Question

In a$∆PQR$, $P$ is the largest angle and$\cos P=\frac{1}{3}$, Further in the circle of the triangle touches the sides$PQ$, $QR$, and $RP$ at $N$, $L$, and $M$ respectively, such that the lengths of $PN$, $QL$, and $RM$ are consecutive even integers, Then, possible length(s) of the side (s) of the triangle is (are)

• A. $16$
• B. $18$
• C. $24$
• D. $22$

Answer 1

Answer: (B), (D)

$22$

Explanation for the correct options:

Step 1: Finding value of the parts of the side of the triangle:

Given,$∆PQR$ in which, $P$ is the largest angle and$\cos P=\frac{1}{3}$,

The circle touches the sides$PQ$, $QR$, and $RP$ at $N$, $L$, and $M$ respectively, such that the lengths of $PN$, $QL$, and $RM$ are consecutive even integers, that is

$PN=2k-2$

$QL=2k$

$RM=2k+2$

Now we know that from an exterior point, the length of two tangents are equal. Hence

$$\begin{gathered} PM=PN \\ =2k-2 \end{gathered}$$

Similarly

$$\begin{gathered} QN=QL \\ =2k \end{gathered}$$

And also

$$\begin{gathered} RL=RM \\ =2k+2 \end{gathered}$$

Step 2: Finding perimeter of triangle in terms of $k$

Now perimeter of a triangle$∆PQR$, for semi perimeter $s$ is

$$\begin{gathered} PQ+QR+RP=2s \\ \Rightarrow \left(PN+NQ\right)+\left(QL+LR\right)+\left(RM+MP\right)=2s \\ \Rightarrow 2\left(2k+2\right)+2\left(2k\right)+2\left(2k-2\right)=2s \\ \Rightarrow 2k+2k+2k-2=s \\ \Rightarrow 6k=s \end{gathered}$$

Step 3: Finding sides of the triangle in terms of $k$

Now we know that

$$\begin{gathered} s-a=2k-2 \\ s-b=2k \\ s-c=2k+2 \end{gathered}$$

Therefore,

$$\begin{gathered} 6k-a=2k-2 \\ \Rightarrow a=4k+2 \end{gathered}$$

Similarly

$$\begin{gathered} 6k-b=2k \\ \Rightarrow b=4k \end{gathered}$$

And also

$$\begin{gathered} 6k-c=2k+2 \\ \Rightarrow c=4k-2 \end{gathered}$$

Step 4: Finding value of $k$ using Cosine formula

Now $\cos P=\frac{1}{3}$

$$\begin{gathered} \Rightarrow \frac{b^2+c^2-a^2}{2bc}=\frac{1}{3}\left[\because \cos \theta =\frac{b^2+c^2-a^2}{2\mathrm{bc}}\right] \\ \Rightarrow \frac{{\left(4k\right)}^2+{\left(4k-2\right)}^2-{\left(4k+2\right)}^2}{2\left(4k-2\right)\left(4k\right)}=\frac{1}{3} \\ \Rightarrow \frac{16k^2-32k}{2\left(4k-2\right)\left(4k\right)}=\frac{1}{3} \\ \Rightarrow \frac{k^2-2k}{\left(2k-1\right)\left(k\right)}=\frac{1}{3} \end{gathered}$$

Simplify to get the values for $k$.

$\begin{array}{rcl}3\left(k^2-2k\right)& =& \left(2k-1\right)\left(k\right)\\ & \Rightarrow & 3k^2-6k=2k^2-k\\ & \Rightarrow & k^2-5k=0\\ & \Rightarrow & \left(k-5\right)k=0\\ \Rightarrow k& =& 0,5\end{array}$

Step 5: Finding the value of the sides of the triangle

Now if we put $k=0$ in $\text{'}c\text{'}$ we get $c=4k-2\Rightarrow c=4\times 0-2=-2$

$c=-2$ is not possible because the length is not negative, hence $k=0$is rejected.

Therefore,$k=5$in $a,b,c$ we get

$\begin{array}{rcl}a& =& 4k+2\\ & =& 4\times 5+2\\ & =& 22\end{array}$

Similarly

$\begin{array}{rcl}b& =& 4\times 5\\ & =& 20\end{array}$

And also

$\begin{array}{rcl}c& =& 4k-2\\ & =& 4\times 5-2\\ & =& 18\end{array}$

Therefore the possible values of sides of a triangle are $22$,$20$ and $18$

Hence, option (B) and (D) is the correct answer.