ABC-with-sides-a-b-c-Z-and-A-3-B-If-its-circumference-is-minimum-then-find-a-b-c-

Question Number 21973 by Joel577 last updated on 08/Oct/17

Δ A B C with sides a, b, c \in Z and ∠ A = 3 ∠ B

If its circumference is minimum, then

find a, b, c

Commented by Rasheed.Sindhi last updated on 08/Oct/17

Let ∠ B = θ \Rightarrow ∠ A = 3 θ, ∠ C = π - 4 θ

Let a, b & c are respective opposite

sides of ∠ A, ∠ B & ∠ C .

sine law : \frac{a}{\sin (3 θ)} = \frac{b}{\sin (θ)} = \frac{c}{\sin (π - 4 θ)} = k (say)

a = k \sin (3 θ), b = k \sin (θ),

c = k \sin (π - 4 θ)

a, b, c \in Z \Rightarrow \frac{a}{b}, \frac{b}{c}, \frac{a}{c} \dots \in Q

\frac{\sin (θ)}{\sin (3 θ)}, \frac{\sin (θ)}{\sin (π - 4 θ)}, \frac{\sin (3 θ)}{\sin (π - 4 θ)} etc

are rational numbers .

⋮

a = \sqrt{b^{2} + c^{2} - 2 b c \cos (3 θ)}

⋮

Answered by ajfour last updated on 09/Oct/17

\frac{\sin 3 θ}{a} = \frac{\sin θ}{b} = \frac{\sin (π - 4 θ)}{c} (∠ B = θ)

p = a + b + c

= \frac{b \sin 3 θ}{\sin θ} + b + \frac{b \sin 4 θ}{\sin θ}

p i s m i n i m u m, \Rightarrow

3 - 4 \sin^{2} θ + 1 + 4 \cos θ \cos 2 θ i s m i n .

\Rightarrow - 8 \sin θ \cos θ - 4 \sin θ \cos 2 θ

- 8 \cos θ \sin 2 θ = 0

\Rightarrow 2 \cos θ + \cos 2 θ + 4 \cos^{2} θ = 0

\Rightarrow 6 \cos^{2} θ + 2 \cos θ - 1 = 0

\cos θ = \frac{- 2 \pm 2 \sqrt{7}}{12}

\cos θ = \frac{- 1 \pm \sqrt{7}}{6}

l e t u s c h o o s e \cos θ = \frac{\sqrt{7} - 1}{6}

\sin θ = \sqrt{1 - \frac{(8 - 2 \sqrt{7})}{36}}

= \frac{\sqrt{28 + 2 \sqrt{7}}}{6}

\dots \dots . .

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