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Question Number 64119 by ajfour last updated on 13/Jul/19

Commented by ajfour last updated on 13/Jul/19

$I n t e r m s o f t h e s i d e s o f △ A B C, f i n d$ $t h e r a d i u s o f t h e c e n t r a l c i r c l e .$
	Answered by ajfour last updated on 14/Jul/19

	$p : r a d i u s o f c i r c l e w i t h c e n t e r A$ $q : c e n t e r B;$ $r : c e n t e r C$ $a n d R : r a d i u s o f c e n t r a l c i r c l e$ $q + r = a$ $r + p = b$ $p + q = c$ $p + q + r = s$ $p = s - a, q = s - b, r = s - c$ $l e t c e n t e r o f s m a l l c i r c l e b e (0, k_{A})$ ${(x + q)}^{2} + y^{2} = {(q + R)}^{2}$ ${(x - r)}^{2} + y^{2} = {(r + R)}^{2}$ $\Rightarrow (2 x + q - r) (q + r) = (q - r) (q + r + 2 R)$ $\Rightarrow 2 x = \frac{(q - r) (q + r + 2 R)}{q + r} - (q - r)$ $x = \frac{(q - r) R}{q + r}$ $y^{2} = {(q + R)}^{2} - {(q + \frac{(q - r) R}{q + r})}^{2}$ $y^{2} = [q + R + q + \frac{(q - r) R}{q + r}] [R - \frac{(q - r) R}{q + r}]$ $y^{2} = \frac{4 q r R}{q + r} [1 + \frac{R}{q + r}]$ $k_{A} = \sqrt{\frac{4 q r R}{q + r} (1 + \frac{R}{q + r})}$ $\Rightarrow k_{A} = \frac{2}{a} \sqrt{q r R (a + R)}$ $l e t A r e a o f △ A B C b e △ .$ $△ = \frac{{a k}_{A} + {b k}_{B} + {c k}_{C}}{2}$ $\Rightarrow \sqrt{q r R (a + R)} + \sqrt{r p R (b + R)} +$ $+ \sqrt{p q R (c + R)} = △$ $. . . .$
	Answered by ajfour last updated on 15/Jul/19

	$l e t c e n t e r o f c e n t r a l c i r c l e b e$ $(h, k) a n d r a d i u s ρ .$ $l e t B b e o r i g i n .$ $h^{2} + k^{2} = {(q + ρ)}^{2}$ ${(h - q - r)}^{2} + k^{2} = {(r + ρ)}^{2}$ $\Rightarrow (q + r) [2 h - (q + r)] = (q - r) (q + r + 2 ρ)$ $2 h = \frac{(q - r) (2 ρ + q + r)}{q + r} + q + r$ $h = \frac{ρ (q - r)}{q + r} + q$ $k^{2} = {(q + ρ)}^{2} - {[\frac{ρ (q - r)}{q + r} + q]}^{2}$ $k^{2} = ρ^{2} + 2 q ρ - \frac{ρ^{2} {(q - r)}^{2}}{{(q + r)}^{2}} - \frac{2 q ρ (q - r)}{q + r}$ $= \frac{4 q r ρ^{2}}{{(q + r)}^{2}} + \frac{4 q r ρ}{(q + r)} = \frac{4 q r ρ [ρ + (q + r)]}{{(q + r)}^{2}}$ $l e t A (x_{A}, y_{A})$ $x_{A} = (p + q) \cos B$ $y_{B} = (p + q) \sin B$ ${[h - (p + q) \cos B]}^{2} + {[k - (p + q) \sin B]}^{2} = {(p + ρ)}^{2}$ $\Rightarrow {(q + ρ)}^{2} + {(p + q)}^{2} - {(ρ + p)}^{2}$ $= 2 (p + q) [h \cos B + k \sin B]$ $\Rightarrow {[\frac{q^{2} + q ρ + p q - p ρ}{p + q} - h \cos B]}^{2}$ $= k^{2} \sin^{2} B$ $\Rightarrow {[\frac{q^{2} + q ρ + p q - p ρ}{p + q} - (\frac{ρ (q - r)}{q + r} + q) \cos B]}^{2}$ $= \frac{4 q r ρ [ρ + (q + r)]}{{(q + r)}^{2}} \sin^{2} B$ $\Rightarrow {[\frac{ρ (q - p)}{c} - \frac{ρ (q - r) \cos B}{a} + q (1 - \cos B)]}^{2}$ $= \frac{4 q r ρ^{2}}{a^{2}} \sin^{2} B + \frac{4 q r ρ}{a} \sin^{2} B$ $\Rightarrow q u a d r a t i c i n ρ;$ $o n e i s r a d i u s o f c i r c u m c i r c l e$ $o f t h e t h r e e c i r c l e s, a n d t h e o t h e r$ $t h e r a d i u s o f r e q u i r e d c e n t r a l o n e .$ $\Rightarrow {[(A - B) ρ + C]}^{2} = D ρ^{2} + E ρ$ $[{(A - B)}^{2} - D] ρ^{2} + [2 C (A - B) - E] ρ + C^{2} = 0$ $ρ = \frac{- 2 C (A - B) + E - \sqrt{- 4 C (A - B) + E^{2} + 4 C^{2} D}}{2 [{(A - B)}^{2} - D]}$ $ρ = \frac{- 2 q (1 - \cos B) [\frac{q - p}{c} - \frac{(q - r)}{a} \cos B] + \frac{4 q r}{a} \sin^{2} B - \sqrt{- 4 q (1 - \cos B) [\frac{q - p}{c} - \frac{(q - r)}{a} \cos B] + \frac{16 q^{2} r^{2}}{a^{2}} \sin^{4} B + 4 q^{2} {(1 - \cos B)}^{2} (\frac{4 q r}{a^{2}} \sin^{2} B)}}{2 {[\frac{q - p}{c} - \frac{(q - r)}{a} \cos B]}^{2} - \frac{8 q r}{a^{2}} \sin^{2} B}$ $w i t h \cos B = \frac{c^{2} + a^{2} - b^{2}}{2 a c}$ $a n d p = s - a, q = s - b, r = s - c$ $◼$
	Commented by ajfour last updated on 15/Jul/19

