Twin-of-Q-3943-Three-circles-are-drawn-in-a-plane-in-such-a-way-that-a-closed-region-is-produced-which-is-not-included-in-any-of-the-circles-Determine-the-area-of-this-region-Circles-Radii-Cente

Question Number 4036 by Rasheed Soomro last updated on 27/Dec/15

You can't use 'macro parameter character #' in math mode

Three circles are drawn in a plane

in such a way that a closed region

is produced which is not included

in any of the circles . Determine the

area of this region .

\underline{Circles ∣ Radii ∣ Centers}

A \boldsymbol r_{1} C_{1}

B \boldsymbol r_{2} C_{2}

C \boldsymbol r_{3} C_{3}

Conditions :

\boldsymbol r_{1} + \boldsymbol r_{2} ⩾ C_{1} C_{2}

\boldsymbol r_{2} + \boldsymbol r_{3} ⩾ C_{2} C_{3}

\boldsymbol r_{3} + \boldsymbol r_{1} ⩾ C_{3} C_{1}

Commented by Yozzii last updated on 27/Dec/15

Commented by Yozzii last updated on 27/Dec/15

A r e a o f r e g i o n G H I i s r e q u i r e d .

Commented by Rasheed Soomro last updated on 27/Dec/15

Yes . Exactly!

In other cases two or all the three

circles are tangent to one another .

Commented by Yozzii last updated on 27/Dec/15

Commented by Yozzii last updated on 27/Dec/15

D i s r e g a r d t h e e x p l i c i t m e a s u r e m e n t s

o f a n g l e s g i v e n i n t h e d i a g r a m .

I o n l y r e q u i r e d t h e i m a g e t o d e l i n e a t e

t h e i m p o r t a n t a n g l e s f o r t h e c a l c u l a t i o n .

Commented by Yozzii last updated on 27/Dec/15

L e t A_{1} b e t h e a r e a o f △ A E C . B y t h e

q u e s t i o n w e k n o w t h e l e n g t h s A E, E C

& C A . S o, {H e r o n}^{'} s f o r m u l a c o u l d

b e u s e d t o f i n d A_{1} .

A_{1} = \sqrt{s (s - C A) (s - A E) (s - E C)}

s = \frac{A E + E C + C A}{2} .

L e t θ_{1} = ∠ A C E, θ_{2} = ∠ C E A a n d θ_{3} = ∠ E A C .

B y c o s i n e r u l e

{A E}^{2} = {E C}^{2} + {C A}^{2} - 2 (E C) (C A) c o s θ_{1}

\Rightarrow θ_{1} = {c o s}^{- 1} {\frac{{E C}^{2} + {C A}^{2} - {A E}^{2}}{2 (E C) (C A)}} .

I n l i k e f a s h i o n, θ_{2} c a n b e f o u n d

a n d θ_{3} = π - θ_{1} - θ_{2} .

L e t r_{1}, r_{2}, r_{3} b e t h e r a d i i o f t h e c i r c l e s

w i t h c e n t r e s C, E a n d A r e s p e c t i v e l y .

T h e a r e a A_{2} o f t h e s e c t o r M C L (f r o m d i a g r a m)

i s g i v e n b y A_{2} = \frac{1}{2} r_{1}^{2} θ_{1} . S i m i l a r l y,

t h e a r e a A_{3} o f t h e s e c t o r O E N i s A_{3} = \frac{1}{2} r_{2}^{2} θ_{2}

a n d t h e a r e a A_{4} o f t h e s e c t o r K A J i s

A_{4} = \frac{1}{2} r_{3}^{2} θ_{3} . N o w, t h e v a l u e o f

s = \underset{i = 2}{\sum^{4}} A_{i} i n c l u d e s d u p l i c a t i o n o f

p a i r w i s e o v e r l a p p i n g o f c i r c l e s . T o

f i n d t h e c o r r e c t v a l u e o f s t o s u b t r a c t

f r o m A_{1} t o g i v e t h e r e q u i r e d a r e a, w o r k

s b y p a r t s . S t a r t i n g w i t h A_{2}, A_{2} + A_{3}

c o n s i d e r s t h e o v e r l a p o f c i r c l e s C a n d E

t w o t i m e s . \Rightarrow s u i t a b l e a r e a,

s_{1} = A_{2} + A_{3} - A r e a (L T N) .

N e x t, A_{2} + A_{3} + A_{4} c o n s i d e r s t h e

o v e r l a p b e t w e e n c i r c l e s A a n d E, a n d a l s o A a n d C, t w i c e .

S o, t r u l y s = \underset{r = 2}{\sum^{4}} A_{r} - A r e a (L T N) - A r e a (O K R) - A r e a (M B J) .

M r . {J a i n}^{'} s a n s w e r t o q u e s t i o n 3877 c a n h e l p

y o u f i n d A r e a s L T N, O K R a n d M B J,

o n c e y o u h a l f e a c h r e s u l t f o r a r e a

f o u n d u s i n g h i s m e t h o d d i r e c t l y .

A f t e r w a r d s,

A r e a (R B T) = A_{1} - s .

Commented by Rasheed Soomro last updated on 27/Dec/15

G REA T!

Y OU H AVE D ONE S UCCESSFULLY!