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?
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?
Hello, fardeen_gen!
I have a start on this problem, but I haven't finished it yet.
. . Maybe this will inspire you . . .
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).