Question

Let $\left({\alpha }_1,{\beta }_1\right),\left({\alpha }_2,{\beta }_2\right),\left({\alpha }_3,{\beta }_3\right)$ be vertices of a $△ABC$ with ${\alpha }_1,{\alpha }_2,{\alpha }_3,{\beta }_1,{\beta }_2,{\beta }_3$ are prime values of $k$ in increasing order for which one root of equation $(k-5)x^2-2kx+k-4=0$ is smaller than 1 and other exceed 2 . If $P(p,q)$ is a point inside the triangle such that area of $△PAC=$ area of $△PAB=$ area of $△PBC$, then find the value of $\left(\frac{p+q}{10}\right)$.

Answer 1

Answer: 3

$\Theta 1$ and 2 lies between roots,
$\therefore (k-5)f(1)<0$
$\Rightarrow (k-5)(k-5-2k+k-4)<0$
$\Rightarrow (k-5)(-9)<0\Rightarrow k-5>0\Rightarrow k>5$
$\text{ and }(k-5)f(2)<0$
$\Rightarrow (k-5)(4(k-5)-4k+k-4)<0$
$\Rightarrow (k-5)(k-24)<0\Rightarrow 5<k<24$
$\therefore k\in (5,24)$
$\text{ prime numbers }=7,11,13,17,19,23$
$\therefore \text{ points will be }=(7,17)(11,19)\text{ and }(13,23)$
$\text{ Clearly, Pis centroid }$
$\therefore P=\left(\frac{31}{3},\frac{59}{3}\right)$
$\frac{p+q}{10}=\frac{\frac{90}{3}}{10}=3$