Three points on sides of equilateral triangle

Question

Let's choose three points on the sides of an equilateral triangle(one point on each side) and construct a triangle with these three points. what is the probability that area of this triangle be at least one half of the area of equilateral triangle?

Answer 1

Place the equilateral triangle s.t its vertices are in the points $A=(0,0), B=(1, \sqrt3), C = (2,0)$ (I'm assuming sidelength $2$ to have nicer constants). Denote by $x_1, x_2$ and $x_3$ the $x$-coordinates of the random points on the sides $AC, AB$ and $BC$ respectively. So

$$x_1 \in [0,2], x_2\in[0,1] x\in[1,2].$$

Then the $y$-coordinates are $y_1 = 0, y_2 = \sqrt3 x_2$ and $y_3 = \sqrt3(2-x_3)$. Now use the formula for the area of a triangle given its vertices to get (see for example here)

$$A(x_1, x_2, x_3) = \frac{\sqrt3}{2} \left|x_1x_2 + x_1x_3-2x_1-2x_2x_3+2_2\right|$$

The area of the original equlateral triangle is $\sqrt3$ so we are lead to the requirement that $$f(x_1, x_2, x_3) \geq 1$$ where $f(x_1, x_2, x_3) = |x_1x_2 + x_1x_3-2x_1-2x_2x_3+2_2|$. So the probability is given by the volume of the region of $X = [0,2]\times[0,1]\times[1,2]$ where this is satisfied, divided by $2$ (which is the total volume of $X$.

I haven't managed to calculate the volume (yet), but maybe we must split it in two parts $f\geq1$ and $f\leq-1$.

EDIT: I believe it can be calculated in $4$ parts depending on whether $f\geq1$ or $f\leq-1$ and whether $x_1 > 2x_2$ or $x_1 < 2x_2$. Let's denote by $\alpha$ the function that cuts the value to the inteval, i.e $\alpha(t) = 1$ if $t<1$, $\alpha(t) = t$ if $t\in[1,2]$ and $\alpha(t) = 2$ if $t>2$. I got the following:

$$I_1 = \int_0^2 { \int_0^{\frac{1}{2}x_1} {\alpha\left( \frac{-1+x_1x_2-2x_2}{x_1-2x_2}\right)} }dx_2dx_1$$ $$I_2 = \int_0^2 { \int_{\frac{1}{2}x_1}^1 {\alpha\left( \frac{1-x_1x_2+2x_1-2x_2}{x_1-2x_2}\right)} }dx_2dx_1$$ $$I_3 = \int_0^2 { \int_0^{\frac{1}{2}x_1} {\alpha\left( \frac{-1-x_1x_2+2x_1-2x_2}{x_1-2x_2}\right)} }dx_2dx_1$$ $$I_4 = \int_0^2 { \int_{\frac{1}{2}x_1}^1 {\alpha\left( \frac{1+x_1x_2-2x_2}{x_1-2x_2}\right)} }dx_2dx_1$$

If my calculations are correct, then the probability is $\frac{1}{2}(I_1+I_2+I_3+I_4)$. Still, I don't really know how to calculate these integrals. Plotting the the constraint equations $f=1$ and $f=-1$ in Wolfram alpha seems to indicate that they are hyperboloids but they are slanted and I dodn't know, maybe some coordinate transformation would be necessary(?), but then again our box is according to the coordinate axis.

EDIT2: Here's my python code I used in the simulation:

import math, random

#----Functions----------------
def area(a, b, c):
    return abs(a[0]*(b[1]-c[1])+b[0]*(c[1]-a[1])+c[0]*(a[1]-b[1]))/2.0

def random_between(a, b):
    t = random.random()
    return (a[0]+t*(b[0]-a[0]), a[1] + t*(b[1]-a[1]))

def mean(a):
    return sum(a)/float(len(a))

def var(a):
    m = mean(a)
    v = 0
    for x in a:
        v += (x-m)**2
    return v/float(len(a))
#-----------------------------


#The verteces of the original triangle
A = (0,0)
B = (2,0)
C = (1,math.sqrt(3))

AREA = area(A,B,C)
HALF_AREA = AREA/2.0

#----- the simulation---------
n = 1000000
count = 0
area_percents = []
for i in xrange(n):
    x = random_between(A, B)
    y = random_between(B, C)
    z = random_between(C, A)
    area_of_triangle = area(x,y,z)
    area_percents.append(area_of_triangle/AREA)
    if area_of_triangle >= HALF_AREA:
        count+=1

print "P(area over half) = "+str(float(count)/n)
print "Mean of the area percentages: " + str(mean(area_percents))
print "Variance of the area percentages: " + str(var(area_percents))

## I got the numbers for the distribution like this
## but there must be a better way to make a graph (with some libraries)
"""
graph = {}
categories = 100
d = 1.0/categories
for i in xrange(categories):
    # i ~ [i*d, (i+1)*d)
    graph[i] = 0

for x in area_percents:
    i = 0
    while x > (i+1)*d:
        i += 1
    graph[i] += 1

for x in graph.keys():
    print graph[x]
"""