Luigi, an Italian cook, drops a spaghetti which breaks into three pieces. What is the probability of making a triangle with the three pieces ?


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Question Number 116983 by Olaf last updated on 08/Oct/20

$Luigi, an Italian cook, drops a spaghetti$ $which breaks into three pieces .$ $What is the probability of making a$ $triangle with the three pieces ?$
Commented by mr W last updated on 08/Oct/20

$i t h i n k {i t}^{'} s \frac{1}{8} .$
Commented by Olaf last updated on 08/Oct/20

$No sir . . .$
Commented by mr W last updated on 08/Oct/20

Commented by mr W last updated on 08/Oct/20

$t h e n \frac{1}{4}$
Commented by Olaf last updated on 08/Oct/20

$Yes sir! Bravo!$ $If x, y and z are the 3 lenghts of$ $spaghetti with 0 ⩽ x, y, z ⩽ l this$ $defines an equilateral triangle in space .$ $The equation is x + y + z = l .$ $The area of this triangle (all possible$ $cases) is \frac{\sqrt{3}}{4} \sqrt{2 l} = \frac{\sqrt{6}}{4} l .$ $But, to make a triangle with the 3$ $pieces we need 0 ⩽ x, y, z ⩽ \frac{l}{2} .$ $This defines a second equilateral$ $triangle in space (all probable cases)$ $with an area a quarter of the first$ $one .$
	Answered by mr W last updated on 09/Oct/20

	Commented by mr W last updated on 09/Oct/20

	$t h i s i s h o w i s o l v e d .$ $s a y t h e l e n g t h o f s p a g e t t i A B i s 1 .$ $i t b r e a k s r a n d o m l y a t p o i n t C & D$ $w i t h A C = x a n d A D = y .$ $0 < x, y < 1$ $t h e p o i n t s i n s q u a r e 1 \times 1 r e p r e s e n t$ $a l l p o s s i b i l i t i e s . w e o n l y n e e d t o t r e a t$ $h a l f o f a l l p o s s i b i l i t i e s : y ⩾ x . t h a t i s$ $t h e o r a n g e t r i a n g l e a r e a .$ $A C = x$ $C D = y - x$ $D B = 1 - y$ $s u c h t h a t t h e t h r e e p i e c e s c a n f o r m$ $a t r i a n g l e, w e m u s t h a v e$ $x + (y - x) > 1 - y \Rightarrow y > \frac{1}{2} . . . (i)$ $x + (1 - y) > y - x \Rightarrow y - x > \frac{1}{2} . . . (i i)$ $(y - x) + (1 - y) > x \Rightarrow x < \frac{1}{2} . . . (i i i)$ $(i), (i i), (i i i) m e a n t h e p o i n t s i n t h e$ $g r e e n t r i a n g l e a r e a .$ $t h e r e q u e s t e d p r o b a b i l i t y i s$ $p = \frac{a r e a o f y e l l o w t r a n g l e}{a r e a o f o r a n g e t r i a n g l e} = \frac{1}{4}$
	Commented by mr W last updated on 09/Oct/20

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