In a equilateral triangle ABC whose side is a, the points M and N are taken on the side BC, such that the triangles ABM, AMN and ANC have the same perimeter. Calculate the distances from vertex A to points M and N. (solve in detail.)


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Question Number 68664 by Maclaurin Stickker last updated on 14/Sep/19

$In a equilateral triangle A B C whose$ $side is a, the points M and N are taken$ $on the side B C, such that the triangles$ $A B M, A M N and A N C have the same$ $perimeter . Calculate the distances from$ $vertex A to points M and N .$ $(solve in detail .)$
	Answered by mr W last updated on 14/Sep/19

	Commented by mr W last updated on 14/Sep/19

	$l e t B M = N C = x$ $\Rightarrow M N = a - 2 x$ $A M = A N = \sqrt{a^{2} + x^{2} - 2 a x \cos 60 °} = \sqrt{a^{2} + x^{2} - a x}$ $a + x + A M = 2 A M + M N$ $a + x = A M + a - 2 x$ $\Rightarrow A M = 3 x$ $\sqrt{a^{2} + x^{2} - a x} = 3 x$ $a^{2} + x^{2} - a x = 9 x^{2}$ $8 x^{2} + a x - a^{2} = 0$ $x = \frac{(\sqrt{33} - 1) a}{16} \approx 0.2965 a$ $\Rightarrow A M = A N = 3 x = \frac{3 (\sqrt{33} - 1) a}{16} \approx 0.8796 a$
	Commented by Rasheed.Sindhi last updated on 15/Sep/19

	$Sir, why have you assumed B M = N C ?$
	Commented by mr W last updated on 15/Sep/19

	$t h e a s s u m p t i o n i s b a s e d o n s y m m e t r y .$ $c e r t a i n l y i t c a n a l s o b e p r o v e d t h a t$ $B M m u s t b e e q u a l t o N C .$
	Commented by mr W last updated on 15/Sep/19

	$l e t B M = x, N C = y \neq x$ $a + x + \sqrt{x^{2} + a^{2} - a x} = a - x - y + \sqrt{x^{2} + a^{2} - a x} + \sqrt{y^{2} + a^{2} - a y}$ $\Rightarrow \sqrt{y^{2} + a^{2} - a y} = 2 x + y$ $\Rightarrow a^{2} - a y = 4 x^{2} + 4 x y . . . (i)$ $a + x + \sqrt{x^{2} + a^{2} - a x} = a + y + \sqrt{y^{2} + a^{2} - a y}$ $\Rightarrow x + \sqrt{x^{2} + a^{2} - a x} = y + \sqrt{y^{2} + a^{2} - a y}$ $\Rightarrow x + \sqrt{x^{2} + a^{2} - a x} = y + 2 x + y$ $\Rightarrow \sqrt{x^{2} + a^{2} - a x} = 2 y + x$ $\Rightarrow a^{2} - a x = 4 y^{2} + 4 x y . . . (i i)$ $(i) - (i i) :$ $a (x - y) = 4 (x^{2} - y^{2})$ $\Rightarrow a (x - y) = 4 (x - y) (x + y)$ $\Rightarrow x - y = 0 o r a = 4 (x + y)$ $\Rightarrow x = y o r x + y = \frac{a}{4}$ $b u t i f x + y = \frac{a}{4} :$ $(i) + (i i) :$ $2 a^{2} - a (x + y) = 4 (x^{2} + y^{2} + 2 x y)$ $2 a^{2} - a (x + y) = 4 {(x + y)}^{2}$ $2 a^{2} - \frac{a^{2}}{4} = \frac{a^{2}}{4}$ $\frac{3 a^{2}}{2} = 0!$ $\Rightarrow x = y i s t h e o n l y p o s s i b i l i t y!$
	Commented by Rasheed.Sindhi last updated on 15/Sep/19

	$T h α n X a l o t s i r . {Y o u}^{'} re v e r y d e e p$ $i n g e o m e t r y a l s o!$
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