Math Question and Answers


Question and Answers Forum

All Questions Topic List
Geometry Questions
Previous in All Question Next in All Question
Previous in Geometry Next in Geometry
Question Number 54661 by ajfour last updated on 08/Feb/19

Commented by ajfour last updated on 08/Feb/19

$G i v e n a a n d b, f i n d c .$
Commented by behi83417@gmail.com last updated on 08/Feb/19

Commented by behi83417@gmail.com last updated on 09/Feb/19
$AN=(√((a+b)^2 −(b−a)^2 ))=2(√(ab)) AP=(√((c+a)^2 −(c−a)^2 ))=2(√(ac)) cosNAB=((2(√(ab)))/(a+b)),cosPAC=((2(√(ac)))/(a+c)) sinNAB=((b−a)/(b+a)),sinPAC=((c−a)/(c+a)) cosCAB=(((a+b)^2 +(a+c)^2 −(b+c)^2 )/(2(a+b)(a+c)))= =((a^2 +ab+ac−bc)/((a+b)(a+c))) CA^� B+NA^� B+PA^� C=90^• ((a^2 +ab+ac−bc)/((a+b)(a+c)))=((2(√(ac)))/((a+b)(a+c)))(b−a)+((2(√(ab)))/((a+b)(a+c)))(c−a) ⇒a^2 +ab+ac−bc=2(b−a)(√(ac))+2(c−a)(√(ab)) ⇒a^2 +ab+ac−bc=2(b−a)(√(ac))+2c(√(ab))−2a(√(ab)) ⇒[a−(b+2(√(ab)))].c−2(b−a)(√a).(√c)+a(a+b+2(√(ab)))=0 △′=a(b−a)^2 −a[a^2 −(b+2(√(ab)))^2 ]= =a[b^2 +a^2 −2ab−[a^2 −(b^2 +4ab+4b(√(ab)))]]= =2ab[b+a+2(√(ab))]=2ab((√a)+(√b))^2 ⇒(√c)=(((b−a)(√a)±((√a)+(√b)).(√(2ab)))/(a−(b+2(√(ab))))) (√c)=(((b−a)(√a)±(a(√(2b))+b(√(2a))))/(a−(b+2(√(ab)))))= =(((b−a)+(√(2b))((√b)+(√a)))/(a−(b+2(√(ab)))))(√a) ⇒c=a.[(((b−a)±(√(2b))((√b)+(√a)))/(a−(b+2(√(ab)))))]^2 . [(b−a)±(√2)(b+(√(ab)))]^2 =(b−a)^2 ±2(√(2b))((√b)−(√a))((√b)+(√a))^2 +2b((√b)+(√a))^2 ]= =(b−a)^2 +2((√a)+(√b))^2 (b±(√(2b))((√b)−(√a))= =(b−a)^2 +2((√b)+(√a))^2 . { (([((√2)+1)b−(√(2ab))])),(([(1−(√2))b+(√(2ab))])) :} [a−(b+2(√(ab)))]^2 =a^2 +b^2 +4ab+4b(√(ab))−2ab−4a(√(ab))= =(a+b)^2 +4(b−a)(√(ab)) ⇒c=a.(((b−a)^2 −2(√b)((√a)+(√b))^2 [(√(2a))−((√2)+1)(√b)])/((a+b)^2 +4(b−a)(√(ab)))) or: c=a.(((b−a)^2 +2(√b)((√a)+(√b))^2 [(√(2a))−((√2)−1)(√b)])/((a+b)^2 +4(b−a)(√(ab)))) .$
$A N = \sqrt{{(a + b)}^{2} - {(b - a)}^{2}} = 2 \sqrt{a b}$ $A P = \sqrt{{(c + a)}^{2} - {(c - a)}^{2}} = 2 \sqrt{a c}$ $c o s N A B = \frac{2 \sqrt{a b}}{a + b}, c o s P A C = \frac{2 \sqrt{a c}}{a + c}$ $s i n N A B = \frac{b - a}{b + a}, s i n P A C = \frac{c - a}{c + a}$ $c o s C A B = \frac{{(a + b)}^{2} + {(a + c)}^{2} - {(b + c)}^{2}}{2 (a + b) (a + c)} =$ $= \frac{a^{2} + a b + a c - b c}{(a + b) (a + c)}$ $C \overset{}{A B} + N \overset{}{A B} + P \overset{}{A C} = 90^{∙}$ $\frac{a^{2} + a b + a c - b c}{(a + b) (a + c)} = \frac{2 \sqrt{a c}}{(a + b) (a + c)} (b - a) + \frac{2 \sqrt{a b}}{(a + b) (a + c)} (c - a)$ $\Rightarrow a^{2} + a b + a c - b c = 2 (b - a) \sqrt{a c} + 2 (c - a) \sqrt{a b}$ $\Rightarrow a^{2} + a b + a c - b c = 2 (b - a) \sqrt{a c} + 2 c \sqrt{a b} - 2 a \sqrt{a b}$ $\Rightarrow [a - (b + 2 \sqrt{a b})] . c - 2 (b - a) \sqrt{a} . \sqrt{c} + a (a + b + 2 \sqrt{a b}) = 0$ $△^{'} = a {(b - a)}^{2} - a [a^{2} - {(b + 2 \sqrt{a b})}^{2}] =$ $= a [b^{2} + a^{2} - 2 a b - [a^{2} - (b^{2} + 4 a b + 4 b \sqrt{a b})]] =$ $= 2 a b [b + a + 2 \sqrt{a b}] = 2 a b {(\sqrt{a} + \sqrt{b})}^{2}$ $\Rightarrow \sqrt{c} = \frac{(b - a) \sqrt{a} \pm (\sqrt{a} + \sqrt{b}) . \sqrt{2 a b}}{a - (b + 2 \sqrt{a b})}$ $\sqrt{c} = \frac{(b - a) \sqrt{a} \pm (a \sqrt{2 b} + b \sqrt{2 a})}{a - (b + 2 \sqrt{a b})} =$ $= \frac{(b - a) + \sqrt{2 b} (\sqrt{b} + \sqrt{a})}{a - (b + 2 \sqrt{a b})} \sqrt{a}$ $\Rightarrow c = a . {[\frac{(b - a) \pm \sqrt{2 b} (\sqrt{b} + \sqrt{a})}{a - (b + 2 \sqrt{a b})}]}^{2} .$ ${[(b - a) \pm \sqrt{2} (b + \sqrt{a b})]}^{2} = {(b - a)}^{2} \pm 2 \sqrt{2 b} (\sqrt{b} - \sqrt{a}) {(\sqrt{b} + \sqrt{a})}^{2} + 2 b {(\sqrt{b} + \sqrt{a})}^{2}] =$ $= {(b - a)}^{2} + 2 {(\sqrt{a} + \sqrt{b})}^{2} (b \pm \sqrt{2 b} (\sqrt{b} - \sqrt{a}) =$ $= {(b - a)}^{2} + 2 {(\sqrt{b} + \sqrt{a})}^{2} . {\begin{cases} [(\sqrt{2} + 1) b - \sqrt{2 a b}] \\ [(1 - \sqrt{2}) b + \sqrt{2 a b}] \end{cases}$ ${[a - (b + 2 \sqrt{a b})]}^{2} = a^{2} + b^{2} + 4 a b + 4 b \sqrt{a b} - 2 a b - 4 a \sqrt{a b} =$ $= {(a + b)}^{2} + 4 (b - a) \sqrt{a b}$ $\Rightarrow c = a . \frac{{(b - a)}^{2} - 2 \sqrt{b} {(\sqrt{a} + \sqrt{b})}^{2} [\sqrt{2 a} - (\sqrt{2} + 1) \sqrt{b}]}{{(a + b)}^{2} + 4 (b - a) \sqrt{a b}}$ $o r :$ $c = a . \frac{{(b - a)}^{2} + 2 \sqrt{b} {(\sqrt{a} + \sqrt{b})}^{2} [\sqrt{2 a} - (\sqrt{2} - 1) \sqrt{b}]}{{(a + b)}^{2} + 4 (b - a) \sqrt{a b}} .$
	Answered by mr W last updated on 09/Feb/19

