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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
	$BC=ai BA=c^� =hi+kj BI=rcot (β/2)i+rj BR=rcot (β/2)(((hi+kj)/( (√(h^2 +k^2 ))))) eq. of JC r_J ^� =ai+ λ{(a−((rh)/( (√(h^2 +k^2 ))))cot (β/2))i−(((rk)/( (√(h^2 +k^2 )))))j} eq. of JA r_J ^� =(hi+kj)+μ{(rcot (β/2)−h)i−kj} r_J ^� =r_J ^� a+λ(a−((rh)/( (√(h^2 +k^2 ))))cot (β/2)) =h+μ(rcot (β/2)−h) & ((λr)/( (√(h^2 +k^2 ))))=μ−1 ⇒ a+λ(a−((rh)/( (√(h^2 +k^2 ))))cot (β/2)) =h+(1+((λr)/( (√(h^2 +k^2 )))))(rcot (β/2)−h) ⇒ λ=((a−h−(rcot (β/2)−h))/((r^2 /( (√(h^2 +k^2 ))))cot (β/2)−a)) IJ=[a+{((a−h−(rcot (β/2)−h))/((r^2 /( (√(h^2 +k^2 ))))cot (β/2)−a))}{(a−((rh)/( (√(h^2 +k^2 ))))cot (β/2))−rcot (β/2)}]i +[{((a−h−(rcot (β/2)−h))/((r^2 /( (√(h^2 +k^2 ))))cot (β/2)−a))}(r/( (√(h^2 +k^2 ))))−r]j h=ccos β and k=csin β ⇒ IJ=[a+{((a−ccos β−(rcot (β/2)−ccos β))/((r^2 /c)cot (β/2)−a))}{(a−rcos βcot (β/2))−rcot (β/2)}]i +[{((a−ccos β−(rcot (β/2)−ccos β))/((r^2 /c)cot (β/2)−a))}(r/c)−r]j r=((2△)/(a+b+c))$
	$B C = a i$ $B A = \bar{c} = h i + k j$ $B I = r \cot \frac{β}{2} i + r j$ $B R = r \cot \frac{β}{2} (\frac{h i + k j}{\sqrt{h^{2} + k^{2}}})$ $e q . o f J C$ ${\bar{r}}_{J} = a i +$ $λ {(a - \frac{r h}{\sqrt{h^{2} + k^{2}}} \cot \frac{β}{2}) i - (\frac{r k}{\sqrt{h^{2} + k^{2}}}) j}$ $e q . o f J A$ ${\bar{r}}_{J} = (h i + k j) + μ {(r \cot \frac{β}{2} - h) i - k j}$ ${\bar{r}}_{J} = {\bar{r}}_{J}$ $a + λ (a - \frac{r h}{\sqrt{h^{2} + k^{2}}} \cot \frac{β}{2})$ $= h + μ (r \cot \frac{β}{2} - h)$ $& \frac{λ r}{\sqrt{h^{2} + k^{2}}} = μ - 1$ $\Rightarrow a + λ (a - \frac{r h}{\sqrt{h^{2} + k^{2}}} \cot \frac{β}{2})$ $= h + (1 + \frac{λ r}{\sqrt{h^{2} + k^{2}}}) (r \cot \frac{β}{2} - h)$ $\Rightarrow λ = \frac{a - h - (r \cot \frac{β}{2} - h)}{\frac{r^{2}}{\sqrt{h^{2} + k^{2}}} \cot \frac{β}{2} - a}$ $I J = [a + {\frac{a - h - (r \cot \frac{β}{2} - h)}{\frac{r^{2}}{\sqrt{h^{2} + k^{2}}} \cot \frac{β}{2} - a}} {(a - \frac{r h}{\sqrt{h^{2} + k^{2}}} \cot \frac{β}{2}) - r \cot \frac{β}{2}}] i$ $+ [{\frac{a - h - (r \cot \frac{β}{2} - h)}{\frac{r^{2}}{\sqrt{h^{2} + k^{2}}} \cot \frac{β}{2} - a}} \frac{r}{\sqrt{h^{2} + k^{2}}} - r] j$ $h = c \cos β a n d k = c \sin β$ $\Rightarrow$ $I J = [a + {\frac{a - c \cos β - (r \cot \frac{β}{2} - c \cos β)}{\frac{r^{2}}{c} \cot \frac{β}{2} - a}} {(a - r \cos β \cot \frac{β}{2}) - r \cot \frac{β}{2}}] i$ $+ [{\frac{a - c \cos β - (r \cot \frac{β}{2} - c \cos β)}{\frac{r^{2}}{c} \cot \frac{β}{2} - a}} \frac{r}{c} - r] j$ $r = \frac{2 △}{a + b + c}$
	Commented by mr W last updated on 04/Dec/23

	$v e r y n i c e!$
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