Examples on Circumcircles Incircles and Excircles Set 1

Example - 55

Prove the following results

(a)\(\begin{align}\frac{1}{{{r_1}}} + \frac{1}{{{r_2}}} + \frac{1}{{{r_3}}} = \frac{1}{r}\end{align}\)   (b) \({r_1} + {r_2} + {r_3} - r = 4R\)    (c)  \({r_1}{r_2} + {r_2}{r_3} + {r_3}{r_1} = {s^2}\)

Solution: (a)

\[\begin{align}&{\text{LHS}} = \frac{1}{{{r_1}}} + \frac{1}{{{r_2}}} + \frac{1}{{{r_3}}} = \frac{{s - a}}{\Delta } + \frac{{s - b}}{\Delta } + \frac{{s - c}}{\Delta }  \\  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\qquad\qquad\quad\;\;= \frac{s}{\Delta }  \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\qquad\qquad\quad\;\;= \frac{1}{r}  \\ \end{align} \]

(b)

\[\begin{align}{\text{LHS}} = {r_1} + {r_2} + {r_3} &= \frac{\Delta }{{s - a}} + \frac{\Delta }{{s - b}} + \frac{\Delta }{{s - c}} - \frac{\Delta }{s} \\&= \frac{\Delta }{{s(s - a)(s - b)(s - c)}}\left\{ \begin{gathered}s(s - b)(s - c) + s(s - c)(s - a) \\  + s(s - a)(s - b) - (s - a)(s - b)(s - c) \\ \end{gathered} \right\} \\&= \frac{{abc}}{\Delta }\text{upon simplificaiton (Verify)}\\&= 4R \end{align} \]

(c)   \[\begin{align}\\&{\text{LHS}} = {r_1}{r_2} + {r_2}{r_3} + {r_3}{r_1}\\&\qquad= \frac{{{\Delta ^2}}}{{(s - a)(s - b)}} + \frac{{{\Delta ^2}}}{{(s - b)(s - c)}} + \frac{{{\Delta ^2}}}{{(s - c)(s - a)}}  \\ &\qquad= \frac{{{\Delta ^2}}}{{(s - a)(s - b)(s - c)}}\;\;\;\;\left\{ {(s - c) + (s - a) + (s - b)} \right\}  \\  &\qquad= \frac{{{\Delta ^2}}}{{(s - a)(s - b)(s - c)}} \cdot s  \\ &\qquad= {s^2} \\ \end{align} \]

Example - 56

Prove that  \(\begin{align}\frac{1}{{{r^2}}} + \frac{1}{{r_1^2}} + \frac{1}{{r_2^2}} + \frac{1}{{r_3^2}} = \frac{{{a^2} + {b^2} + {c^2}}}{{{\Delta ^2}}}\end{align}\)

Solution:

\[\begin{align}  {\text{LHS}} = \frac{1}{{{r^2}}} + \frac{1}{{r_1^2}} + \frac{1}{{r_2^2}} + \frac{1}{{r_3^2}} &= \frac{{{s^2} + {{(s - a)}^2} + {{(s - b)}^2} + {{(s - c)}^2}}}{{{\Delta ^2}}}  \\  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= \frac{{4{s^2} - 2s(a + b + c) + ({a^2} + {b^2} + {c^2})}}{{{\Delta ^2}}}  \\ \end{align} \]

Since  \(a + b + c = 2s\)  , the expression above reduces to the RHS.

Example - 57

Prove that  \(\begin{align}\frac{{{r_1}}}{{bc}} + \frac{{{r_2}}}{{ca}} + \frac{{{r_3}}}{{ab}} = \frac{1}{r} - \frac{1}{{2R}}\end{align}\)

Solution: Let us focus on one term of the LHS.

\[\begin{align}&\frac{{{r_1}}}{{bc}} = \frac{{\left( {4R\sin \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}} \right)}}{{(2R\sin B)(2R\sin C)}}  \\  \,\,\,\,\,\,\, &\quad\;= \frac{{{{\sin }^2}\frac{A}{2}}}{{4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}}} = \frac{{{{\sin }^2}\frac{A}{2}}}{r}  \\ \end{align} \]

Thus, using similar expressions for the other terms, the LHS reduces to

\[\begin{align}&{\text{LHS}} = \frac{{{{\sin }^2}\frac{A}{2} + {{\sin }^2}\frac{B}{2} + {{\sin }^2}\frac{C}{2}}}{r} = \frac{1}{{2r}}\left[ {3 - \left( {\cos A + \cos B + \cos C} \right)} \right]  \\  \,\,\,\,\,\,\,\,\,\,\,&\quad\;\;\; = \frac{1}{{2r}}\left[ {3 - \left( {1 + 4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}} \right)} \right] \\  \,\,\,\,\,\,\,\,\,\,\, &\quad\;\;\;= \frac{1}{{2r}}\left[ {2 - \frac{r}{R}} \right]  \\  \,\,\,\,\,\,\,\,\,\, &\quad\;\;\;= \frac{1}{r} - \frac{1}{{2R}} \\ \end{align} \]

Example - 58

Prove that \(\begin{align}\Delta  = 4Rr\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}\end{align}\)

Solution: Starting with the RHS, and using  \(\begin{align}r = 4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2},\end{align}\)  we have

\[\begin{align}&{\text{RHS}} = 16{R^2}\left( {\sin \frac{A}{2}\cos \frac{A}{2}} \right)\left( {\sin \frac{B}{2}\cos \frac{B}{2}} \right)\left( {\sin \frac{C}{2}\cos \frac{C}{2}} \right)  \\  \,\,\,\,\,\,\,\,\,\,\,\, &\qquad= 2{R^2}\sin A\sin B\sin C  \\ \end{align} \]

Now, using  \(\begin{align}R = \frac{{abc}}{{4\Delta }}\end{align}\), we have

\[\begin{align}&{\text{RHS}} = 2{\left( {\frac{{abc}}{{4\Delta }}} \right)^2}\left( {\frac{{2\Delta }}{{bc}}} \right)\left( {\frac{{2\Delta }}{{ca}}} \right)\left( {\frac{{2\Delta }}{{ab}}} \right)  \\  \,\,\,\,\,\,\,\,\,\,\,\, &\qquad= \Delta  = {\text{LHS}}  \\ \end{align} \]