Take some point V and draw two rays from it. Choose some other point W in between those two rays. Then, construct a line that touches both rays and passes through W.

Now, this line forms a closed triangle together with the two rays. The point W divides this line into two segments (x1, x2). What is the ratio of these two segments such that the area of the enclosed triangle is minimal?

Does this minimal area even exist?

Place the angle on a coordinate system so that V is at the origin and W is at (0,1). Then one ray is in the first quadrant and the other ray is in the second quadrant.

Let the equation of the first ray be y=m*x and let the equation of the second ray be y=-n*x. Let k be the slope of the line through W, its equation will be y=k*x+1. The valid range of k is m>k>-n.

Let A be the intersection in the first quadrant: (1/(m-k), m/(m-k)). Then x1=AW.

Let B be the intersection in the second quadrant: (-1/(n+k), n/(n+k)). Then x2=WB.

Then the area of ABV = area AWV + area BWV = (1/2)*1*(1/(m-k)) + (1/2)*1*(1/(n+k))

Simplfied, area ABV = (1/2)*[1/(m-k) + 1/(n+k)]

Differentiating with respect to k yields: 0 = dArea/dk = (1/2) * [1/(m-k)^2 - 1/(n+k)^2]

Simplfying the equation yields: (m-k)^2 - (n+k)^2 = 0

Which makes k=(m-n)/2

This point is the only extrema. It is a minimum since the limit of the area of ABV as k approaches either -n or m tends to infinity.

Substituting back into points A and B gives A(2/(m+n), 2*m/(m+n)) and B(-2/(m+n), 2*n/(m+n))

The midpoint of AB is (0,1), which is W. That means AW=WB (ie. x1/x2=1) occurs when the area of the triangle ABV is minimized.