Question

An equilateral triangle $ABC$ is inscribed in the parabola $y=x^2$ and one of the side of the equilateral triangle has the gradient 2 . If the sum of $x$-coordinates of the vertices of the triangle is a rational in the form $\frac{p}{q}$ where $p$ and $q$ are coprime, then find the value of $(p+q)$.

Answer 1

Answer: 14

$y=x^2$
To find: $t_1+t_2+t_3=$ ?
$m_1=\frac{t_2^2-t_1^2}{t_2-t_1}=t_2+t_1+t_3$
|||$ly{m}^2=t^2+t$
and $m^3=t^3+t^1$
hence $\sum t_i=\frac{m_1+m_2+m_3}{2}$
Now $\tan 60^{\circ }=\left|\frac{m-2}{1+2m}\right|$
$\Rightarrow \pm \sqrt{3}(1+2m)=m-2$
Taking + ve sign, we get
$\sqrt{3}(1+2m)=m-2\Rightarrow m(2\sqrt{3}-1)=-(2+\sqrt{3})\Rightarrow m=\frac{-(2+\sqrt{3})}{2\sqrt{3}-1}$
Taking - ve sign, we get
$m_{-2}=-2\sqrt{3}m-\sqrt{3}\Rightarrow m(2\sqrt{3}+1)=2-\sqrt{3}\Rightarrow m=\frac{2-\sqrt{3}}{2\sqrt{3}+1}$
$\therefore m_1=\frac{-(2+\sqrt{3})}{2\sqrt{3}-1};m_2=\frac{2-\sqrt{3}}{2\sqrt{3}+1}\text{ and }{m}_3=2$
$\therefore \sum _{i=1}^3{t}_i=\frac{m_1+m_2+m_3}{2}$
$=\frac{\frac{-(2+\sqrt{3})}{2\sqrt{3}-1}+\frac{2-\sqrt{3}}{2\sqrt{3}+1}+2}{2}=\frac{\frac{-4\sqrt{3}-2-6-\sqrt{3}+4\sqrt{3}-6-2+\sqrt{3}+22}{11}}{2}$
$=\frac{6}{22}=\frac{3}{11}=\frac{p}{q}\Rightarrow (p+q)=14$