Let D be a point in the plane of ΔABC such that
ray AD intersects the interior of side BC.
Construct a line through D intersecting lines AB and
AC at points B' and C' respectively such that the
perimeter of ΔAB'C' equals BC.
I didn't know where to begin so I simplified things by making the triangle isosceles and right with A=(0,0), B=(1,0), C=(0,1)
That helped me see the problem is not possible unless D is inside the triangle and close to A.
So I chose to ignore D and focus on where the line B'C' can be.
It turns out if B'=(x,0) then to make the perimeter equal to BC, C'=(0,(√2x1)/(x√2)).
Graphing all of these lines B'C' seems to form the exterior of a quarter circle with center (√2/2,√2/2) and radius √2/2.
I was able to prove this.
So for any D inside this region we can find which line we are on and determine B' and C'.

Posted by Jer
on 20130509 13:10:44 