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Question Number 202868 by hardmath last updated on 04/Jan/24

Commented by mr W last updated on 04/Jan/24

$w h a t i s p ? p e r i m e t e r o r$ $s e m i - p e r i m e t e r ?$
Commented by hardmath last updated on 04/Jan/24

$My dear professor, semi - perimeter$
	Answered by mr W last updated on 05/Jan/24

	Commented by hardmath last updated on 05/Jan/24

	$My darling professor, you are perfect . . .$
	Commented by mr W last updated on 05/Jan/24

	$p = \frac{a + b + c}{2}$ ${A T}_{B} = {A T}_{C} = p - a$ ${B T}_{A} = {B T}_{C} = p - b$ ${C T}_{A} = {C T}_{B} = p - c$ $S = a r e a o f Δ A B C$ $= p r = \frac{a b c}{4 R} = \sqrt{p (p - a) (p - b) (p - c)}$ $i n Δ A B C :$ ${a A T}_{A}^{2} = (p - b) b^{2} + (p - c) c^{2} - a (p - b) (p - c)$ ${A T}_{A}^{2} = \frac{(p - b) b^{2} + (p - c) c^{2}}{a} - (p - b) (p - c)$ ${A T}_{A}^{2} = \frac{(p - a) a^{2} + (p - b) b^{2} + (p - c) c^{2} - a b c}{a} + {(p - a)}^{2}$ ${A T}_{A}^{2} = \frac{p (a^{2} + b^{2} + c^{2}) - (a^{3} + b^{3} + c^{3}) - a b c}{a} + {(p - a)}^{2}$ ${A T}_{A}^{2} = \frac{p (2 p^{2} - 2 r^{2} - 8 R r) - (2 p^{3} - 6 r^{2} p - 12 R r p) - a b c}{a} + {(p - a)}^{2}$ ${A T}_{A}^{2} = \frac{4 r^{2} p + 4 R r p - a b c}{a} + {(p - a)}^{2}$ ${A T}_{A}^{2} = \frac{4 {p r}^{2} + 4 R S - 4 R S}{a} + {(p - a)}^{2}$ ${A T}_{A}^{2} = \frac{4 {p r}^{2}}{a} + {(p - a)}^{2}$ $\Rightarrow {A T}_{A} = \sqrt{\frac{4 {p r}^{2}}{a} + {(p - a)}^{2}}$ $s i m i l a r l y$ ${B T}_{B} = \sqrt{\frac{4 {p r}^{2}}{b} + {(p - b)}^{2}}$ ${C T}_{C} = \sqrt{\frac{4 {p r}^{2}}{c} + {(p - c)}^{2}}$ $\frac{{A G}_{e}}{G_{e} T_{A}} \times \frac{{B T}_{A}}{B C} \times \frac{{C T}_{B}}{T_{B} A} = 1$ $\frac{{A G}_{e}}{G_{e} T_{A}} \times \frac{p - b}{a} \times \frac{p - c}{p - a} = 1$ $\Rightarrow \frac{{A G}_{e}}{G_{e} T_{A}} = \frac{a (p - a)}{(p - b) (p - c)}$ $\Rightarrow {A G}_{e} = \frac{a (p - a)}{a (p - a) + (p - b) (p - c)} \times {A T}_{A}$ $= \frac{4 a (p - a)}{2 (a b + b c + c a) - (a^{2} + b^{2} + c^{2})} \times {A T}_{A}$ $= \frac{4 a (p - a)}{2 (p^{2} + r^{2} + 4 R r) - (2 p^{2} - 2 r^{2} - 8 R r)} \times {A T}_{A}$ $= \frac{a (p - a)}{r^{2} + 4 R r} \times {A T}_{A}$ $\Rightarrow G_{e} T_{A} = \frac{(p - b) (p - c)}{a (p - a) + (p - b) (p - c)} \times {A T}_{A}$ $= \frac{(p - b) (p - c)}{r^{2} + 4 R r} \times {A T}_{A}$ $i n Δ O B C :$ ${a O T}_{A}^{2} = (p - b) R^{2} + (p - c) R^{2} - a (p - b) (p - c)$ $\Rightarrow {O T}_{A}^{2} = R^{2} - (p - b) (p - c)$ $i n Δ {O A T}_{A} :$ ${A T}_{A} \times {O G}_{e}^{2} = {A G}_{e} \times {O T}_{A}^{2} + G_{e} T_{A} \times {O A}^{2} - {A T}_{A} \times {A G}_{e} \times G_{e} T_{A}$ ${O G}_{e}^{2} = \frac{{A G}_{e}}{{A T}_{A}} \times {O T}_{A}^{2} + \frac{G_{e} T_{A}}{{A T}_{A}} \times {O A}^{2} - {A G}_{e} \times G_{e} T_{A}$ ${O G}_{e}^{2} = \frac{a (p - a)}{r^{2} + 4 R r} \times [R^{2} - (p - b) (p - c)] + \frac{(p - b) (p - c) R^{2}}{r^{2} + 4 R r} - \frac{a (p - a)}{r^{2} + 4 R r} \times \frac{(p - b) (p - c)}{r^{2} + 4 R r} \times {A T}_{A}^{2}$ ${O G}_{e}^{2} = \frac{[a (p - a) + (p - b) (p - c)] R^{2}}{r^{2} + 4 R r} - \frac{a (p - a) (p - b) (p - c)}{r^{2} + 4 R r} - \frac{a (p - a) (p - b) (p - c)}{{(r^{2} + 4 R r)}^{2}} \times {A T}_{A}^{2}$ ${O G}_{e}^{2} = \frac{(r^{2} + 4 R r) R^{2}}{r^{2} + 4 R r} - \frac{{a S}^{2}}{(r^{2} + 4 R r) p} - \frac{{a S}^{2}}{{(r^{2} + 4 R r)}^{2} p} \times {A T}_{A}^{2}$ ${O G}_{e}^{2} = R^{2} - \frac{{a S}^{2}}{{(r^{2} + 4 R r)}^{2} p} (r^{2} + 4 R r + {A T}_{A}^{2})$ ${O G}_{e}^{2} = R^{2} - \frac{{a p r}^{2}}{{(r^{2} + 4 R r)}^{2}} [r^{2} + 4 R r + \frac{4 {p r}^{2}}{a} + {(p - a)}^{2}]$ ${O G}_{e}^{2} = R^{2} - \frac{a p}{{(r + 4 R)}^{2}} (p^{2} + r^{2} + 4 R r + \frac{4 {p r}^{2}}{a} + a^{2} - 2 p a)$ ${O G}_{e}^{2} = R^{2} - \frac{a p}{{(r + 4 R)}^{2}} (a b + b c + c a + \frac{4 {p r}^{2}}{a} + a^{2} - a^{2} - a b - c a)$ ${O G}_{e}^{2} = R^{2} - \frac{a p}{{(r + 4 R)}^{2}} (b c + \frac{4 {p r}^{2}}{a})$ ${O G}_{e}^{2} = R^{2} - \frac{p}{{(r + 4)}^{2}} (a b c + 4 {p r}^{2})$ ${O G}_{e}^{2} = R^{2} - \frac{p}{{(r + 4 R)}^{2}} (4 R p r + 4 {p r}^{2})$ ${O G}_{e}^{2} = R^{2} - \frac{4 p^{2} r (R + r)}{{(4 R + r)}^{2}} ✓$
	Commented by hardmath last updated on 05/Jan/24

	$My darling professor, this is excellent$ $evidence, thank you so much . . .$
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