	${[\sqrt{{(a + c)}^{2} - {(c - a)}^{2}} - (b - a)]}^{2} + {[\sqrt{{(a + b)}^{2} - {(b - a)}^{2}} - (c - a)]}^{2} = {(b + c)}^{2}$ ${[2 \sqrt{a c} - (b - a)]}^{2} + {[2 \sqrt{a b} - (c - a)]}^{2} = {(b + c)}^{2}$ $4 a c - 4 (b - a) \sqrt{a c} + b^{2} - 2 a b + a^{2} + 4 a b - 4 (c - a) \sqrt{a b} - 2 (a + b) c + a^{2} - b^{2} = 0$ $\Rightarrow (a - b - 2 \sqrt{a b}) c - 2 (b - a) \sqrt{a c} + a (a + b + 2 \sqrt{a b}) = 0$ $\Rightarrow \sqrt{c} = \frac{(a - b) \sqrt{a} \pm \sqrt{2 a b (a + b + 2 \sqrt{a b})}}{b - a + 2 \sqrt{a b}}$ $(t a k e ‘ ‘ +^{″} f o r t h e u p p e r g r e e n c i r c l e$ $a n d ‘ ‘ -^{″} f o r t h e l o w e r b l u e c i r c l e)$ $\Rightarrow c = a {[\frac{a - b \pm \sqrt{2 b (a + b + 2 \sqrt{a b})}}{b - a + 2 \sqrt{a b}}]}^{2}$
	Commented by ajfour last updated on 09/Feb/19

	$w h e n a = b, {i s n}^{'} t c = 2 a t h e n, w h y n o t ?$ $T h a n k s b o t h S i r s .$
	Commented by mr W last updated on 09/Feb/19

	Commented by mr W last updated on 09/Feb/19

	Commented by mr W last updated on 09/Feb/19

	Commented by ajfour last updated on 09/Feb/19

	$G r e a t S i r, t h a n k s f o r t h e i m a g e s .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum