Examples on Circumcircles Incircles and Excircles Set 6

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Example -72

Tangents are drawn to the incircle of   \(\Delta ABC\)  which are parallel to its sides. Let  \(x,\;y,\;z\) be the lengths of the tangents opposite  \(a,\;b,\;c\)  to respectively. Find the value of  \(\begin{align}\frac{x}{a} + \frac{y}{b} + \frac{z}{c}\end{align}\)  .

Solution:

Now, observe that the incircle of  \(\Delta ABC\) is the ex-circle opposite to A for  \(\Delta ADE\) . The semiperimeter of   \(\Delta ADE\)  is

\[\begin{align}&{s_{\Delta ADE}} = \frac{1}{2}\left( {x + \frac{{bx}}{a} + \frac{{cx}}{a}} \right) = \frac{{sx}}{a}  \\   &\Rightarrow\quad   r = {s_{\Delta ADE}}\tan \frac{A}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {how?} \right)  \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\quad= \frac{{sx}}{a}\tan \frac{A}{2} \\\end{align}\]

Since r itself equals \((s - a)\tan \frac{A}{2}\) , we have

\[\begin{align}&\qquad\;\;\; s - a = \frac{{sx}}{a}  \\   &\Rightarrow\quad   \frac{x}{a} =   1 - \frac{a}{s} \\   &\Rightarrow\quad   \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 3 - \left( {\frac{{a + b + c}}{s}} \right) \\ &\qquad\qquad\qquad\qquad= 3 - 2  \\  &\qquad\qquad\qquad\qquad= 1 \\ \end{align}\]

Example -73

If the incricle of  \(\Delta ABC\) passes through its circumcenter, find the value of \(\cos A + \cos B + \cos C\)   .

Solution: We have  \(IO = r\)  . Using the result of Example - 71 Part (b), we have

\[\begin{align}&\qquad\quad\;\;\,{R^2} - 2Rr = {r^2} \\ &\Rightarrow\qquad  {R^2}\left( {1 - 8\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}} \right) = 16{R^2}{\sin ^2}\frac{A}{2}{\sin ^2}\frac{B}{2}{\sin ^2}\frac{C}{2} \\&\Rightarrow \qquad  {\left( {1 + 4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}} \right)^2} = 2 \\&\Rightarrow \qquad \cos A + \cos B + \cos C = 1 + 4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2} = \sqrt 2 \\ \end{align} \]

Example -74

Let In be the area of a n-sided regular polygon in a circle of unit radius. Let On be the area of a n-sided regular polygon circumscribing the given circle. Show that

\[{I_n} = \frac{{{O_n}}}{2}\left\{ {1 + \sqrt {1 - {{\left( {\frac{{2{I_n}}}{n}} \right)}^2}} } \right\}\]

Solution:

Also,

\[\begin{align}  \cos \theta  = \frac{{OY}}{{O{Q_1}}} \quad\Rightarrow \quad  &O{Q_1} = \sec \theta  = \sec \frac{\pi }{n}  \\  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow\quad & {\text{area}}\,(\Delta O{Q_1}{Q_2}) = \frac{1}{2}{\sec ^2}\frac{\pi }{n}\;\sin \frac{{2\pi }}{n} = \tan \frac{\pi }{n}  \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow\quad  &{O_n} = n\tan \frac{\pi }{n} \\ \end{align} \]

Thus,

\[\begin{align}&\;\; \frac{{{I_n}}}{{{O_n}}} = \frac{{\sin \frac{{2\pi }}{n}}}{{2\tan \frac{\pi }{n}}} = 2{\cos ^2}\frac{\pi }{n} = 1 + \cos \frac{{2\pi }}{n} \\ &\qquad\qquad\qquad\quad = 1 + \sqrt {1 - {{\sin }^2}\frac{{2\pi }}{n}} \\ &\qquad\qquad\qquad\quad = 1 + \sqrt {1 - {{\left( {\frac{{2{I_n}}}{n}} \right)}^2}} \\& \quad\quad\;\;  \Rightarrow \qquad  {I_n} = \frac{{{O_n}}}{2}\left\{ {1 + \sqrt {1 - {{\left( {\frac{{2{I_n}}}{n}} \right)}^2}} } \right\} \\ \end{align} \]

Example -75

In a  \(\Delta ABC\) ,  \(a,\;c,\;\angle A\) are fixed. The third side may have two possible value say  \({b_1}\) and  \({b_2}\) . It is given that  \({b_2} = 2{b_1}\) . Find the value of  \(\begin{align}\frac{c}{a}\sqrt {1 + 8{{\sin }^2}A} \end{align}\) .

Solution: Using the cosine rule, we get a quadratic in b.

\[\begin{align}&{b^2} - 2bc\cos A + \left( {{c^2} - {a^2}} \right) = 0  \\   \Rightarrow\qquad &{b_1} + {b_2} = 2c\cos A,\;{b_1}{b_2} = {c^2} - {a^2}  \end{align} \]

Using  \({b_2} = 2{b_1},\)  we have

\[{b_1} = \frac{{2c}}{3}\cos A,\;\;\;b_1^2 = \frac{{{c^2} - {a^2}}}{2}\]

Thus,

\[\begin{align}&\frac{{4{c^2}}}{9}{\cos ^2}A = \frac{{{c^2} - {a^2}}}{2}  \\   \Rightarrow\qquad   &9({c^2} - {a^2}) = 8{c^2}{\cos ^2}A  \\   \Rightarrow \qquad &9{a^2} = 9{c^2} - 8{c^2}{\cos ^2}A \\  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\;\;\;\;\;= {c^2}(1 + 8{\sin ^2}A)  \\   \Rightarrow\qquad   &\frac{c}{a}\sqrt {1 + 8{{\sin }^2}A}  = 3  \\ \end{align} \]

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