Question-201298

Question Number 201298 by ajfour last updated on 03/Dec/23

Commented by ajfour last updated on 03/Dec/23

F i n d I J i n t e r m s o f a, b, c .

Commented by mr W last updated on 05/Jan/24

I J = r \sqrt{1 - \frac{3 p^{2}}{{(4 R + r)}^{2}}}

s e e Q 202925

Answered by ajfour last updated on 03/Dec/23

B C = a i

B A = \bar{c} = h i + k j

B I = r \cot \frac{β}{2} i + r j

B R = r \cot \frac{β}{2} (\frac{h i + k j}{\sqrt{h^{2} + k^{2}}})

e q . o f J C

{\bar{r}}_{J} = a i +

λ {(a - \frac{r h}{\sqrt{h^{2} + k^{2}}} \cot \frac{β}{2}) i - (\frac{r k}{\sqrt{h^{2} + k^{2}}}) j}

e q . o f J A

{\bar{r}}_{J} = (h i + k j) + μ {(r \cot \frac{β}{2} - h) i - k j}

{\bar{r}}_{J} = {\bar{r}}_{J}

a + λ (a - \frac{r h}{\sqrt{h^{2} + k^{2}}} \cot \frac{β}{2})

= h + μ (r \cot \frac{β}{2} - h)

& \frac{λ r}{\sqrt{h^{2} + k^{2}}} = μ - 1

\Rightarrow a + λ (a - \frac{r h}{\sqrt{h^{2} + k^{2}}} \cot \frac{β}{2})

= h + (1 + \frac{λ r}{\sqrt{h^{2} + k^{2}}}) (r \cot \frac{β}{2} - h)

\Rightarrow λ = \frac{a - h - (r \cot \frac{β}{2} - h)}{\frac{r^{2}}{\sqrt{h^{2} + k^{2}}} \cot \frac{β}{2} - a}

I J = [a + {\frac{a - h - (r \cot \frac{β}{2} - h)}{\frac{r^{2}}{\sqrt{h^{2} + k^{2}}} \cot \frac{β}{2} - a}} {(a - \frac{r h}{\sqrt{h^{2} + k^{2}}} \cot \frac{β}{2}) - r \cot \frac{β}{2}}] i

+ [{\frac{a - h - (r \cot \frac{β}{2} - h)}{\frac{r^{2}}{\sqrt{h^{2} + k^{2}}} \cot \frac{β}{2} - a}} \frac{r}{\sqrt{h^{2} + k^{2}}} - r] j

h = c \cos β a n d k = c \sin β

\Rightarrow

I J = [a + {\frac{a - c \cos β - (r \cot \frac{β}{2} - c \cos β)}{\frac{r^{2}}{c} \cot \frac{β}{2} - a}} {(a - r \cos β \cot \frac{β}{2}) - r \cot \frac{β}{2}}] i

+ [{\frac{a - c \cos β - (r \cot \frac{β}{2} - c \cos β)}{\frac{r^{2}}{c} \cot \frac{β}{2} - a}} \frac{r}{c} - r] j

r = \frac{2 △}{a + b + c}

Commented by mr W last updated on 04/Dec/23

v e r y n i c e!

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