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


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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

$w o w . . . . {y o u}^{'} r e a m a z i n g s i r .$ $I t f e e l s g o o d t o b e h e r e . I^{'} l l l e a r n$ $a l o t f r o m y o u a l l . G r a c i a s . . . . .$
	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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