Question

An altitude $BD$ and angular bisector $BE$ are drawn in $△ABC$ from the vertex $B.$ It is known that the length of side $AC=1$ and the magnitude of angle $BEC,ABD,$ $ABE,BAC$ form an arithmetic progression. Let $B^{\prime }$ be the image of point $B$ with respect to side $AC$ of $△ABC$. If the length $B{B}^{\prime }$ is equal to $\sqrt{\frac{a}{b}};$ where $a,b\in N,$ then the least possible value of $a+b$ is

Answer 1

Let $\angle BEC=\alpha -3\beta$
$\angle ABD=\alpha -\beta$
$\angle ABE=\alpha +\beta$
$\angle BAC=\alpha +3\beta$


$\alpha +3\beta +\alpha +\beta =\alpha -3\beta$
$\Rightarrow \alpha =-7\beta$
$\alpha +3\beta +\alpha -\beta =\frac{\pi }{2}$
$\Rightarrow 2\alpha +2\beta =\frac{\pi }{2}$
$\Rightarrow -14\beta +2\beta =\frac{\pi }{2}$
$\Rightarrow \beta =-\frac{\pi }{24},\alpha =\frac{7\pi }{24}$
$\Rightarrow \angle B=2(\alpha +\beta )=\frac{\pi }{2}$


In $△ABD,$
$\tan \frac{\pi }{6}=\frac{BD}{\lambda }...\left(1\right)$
In $△BDC,$
$\tan \frac{\pi }{3}=\frac{BD}{1-\lambda }...\left(2\right)$
On dividing equation $\left(1\right)$ by $\left(2\right),$ we get
$\Rightarrow \frac{\frac{1}{\sqrt{3}}}{\sqrt{3}}=\frac{1-\lambda }{\lambda }$
$\Rightarrow \lambda =3-3\lambda$
$\Rightarrow \lambda =\frac{3}{4}$
$\therefore BD=\frac{\sqrt{3}}{4}$
$\therefore B{B}^{\prime }=\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{\sqrt{4}}$
Thus, $a+b=3+4=7$