A piece of paper has the precise shape of a triangle with the three side lengths AB, AC and BC being respectively 2, 4/3 and √10/3. The paper is folded along a line perpendicular to the side AB.

Determine the largest possible overlapped area.

I'll leave out most of the algebra

Placing A and B on the x-axis and C on the y-axis we can coordinatize the points using law of cosine.  cos(A)=7/8
A=(-7/6,0)
B=(5/6,0)
C=(0,√15/6)
Clearly we want to fold AC over BC

The line containing BC is y=-√15x/5+√15/6
the fold becomes a vertical line x=-t
the reflected section of AC is on the line y=-√15x/5-2√15t/7+√15/6

the intersection of these lines is x=5t
y=-√15t+√15/6
drawing this vertical line splits the overlap region into a trapezoid and a right triangle
Area = .5(6t)(-√15t/7+√15/6+-√15t+√15/t) + .5(5/6-5t)(-√15t+√15/t)
=-13√15tē/14 + √15t/6 + 5√15/72
which is a quadratic, so its maximum is at t=-b/(2a)=7/78
And the area here is √15/13

Posted by Jer on 2014-02-12 17:03:47