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

$ 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 $
|||$ly \,\,\,m^2 = t^2+ t%3$
and $m^3 = t^3 + t^1$
hence $\sum t _i=\frac{ m _1+ m _2+ m _3}{2}$
Now $\tan 60^{\circ}=\left|\frac{m-2}{1+2 m}\right|$
$\Rightarrow \pm \sqrt{3}(1+2 m )= m -2$
Taking + ve sign, we get
$\sqrt{3}(1+2 m )= m -2 \Rightarrow m (2 \sqrt{3}-1)=-(2+\sqrt{3}) \Rightarrow m =\frac{-(2+\sqrt{3})}{2 \sqrt{3}-1}$
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} $
$\therefore m _1=\frac{-(2+\sqrt{3})}{2 \sqrt{3}-1} ; m _2=\frac{2-\sqrt{3}}{2 \sqrt{3}+1} \text { and } m _3=2$
$\therefore \displaystyle\sum_{i=1}^3 t _i=\frac{ m _1+ m _2+ m _3}{2}$
$=\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 $