ali1254 said:
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.
