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Question Number 18320 by Tinkutara last updated on 18/Jul/17

In a triangle A B C with fixed base B C,

the vertex A moves such that \cos B +

\cos C = 4 \sin^{2} \frac{A}{2} . If a, b and c denote

the lengths of the sides of the triangle

opposite to the angles A, B and C

respectively, then

(1) b + c = 4 a

(2) b + c = 2 a

(3) Locus of point A is an ellipse

(4) Locus of point A is a pair of straight

lines

Answered by ajfour last updated on 19/Jul/17

Commented by ajfour last updated on 19/Jul/17

Given : \cos B + \cos C = 4 \sin^{2} \frac{A}{2}

and a = constant .

\Rightarrow \cos B + \cos C = 2 (1 - \cos A)

\cos B = \frac{x}{c}, \cos C = \frac{a - x}{b},

\cos A = \frac{b^{2} + c^{2} - a^{2}}{2 bc}

x^{2} + y^{2} = c^{2}

{(a - x)}^{2} + y^{2} = b^{2}

subtracting we get

a (2 x - a) = - (b + c) (b - c)

\Rightarrow x = \frac{a}{2} - \frac{(b + c) (b - c)}{2 a}

= \frac{a^{2} - (b + c) (b - c)}{2 a} \dots . . (i)

from given condition

\frac{x}{c} + \frac{a - x}{b} = 2 (1 - \cos A)

\Rightarrow \frac{ac + x (b - c)}{bc} = 2 (1 - \frac{b^{2} + c^{2} - a^{2}}{2 bc})

replacing x from (i) :

\frac{a}{b} + (\frac{b - c}{2 abc}) [a^{2} - (b + c) (b - c)]

= 2 [\frac{2 bc - (b^{2} + c^{2}) + a^{2}}{2 bc}]

{2 a}^{2} c + a^{2} (b - c) - {(b - c)}^{2} (b + c)

= - 2 a {(b - c)}^{2} + {2 a}^{3}

\Rightarrow a^{2} (2 c + b - c - 2 a)

= {(b - c)}^{2} (b + c - 2 a)

a^{2} (b + c - 2 a) = {(b - c)}^{2} (b + c - 2 a)

\Rightarrow 2 a = b + c or a =∣ b - c ∣

If we accept : b + c = 2 a = constant,

then locus of A is an ellipse with

foci at B and C .

If we accept a =∣ b - c ∣ then A lies

on produced BC and on either side .

Commented by Tinkutara last updated on 19/Jul/17

Thanks Sir!