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 xaxis and C on the yaxis 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/52√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(5/65t)(√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 20140212 17:03:47 