Examples on Circumcircles Incircles and Excircles Set 2
Example - 59
Prove that \(\begin{align}\cos A + \cos B + \cos C = 1 + \frac{r}{R}\end{align}\)
Solution: We’ll use pre-determined results:
\[\begin{align}& \cos A+\cos B+\cos C=1+4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\,\,\,(\text{Section-2,Result}\,\,iv) \\ &\qquad \qquad\qquad\qquad\quad\;=1+\frac{r}{R}\,\,\,\,\,\text{ }\!\!~\!\!\text{ }(\text{Inradius-Circumradius relation}) \end{align}\]
Example - 60
Evalutate \(\begin{align}\left( {r + {r_1}} \right)\tan \left( {\frac{{B - C}}{2}} \right) + (r + {r_2})\tan \left( {\frac{{C - A}}{2}} \right) + (r + {r_3})\tan \left( {\frac{{A - B}}{2}} \right)\end{align}\)
Solution: Lets focus on the first term.
Also, using Napier’s law,
\[\tan \left( {\frac{{B - C}}{2}} \right) = \left( {\frac{{b - c}}{{b + c}}} \right)\cot \frac{A}{2}\]
Thus, the first term becomes
\[(r + {r_1})\tan \left( {\frac{{B - C}}{2}} \right) = b - c\]
Similarly, the second and third terms are (c – a) and (a – b), so that the required sum is zero.
Example - 61
Let \({h_1},\;{h_2},\;{h_3}\) be the lengths of the altitudes in \(\Delta ABC\) . Prove that
(a)\(\begin{align}\frac{1}{{{h_1}}} + \frac{1}{{{h_2}}} + \frac{1}{{{h_3}}} = \frac{1}{r}\end{align}\) (b) \({h_1}\;{h_2}\;{h_3} = \frac{{2{\Delta ^2}}}{R}\) (c) \(\begin{align}\frac{{\cos A}}{{{h_1}}} + \frac{{\cos B}}{{{h_2}}} + \frac{{\cos C}}{{{h_3}}} = \frac{1}{R}\end{align}\)
Solution: Using the fact that
\[\Delta = \frac{1}{2}a{h_1} = \frac{1}{2}b{h_2} = \frac{1}{2}c{h_3}\]
these results are easy to justfy
(a) \(\begin{align}\frac{1}{{{h_1}}} + \frac{1}{{{h_2}}} + \frac{1}{{{h_3}}} = \frac{{a + b + c}}{{2\Delta }} = \frac{s}{\Delta } = \frac{1}{r}\end{align}\)
(b) \(\begin{align}{h_1}\;{h_2}\;{h_3} = \frac{{8{\Delta ^3}}}{{abc}} = \frac{{8{\Delta ^3}}}{{4R\Delta }} = \frac{{2{\Delta ^2}}}{R}\end{align}\)
(c) \(\begin{align}\frac{{\cos A}}{{{h_1}}} + \frac{{\cos B}}{{{h_2}}} + \frac{{\cos C}}{{{h_3}}} = \frac{1}{{2\Delta }}(a\cos A + b\cos B + c\cos C)\end{align}\)
\(\begin{align}&= \frac{R}{{2\Delta }}(\sin 2A + \sin 2B + \sin 2C)\,\,\,\,\,\,\,\,\,\left( {{\text{How?}}} \right) \\ &= \frac{R}{{2\Delta }}(4\sin A\sin B\sin C) \\ &= \frac{{2R}}{\Delta } \cdot \frac{a}{{2R}} \cdot \frac{b}{{2R}} \cdot \frac{c}{{2R}} \\ &= \frac{1}{R} \end{align} \)
Example - 62
If \(A,\;{A_1},\;{A_2},\;{A_3}\) are the areas of the incircle and the three ex-circles respectively of \(\Delta ABC\) , show that
\[\frac{1}{{\sqrt A }} = \frac{1}{{\sqrt {{A_1}} }} + \frac{1}{{\sqrt {{A_2}} }} + \frac{1}{{\sqrt {{A_3}} }}\]
Solution:
\[\begin{align}&{\text{RHS}} = \frac{1}{{\sqrt \pi }}\left( {\frac{1}{{{r_1}}} + \frac{1}{{{r_2}}} + \frac{1}{{{r_3}}}} \right) \\ \,\,\,\,\,\,\,\,\,\,\, &\qquad= \frac{1}{{\sqrt \pi }} \cdot \frac{1}{r} ({\text{Example}} - 55,\;{\text{Part}} - {\text{a}}) \\ \,\,\,\,\,\,\,\,\,\,\, &\qquad= \frac{1}{{\sqrt A }}\\ \end{align} \]
Example - 63
What can we say about a \(\Delta ABC\) for which the relation
\[\frac{{a\cos A + b\cos B + c\cos C}}{{a + b + c}} = \frac{{a\sin B + b\sin C + c\sin A}}{{9R}}\]
Solution:
\[\begin{align}&{\text{LHS}} = \frac{{4R\sin A\sin B\sin C}}{{2R(\sin A + \sin B + \sin C)}} \qquad({\text{how}}?) \\ &\quad\;\;\;= \frac{{2\sin A\sin B\sin C}}{{4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}}} \\ \,\,\,\,\,\,\,\,\, &\quad\;\;\;= 4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2} = \frac{r}{R} = \frac{1}{s} \cdot \frac{\Delta }{R} \\ {\text{RHS}} &\quad\;\;\;= \frac{{a\left( {\frac{{2\Delta }}{{ca}}} \right) + b\left( {\frac{{2\Delta }}{{ab}}} \right) + c\left( {\frac{{2\Delta }}{{bc}}} \right)}}{{9R}}\qquad ({\text{How}}?) \\ \,\,\,\,\,\,\,\,\, &\quad\;\;\;= \frac{{2(ab + bc + ac)}}{{9\;abc}} \cdot \frac{\Delta }{R}\\ \end{align} \]
Equating the expressions for LHS and RHS, we obtain
\[\begin{align}&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{1}{s} = \frac{{2(ab + bc + ac)}}{{9\;abc}} \\ &\Rightarrow\quad 9\;abc = (a + b + c)(ab + bc + ca)\\ &\Rightarrow\quad a({b^2} + {c^2} - 2bc) + b({c^2} + {a^2} - 2ca) + c({a^2} + {b^2} - 2ab) = 0 \\ &\Rightarrow \quad a{(b - c)^2} + b{(c - a)^2} + c{(a - b)^2} = 0 \\\end{align} \]
This can only happen if \(a = b = c\) . Thus, the triangle must be equilateral.