Arithmetic sequence and length of right-angled triangle

[cccc]

Does anyone have any advice on how to tackle this question. 

So far what i have done is to say 

1/x+ d = 1/y 

So (1+xd)/x = 1/y 

and that (1+2xd)/x = 1/z 

z=x/(1+2xd) 

So it means that I can use the pythagoras theorem to solve it, but i don't know which one is longest? 

and also there are too many variables involved?

[kw0573]

"...." are added for indentation purposes.
Make the substitution a, b, c, such that
a = 1/x, b = 1/y, c = 1/z; a, b, c, is an increasing arithmetic progression
let d = difference = c - b = b - a = (c - a) / 2
x - z ...... = 1/a - 1/c = (c-a) / (ac) = (bc - ab) / (abc) 
y ........... = 1/b = ac / (abc)
x - y + z  = 1/a - 1/b + 1/c = (bc - ac + ab) / (abc) = (bc - a(c - b)) / (abc) = (b(a + c) - ac) / (abc)
If these form a right triangle, so does {bc - ab, ac, bc - ac + ab}; now we have to decide which is the biggest. Once we do that, apply Pythagorean and compare! Here we are almost half way. Expand to see the full solution, or pause here to try on your own.

Spoiler

 

i) compare bc - a(b) with
..................bc - a(c -b) or equivalently bc - ad
b > d because b - d = a > 0
We can also write bc - ab as b(c-a) = b(2d)

ii) compare ac with bc - ac + ab, or equivalently b(a + c) - ac
Make substitution a = b - d, and c = b + d
b(a + c) - ac = b(b - d + b + d) - (b-d)(b+d) = b(2b) - (b² - d²) = b² + d².
ac ................= (b - d)(b + d) = b² - d² < b² + d². So bc - ac + ab is the biggest term. 

At this time, we would rewrite {bc - ab, ac, bc - ac + ab} as {b(2d), b² - d², b² + d²)

Squaring each term, we get

{4b²d², b⁴ - 2b²d² + d⁴, b⁴ + 2b²d² + d⁴}

Because (4b²d²) + (b⁴ - 2b²d² + d⁴) = b⁴ + 2b²d² + d⁴, the proposed relationship is true. QED

 

 

[cccc]

@kw0573

16 hours ago, kw0573 said:

let d = difference = c - b = b - a = (c - a) / 2
x - z ...... = 1/a - 1/c = (c-a) / (ac) = (bc - ab) / (abc) 
y ........... = 1/b = ac / (abc)
x - y + z  = 1/a - 1/b + 1/c = (bc - ac + ab) / (abc) = (bc - a(c - b)) / (abc) = (b(a + c) - ac) / (abc)
 

  Reveal hidden contents

 

i) compare bc - a(b) with
..................bc - a(c -b) or equivalently bc - ad
b > d because b - d = a > 0
We can also write bc - ab as b(c-a) = b(2d)

ii) compare ac with bc - ac + ab, or equivalently b(a + c) - ac
Make substitution a = b - d, and c = b + d
b(a + c) - ac = b(b - d + b + d) - (b-d)(b+d) = b(2b) - (b² - d²) = b² + d².
ac ................= (b - d)(b + d) = b² - d² < b² + d². So bc - ac + ab is the biggest term. 

At this time, we would rewrite {bc - ab, ac, bc - ac + ab} as {b(2d), b² - d², b² + d²)

Squaring each term, we get

{4b²d², b⁴ - 2b²d² + d⁴, b⁴ + 2b²d² + d⁴}

Because (4b²d²) + (b⁴ - 2b²d² + d⁴) = b⁴ + 2b²d² + d⁴, the proposed relationship is true. QED

 

 

 

Thanks for the reply! But how did you get from c-a / ac to (bc-ab)/abc. 

and same for all the remaining two lines as well? 

I know you rearrange the first equation, but don't you get (b-a)-(c-b) = (c-a)/2 

So hence

2*(2b-a-c) = (c-a)? 

[kw0573]

@cccc
1)  (c-a) / (ac) I just multiplied numerator and denominator by b. b(c-a) / (abc) = (bc - ab) / abc. A common denominator will help comparing between the three expressions.
2) (b - a) - (c - b) = 0 because it's basically d - d = 0. Which line are you referring to? We can continue in a chat; shoot me a message.