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Q.
Suppose that the side lengths of a triangle are three consecutive integers and one of the angles is twice another. The number of such triangles is/are
NTA AbhyasNTA Abhyas 2020
Solution:
Let $\text{B}=2\text{A}$ and
CD = $\frac {ab}{a+c}$ & $AD = \frac {bc}{a+c}$
Now, $\Delta \text{ABC} \, and \, \Delta \text{BDC}$ are similar, so
$\frac{\text{BC}}{\text{AC}} = \frac{\text{CD}}{\text{BC}} \Rightarrow \left(\text{a}\right)^{2} = \frac{\text{ab}}{\text{a} + \text{c}} \text{b} \, ⇒ \, \left(\text{b}\right)^{2} = \text{a} \left(\text{a} + \text{c}\right)$ ......(i)
Since, $\text{b} > \text{a} \, \, \Rightarrow $ Either $\text{b} = \text{a} + 1$ or $\text{b} = \text{a} + 2,$ if $\text{b} = \text{a} + 1,$ then [From Eq.(i)]
$\left(\text{a} + 1\right)^{2} = \left(\text{a} + \text{c}\right) \text{a} \Rightarrow \text{c} = 2 + \frac{1}{\text{a}}$
$\text{c}$ is integer $\Rightarrow $ $\text{a} = 1 , \text{b} = 2 , \text{c} = 3$ but then, no triangle will form.
If $\text{b} = \text{a} + 2 ,$ then obviously $\text{c} = \text{a} + 1 ,$ and then, [from Eq. (i)]
$\left(\text{a} + 2\right)^{2}=\text{a}\left(2 \text{a} + 1\right)$
$\Rightarrow \text{a}^{2}-3a-4=0 \, \, \text{or} \, \text{a}=4$
$\therefore \, \, \text{a}=4, \, \text{b}=6,\text{c}=5$ is the only possible solution.
