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

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Since the three triangles $ABP,ACP$ and $BCP$ have equal bases,their areas are proportional to their length of altitudes.
But i dont know how to solve further and find the required probability.Please help me.Thanks.

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  • $\begingroup$ Um, I'm kind of perplexed. Some triangle must be the largest and as ABC is equilateral it's perfectly symmetric so any triangle is as likely as any other. So the probability is 1/3. How could it possibly be anything else? $\endgroup$ – fleablood Nov 25 '15 at 5:38
  • $\begingroup$ Looks logically true $\endgroup$ – Ekaveera Kumar Sharma Nov 25 '15 at 6:57
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There are six equally likely outcomes:

$ABP>BCP>ACP$

$ABP>ACP>BCP$

$BCP>ACP>ABP$

$BCP>ABP>ACP$

$ACP>BCP>ABP$

$ACP>ABP>BCP$

Each of them are mutually exclusive and their sum is $1$. Hence $BCP$ being the largest has probability $1\over3$. (The equal area case has probability $0$ and can be ignored because we have a dense space, that is, $>$ sign has same probability as $\geq$ sign.)

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  • $\begingroup$ Why does the equal area case has probability $0?$I do not understand. $\endgroup$ – Vinod Kumar Punia Nov 25 '15 at 4:57
  • $\begingroup$ This is a current problem in our probability model. You might want to look at this: math.stackexchange.com/questions/1479453/what-is-randomness/… $\endgroup$ – cr001 Nov 25 '15 at 5:01
  • $\begingroup$ Equal area case requires P to be in precise locations. This locations (which I believe lie on the straight line angle bisectors) together have no measurable area and have 0 probability of occurring. This is the same reason choosing a random real number an [0,1] will have a 50% chance of being less than 1/2, a 50% chance of being less than or equal to 1/2, a 50% chance of being larger than 1/2 and a 50% chance of being larger or equal to 1/2 but an absolute 0 probability of being equal to 1/2. $\endgroup$ – fleablood Nov 25 '15 at 5:42
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Let distance from P to AB, AC and BC to be $L_{ab}$, $L_{ac}$ and $L_{bc}$, respectively.

Then $L_{ac}+L_{ac}+L_{bc}$ is constant and equal to the distance from A to BC (let's call this L), because sum of areas of three triangles ABP, ACP and BCP should be equal to the area of ABC.

So the condition becomes

$$L_{ab} > L_{ac}$$ $$L_{ab} > L_{bc}$$

So, P should exist within the region defined by

  1. Center of ABC
  2. C
  3. Mid point of A-C
  4. Mid point of B-C

And the area of the region is $\frac{1}{3}$ ABC. $$\therefore Probability = \frac{1}{3}$$

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  • $\begingroup$ Kay K,i think the correct probability is $\frac{1}{3},$not $\frac{1}{4}?$ $\endgroup$ – Vinod Kumar Punia Nov 25 '15 at 5:05
  • $\begingroup$ P should exist within an equilateral triangle of which corners are C, mid point between A-C, and mid point between B-C. It's area is 1/4 of ABC. $\endgroup$ – Kay K. Nov 25 '15 at 5:08
  • $\begingroup$ Sorry I misinterpreted the question. I thought ABP > ACP+BCP. It seems like ABP > ACP, and ABP > BCP. I will correct my solution. $\endgroup$ – Kay K. Nov 25 '15 at 5:12

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