ABCD-is-a-square-with-side-length-1-E-is-a-moving-point-between-B-amp-C-F-is-a-moving-point-between-C-amp-D-Find-the-maximum-radius-of-inscribed-circle-in-AEF-

Question Number 90456 by Tony Lin last updated on 23/Apr/20

A B C D i s a s q u a r e w i t h s i d e l e n g t h = 1

E i s a m o v i n g p o i n t b e t w e e n B & C

F i s a m o v i n g p o i n t b e t w e e n C & D

F i n d t h e m a x i m u m r a d i u s o f i n s c r i b e d

c i r c l e i n △ A E F

Answered by mr W last updated on 23/Apr/20

r = \frac{\sqrt{2}}{2} (1 - \tan θ) \tan (\frac{π}{8} + \frac{θ}{2})

r_{m a x} = 0.3032 a t θ = 9.3719 °

Commented by Tony Lin last updated on 24/Apr/20

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

h o w t o d o i t ?

Answered by ajfour last updated on 24/Apr/20

Commented by ajfour last updated on 24/Apr/20

∠ E A G = \frac{π}{4} - θ, A E = \sec θ

2 ϕ = \frac{π}{4} + θ \Rightarrow ϕ = \frac{π}{8} + \frac{θ}{2}

E G = A E \cos (2 ϕ)

= \sec θ \cos (\frac{π}{4} + θ)

r = E G \tan ϕ

r = \frac{1}{\sqrt{2}} (1 - \tan θ) \tan (\frac{π}{8} + \frac{θ}{2})

\tan \frac{π}{8} = \sqrt{2} - 1

l e t \tan \frac{θ}{2} = t

r = \frac{1}{\sqrt{2}} (1 - \frac{2 t}{1 - t^{2}}) [\frac{t + \sqrt{2} - 1}{1 - t (\sqrt{2} - 1)}]

s a y \sqrt{2} - 1 = a

r \sqrt{2} = \frac{(1 - t^{2} - 2 t) (t + a)}{(1 - t^{2}) (1 - a t)}

= \frac{- {t^{3} + (a + 2) t^{2} + (2 a - 1) t - a}}{{a t}^{3} - t^{2} - a t + 1}

\frac{d (r \sqrt{2})}{d t} = 0

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

+ (3 {a t}^{2} - 2 t - a) {t^{3} + (a + 2) t^{2} + (2 a - 1) t - a} = 0

\Rightarrow

{(a + 1)}^{2} t^{4} + 4 a^{2} t^{3} - 2 (a^{2} + 1) t^{2}

- 4 t + {(a - 1)}^{2} = 0

w i t h a = \sqrt{2} - 1,

a p p r o p r i a t e t = 0.08197

\Rightarrow θ_{0} \approx 9.3721 °

r = \frac{1}{\sqrt{2}} (1 - \tan θ_{0}) \tan (\frac{π}{8} + \frac{θ_{0}}{2})

\Rightarrow r_{m a x} \approx 0.3032

Commented by Tony Lin last updated on 24/Apr/20

t h a n k s s i r, I g o t i t