all be even integers. So, we may put x=2m, y=2n, z=2k to get,

mnk=4(m+n+k)

⇒ mnk-4m = 4(n+k)

⇒ m = 4(n+k)/(nk-4)

Without loss of generality assume m≥n≥k, then we have 2m≥2n≥n+k ⇒ nk≤12.

Since 4<nk≤12 we may test integral values of n and k to find all such triangles.

Finally, we see that there is only 5 such triangles:

(a,b,c) = (5,12,13), (6,8,10), (6,25,29), (7,15,20), (9,10,17)

..................................................................................................................................

I followed pretty easily to the three little lines starting with mnk=4(m+n+k) but had a little

difficulty figuring out why 4<nk≤12. So I went back to the line mnk=4(m+n+k) and got it

from there. First rewrite mnk=4(m+n+k) as nk = 4(m+n+k)/m.

Since m≥n≥k then 3m = (m+m+m) ≥ (m+n+k). Therefore

4(3m) 4(m+m+m) 4(m+n+k)

12 = 3*4 = _____ = __________ ≥ ________ = nk because m≥n≥k.

m m m

implies (m+m+m) ≥ (m+n+k).

Also rewriting mnk=4(m+n+k) as 4 = nk*[m/(m+n+k)] we see that nk is being multiplied

by the quantity [m/(m+n+k)] which is less than one. Hence nk>4.

So now we have the 4<nk≤12.

Testing all n≥k being bigger than four but less than or equal to 12 gives us the combinations

(n,k) ∈ {(12,1),(6,2),(4,3),(3,2),(4,2),(5,2),(3,3),(6,1),(8,1),(5,1),(7,1), (9,1),(10,1),(11,1)}

to test (14 of them).

Testing means substitute the n and k into mnk=4(m+n+k) and solve for m. If the result is a

positive integer then double the m, n, and k to get the x, y, and z. Then put these values

into the equations x=-a+b+c

y= a-b+c

z= a+b-c and solve this system for a, b and c (the sides of the triangles).

Example: n=4 and k=3 gives m*4*3=4(m+4+3); that is, 12m=4m+28 so m=28/8 = 7/2

so this case doesn't give an integral value of m. Case fails.

Example: n=6 and k=1 gives m*6*1=4(m+6+1); that is, 6m=4m+28 so m=14

This case gives an integral value for m so x=2m=28, y=2n=12 and z=2k=2.

Substituting these in to the three equations involving a, b, and c we obtain

28 = -a + b + c

12 = a - b + c

2 = a + b - c

Solving this system by addition/subtraction we obtain a=7, b=15, c=20 which is one of the

triangles that works. (Adding each pair of equations eliminates two variables leaving the

third which is easy to solve.)

A nice solution to the problem, but Dan leaves out a few steps that one must scratch the

head about to fill in the gaps. gotta

Writing "pretty" math (two dimensional) is easier to read and grasp than LaTex (one dimensional).

LaTex is like painting on many strips of paper and then stacking them to see what picture they make.

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Sorry for my late reply.

In the first place, how do you know that if 16(x+y+z) = xyz then x, y and z are even?

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'You have made another human being happy. There is no greater accomplishment.' -bobbym

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,229

Hi Agnishom;

The statement 16(x+y+z)=xyz is not generally true. Just look at x = 6, y = 7 and z =8.

He says above, there might something else that implies that they are all even.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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Should I post my next question?

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'You have made another human being happy. There is no greater accomplishment.' -bobbym

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,229

Hi Agnishom;

You can but since Dan did not explain you have to consider the possibility that his proof is flawed. I am not saying that it is, just that it is possible.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**noelevans****Member**- Registered: 2012-07-20
- Posts: 236

Hi!

I believe you all are right. The x, y an z do not have to be even, only the product xyz. So only

one of the three x, y, z has to be even; that is, one or more of x, y, z are even. But I think his

proof essentially holds if his evenness assumption is ignored.

Starting with the 16(x+y+z)=xyz and reasoning in a similar fashion as before for the

4<nk≤12 we can arrive at 16<yz≤48 (assuming z≤y≤x without loss of generality).

Then for y=6 and z=4, 16<yz≤48 and substituting these into 16(x+y+z)=xyz we obtain x=20.

Putting these values for x, y and z into the system of equations involving a, b, and c

we obtain a=5, b=12 and c=13 which is one of the triangles that works.

I'll leave it to you to check out the other possibilities for yz to see if the other 4 triangles are

obtained.

Edit: P. S. Actually all three of x, y and z must be odd or all three must be even to make a, b

and c come out to be integers when solving for a, b and c.

If Dan had just left out the sentence: "We see from above that x,y,z must all be even integers."

then he could have still made the substitutions x=2m, y=2n and z=2k but readers might wonder

why he did so. It appears that all that does is change the limits from '16 to 48' to '4 to 12' which

makes for different pairs to try for solving for x and m, respectively.

*Last edited by noelevans (2012-12-15 06:29:47)*

Writing "pretty" math (two dimensional) is easier to read and grasp than LaTex (one dimensional).

LaTex is like painting on many strips of paper and then stacking them to see what picture they make.

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Do you know Dan?

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'You have made another human being happy. There is no greater accomplishment.' -bobbym

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,229

Hi Agnishom;

I do not know him.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**noelevans****Member**- Registered: 2012-07-20
- Posts: 236

I don't know him either.

LaTex is like painting on many strips of paper and then stacking them to see what picture they make.

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Anyone saw this: http://jwilson.coe.uga.edu/emt668/EMAT6 … ngles.html?

BTW, Someone please explain me the x+y, y+z, x+z theorem about the sides of triangles?

*Last edited by Agnishom (2012-12-15 23:43:13)*

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'You have made another human being happy. There is no greater accomplishment.' -bobbym

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,229

Hi Agnishom;

Thanks for the link! I did not know that one.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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I think I understand this proof

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'You have made another human being happy. There is no greater accomplishment.' -bobbym

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,229

Hi Agnishom;

I did a lot of searching for a solution to that. How did you find it?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

**Online**

Just googling!

Hope it is correct

Please explain me this thing: Why do they assume the sides are in the form of (x+y), (y+z) and (x+z)?

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'You have made another human being happy. There is no greater accomplishment.' -bobbym

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,229

Hi Agnishom;

Looks like a clever trick. But the sides can be labelled anything as long as the triangle inequality holds. And it does hold,

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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That means every triangle can be put into that form.

How can you say?

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'You have made another human being happy. There is no greater accomplishment.' -bobbym

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,229

Hi;

Yes, as long as x, y and z are greater than 0.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

**Online**

Question 9: **A triangle has sides of length at most 2, 3 & 4. What is the maximum area the triangle can have?**

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'You have made another human being happy. There is no greater accomplishment.' -bobbym

**Online**