What's the Probability That Area of MNP is ≥ Half of ABC?

[ali1254]

Hi. Choose 3 random points M,N,P on sides of an equilateral triangle ABC . What's the probability that area of MNP is greater than or equal to half of area of ABC? What's the probability if points are chosen inside ABC?

 

[ali1254]

Hello and welcome to MHB, ali1254! :D

Since you are a new member of our community, I feel I should point out our guideline for posting in this particular forum:

http://mathhelpboards.com/challenge-questions-puzzles-28/guidelines-posting-answering-challenging-problem-puzzle-3875.html

If you are instead asking for help with this problem, then let me know and I can move the thread to the appropriate forum. If not, and you already have the solution and are posting it as a challenge then I will simply delete this post so it doesn't clutter your thread. :D

hi. i really do not know the answer. i'll be glad if you can help me solve it. thanks a lot.

 

[MarkFL]

hi. i really do not know the answer. i'll be glad if you can help me solve it. thanks a lot.

Okay, I have moved the thread here. :D

Are you certain the problem is correctly stated?

As I see it, there is only 1 way for the area of $\triangle MNP$ to be greater than or equal to $\triangle ABC$, and that is if their areas are equal because they are the same triangle. The area of $\triangle MNP$ can only become smaller as each of $M,\,N,\,P$ move away from the vertices of $\triangle ABC$.

 

[ali1254]

Okay, I have moved the thread here. :D

Are you certain the problem is correctly stated?

As I see it, there is only 1 way for the area of $\triangle MNP$ to be greater than or equal to $\triangle ABC$, and that is if their areas are equal because they are the same triangle. The area of $\triangle MNP$ can only become smaller as each of $M,\,N,\,P$ move away from the vertices of $\triangle ABC$.

oh. sorry. i corrected the question.

 

[MarkFL]

oh. sorry. i corrected the question.

Okay, I suspected that would be the case, but I wanted to be sure.

Can you show what you have tried so far so our helpers can see where you are stuck and/or may be going astray?

 

[ali1254]

Okay, I suspected that would be the case, but I wanted to be sure.

Can you show what you have tried so far so our helpers can see where you are stuck and/or may be going astray?

i have no idea . i can't find when the area is divided into half.

 

[Opalg]

Choose 3 random points M,N,P on sides of an equilateral triangle ABC . What's the probability that area of MNP is greater than or equal to half of area of ABC?

Suppose that the vertices of a triangle are given by the vectors $\vec{0}$, $\vec{a}$, $\vec{b}$ in 3-dimensional space. The area of the triangle is given by half the magnitude of the cross product, $\frac12|\vec{a}\times \vec{b}|$. Now suppose that points are randomly chosen on the three sides of the triangle, given by vectors $x\vec{a}$, $(1-y)\vec{b}$, $(1-z)\vec{a} + z\vec{b}$, where each of $x,y,z$ is chosen randomly in the unit interval. The area of the triangle formed by those three points is $$\tfrac12\bigl|\bigl((1-z)\vec{a} + z\vec{b} - x\vec{a})\bigr) \times \bigl((1-z)\vec{a} + z\vec{b} - (1-y)\vec{b})\bigr)\bigr|,$$ which simplifies to $\frac12|\vec{a}\times \vec{b}|\bigl(xyz + (1-x)(1-y)(1-z)\bigr).$

So the problem is to find the volume of the region in the unit cube $\{(x,y,z):0\leqslant x,y,z \leqslant 1\}$ in which $f(x,y,z) \geqslant \frac12$, where $f(x,y,z) = xyz + (1-x)(1-y)(1-z).$ I tried to find that volume by calculus, and ended up with a complicated integral, which Wolfson Alpha evaluated as approximately $0.068$. That is my best guess for the answer to the problem. I should be very pleased if someone has a better method to tackle it.

Notice that the answer to the problem is completely independent of the shape of the original triangle. It does not have to be equilateral. In fact, it might be easier to tackle the problem using some other shape, such as an isosceles right-angled triangle.