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# Hi y'all this is a very similar question, but it's different. Sorry!!

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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\$?

Apr 19, 2015

#1
+95177
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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 :)

Apr 20, 2015

#1
+95177
+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