# Examples on Circumcircles Incircles and Excircles Set 5

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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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Trigonometry
grade 11 | Questions Set 1
Trigonometry