A piece of paper has the precise shape of a triangle (which we will denote by triangle ABC), where the respective lengths of the crease whenever the paper is folded such that; the vertex A is joined onto vertex B, the vertex A is joined onto vertex C and, the vertex B is joined onto vertex C are 7, 8 and 9.

Determine the lengths AB, AC and BC.

Call A=(0,0), B=(a,b), C=(c,b)

Joining vertex A to B creates a crease along y-b/2 = -a/b(x-a/2) which has endpoints (-b^2/2a+a/2,b) and (a/2,b/2)

Joining vertex A to C creates a crease along y-b/2 = -c/b(x-c/2)
which has endpoints (-b^2/2c+c/2,b) and (c/2,b/2)

Joining vertex B to C creates a crease along x=(a+c)/2 which has endpoints ((a+c)/2,b(a+c)/2a) and ((a+c)/2,b)

Setting the first two crease lengths to 7 and 9 and the last to 8 we get the system:
b^4/4a^2 + b^2/4 = 49
b^4/4c^2 + b^2/4 = 81
b - b(a+c)/2a = 8

Solving the first two for a and c respectively and substituting into the third we get (after simplifying a lot) the cubic

b^3 - 4^2 - 324b + 2592 = 0

Of the three solutions, one gives imaginary coordinates for a and c.
The two real solutions are

Case1
a=8.993908283
b=9.5815351
c=-6.024826225
Side lengths 15.01873451, 13.14139266, 11.31831904

Case2
a=111.097725
b=13.891819
c=-16.85985545
Side lengths 127.9575805, 111.9628828, 21.845783

Posted by Jer on 2012-05-21 13:41:55