Question

In a scalene acute $\Delta ABC,$ it is known that the line joining circumcentre and ortho centre is parallel to $BC$ then the angle $A$ lies in which of the following intervals:

• A. $(\frac{\pi }{4},\frac{\pi }{3})$
• B. $(\frac{\pi }{6},\frac{\pi }{4})$
• C. $(\frac{\pi }{3},\frac{\pi }{2})$
• D. $(\frac{\pi }{6},\frac{\pi }{3})$

Answer 1

The correct option is C $(\frac{\pi }{3},\frac{\pi }{2})$

In the adjacent figure $A{A}_1$ and $B{B}_1$ are altitudes drawn from $A$ and $B$ on the sides $BC$ and $AC$ respectively. Let $P,O$ be the orthocentre and circum centre of the $\Delta$


We have been given that $P{A}_1=O{O}_1$
In $\Delta O{O}_{1}C,$
$\angle O_{1}OC=A$ and $OC=R$
$\Rightarrow O{O}_1=R\cos A$
Also, $B{A}_1=AB\cos B=c\cos B$
and
In right angle $△B{B}_{1}C:\angle B_{1}BC={90}^{\circ }-\angle B_{1}CB$
and so for right angle $△P{A}_{1}B:\angle BP{A}_1=\angle B_{1}CB=\angle C$
$\Rightarrow P{A}_1=B{A}_1\cot C\Rightarrow P{A}_1=c\cos B\cot C=\frac{c\cos B\cos C}{\sin C}\Rightarrow P{A}_1=2R\cos B\cos C$
Thus $R\cos A=2R\cos B\cos C\Rightarrow 2\cos B\cos C+\cos (B+C)=0\Rightarrow \sin B\sin C=3\cos B\cos C\Rightarrow \tan B\tan C=3$
Now, we know in any triangle
$\sum \tan A=\tan A\tan B\tan C\Rightarrow \tan A+\tan B+\tan C=3\tan A\Rightarrow \tan B+\tan C=2\tan A$
Now, $(\tan B-\tan C{)}^2>0$ as $\angle B\ne \angle C\Rightarrow (\tan B+\tan C{)}^2>4\tan B\tan C\Rightarrow (4{\tan }^{2}A)>12\Rightarrow \tan A>\sqrt{3}$
$\therefore A\in (\frac{\pi }{3},\frac{\pi }{2})$