Examples on Circumcircles Incircles and Excircles Set 3

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  \(\frac{3}{4}\) .