Question-169585

Question Number 169585 by Shrinava last updated on 03/May/22

Commented by mr W last updated on 03/May/22

p l e a s e c h e c k t h e q u e s t i o n! o n e {c a n}^{'} t

i m a g e h o w i t l o o k s l i k e . m a y b e

D o n B C, E o n C A a n d F o n A B .

Commented by Shrinava last updated on 03/May/22

Sorry professor, yes will be as you say

Answered by mr W last updated on 03/May/22

Commented by mr W last updated on 03/May/22

{i t}^{'} s e a s y t o s e e t h a t

Δ E F D \sim Δ A B C

\Rightarrow \frac{p}{c} = \frac{q}{a} = \frac{r}{b} = k, s a y

\Rightarrow p = k c, q = k a, r = k b

a = \frac{q}{\tan B} + \frac{r}{\sin C}

\Rightarrow a (\frac{1}{k} - \frac{1}{\tan B}) = \frac{b}{\sin C} \dots (i)

b = \frac{r}{\tan C} + \frac{p}{\sin A}

\Rightarrow b (\frac{1}{k} - \frac{1}{\tan C}) = \frac{c}{\sin A} \dots (i i)

c = \frac{p}{\tan A} + \frac{q}{\sin B}

\Rightarrow c (\frac{1}{k} - \frac{1}{\tan A}) = \frac{a}{\sin B} \dots (i i i)

(i) \times (i i) \times (i i i) :

(\frac{1}{k} - \frac{1}{\tan A}) (\frac{1}{k} - \frac{1}{\tan B}) (\frac{1}{k} - \frac{1}{\tan C}) = \frac{1}{\sin A \sin B \sin C}

(\frac{1}{k} - \frac{\cos A}{\sin A}) (\frac{1}{k} - \frac{\cos B}{\sin B}) (\frac{1}{k} - \frac{\cos C}{\sin C}) = \frac{1}{\sin A \sin B \sin C}

(\sin A - k \cos A) (\sin B - k \cos B) (\sin C - k \cos C) = k^{3}

(1 + \cos A \cos B \cos C) k^{3} - (\sin A \sin B \sin C) k^{2} + (1 + \cos A \cos B \cos C) k - \sin A \sin B \sin C = 0

(k - \frac{\sin A \sin B \sin C}{1 + \cos A \cos B \cos C}) (k^{2} + 1) = 0

\Rightarrow k = \frac{\sin A \sin B \sin C}{1 + \cos A \cos B \cos C}

\frac{p}{a} + \frac{q}{b} + \frac{r}{c} = k (\frac{c}{a} + \frac{a}{b} + \frac{b}{c}) ⩾ 3 k

m a x i m u m i f \frac{c}{a} = \frac{a}{b} = \frac{b}{c}, i . e . a = b = c,

i . e . A = B = C = \frac{π}{3} a n d

k = \frac{\sin A \sin B \sin C}{1 + \cos A \cos B \cos C} = \frac{{(\frac{\sqrt{3}}{2})}^{3}}{1 + {(\frac{1}{2})}^{3}} = \frac{\sqrt{3}}{3}

\Rightarrow \frac{p}{a} + \frac{q}{b} + \frac{r}{c} ⩾ 3 k = 3 \times \frac{\sqrt{3}}{3} = \sqrt{3} ✓

Commented by Shrinava last updated on 04/May/22

As always, the perfect solution,

thank you so much Professor

Commented by Tawa11 last updated on 04/May/22

Great sir

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