Question-111432

Question Number 111432 by ajfour last updated on 03/Sep/20

Commented by ajfour last updated on 03/Sep/20

F i n d p e r i m e t e r o f △ O A B .

Commented by ajfour last updated on 04/Sep/20

m r W S i r, c a n y o u p l e a s e c o n f i r m

a n s w e r t o t h i s q u e s t i o n ?

Answered by ajfour last updated on 04/Sep/20

Commented by ajfour last updated on 04/Sep/20

\cos α = \frac{1}{a + 1}

\sin β = \frac{a}{a + 1} = 1 - \cos α = \frac{b}{1 + 2 a + b}

α + β = 60 °

\Rightarrow \sin β = \frac{a}{a + 1} = \frac{\sqrt{3}}{2} (\frac{1}{a + 1}) - \frac{1}{2} \sqrt{1 - \frac{1}{{(a + 1)}^{2}}}

\Rightarrow 2 a = \sqrt{3} - \sqrt{{(a + 1)}^{2} - 1}

\Rightarrow 3 + 4 a^{2} - 4 \sqrt{3} a = a^{2} + 2 a

\Rightarrow 3 a^{2} - (4 \sqrt{3} + 2) a + 3 = 0

\Rightarrow a = \frac{2 \sqrt{3} + 1}{3} - \frac{\sqrt{{(2 \sqrt{3} + 1)}^{2} - 9}}{3}

\Rightarrow a = \frac{2 \sqrt{3} + 1 - 2 \sqrt{1 + \sqrt{3}}}{3} \approx 0.38611

1 + \frac{1}{a} = 1 + \frac{2 a + 1}{b}

\Rightarrow b = a (2 a + 1) \approx 0.68425

\dots . .

Commented by mr W last updated on 06/Feb/25

\sin β = \frac{a}{a + 1} = 1 - \cos α \neq \frac{b}{1 + 2 a + b}

t h e y a r e n o t c o l l i n e a r!

Answered by 1549442205PVT last updated on 05/Sep/20

(See the figure below)

HJ = HQ = x; EP = EG = EF = y

Denote by x the radius of the circle

(B), by y - the radius of small circle

Apply {Steward}^{'} s theorem we have :

{(1 + x)}^{2} (1 + y) + 2^{2} (x + y) - {(1 + y)}^{2} (1 + x + 2 y)

= (1 + x + 2 y) (1 + y) (x + y)

\Leftrightarrow 1 + 2 x + x^{2} + y + 2 xy + x^{2} y + 4 x + 4 y - 1

- 2 y - y^{2} - x - 2 xy - {xy}^{2} - 2 y - {4 y}^{2} - {2 y}^{3}

= x + x^{2} + 4 xy + x^{2} y + {3 xy}^{2} + {3 y}^{2} + {2 y}^{3}

\Leftrightarrow {4 y}^{3} + {8 y}^{2} + {4 xy}^{2} + 4 xy - 4 x = 0 (1)

Since NE / / HQ, by {Thalet}^{'} theorem

we have : \frac{NE}{HQ} = \frac{OE}{OH} \Leftrightarrow \frac{y}{x} = \frac{1 + y}{1 + x + y}

\Leftrightarrow x + xy = y + xy + y^{2} \Leftrightarrow x = {2 y}^{2} + y (2)

Replace (2) into (1) we get :

{4 y}^{3} + {8 y}^{2} + {4 y}^{2} ({2 y}^{2} + y) - 4 y ({2 y}^{2} + y) = 0

\Leftrightarrow {8 y}^{4} + {16 y}^{3} + {4 y}^{2} - 4 y = 0

\Leftrightarrow {2 y}^{3} + {4 y}^{2} + y - 1 = 0

\Leftrightarrow (y + 1) ({2 y}^{2} + 2 y - 1) = 0

\Leftrightarrow {2 y}^{2} + 2 y - 1 = 0; Δ^{'} = 1 + 2 = 3

\Rightarrow y = \frac{- 1 + \sqrt{3}}{2} . Replace into (2) we get

x = ({2 y}^{2} + y) = 2 . \frac{2 - \sqrt{3}}{2} + \frac{\sqrt{3} - 1}{2} = \frac{3 - \sqrt{3}}{2}

Thus, x = \frac{3 - \sqrt{3}}{2} \approx 0.6339

y = \frac{\sqrt{3} - 1}{2} \approx 0.3660 . Consequently, the

perimeter of triangle OAB equal to

4 + 2 x + 2 y = 4 + 3 - \sqrt{3} + \sqrt{3} - 1 = 6

Commented by 1549442205PVT last updated on 04/Sep/20

Commented by ajfour last updated on 04/Sep/20

q u e s t i o n i s a l r i g h t s i r .

Commented by 1549442205PVT last updated on 05/Sep/20

Thank you, Sir . I mistaked and corrected

E is vertex of isosceles triangle . But

my solution {don}^{'} t use this hypothesis