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

#1**+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**+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