Nom d'utilisateurMelody
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(1) A line segment is broken at two random points along its length. What is the probability that the three new segments can be arranged to form a triangle?


I have used a Boolean contour map to solve this.

I have allowed the line to be 3 units long just because it is easier to use numbers.


I have allowed the line to split into 3 peices with lengths   x,   y-x,  and 3-y

Now all these lengths must between 0 and 3 units long

And x and y must be between 0 and 3 units long.

and x


I am gong to arbitrarily say

x <  y-x < 3-y

This does not have to be true of course but I will think about the consequences of saying this later.


So we have



So it is inside a 3 by 3 box.




y>x       It is the above the line y=x


y-x > x

y >2x




y-x < 3-y

2y < 3 +x

y< 0.5x +1.5


But on top of all these constraints we have the main constraint which is that the sum of the 2 short sides must be longer than the long side, otherwise a triangle cannot exist.


x + y-x >3-y

2y > 3

y > 1.5


Here is the Boolean contour map:



So the entire area is 9 units squared, and the bit that fits the constraints is the darkest little triangle and it has an area of 

1/2 * 0.75* 0.5 = 0.1875 units squared


Now the bit that is confusing me a bit is that the ribbon does not need to be cut from smallest to biggest so this is not the only answer

It coudl be

S,M, B    that is the one I have done


M, S, B

M, B, S

B, S,M

B, M, S

That is 6 possibilities.

So I think I need to multiply  0.1875 by 6 = 1.125      (I admit I am a little confused here )


So I think the probability of being able to make a triangle is    1.125/9 = 0.125 =   1/8

Melody 3 hours ago
Melody 9 hours ago

Here is one, by GingerAle and Hectictar, that Hectictar just asked me to include.  Thanks Girls :)


This is the original thread but I will includ most of the salient coding underneath.








\(\begin{array}{|rcll|} \hline \angle {ABC} &=& 27° \\ \measuredangle{ABC} &=& 27° \\ \stackrel {\; \frown} {ABC} &=& 27° \\ \stackrel { \hspace{.1em} \wedge} {AC} &=& 27° \\ \stackrel {\, \hspace{.1em} \frown} {AC} &=& 27° \\ \hline \end{array}\\\ \overset{\; \frown}{AC} = 27° \\\)






\angle {ABC} &=& 27° \\

\measuredangle{ABC} &=& 27° \\

\stackrel {\; \frown} {ABC} &=& 27° \\

\stackrel { \hspace{.1em} \wedge} {AC} &=& 27° \\

\stackrel {\, \hspace{.1em}  \frown} {AC} &=& 27° \\

\hline \end{array}\\\

\overset{\; \frown}{AC} = 27° \\



I only just realised that Ginger has not used the \boxed command.

I wonder how she got the vertical side lines on her box??   I'll have to work it out ://

Arr got it, it is a part of the array command. 

Thanks Ginger :)



Oh Hectictar also mentioned that there is a number of ways to get the degree sign.

These are Hectictars words:

"I have noticed that there are actually two signs that look like degree signs in the "special character" selection. One of them is a masculine ordinal indicator: º  and the other is the degree sign: °  . (If you hover over the symbol it tells the name of it.)

I just use the command   ^\circ     Like this     5^\circ  displays as  \(5^\circ\)

Melody 25 sept. 2017