	$i f b = c = a t h e n B = 60 ° a n d$ $p = q = r = \frac{a}{2}$ $\Rightarrow ρ = \frac{\frac{3 a}{4} - \sqrt{\frac{9 a^{2}}{16} + \frac{3 a^{2}}{16}}}{- \frac{3}{2}}$ $\Rightarrow ρ = \frac{a}{\sqrt{3}} - \frac{a}{2} (c o r r e c t i n d e e d) .$ $ρ_{o u t e r} = \frac{a}{\sqrt{3}} + \frac{a}{2} .$
	Commented by mr W last updated on 15/Jul/19

	$i n t h i s s p e c i a l c a s e :$ $s = \frac{a}{2 (2 \sqrt{3} \pm 3)} = \frac{(2 \sqrt{3} \mp 3) a}{6}$ $\pm o n e f o r t h e s m a l l i n s c r i b e d, a n d o n e$ $f o r t h e b i g c i r c u m c i r c l e .$ $i . e . y o u r r e s u l t i s c o r r e c t f o r t h e$ $s m a l l c i r c l e .$
	Commented by ajfour last updated on 15/Jul/19

	Commented by ajfour last updated on 15/Jul/19

	$T h a n k s s i r, i t t u r n e d r i g h t, m y$ $f a i t h i n m y s e c o n d s o l u t i o n;$ $f o r i t w a s j u s t a q u a d r a t i c .$ $c a n y o u t r y c o m p r e s s i n g m y$ $f o u n d r e s u l t, S i r ?$
	Commented by mr W last updated on 16/Jul/19

