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Question Number 197057 by ajfour last updated on 07/Sep/23

Commented by ajfour last updated on 07/Sep/23

$T h e a n t h a s t o c l i m b u p t h e p l a n e$ $a n d s u r m o u n t t h e w a l l o f h e i g h t c,$ $a n d d e s c e n d t h e n r e a c h B . F i n d t h e$ $s h o r t e s t l e n g t h o f p a t h .$
Commented by Frix last updated on 08/Sep/23

$Depending on the values of a, b, c the path$ $along the diagonal of the rectangle a b and$ $the rectangle b c on the right side can be$ $shorter . We have$ $\sqrt{b^{2} + {(a + b + c)}^{2}} ⋚ \sqrt{a^{2} + b^{2}} + \sqrt{b^{2} + c^{2}}$ $Let b = p a \land c = q a; p, q > 0$ $a \sqrt{p^{2} + {(1 + p + q)}^{2}} ⋚ a \sqrt{1 + p^{2}} + a \sqrt{p^{2} + q^{2}}$ $p^{2} + {(p + q + 1)}^{2} ⋚ 2 p^{2} + q^{2} + 1 + 2 \sqrt{(p^{2} + q^{2}) (p^{2} + 1)}$ $p + p q + q ⋚ \sqrt{(p^{2} + q^{2}) (p^{2} + 1)}$ $p (p^{3} - 2 p q - 2 q (q + 1)) ⋛ 0$ $p^{3} - 2 p q - 2 q (q + 1) ⋛ 0$ $Let q = α p; α > 0$ $p (p^{2} - 2 α (α + 1) p - 2 α) ⋛ 0$ $p^{2} - 2 α (α + 1) p - 2 α ⋛ 0$ $p^{2} - 2 α (α + 1) p - 2 α = 0$ $p = α (α + 1) \pm \sqrt{α (α + 2) (α^{2} + 1)}$ $p_{1} = α (α + 1) - \sqrt{α (α + 2) (α^{2} + 1)} < 0 \forall α > 0$ $p_{2} = α (α + 1) + \sqrt{α (α + 2) (α^{2} + 1)} > 0 \forall α > 0$ $We have p > 0 \Rightarrow$ $(p - p_{1}) (p - p_{2}) < 0; 0 < p < p_{2}$ $\sqrt{b^{2} + {(a + b + c)}^{2}} > \sqrt{a^{2} + b^{2}} + \sqrt{b^{2} + c^{2}}$ $(p - p_{1}) (p - p_{2}) = 0; p = p_{2}$ $\sqrt{b^{2} + {(a + b + c)}^{2}} = \sqrt{a^{2} + b^{2}} + \sqrt{b^{2} + c^{2}}$ $(p - p_{1}) (p - p_{2}) > 0; p > p_{2}$ $\sqrt{b^{2} + {(a + b + c)}^{2}} < \sqrt{a^{2} + b^{2}} + \sqrt{b^{2} + c^{2}}$
	Answered by Frix last updated on 07/Sep/23

	$\sqrt{b^{2} + {(a + b + c)}^{2}}$ $The shortest path is the diagonal of the$ $rectangle we get by unfolding the object .$ $Its sides are b and (a + b + c)$
	Answered by mahdipoor last updated on 07/Sep/23

	$A B = A M + M N + N B \Rightarrow$ $p (x, y) = \sqrt{a^{2} + x^{2}} + \sqrt{c^{2} + {(b - x - y)}^{2}} + \sqrt{y^{2} + b^{2}}$ $0 ⩽ x, y ⩽ b$ $\frac{d p}{d x} = 0 = \frac{x}{\sqrt{x^{2} + a^{2}}} + \frac{(- b + x + y)}{\sqrt{c^{2} + {(b - x - y)}^{2}}}$ $\frac{d p}{d y} = 0 = \frac{y}{\sqrt{y^{2} + b^{2}}} + \frac{(- b + x + y)}{\sqrt{c^{2} + {(b - x - y)}^{2}}}$ $\Rightarrow \frac{x}{\sqrt{x^{2} + a^{2}}} = \frac{y}{\sqrt{y^{2} + b^{2}}} \Rightarrow b x = a y$ $\frac{x}{\sqrt{x^{2} + a^{2}}} = \frac{b - (1 + \frac{b}{a}) x}{\sqrt{c^{2} + {(b - (1 + \frac{b}{a}) x)}^{2}}} \Rightarrow$ $x^{2} (c^{2} + {(b - m x)}^{2}) = (x^{2} + a^{2}) {(b - m x)}^{2}$ $\Rightarrow x^{2} c^{2} = a^{2} {(b - m x)}^{2} \Rightarrow x c = a b - (a + b) x$ $\Rightarrow x = \frac{a b}{(a + b) + c} & y = \frac{b^{2}}{(a + b) + c} 0 ⩽ x, y ⩽ b$
	Commented by Frix last updated on 07/Sep/23

	$You {don}^{'} t need this . Look at my explanation .$
	Commented by mahdipoor last updated on 07/Sep/23

	$Thats right your solution is logical and$ $faster, the answer is the same for both$
	Commented by Frix last updated on 07/Sep/23

	$Yes .$
	Answered by ajfour last updated on 08/Sep/23

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