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 $\displaystyle ABC$, length of the sides $\displaystyle a,b,c$ are in Arithmetic Progression,
then it is necessary that:
. . $\displaystyle \begin{array}{cccccc}(A)& \dfrac{2}{3} & < & \dfrac{b}{c} & < & 2 \\ \\[-2mm]
(B)& \dfrac{1}{3} &<& \dfrac{b}{c} &<& \dfrac{2}{3} \\ \\[-2mm]
(C)& \dfrac{2}{3} &<& \dfrac{b}{a} &<& 2 \\ \\[-2mm]
(D)& \dfrac{1}{3} &<& \dfrac{b}{a} &<& \dfrac{2}{3} \end{array}$
More than one option is correct. How to solve this problem?
Assume that $\displaystyle a,b,c$ are ordered from smallest to largest: .$\displaystyle a \,<\, b \,<\, c$
Since the form an arithmetic progression, there is a common difference $\displaystyle d.$
Then the three sides are: . $\displaystyle b-d,\;b,\;b+d$
The Triangle Inequality says: .$\displaystyle a + b \:>\:c$
. . which gives us: .$\displaystyle (b-d) + b \:>\:b+d\quad\Rightarrow\quad b \:>\:2d \quad\rightarrow\quad d \:<\:\frac{b}{2} $
Add $\displaystyle b$ to both sides: .$\displaystyle b + d \:< \:b + \frac{b}{2} \quad\Rightarrow\quad b+d \:<\:\frac{3b}{2}$
Take reciprocals: .$\displaystyle \frac{1}{b+d} \:>\:\frac{2}{3b}$
$\displaystyle \text{Multiply by }b\!:\;\;\frac{b}{\underbrace{b+d}_{\text{This is }c}} \:<\:\frac{2}{3}$
. . Therefore: .$\displaystyle \frac{b}{c}\:<\:\frac{2}{3}$
We have half of statement (B); I'm still working on the other half.
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We have: .$\displaystyle d \:<\:\frac{b}{2}$
Multiply by -1: .$\displaystyle -d \:>\:-\frac{b}{2}$
Add $\displaystyle b$ to both sides: .$\displaystyle b - d \:>\:b -\frac{b}{2} \quad\Rightarrow\quad b - d \:>\:\frac{b}{2}$
Take reciprocals: .$\displaystyle \frac{1}{b-d} \:<\:\frac{2}{b}$
$\displaystyle \text{Multiply by }b\!:\;\;\frac{b}{\underbrace{b-d}_{\text{This is }a}} \:<\:2$
. . Therefore: .$\displaystyle \frac{b}{a} \:<\:2$
And we have half of statment (C).