	$t h e f i n a l r e s u l t i s$ $ρ_{i} = \frac{1}{2 \sqrt{\frac{1}{p q} + \frac{1}{q r} + \frac{1}{r p}} + (\frac{1}{p} + \frac{1}{q} + \frac{1}{r})}$ $ρ_{o} = \frac{1}{2 \sqrt{\frac{1}{p q} + \frac{1}{q r} + \frac{1}{r p}} - (\frac{1}{p} + \frac{1}{q} + \frac{1}{r})}$
	Answered by mr W last updated on 15/Jul/19
	$a=q+r b=r+p c=p+q s=a+b+c ⇒p=s−a ⇒q=s−b ⇒r=s−c center of inscribed circle M MA=p+R MB=q+R MC=r+R cos θ_A =((MB^2 +MC^2 −a^2 )/(2MB×MC))=(((q+R)^2 +(r+R)^2 −a^2 )/(2(q+R)(r+R))) cos θ_B =(((r+R)^2 +(p+R)^2 −b^2 )/(2(r+R)(p+R))) cos θ_C =(((p+R)^2 +(q+R)^2 −c^2 )/(2(p+R)(q+R))) θ_A +θ_B +θ_C =2π cos (θ_A +θ_B )=cos (2π−θ_C ) (((q+R)^2 +(r+R)^2 −a^2 )/(2(q+R)(r+R)))×(((r+R)^2 +(p+R)^2 −b^2 )/(2(r+R)(p+R)))−((√({[2(q+R)(r+R)]^2 −[(q+R)^2 +(r+R)^2 −a^2 ]^2 }{[2(r+R)(p+R)]^2 −[(r+R)^2 +(p+R)^2 −b^2 ]^2 }))/(2(q+R)(r+R)2(r+R)(p+R)))=(((p+R)^2 +(q+R)^2 −c^2 )/(2(p+R)(q+R))) (({2R^2 +2(q+r)R+q^2 +r^2 −a^2 }{2R^2 +2(r+p)R+r^2 +p^2 −b^2 }−(√({[2(q+R)(r+R)]^2 −[(q+R)^2 +(r+R)^2 −a^2 ]^2 }{[2(r+R)(p+R)]^2 −[(r+R)^2 +(p+R)^2 −b^2 ]^2 })))/(2(r+R)^2 ))=2R^2 +2(p+q)R+p^2 +q^2 −c^2 {2R^2 +2(q+r)R+q^2 +r^2 −a^2 }{2R^2 +2(r+p)R+r^2 +p^2 −b^2 }−(√({(q+r+2R)^2 −a^2 }{a^2 −(q−r)^2 }{(r+p+2R)^2 −b^2 }{b^2 −(r−p)^2 }))=2(r+R)^2 {2R^2 +2(p+q)R+p^2 +q^2 −c^2 } {2R^2 +2(q+r)R+q^2 +r^2 −a^2 }{2R^2 +2(r+p)R+r^2 +p^2 −b^2 }−(√((q+r+2R+a)(q+r+2R−a)4qr(r+p+2R+b)(r+p+2R−b)4rp))=2(r+R)^2 {2R^2 +2(p+q)R+p^2 +q^2 −c^2 } {2R^2 +2(q+r)R−2qr}{2R^2 +2(r+p)R−2rp}−16Rr(√(pq(R+a)(R+b)))=2(r+R)^2 {2R^2 +2(p+q)R−2pq} {R^2 +(q+r)R−qr}{R^2 +(r+p)R−rp}−4Rr(√(pq(R+a)(R+b)))=(R^2 +2rR+r^2 ){R^2 +(p+q)R−pq} R^4 +(q+r)R^3 −qrR^2 +(r+p)R^3 +(q+r)(r+p)R^2 −qr(r+p)R−rpR^2 −rp(q+r)R+pqr^2 −4Rr(√(pq(R+a)(R+b)))=R^4 +2rR^3 +r^2 R^2 +(p+q)R^3 +2r(p+q)R^2 +r^2 (p+q)R−pqR^2 −2pqrR−pqr^2 (pq−qr−rp)R^2 −(p+q)r^2 R+pqr^2 =2Rr(√(pq(R+q+r)(R+r+p))) .....$
	$a = q + r$ $b = r + p$ $c = p + q$ $s = a + b + c$ $\Rightarrow p = s - a$ $\Rightarrow q = s - b$ $\Rightarrow r = s - c$ $c e n t e r o f i n s c r i b e d c i r c l e M$ $M A = p + R$ $M B = q + R$ $M C = r + R$ $\cos θ_{A} = \frac{{M B}^{2} + {M C}^{2} - a^{2}}{2 M B \times M C} = \frac{{(q + R)}^{2} + {(r + R)}^{2} - a^{2}}{2 (q + R) (r + R)}$ $\cos θ_{B} = \frac{{(r + R)}^{2} + {(p + R)}^{2} - b^{2}}{2 (r + R) (p + R)}$ $\cos θ_{C} = \frac{{(p + R)}^{2} + {(q + R)}^{2} - c^{2}}{2 (p + R) (q + R)}$ $θ_{A} + θ_{B} + θ_{C} = 2 π$ $\cos (θ_{A} + θ_{B}) = \cos (2 π - θ_{C})$ $\frac{{(q + R)}^{2} + {(r + R)}^{2} - a^{2}}{2 (q + R) (r + R)} \times \frac{{(r + R)}^{2} + {(p + R)}^{2} - b^{2}}{2 (r + R) (p + R)} - \frac{\sqrt{{{[2 (q + R) (r + R)]}^{2} - {[{(q + R)}^{2} + {(r + R)}^{2} - a^{2}]}^{2}} {{[2 (r + R) (p + R)]}^{2} - {[{(r + R)}^{2} + {(p + R)}^{2} - b^{2}]}^{2}}}}{2 (q + R) (r + R) 2 (r + R) (p + R)} = \frac{{(p + R)}^{2} + {(q + R)}^{2} - c^{2}}{2 (p + R) (q + R)}$ $\frac{{2 R^{2} + 2 (q + r) R + q^{2} + r^{2} - a^{2}} {2 R^{2} + 2 (r + p) R + r^{2} + p^{2} - b^{2}} - \sqrt{{{[2 (q + R) (r + R)]}^{2} - {[{(q + R)}^{2} + {(r + R)}^{2} - a^{2}]}^{2}} {{[2 (r + R) (p + R)]}^{2} - {[{(r + R)}^{2} + {(p + R)}^{2} - b^{2}]}^{2}}}}{2 {(r + R)}^{2}} = 2 R^{2} + 2 (p + q) R + p^{2} + q^{2} - c^{2}$ ${2 R^{2} + 2 (q + r) R + q^{2} + r^{2} - a^{2}} {2 R^{2} + 2 (r + p) R + r^{2} + p^{2} - b^{2}} - \sqrt{{{(q + r + 2 R)}^{2} - a^{2}} {a^{2} - {(q - r)}^{2}} {{(r + p + 2 R)}^{2} - b^{2}} {b^{2} - {(r - p)}^{2}}} = 2 {(r + R)}^{2} {2 R^{2} + 2 (p + q) R + p^{2} + q^{2} - c^{2}}$ ${2 R^{2} + 2 (q + r) R + q^{2} + r^{2} - a^{2}} {2 R^{2} + 2 (r + p) R + r^{2} + p^{2} - b^{2}} - \sqrt{(q + r + 2 R + a) (q + r + 2 R - a) 4 q r (r + p + 2 R + b) (r + p + 2 R - b) 4 r p} = 2 {(r + R)}^{2} {2 R^{2} + 2 (p + q) R + p^{2} + q^{2} - c^{2}}$ ${2 R^{2} + 2 (q + r) R - 2 q r} {2 R^{2} + 2 (r + p) R - 2 r p} - 16 R r \sqrt{p q (R + a) (R + b)} = 2 {(r + R)}^{2} {2 R^{2} + 2 (p + q) R - 2 p q}$ ${R^{2} + (q + r) R - q r} {R^{2} + (r + p) R - r p} - 4 R r \sqrt{p q (R + a) (R + b)} = (R^{2} + 2 r R + r^{2}) {R^{2} + (p + q) R - p q}$ $R^{4} + (q + r) R^{3} - {q r R}^{2} + (r + p) R^{3} + (q + r) (r + p) R^{2} - q r (r + p) R - {r p R}^{2} - r p (q + r) R + {p q r}^{2} - 4 R r \sqrt{p q (R + a) (R + b)} = R^{4} + 2 {r R}^{3} + r^{2} R^{2} + (p + q) R^{3} + 2 r (p + q) R^{2} + r^{2} (p + q) R - {p q R}^{2} - 2 p q r R - {p q r}^{2}$ $(p q - q r - r p) R^{2} - (p + q) r^{2} R + {p q r}^{2} = 2 R r \sqrt{p q (R + q + r) (R + r + p)}$ $. . . . .$
	Commented by ajfour last updated on 15/Jul/19

	$T h i s i s i n j u s t i c e S i r .$ $Y o u s h o u l d n o t e f f o r t l i m i t l e s s . .$
	Commented by mr W last updated on 14/Jul/19

	$t h i s s e e m s t o b e a t o u g h w a y t o o b t a i n$ $t h e r a d i u s . i t h i n k y o u r f i r s t m e t h o d$ $w i t h u s e o f Δ s h o u l d b e a b l e t o r e a c h$ $t h e g o a l .$
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