There exist infinitely many non similar triangles

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There exist infinitely many non similar triangles

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Source: Transylvanian Hungarian MC (Romania) 2012 - G11 - P6

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5452 posts

#1 h Feb 17, 2012, 10:02 AM • 1 Y

Y by Adventure10

Show that there exist infinitely many non similar triangles such that the side-lengths are positive integers and the areas of squares constructed on their sides are in arithmetic progression.

Proposed by Ferenc Olosz

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mlm95

#2 h Feb 20, 2012, 1:45 PM • 2 Y

Y by Amir Hossein, Adventure10

let $a=(a_{n}+2)^{2}-2$

$b=(a_{n})^{2}-2$

$c=(a_{n}+1)^{2}+1$

then $a^{2}+b^{2}=2c^{2}$

where $(a_{n})(n\geq1)$ is the sequence such that for every $i$ , $a_{i}\in \mathbb{N}$

and $a_{1}\geq3$ and for every $i\geq1$ , $a_{i+1}>2(a_{i})^{2}+2a_{i}$

some one can this satisfies the conditions

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#3 h Aug 19, 2012, 3:43 AM • 2 Y

Y by Adventure10, Mango247

In the solution by mim95, it is true that $a + b = 2c$ , but not that $a^2 + b^2 = 2c^2$ . I don't see how the solution works with this issue.

EDIT: There are many things wrong with this post of mine. I copied down $c$ incorrectly onto the paper I did my work in, which left everything messed up no matter how many times I checked my work. The solution above seems to work.

I really should be more careful...

This post has been edited 1 time. Last edited by Duelist, Aug 19, 2012, 4:31 AM

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#4 h Aug 19, 2012, 4:02 AM • 2 Y

Y by Adventure10, Mango247

Duelist wrote:

In the solution by mim95, it is true that $a + b = 2c$ , but not that $a^2 + b^2 = 2c^2$ . I don't see how the solution works with this issue.

No $a+b=2c$ is not true. Another construction

$a=u^2-v^2+2uv,b=u^2-v^2-2uv,c=u^2+v^2,u>4v$

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#5 h Aug 19, 2012, 1:47 PM • 2 Y

Y by Adventure10, Mango247

let the sides be $a,b,c$ ,
let $a \ge b\ge c$
so , we have $a^2+c^2=2b^2$ . put $a/b=x ,c/b=y, x,y>0$
so , we get $x^2+y^2=2$ ....(*) ,
with $x,y\in Q, x,y>0$
now , consider a particular solution $(p,q)$ .
consider the slope (k) of the line joining $(p,q)$ & $(x,y)$ .
$k=\frac{y-q}{x-p}$ , now express $y$ in terms of $k,p,q$ and put that in (*). then appying Vieta's relation , we can easily solve the problem.we must keep in mind the problem conditions for choosing $k$ carefully.
a such set of solution is given by $(a,b,c)=(k^2-1+2k,k^2+1,k^2-2k-1)$ .
we must choose $k\in N , [k\ge5]$ carefully such that , they are consistent with the problem condition.
it is quite obvious that , we can make infinitely many non-similar triangles satisfying the condition from this family with suitable values of $k$
[the proof of the last sentence is trivial ,so i am not posting it. it uses basic divisibility and triangle inequality ]

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#6 h Jan 24, 2018, 9:54 PM • 2 Y

Y by Adventure10, Mango247

mlm95 wrote:

let $a=(a_{n}+2)^{2}-2$

$b=(a_{n})^{2}-2$

$c=(a_{n}+1)^{2}+1$

Why these expressions?
Thanks (I know I'm bumping a 6 year old post.)

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#7 h Oct 31, 2021, 3:53 AM

Y by

After 1 hour with this problem I managed to solve it

. First; I tried to see what possible triangles could be (to study their properties) I found $(1,5,7); (7,17,23); (23,37,47).$ Notice that $5 = 1 ^ 2 + 2 ^ 2; 17 = 1 ^ 2 + 4 ^ 2; 37 = 1 ^ 2 + 6 ^ 2$
Also: $1 = 2 ^ 2-2 (2) +1; 7 = 2 ^ 2 + 2 (2) + 1$ similarly the other cases. Therefore the triples that satisfy are: $(a, b, c) = (x ^ 2-2x-1, x ^ 2 + 1, x ^ 2 + 2x-1)$ for all $x>4$ positive integer $\blacksquare$

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#8 h Jun 16, 2022, 8:27 PM

Y by

Albert123 wrote:

After 1 hour with this problem I managed to solve it

. First; I tried to see what possible triangles could be (to study their properties) I found $(1,5,7); (7,17,23); (23,37,47).$ Notice that $5 = 1 ^ 2 + 2 ^ 2; 17 = 1 ^ 2 + 4 ^ 2; 37 = 1 ^ 2 + 6 ^ 2$
Also: $1 = 2 ^ 2-2 (2) +1; 7 = 2 ^ 2 + 2 (2) + 1$ similarly the other cases. Therefore the triples that satisfy are: $(a, b, c) = (x ^ 2-2x-1, x ^ 2 + 1, x ^ 2 + 2x-1)$ for all $x>4$ positive integer $\blacksquare$

Bro (1,5,7) is not triangle

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