Question

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

• A. $1$
• B. $0$
• C. $4$
• D. $2$

Answer 1

Answer: (A)

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
$\frac{\text{BC}}{\text{AC}}=\frac{\text{CD}}{\text{BC}}\Rightarrow {\left(\text{a}\right)}^2=\frac{\text{ab}}{\text{a}+\text{c}}\text{b}\Rightarrow {\left(\text{b}\right)}^2=\text{a}\left(\text{a}+\text{c}\right)$ ......(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.