A few questions

bobbym · 2012-12-11 04:28:49

Hi;

Not necessary Agnishom found the link below.

Agnishom · 2012-12-11 04:31:59

Hello Bobbym,
Would you mind if I go to bed now?

bobbym · 2012-12-11 04:35:30

Hi;

Of course you can get some sleep see you later.

Agnishom · 2012-12-11 04:35:32

Well, are you seeing this link?

bobbym · 2012-12-11 04:39:32

Hi;

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

noelevans · 2012-12-11 08:53:38

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)

Agnishom · 2012-12-11 15:07:26

Hi noelevans!
Will you explain the proof?

noelevans · 2012-12-12 19:28:10

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

Agnishom · 2012-12-15 01:01:23

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?

bobbym · 2012-12-15 01:13:07

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.

Agnishom · 2012-12-15 01:44:43

Should I post my next question?

bobbym · 2012-12-15 01:50:38

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.

noelevans · 2012-12-15 03:32:38

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)

Agnishom · 2012-12-15 17:07:05

Do you know Dan?

bobbym · 2012-12-15 17:44:48

Hi Agnishom;

I do not know him.

noelevans · 2012-12-15 18:17:44

I don't know him either.

Agnishom · 2012-12-15 23:39:16

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)

bobbym · 2012-12-16 00:05:09

Hi Agnishom;

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

Agnishom · 2012-12-16 00:11:28

I think I understand this proof

bobbym · 2012-12-16 11:00:08

Hi Agnishom;

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

Agnishom · 2012-12-17 00:46:49

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)?

bobbym · 2012-12-17 00:59:50

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,

Agnishom · 2012-12-17 01:03:09

That means every triangle can be put into that form.
How can you say?

bobbym · 2012-12-17 01:07:07

Hi;

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

Agnishom · 2012-12-17 01:27:23

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

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