# Problem Study (Geometry applying Trigonometry and Calculus)

(a)    Let be a triangle with right angle and hypotenuse $a=BC$.
(See the figure.)
If the inscribed circle touches the hypotenuse at D,
show that $CD&space;=&space;\frac{1}{2}\left(&space;{BC&space;+&space;AC-AB}&space;\right)$.
(b)   If  $\theta&space;=&space;\frac{1}{2}\angle&space;C$, express the radius $r$ of the inscribed circle in terms  of $a$ and $\theta$
(c)   If  $a$ is fixed and  $\theta$ varies, find the maximum value of $r$.
Solution
Let $O$  be the centre of the circle and $E$ and $F$ be points of
tangency of $AC$ and $AB$ respectively.
$BC&space;=&space;a$ (given)
Draw $OE$ and $OF$
$\therefore&space;OE&space;=&space;OF&space;=&space;r$
Since $OE&space;\bot&space;AB$ and $OF&space;\bot&space;AC$, $AEOF$ is a square.
Therefore $OE&space;=&space;OF&space;=&space;AE&space;=&space;AF&space;=&space;r.$
Let $CD&space;=&space;CF&space;=&space;x$.
$\therefore&space;\&space;BD&space;=&space;BE&space;=&space;a-x.$
(a)      $\frac{1}{2}\left(&space;{BC&space;+&space;AC-AB}&space;\right)$
$=\frac{1}{2}(a&space;+&space;x&space;+&space;r-(r&space;+&space;a-x))$
$=\frac{1}{2}&space;\&space;(a&space;+&space;x&space;+&space;r&space;-r&space;-&space;a&space;+&space;x)$
$=&space;x$
$\therefore&space;\&space;\&space;CD&space;=&space;\frac{1}{2}\left(&space;{BC&space;+&space;AC-AB}&space;\right).$
(b)   Draw $OC$. Since $O$ is incentre, $OC$ bisects $\angle&space;C$.
In right $\Delta&space;\&space;COF$, $\frac{x}{r}=\cot\theta$
$\therefore&space;\&space;x=r\cot\theta$
In right $\Delta&space;\&space;ABC$ , By Pythagoras Theorem,
$AB^2+AC^2=BC^2$
$\therefore&space;\&space;(r+a-x)^2+(r+x)^2=a^2$
$\therefore&space;\&space;(r-x+a)^2+(r+x)^2=a^2$
$\therefore&space;\&space;(r-x)^2+2a(r-x)+a^2+(r+x)^2=a^2$
$\therefore&space;\&space;(r-x)^2+2a(r-x)+(r+x)^2=0$
$\therefore&space;\&space;(r-x)^2+(r+x)^2=2a(x-r)$
$\therefore&space;\&space;r^2-2xr+x^2+r^2+2xr+x^2&space;=&space;2a(x-r)$
$\therefore&space;\&space;2(r^2+x^2)&space;=&space;2a(x-r)$
$\therefore&space;\&space;(r^2+x^2)&space;=&space;a(x-r)$
$\therefore&space;\&space;(r^2+r^2\cot^2\theta&space;)&space;=&space;a(r\cot\theta-r)$
$\therefore&space;\&space;r^2(1+\cot^2\theta&space;)&space;=&space;ar(\cot\theta-1)$
$\therefore&space;\&space;r\csc^2\theta&space;=&space;a\left&space;(&space;\frac{\cos\theta}{\sin\theta}&space;-1\right&space;)$
$\therefore&space;\&space;\frac{r}{\sin^2\theta}&space;=&space;a\left&space;(&space;\frac{\cos\theta-\sin\theta}{\sin\theta}&space;\right&space;)$
$\therefore&space;\&space;r=a(\sin\theta\cos\theta-\sin^2\theta)$
$\therefore&space;\&space;r=\frac{a}{2}(2\sin\theta\cos\theta-2\sin^2\theta)$
$\therefore&space;\&space;r=\frac{a}{2}(\sin2\theta-2\sin^2\theta)$
(c)   Since $a$ is fixed and  $\theta$ varies, $r$ is a function of $\theta$ and the rate of
change of $r$ with respect to $\theta$ is
$\frac{dr}{d\theta}=\frac{a}{2}(2\cos2\theta-4\sin\theta\cos\theta)$
$\therefore&space;\&space;\frac{dr}{d\theta}=a(\cos2\theta-\sin2\theta)$
$r$ has stationary value when  $\frac{dr}{d\theta}=0$.
$\therefore&space;\&space;a(\cos2\theta-\sin2\theta)=0$
$\therefore&space;\&space;\cos2\theta=\sin2\theta$
$\therefore&space;\&space;\tan2\theta=1$
$\therefore&space;\&space;2\theta=\frac{&space;\pi}{4}$  and $\theta=\frac{&space;\pi}{8}$.
$\frac{d^2r}{d\theta^2}=a(-2\sin2\theta-2\cos2\theta)$
$=-2a(\sin2\theta+\cos2\theta)$
When $\theta=\frac{&space;\pi}{8}$,
$\frac{d^2r}{d\theta^2}=-2a\left&space;(&space;\sin\frac{\pi&space;}{4}+\cos\frac{\pi&space;}{4}&space;\right&space;)=-2\sqrt&space;2&space;a<0$.
$r$ will be maximum value when $\theta=\frac{&space;\pi}{8}$.
$\sin\frac{\pi}{8}=\sin\frac{\frac{\pi}{4}}{2}$
$=\&space;\sqrt\frac{1-\cos\frac{\pi}{4}}{2}$
$=\&space;\sqrt\frac{1-\frac{\sqrt2}{2}}{2}$
$=\&space;\frac{{\sqrt&space;{2&space;-&space;\sqrt&space;2&space;}&space;}}{2}$
$\therefore&space;\&space;\sin^2\frac{\pi}{8}=&space;\frac{{2&space;-&space;\sqrt&space;2&space;}}{4}$
$r=\frac{a}{2}(\sin2\theta-2\sin^2\theta)$
$=\frac{a}{2}(\sin\frac{\pi}{4}-2\sin^2\frac{\pi}{8})$
$=\frac{a}{2}\left&space;(\frac{\sqrt2}{2}-2\left&space;(\frac{2-\sqrt2}{4}&space;\right&space;)&space;\right&space;)$
$=\frac{a}{2}\left&space;(\frac{\sqrt2}{2}-\frac{2-\sqrt2}{2}&space;\right&space;)$
$=\frac{a}{2}\left&space;(\sqrt2-1\right&space;)$ $\Leftarrow$