Question

Let $XY$ be the diameter of a semicircle with center $O$. Let $A$ be a variable point on the semicircle and $B$ another point on the semicircle such that $AB$ is parallel to $XY$. The value of $\angle BOY$ for which the inradius of triangle $AOB$ is maximum, is

• A. ${\cos }^{-1}\left(\frac{\sqrt{5}-1}{2}\right)$
• B. ${\sin }^{-1}\left(\frac{\sqrt{5}-1}{2}\right)$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{5}$

Answer 1

The correct option is A ${\cos }^{-1}\left(\frac{\sqrt{5}-1}{2}\right)$

Let the angle $\angle BOY$ be $\theta$ (as shown in the figure), when the inradius of $△AOB$ is maximum.

We know that, area of triangle is $A=\frac{1}{2}\text{base}\times \text{height}=\text{inradius}\times \text{semi-perimeter}$

Height of $△AOB=R\cos (90-\theta )=R\sin \theta$

Base is $AB=2R\sin (90-\theta )=2R\cos \theta$

Area of $△AOB=\frac{1}{2}R\sin \theta \times 2R\cos \theta =R^2\sin \theta \cos \theta$

Semi-perimeter $s=\frac{1}{2}(R+R+2R\cos \theta )=R(1+\cos \theta )$

Thus, Inradius $r=\frac{\Delta }{s}=\frac{R^2\sin \theta \cos \theta }{R(1+\cos \theta )}$

If inradius is maximum, then $\frac{dr}{d\theta }=0$

$\Rightarrow \frac{(1+\cos \theta )({\cos }^2\theta -{\sin }^2\theta )-\sin \theta \cos \theta (-\sin \theta )}{(1+\cos \theta {)}^2}=0$

$\Rightarrow (1+\cos \theta )(2{\cos }^2\theta -1)+\cos \theta (1-{\cos }^2\theta )=0$

Let $\cos \theta =C$

Then, $(1+C)(2C^2-1)+C(1-C^2)=0\Rightarrow C^3+2C^2-1=0$

$C=-1,\frac{\sqrt{5}-1}{2},-\frac{\sqrt{5}+1}{2}$

Thus, $\cos \theta =\frac{\sqrt{5}-1}{2}$