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Question Number 102836 by ajfour last updated on 11/Jul/20

Commented by ajfour last updated on 11/Jul/20

$△ A B C i s e q u i l a t e r a l w i t h s i d e a .$ $I f r e g i o n s 1, 2, 3 h a v e e q u a l$ $p e r i m e t e r s, f i n d x / a .$
	Answered by mr W last updated on 11/Jul/20

	$C D = p$ $C E = q$ $D E = \sqrt{p^{2} + q^{2} - 2 p q \cos 60 °} = \sqrt{p^{2} + q^{2} - p q}$ $B E = \sqrt{a^{2} + {(a - q)}^{2} - 2 a (a - q) \cos 60 °} = \sqrt{a^{2} + {(a - q)}^{2} - a (a - q)}$ $P_{1} = p + q + \sqrt{p^{2} + q^{2} - p q}$ $P_{2} = a - p + \sqrt{p^{2} + q^{2} - p q} + \sqrt{a^{2} + {(a - q)}^{2} - a (a - q)}$ $P_{3} = 2 a - q + \sqrt{a^{2} + {(a - q)}^{2} - a (a - q)}$ $P_{3} = P_{2} :$ $2 a - q = a - p + \sqrt{p^{2} + q^{2} - p q}$ $a + p - q = \sqrt{p^{2} + q^{2} - p q}$ $a^{2} + 2 a (p - q) - p q = 0$ $w i t h α = \frac{p}{a}, β = \frac{q}{a}$ $\Rightarrow 2 (α - β) - α β + 1 = 0 . . . (i)$ $P_{1} = P_{2} :$ $p + q = a - p + \sqrt{a^{2} + {(a - q)}^{2} - a (a - q)}$ $2 p + q - a = \sqrt{a^{2} + {(a - q)}^{2} - a (a - q)}$ $4 p^{2} + 4 p q - 4 a p - a q = 0$ $\Rightarrow 4 α^{2} + 4 α β - 4 α - β = 0 . . . (i i)$ $f r o m (i) a n d (i i) :$ $β = \frac{2 α + 1}{α + 2} = \frac{4 α (1 - α)}{4 α - 1}$ $\Rightarrow 4 α^{3} + 12 α^{2} - 6 α - 1 = 0$ $w e g e t :$ $α = 0.5504, β = 0.8237$ $\frac{x}{a} = 1 - α = 0.4496$
	Commented by ajfour last updated on 11/Jul/20

	$SUPERB! S i r, G r e a t s o l u t i o n,$ $t h a n k s a l o t .$
	Answered by 1549442205 last updated on 11/Jul/20

	Commented by 1549442205 last updated on 12/Jul/20

	$Putting BD = m, CE = n, BC = aWe need$ $find \frac{x}{a} = \frac{m}{a} . From the cosine theorem we$ $get DE = \sqrt{n^{2} + {(a - m)}^{2} - n (a - m)}$ $BE = \sqrt{a^{2} + {(a - n)}^{2} - a (a - n)} . From$ $the hypothesis p (Δ ABE) = p (Δ BDE)$ $we get AB + AE = BD + DE$ $\Leftrightarrow a + a - n = m + \sqrt{{(a - m)}^{2} + n^{2} - n (a - m)}$ $\Leftrightarrow {(2 a - n - m)}^{2} = {(a - m)}^{2} + n^{2} - n (a - m)$ $\Leftrightarrow {4 a}^{2} + n^{2} + m^{2} - 4 an - 4 am + 2 mn =$ $a^{2} - 2 am + m^{2} + n^{2} - an + mn$ $\Leftrightarrow {3 a}^{2} - 3 an - 2 am + mn = 0 (1)$ $p (Δ BDE) = p (Δ CDE) \Leftrightarrow BD + BE = CD + CE$ $\Leftrightarrow m + \sqrt{a^{2} + {(a - n)}^{2} - a (a - n)} = a - m + n$ $\Leftrightarrow a^{2} + {(a - n)}^{2} - a (a - n) = {(a + n - 2 m)}^{2}$ $\Leftrightarrow a^{2} - an + n^{2} = a^{2} + n^{2} + {4 m}^{2} + 2 an - 4 am - 4 mn$ $\Leftrightarrow {4 m}^{2} + 3 an - 4 am - 4 mn = 0 (2)$ $From (1) we get n = \frac{{3 a}^{2} - 2 am}{3 a - m} () . Replace$ $into (2) we get$ ${4 m}^{2} + (3 a - 4 m) \times \frac{{3 a}^{2} - 2 am}{3 a - m} - 4 am = 0$ $\Leftrightarrow {12 am}^{2} - {4 m}^{3} + {9 a}^{3} - {18 a}^{2} m + {8 am}^{2} - {12 a}^{2} m + {4 am}^{2} = 0$ $\Leftrightarrow {4 m}^{3} + {30 a}^{2} m - {2 am}^{2} - {9 a}^{3} = 0$ $\Leftrightarrow 4 {(\frac{m}{a})}^{3} - 24 {(\frac{m}{a})}^{2} + 30 (\frac{m}{a}) - 9 = 0$ $\Leftrightarrow \frac{x}{a} = \frac{m}{a} = 0.4495847837$ $From () we get :$ $y = \frac{n}{a} = 2 - \frac{3}{3 - \frac{m}{a}} = 0.823720452$
	Commented by ajfour last updated on 12/Jul/20

	$t h a n k s S i r, e x c e l l e n t .$
	Commented by 1549442205 last updated on 13/Jul/20

	$You are welcome sir .$
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