Resolution of triangles?

**fardeen_gen** · July 18th 2008, 09:17 AM

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?

**Moo** · July 18th 2008, 10:50 AM

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)

**Soroban** · July 18th 2008, 03:30 PM

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

. . $\begin{array}{cccccc}(A)& \dfrac{2}{3} & < & \dfrac{b}{c} & < & 2 \\ \\[-2mm]<br /> (B)& \dfrac{1}{3} &<& \dfrac{b}{c} &<& \dfrac{2}{3} \\ \\[-2mm]<br /> (C)& \dfrac{2}{3} &<& \dfrac{b}{a} &<& 2 \\ \\[-2mm]<br /> (D)& \dfrac{1}{3} &<& \dfrac{b}{a} &<& \dfrac{2}{3} \end{array}$

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

Assume that $a,b,c$ are ordered from smallest to largest: . $a \,<\, b \,<\, c$

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

Then the three sides are: . $b-d,\;b,\;b+d$

The Triangle Inequality says: . $a + b \:>\:c$

. . which gives us: . $(b-d) + b \:>\:b+d\quad\Rightarrow\quad b \:>\:2d \quad\rightarrow\quad d \:<\:\frac{b}{2}$

Add $b$ to both sides: . $b + d \:< \:b + \frac{b}{2} \quad\Rightarrow\quad b+d \:<\:\frac{3b}{2}$

Take reciprocals: . $\frac{1}{b+d} \:>\:\frac{2}{3b}$

$\text{Multiply by }b\!:\;\;\frac{b}{\underbrace{b+d}_{\text{This is }c}} \:<\:\frac{2}{3}$

. . Therefore: . $\frac{b}{c}\:<\:\frac{2}{3}$

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

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

We have: . $d \:<\:\frac{b}{2}$

Multiply by -1: . $-d \:>\:-\frac{b}{2}$

Add $b$ to both sides: . $b - d \:>\:b -\frac{b}{2} \quad\Rightarrow\quad b - d \:>\:\frac{b}{2}$

Take reciprocals: . $\frac{1}{b-d} \:<\:\frac{2}{b}$

$\text{Multiply by }b\!:\;\;\frac{b}{\underbrace{b-d}_{\text{This is }a}} \:<\:2$

. . Therefore: . $\frac{b}{a} \:<\:2$

And we have half of statment (C).

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