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Question Number 16595 by ajfour last updated on 24/Jun/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 25/Jun/17

$y o u a r e w e l l c o m e d e a r m r A j f o u r . p l e a s e$ $p o s t y o u r a n s w e r . i a m w a i t i n g f o r i t .$
Commented by ajfour last updated on 24/Jun/17

$A point M lies inside an ∠ AOB;$ $∠ MOA = α, ∠ MOB = β, OM = a$ $(α + β < π) . Find radius of a circle$ $passing through M and cutting off,$ $on sides OA and OB of the given$ $angle, chords of length 2 a .$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 26/Jun/17

$o C = x, o D = y,^{'} P^{'} c e n t e r o f c i r c l e . α + β = φ$ $C, i s t h e i n t e r s e c t o f o B w i t h c i r c l e .$ $D, i s t h e i n t e r s e c t o f o A w i t h c i r c l e .$ $N, i s t h e i n t e r s e c t o f o M w i t h c i r c l e$ $w h e n c o n t i n u e d .$ ${o p}^{2} - R^{2} = o D . o A = y (2 a + y)$ ${o p}^{2} - R^{2} = o C . o . B = x (2 a + x)$ $\Rightarrow y (2 a + y) = x (2 a + x) \Rightarrow x^{2} - y^{2} + 2 a (x - y) = 0$ $\Rightarrow (x - y) (x + y + 2 a) = 0 \Rightarrow x = y = o C = o D .$ ${C D}^{2} = 2 x^{2} - 2 x^{2} c o s φ = 4 x^{2} {s i n}^{2} \frac{φ}{2} \Rightarrow$ $C D = 2 x . s i n \frac{φ}{2}$ ${A B}^{2} = {(x + 2 a)}^{2} + {(y + 2 a)}^{2} - 2 (x + 2 a) (y + 2 a) c o s φ u \Rightarrow$ $A B = 2 (x + 2 a) s i n \frac{φ}{2}$ ${B D}^{2} = x^{2} + {(x + 2 a)}^{2} - 2 x (x + 2 a) c o s φ =$ $= 2 x^{2} + 4 a^{2} + 4 a x - 2 x^{2} c o s φ - 4 a x c o s φ =$ $= 2 (x^{2} - 2 a x) (1 - c o s φ) + 4 a^{2} =$ $= 4 (x^{2} + 2 a x) {s i n}^{2} \frac{φ}{2} + 4 a^{2}$ $\Rightarrow B D = 2 \sqrt{a^{2} + (x^{2} + 2 a x) {s i n}^{2} \frac{φ}{2}}$ $C A = 2 \sqrt{a^{2} + (x^{2} + 2 a x) {s i n}^{2} \frac{φ}{2}}$ ${D M}^{2} = a^{2} + x^{2} - 2 a x c o s α$ ${C M}^{2} = a^{2} + x^{2} - 2 a x c o s β$ ${A M}^{2} = a^{2} + {(2 a + x)}^{2} - 2 x (2 a + x) c o s α$ $(x + 2 a) ({D M}^{2} + 2 a x) = 2 a . a^{2} + x . {A M}^{2}$ $(x + 2 a) (a^{2} + x^{2} - 2 a x c o s α + 2 a x) = 2 a^{3} + x [a^{2} + {(2 a + x)}^{2} - 2 x (2 a + x) c o s α] \Rightarrow$ ${x a}^{2} + x^{3} - 2 {a x}^{2} c o s α + 2 {a x}^{2} + 2 a^{3} + 2 {a x}^{2} - 4 a^{2} x c o s α + 4 a^{2} x =$ $= 2 a^{3} + {x a}^{2} + x {(2 a + x)}^{2} - 2 x^{2} (2 a + x) c o s α$ $\Rightarrow - 2 {a x}^{2} c o s α - 2 x^{3} c o s α + 4 a^{2} x c o s α = 0$ $\Rightarrow x^{2} + a x - 2 a^{2} = 0 \Rightarrow x = \frac{- a \pm \sqrt{a^{2} + 8 a^{2}}}{2}$ $\Rightarrow x = y = \frac{- a + 3 a}{2} = a$ $C D = 2 a s i n \frac{φ}{2}, A B = 6 a s i n \frac{φ}{2},$ $B D = C A = 2 a \sqrt{1 + 3 {s i n}^{2} \frac{φ}{2}}$ $D M = 2 a s i n \frac{α}{2}, C M = 2 a s i n \frac{β}{2}$ $a^{2} = a^{2} + {D M}^{2} - 2 a . D M . c o s (o D M)$ $\Rightarrow c o s (o D M) = \frac{D M}{2 a} = \frac{2 a s i n \frac{α}{2}}{2 a} = s i n \frac{α}{2}$ $\Rightarrow o D M = \frac{π}{2} - \frac{α}{2}$ ${M B}^{2} = {M D}^{2} + {B D}^{2} - 2 M D . B D . c o s (M D B)$ ${M B}^{2} = a^{2} + 9 a^{2} - 6 a^{2} c o s β = 2 a^{2} (5 - 3 c o s β)$ $10 a^{2} - 6 a^{2} c o s β = 4 a^{2} {s i n}^{2} \frac{α}{2} + 4 a^{2} (1 + 3 {s i n}^{2} \frac{φ}{2}) - 8 a^{2} s i n \frac{α}{2} \sqrt{1 + 3 {s i n}^{2} \frac{φ}{2}} . c o s (M D B)$ $2 a^{2} - 6 a^{2} c o s β + 2 a^{2} c o s α - 6 a^{2} c o s (α + β) = - 8 a^{2} s i n \frac{α}{2} \sqrt{1 + 3 {s i n}^{2} \frac{α + β}{2}} . c o s (M D B)$ $\overset{}{\Rightarrow} c o s (M D B) = \sqrt{2} . \frac{3 c o s (α + β) + 3 c o s β - c o s α - 1}{2 s i n \frac{α}{2} . \sqrt{5 - 3 c o s (α + β)}}$
Commented by ajfour last updated on 25/Jun/17

$huge will and effort Sir, should$ $i mention the answer ?$
	Answered by ajfour last updated on 25/Jun/17

	$R = a \sqrt{1 + 4 \frac{\cos^{2} (\frac{α - β}{2}) \sin^{2} (\frac{α + β}{2})}{\cos^{4} (\frac{α + β}{2})}}$
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