In-rectangle-ABCD-AB-8-BC-20-P-is-a-point-on-AD-so-that-BPC-90-If-r-1-r-2-r-3-are-the-radii-of-the-incircles-of-APB-BPC-and-CPD-find-r-1-r-2-r-3-

Question Number 20131 by NECC last updated on 22/Aug/17

I n r e c t a n g l e A B C D, A B = 8,

B C = 20 . P i s a p o i n t o n A D s o

t h a t ∠ B P C = 90 ° . I f r_{1}, r_{2}, r_{3} a r e t h e

r a d i i o f t h e i n c i r c l e s o f A P B,

B P C, a n d C P D . f i n d r_{1} + r_{2} + r_{3}

Commented by ajfour last updated on 22/Aug/17

Commented by NECC last updated on 22/Aug/17

p l e a s e h e l p

Commented by ajfour last updated on 22/Aug/17

r_{1} + r_{2} + r_{3} = r_{1} (1 + \sqrt{5} + 2) = (3 + \sqrt{5}) r_{1}

(a s s h a l l b e p r o v e d b e l o w)

r_{1} + r_{1} \cot θ = A P (f r o m f i g u r e)

O P = \frac{B C}{2} = 10; P Q = A B = 8

O Q = \sqrt{{O P}^{2} - {P Q}^{2}} = 6

B Q = A P = O B - O Q = 10 - 6 = 4

B P = \sqrt{{P Q}^{2} + {B Q}^{2}} = \sqrt{64 + 16} = 4 \sqrt{5}

C D = A B = 8

\tan 2 θ = \frac{P Q}{B Q} = \frac{8}{4} = 2

\frac{2 \tan θ}{1 - \tan^{2} θ} = 2 \Rightarrow \tan^{2} θ + \tan θ - 1 = 0

\Rightarrow \tan θ = \frac{\sqrt{5} - 1}{2}

r_{1} (1 + \cot θ) = A P = 4

r_{1} (1 + \frac{2}{\sqrt{5} - 1}) = 4 \Rightarrow r_{1} = \frac{4 (\sqrt{5} - 1)}{\sqrt{5} + 1}

\Rightarrow r_{1} = 6 - 2 \sqrt{5} .

T h e t h r e e t r i a n g l e s a r e s i m i l a r, s o

t h e i r i n r a d i i a r e i n t h e r a t i o o f

t h e i r s i d e s;

r_{1} : r_{2} : r_{3} = A P : B P : C D

= 4 : 4 \sqrt{5} : 8 = 1 : \sqrt{5} : 2

s o, r_{1} + r_{2} + r_{3} = r_{1} (1 + \sqrt{5} + 2)

= (6 - 2 \sqrt{5}) (3 + \sqrt{5})

= 2 (3 - \sqrt{5}) (3 + \sqrt{5})

= 2 (9 - 5) = 8 .

Commented by NECC last updated on 22/Aug/17

w o w \dots . {y o u}^{'} r e a m a z i n g s i r .

I t f e e l s g o o d t o b e h e r e . I^{'} l l l e a r n

a l o t f r o m y o u a l l . G r a c i a s \dots . .

Answered by Tinkutara last updated on 22/Aug/17

Commented by Tinkutara last updated on 22/Aug/17

In a right triangle where c is hypotenuse,

radius of incircle = \frac{1}{2} (a + b - c)

So r_{1} = \frac{1}{2} (x + 8 - a)

r_{2} = \frac{1}{2} (a + b - 20)

r_{3} = \frac{1}{2} (y + 8 - 20)

r_{1} + r_{2} + r_{3} = \frac{1}{2} (2 \times 8) = 8

Commented by ajfour last updated on 22/Aug/17

t h a n k s f o r t h e c o n f i r m a t i o n .

Commented by NECC last updated on 22/Aug/17

i m m o s t g r a t e f u l s i r

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