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A point $P$ is chosen at random in the interior of equilateral triangle $ABC$. What is the probability that $\triangle ABP$ has a greater area than each of $\triangle ACP$ and $\triangle BCP$?

Mellie  Apr 19, 2015

Best Answer 

 #1
avatar+93677 
+5

Mmm,

I was thinking about this and I think one of the triangles has to be the biggest and they are all equally likely do the probability of ABP being the biggest is 1/3

 

I also decided to follow Chris and try a visual approach.

You want the probability of triangle 1 being the biggest.

If P is in 1b or 2a then   triangle 3 is the biggest

If P is in  2b or 3a then triangle 1 is the biggest

If P is in 3b or 1a then triangle 2 is the biggest.

So

The probability that triangle 1 is the biggest is 1/3

 

That is what I think anyway :)

Could another mathematician please comment on this :)

Melody  Apr 20, 2015
 #1
avatar+93677 
+5
Best Answer

Mmm,

I was thinking about this and I think one of the triangles has to be the biggest and they are all equally likely do the probability of ABP being the biggest is 1/3

 

I also decided to follow Chris and try a visual approach.

You want the probability of triangle 1 being the biggest.

If P is in 1b or 2a then   triangle 3 is the biggest

If P is in  2b or 3a then triangle 1 is the biggest

If P is in 3b or 1a then triangle 2 is the biggest.

So

The probability that triangle 1 is the biggest is 1/3

 

That is what I think anyway :)

Could another mathematician please comment on this :)

Melody  Apr 20, 2015

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