If in a triangle ABC, length of the sides a,b & c are in Arithmetic Progression, then it is necessary that:

A)2/3 < b/c < 2

B)1/3 < b/c < 2/3

C)2/3 < b/a < 2

D)1/3 < b/a < 2/3

More than one options are correct. How to solve this problem?

Printable View

- July 18th 2008, 10:17 AMfardeen_genResolution of triangles?
If in a triangle ABC, length of the sides a,b & c are in Arithmetic Progression, then it is necessary that:

A)2/3 < b/c < 2

B)1/3 < b/c < 2/3

C)2/3 < b/a < 2

D)1/3 < b/a < 2/3

More than one options are correct. How to solve this problem? - July 18th 2008, 11:50 AMMoo
Hello,

What about something using the law of sines ?

And is there more information about a,b and c ? Because propositions A) and C) are equivalent, and so are B) and D) - July 18th 2008, 04:30 PMSoroban
Hello, fardeen_gen!

I have a start on this problem, but I haven't finished it yet.

. . Maybe this will inspire you . . .

Quote:

If in a triangle , length of the sides are in Arithmetic Progression,

then it is necessary that:

. .

More than one option is correct. How to solve this problem?

Assume that are ordered from smallest to largest: .

Since the form an arithmetic progression, there is a common difference

Then the three sides are: .

The Triangle Inequality says: .

. . which gives us: .

Add to both sides: .

Take reciprocals: .

. . Therefore: .

We have half of statement (B); I'm still working on the other half.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

We have: .

Multiply by -1: .

Add to both sides: .

Take reciprocals: .

. . Therefore: .

And we have half of statment (C).