In a ∆ P Q R , P is the largest angle and cos P = 1 3 . Further the incircle of the triangle touches the sides P Q ,   Q R and R P at N ,   L and M respectively, such that the lengths of P N ,   Q L and R M are consecutive even integers. Then possible length(s) of the side(s)…

Question

In a &#8710; P Q R , P is the largest angle and cos P = 1 3 . Further the incircle of the triangle touches the sides P Q , &#160; Q R and R P at N , &#160; L and M respectively, such that the lengths of P N , &#160; Q L and R M are consecutive even integers. Then possible length(s) of the side(s) of the triangle is (are)

Aicharya · Accepted Answer

Consider the following figure:

We assumed the value of sides in terms of $n$ , which satisfies our given condition.

Now applying cosine rule,

$\cos P = \frac{{(2 n + 2)}^{2} + {(2 n + 4)}^{2} - {(2 n + 6)}^{2}}{2 (2 n + 2) (2 n + 4)}$

$= \frac{4 n^{2} - 16}{8 (n + 1) (n + 2)} = \frac{1}{3}$

$= \frac{n^{2} - 4}{2 (n + 1) (n + 2)} = \frac{1}{3}$

$\Rightarrow \frac{n - 2}{2 (n + 1)} = \frac{1}{3}$

$\Rightarrow 3 n - 6 = 2 n + 2 \Rightarrow n = 8$

Length of the sides will be $18, 20$ and $22$ .

In a ∆ P Q R , P is the largest angle and cos P = 1 3 . Further the incircle of the triangle touches…

Solution

In a &#8710; P Q R , P is the largest angle and cos P = 1 3 . Further the incircle of the triangle touches…

Solution

In a ∆ P Q R , P is the largest angle and cos P = 1 3 . Further the incircle of the triangle touches…