+1833 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 $\triangle BCP$?
+118608 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 :)
+118608 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 :)
