Question-191305

Question Number 191305 by Mingma last updated on 22/Apr/23

Commented by mr W last updated on 22/Apr/23

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

i s a r_{m i n} .

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

Commented by mr W last updated on 23/Apr/23

O A = a = 12

O B = b = 5

s a y P (p, r), Q (r, q)

p = r + 2 r \cos θ

q = r + 2 r \sin θ

M = m i d p o i n t o f P Q

M (\frac{p + r}{2}, \frac{q + r}{2}) = [(1 + \cos θ) r, (1 + \sin θ) r]

Q M = \sqrt{3} r

x_{R} = (1 + \cos θ + \sqrt{3} \sin θ) r

y_{R} = (1 + \sin θ + \sqrt{3} \cos θ) r

e q n . o f A B :

\frac{x}{a} + \frac{y}{b} = 1 \Rightarrow b x + a y - a b = 0

d i s t a n c e f r o m R t o A B i s r :

\frac{∣ b (1 + \cos θ + \sqrt{3} \sin θ) r + a (1 + \sin θ + \sqrt{3} \cos θ) r - a b ∣}{\sqrt{a^{2} + b^{2}}} = r

r = \frac{a b}{a + b + \sqrt{a^{2} + b^{2}} + (a + \sqrt{3} b) \sin θ + (\sqrt{3} a + b) \cos θ}

r = \frac{a b}{a + b + \sqrt{a^{2} + b^{2}} + 2 \sqrt{a^{2} + b^{2} + \sqrt{3} a b} \sin (θ + \tan^{- 1} \frac{\sqrt{3} a + b}{a + \sqrt{3} b})}

a t θ + \tan^{- 1} \frac{\sqrt{3} a + b}{a + \sqrt{3} b} = \frac{π}{2}, i . e . θ = \frac{π}{2} - \tan^{- 1} \frac{\sqrt{3} a + b}{a + \sqrt{3} b} :

r_{m i n} = \frac{a b}{a + b + \sqrt{a^{2} + b^{2}} + 2 \sqrt{a^{2} + b^{2} + \sqrt{3} a b}}

w i t h a = 12, b = 5 :

r_{m i n} = \frac{30}{15 + \sqrt{169 + 60 \sqrt{3}}} \approx 0.952

Commented by Mingma last updated on 23/Apr/23

Great Work, sir!

Commented by mr W last updated on 23/Apr/23

w i t h r_{m i n} :

Commented by mr W last updated on 23/Apr/23

Commented by mr W last updated on 23/Apr/23

e x a m p l e s f o r o t h e r p o s s i b i l i t i e s :

Commented by mr W last updated on 23/Apr/23

Commented by mr W last updated on 23/Apr/23

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