# Examples on Circumcircles Incircles and Excircles Set 3

Go back to  'Trigonometry'

Example - 64

In an acute-angled triangle ABC, the circle on the altitude AD as diameter cuts AB at P and AC at Q.

Prove that  \begin{align}PQ = \frac{\Delta }{R}\end{align}

Solution: Let us redraw the figure above with all the angles marked:

Apply the sine rule in  $$\angle PQD:$$

$\frac{{PQ}}{{\sin \left( {B + C} \right)}} = \frac{{DQ}}{{\cos C}} = \frac{{PD}}{{\cos B}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{How?}}} \right)$

But                                                                                                 \begin{align}\frac{{DQ}}{{\cos C}} = \frac{{PQ}}{{\cos B}} = AD,\;{\text{so that}}\end{align}

\begin{align}&PQ = AD\sin \left( {B + C} \right) = AD\sin A \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\qquad\qquad\quad\;= c\sin B\sin A\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{How?}}} \right) \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\qquad\qquad\quad\;= c\left( {\frac{b}{{2R}}} \right)\left( {\frac{a}{{2R}}} \right) \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\qquad\qquad\quad\;= \frac{{abc}}{{4{R^2}}} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\qquad\qquad\quad\;= \frac{\Delta }{R} \\ \end{align}

Example - 65

Three circles whose radii are $$a,\;b,\;c$$  touch each other externally, and the tangents at their points of contact meet in a point. Find the distance of this point from any of the points of contact.

Solution: Observe that the distance of this intersection point will be the same for all the three points of contact:

Note that O is simply the incentre of $$\Delta \,ABC$$ . The distance of O from any of the points of contact is simply the inradius r of $$\Delta \,ABC$$ . We have

\begin{align}&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,s = a + b + c \\ &\Rightarrow\quad \Delta = \sqrt {s\left( {s - (a + b)} \right)\left( {s - (b + c)} \right)\left( {s - (c + a)} \right)} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\quad\quad\;\;\;\;= \sqrt {abc(a + b + c)} \\ &\Rightarrow\quad r = \frac{\Delta }{s} = \sqrt {\frac{{abc}}{{a + b + c}}}\\ \end{align}

Example - 66

Find the

(a) minimum value of \begin{align}\frac{R}{r}\end{align}

(b) maximum value of  \begin{align} S = \frac{{a{{\cos }^2}\frac{A}{2} + b{{\cos }^2}\frac{B}{2} + c{{\cos }^2}\frac{C}{2}}}{{a + b + c}}\end{align}

Solution: (a)

\begin{align}&\frac{r}{R} = 4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2} \\ \,\,\,\,\,\, &\quad\;= 2\left\{ {\cos \frac{{A - B}}{2} - \cos \frac{{A + B}}{2}} \right\}\sin \frac{C}{2} \\ \,\,\,\,\, &\quad\;= 2\left\{ {\cos \left( {\frac{{A - B}}{2}} \right) - \cos \frac{C}{2}} \right\}\sin \frac{C}{2} \\ \,\,\,\,\,\,\,\, &\quad\;\leqslant \;\,2\left( {1 - \sin \frac{C}{2}} \right)\sin \frac{C}{2} \leqslant \frac{1}{2} \\ \end{align}

where the maximum holds when

\begin{align}&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cos \left( {\frac{{A - B}}{2}} \right) = 1\;\;{\text{and}}\;\;\sin \frac{C}{2} = \frac{1}{2} \\ &\qquad\Rightarrow\qquad A = B \qquad\Rightarrow\qquad C = \frac{\pi }{3} \\ &\qquad\Rightarrow\qquad A = B = C = \frac{\pi }{3}\\ \end{align}

Thus,  \begin{align}\frac{R}{r}\end{align}  has a minimum value of 2, when the triangle is equilateral. This could have also been proved using symmetry arguments.

(b)

\begin{align}&S = \frac{{\frac{a}{2}(1 + \cos A) + \frac{b}{2}(1 + \cos B) + \frac{c}{2}(1 + \cos C)}}{{a + b + c}} \\ \,\,\,\,\, &\quad= \frac{1}{2} + \frac{{a\cos A + b\cos B + c\cos C}}{{a + b + c}} \\ \end{align}

Using $$a = 2R\sin A,\;b = 2R\sin B,\;c = 2R\sin C,$$  this can be reduced to

\begin{align}S = \frac{1}{2} + \frac{r}{{2R}} ({\text{verify}})\end{align}

$$\leqslant \;3/4$$  (from the result of part - a)

Therefore, the maximum value of S is  $$\frac{3}{4}$$ .