Examples on Circumcircles Incircles and Excircles Set 5

Go back to  'Trigonometry'

Example -70

Show that area \(\begin{align}(\Delta {I_1}{I_2}{I_3}) = \frac{{abc}}{{2r}}\end{align}\)

Solution: Note a very interesting fact from Fig.30 If you extend  \({I_2}A\) , it will pass through \({I_3}\) , also  \(\begin{align}\angle \,{I_1}A{I_2} = \angle \,{I_1}A{I_3} = \frac{\pi }{2},\;\;{\text{so}}\;{I_1}A\end{align}\)  , is the altitude to  \({I_2}\;{I_3}\) . Similarly  \({I_2}B\) , and  \({I_3}C\)  are altitudes to  \({I_3}{I_1}\)  and  \({I_1}{I_2}\) respectively.

\[\begin{align}& {\Delta }'=\text{area}(\Delta {{I}_{1}}{{I}_{2}}{{I}_{3}})=\frac{1}{2}\times {{I}_{1}}{{I}_{2}}\times A{{I}_{1}} \\  &\quad =\frac{1}{2}\times a\ \text{cosec}\frac{A}{2}\times {{r}_{1}}\ \text{cosec}\frac{A}{2}\text{  }\!\!~\!\!\text{ }(\text{Using the result of the last example}) \\ 
\end{align}\]

Now, substituting for  \(a\;{\text{and}}\;{r_1}\)  , we have

\[\begin{align}&\Delta ' = \frac{1}{2} \times \frac{{2R\sin A}}{{{\text{cosec}}\frac{A}{2}}} \times \frac{{4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}}}{{\sin \frac{A}{2}}}  \\  \,\,\,\,\,\, &\quad= 8{R^2}\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2} \\  \,\,\,\,\,\,&\quad= 8{R^2}\sqrt {\frac{{s(s - a)}}{{bc}}} \sqrt {\frac{{s(s - b)}}{{ca}}} \sqrt {\frac{{s(s - c)}}{{ab}}}  \\  \,\,\,\,\,\, &\quad= \frac{{8{R^2}s\Delta }}{{abc}}\\ \end{align}\]

Using  \(\begin{align}R = \frac{{abc}}{{4\Delta }}\;{\text{and}}\;r = \frac{\Delta }{s}  ,\text{ we have}\;\;  \Delta ' = \frac{{abc}}{{2r}}\end{align}\)

Example -71

In a given  \(\Delta ABC\), find the distance between

(a) circumcenter and orthocenter

(b) circumcenter and incenter

(c) incenter and orthocenter

(d) circumcenter and centroid

Solution: (a)

\[\begin{align}\Rightarrow\quad &\angle HAO = \angle A - 2\left( {\frac{\pi }{2} - \angle B} \right) = \angle A + 2\angle B - \pi 3 \\ &\qquad\quad= \angle A + 2\angle B - (\angle A + \angle B + \angle C) \\  &\qquad\quad= \angle B - \angle C\end{align}\]

Also, recall that  \(AH = 2R\;\cos A\) . Now, applying the cosine rule in  \(\Delta OAH\) , we have

\[\begin{align}  O{H^2}&= A{H^2} + O{A^2} - 2(AH)(OA)\cos (B - C) \\   &= 4{R^2}{\cos ^2}A + {R^2} - 4{R^2}\cos A\cos (B - C) \\   &= {R^2}(1 + 4\cos A(\cos A + \cos (B - C)))  \\   &= {R^2}(1 - 8\cos A\cos B\cos C)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{Verify!}}} \right)  \\   \Rightarrow\quad  &OH = R\sqrt {1 - 8\cos A\cos B\cos C} \\ \end{align}\]

(b)

We have

\[\begin{align}  \angle OAI& = \angle A - \left( {\frac{1}{2}\angle A + \frac{\pi }{2} - \angle B} \right)  \\  \,\,\,\,\,\,\,\,\,\,\,\,\,\, &= \;\left| {\;\frac{{(\angle B - \angle C)}}{2}\;} \right| \\ \end{align}\]

Using the cosine rule in  \(\Delta OAI\) , we have

\[O{I^2} = {R^2} + {r^2}\,{\text{cose}}{{\text{c}}^2}\frac{A}{2} - 2Rr\;{\text{cosec}}\frac{A}{2}\cos \left( {\frac{{C - B}}{2}} \right)\]

Using \(\begin{align}r = 4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2},\;{\text{we}}\;{\text{have}}\end{align}\)

\[O{I^2} = {R^2} + 4{R^2}{\sin ^2}\frac{B}{2}{\sin ^2}\frac{C}{2} - 8{R^2}\sin \frac{B}{2}\sin \frac{C}{2}\left( {\cos \frac{B}{2}\cos \frac{C}{2} + \sin \frac{B}{2}\sin \frac{C}{2}} \right)\]

Upon simplification, this reduces to

\[\begin{align}  OI &= R\sqrt {1 - 8\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}}\qquad\qquad\text{ (Verify!)} \\  \,\,\,\,\,\,\, &= R\sqrt {1 - \frac{{2r}}{R}}\\  \,\,\,\,\,\,\, &= \sqrt {{R^2} - 2Rr}\\ \end{align}\]

(c)

 

Also, using  \(HA = 2R\cos A\) ,  \(\begin{align}IA = r\;{\text{cosec}}\frac{A}{2} = 4R\sin \frac{B}{2}\sin \frac{C}{2}\end{align}\) and applying the cosine rule to  \(\Delta IAH\) and simplifying, we have

\[IH = \sqrt {2{r^2} - 4{R^2}\cos A\cos B\cos C}\qquad\qquad (Verify!)\]

(d)

In  \(\Delta OAD\) , we have the three sides, and we know the ratio in which G divides AD. To find OG, we apply the cosine rule, once in  \(\Delta ODG\)  and once on  \(\angle ODA:\)

\[\cos \theta  = \frac{{O{D^2} + G{D^2} - O{G^2}}}{{2OD \cdot DG}} = \frac{{O{D^2} + A{D^2} - O{A^2}}}{{2\,OD \cdot AD}}\]

From this equality, we can separate OG2:

\[\begin{align}\frac{{{R^2}{{\cos }^2}A + \frac{{A{D^2}}}{9} - O{G^2}}}{{2R\cos A \cdot \frac{{AD}}{9}}} = \frac{{{R^2}{{\cos }^2}A + \frac{{2{b^2} + 2{c^2} - {a^2}}}{4} - {R^2}}}{{2R\cos A \cdot AD}}\end{align}\]

Cross-multiplying and simplifying, we have

\[O{G^2} = {R^2} - \frac{1}{9}\left( {{a^2} + {b^2} + {c^2}} \right)\]

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