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[quote=noelevans]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 Heron’s 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. :up  :wave  gotta :sleep[/quote]

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Topic review (newest first)

Agnishom: 2013-03-17 15:06:15

Thank you

Nehushtan: 2013-03-17 03:09:30

Here you go.

Agnishom: 2013-03-16 12:54:42

Please upload a diagram, I really do not understand from "Applying Pythagoras Theorem..."

Agnishom: 2013-03-15 18:18:18

Thanks Nehushtan
I shall ask you if I have any doubts

bobbym: 2013-03-15 17:19:41

Hi Agnishom;

Thanks for the drawing.

Nehushtan: 2013-03-15 13:45:43

Agnishom wrote:
Please help me with the problem attached

This is what I got.

Let be the centre of and the point of contact of with . Let be the centre of and the point of contact of with .

Let be the point on such that is a rectangle. Applying Pythagoras’s theorem to gives .

Let be the point on such that is a rectangle. Applying Pythagoras’s theorem to gives .

Let be the point on such that is a rectangle. Applying Pythagoras’s theorem to gives .

Then

Hence:

Agnishom: 2013-03-15 13:12:12

This is the only diagram I can imagine.

A rough sketch though...

1001 Words ≈ 7 KB
1 picture > 100 KB

Obviously, a picture is worth more....

Agnishom: 2013-03-15 12:59:39

Bob, if you want I can give a diagram

bobbym: 2013-03-15 09:42:50

Hi Agnishom;

bobbym wrote:
A picture is worth a thousand words.

For people like me who are geometrically challenged this problem is a nightmare. So far I have come up with 10^16 different possible drawings that to me fit those constraints.

Now I am busy looking up externally tangent, internally tangent, tangerine tangent... in the hopes of what externally tangent to the line means. Intersects at one point?

I see that bob has got his circles with their centers on the line, I have lifted my circles and it was hard to do, after all there are an infinite amount of them, to roll on the line. That takes care of externally tangent. Now this second bunch of circles, where do they go?

John McCarthy wrote:
As the Chinese say, 1001 words is worth more than a picture.

Maybe I shouldn't need the diagram...

anonimnystefy: 2013-03-15 07:05:57

I am getting 1/2.

bob bundy: 2013-03-15 04:07:38

hi Agnishom,

I'm not getting this yet. I've drawn a line L . Now it says 'on a straight line l' so I assumed that the circles had their centres on the line, and that each circle touches the ones on either side of it. That covers the 'externally tangential to circle(n-1) and circle(n+1) but apparently also tangential to L itself. How does that happen? And I haven't even got to the second sentence yet.

Bob

Agnishom: 2013-03-15 03:32:52

Please help me with the problem attached

bobbym: 2013-02-12 01:07:24

Please hold on the latex maker is not holding up, I am aware of that and am editing the original post.

All done!

anonimnystefy: 2013-02-12 01:03:23

Hi bobbym

$http://latex.codecogs.com/gif.latex?\frac{\ln{\left%20(1+\frac{2}{3}%20\right%20)}}{1-\frac{2}{3}}\neq%20\ln{5}$

Agnishom: 2013-02-12 00:54:28

Let me try it out

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bobbym wrote:

John McCarthy wrote:

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