Last week we asked you to solve the following mathematical problem.....
A SQUARE piece of toast ABCD of side length 1 and centre O is cut in half to form two equal pieces ABC and CDA.
If the triangle ABC has to be cut into two parts of equal area, one would usually cut along the line of symmetry BO.
However, there are other ways of doing this.
Find, with justification, the length and location of the shortest straight cut which divides the triangle ABC into two parts of equal area.
And this week we reveal the answer.......
This is fairly straightforward using the sine/cosine rules and
the AM-GM inequality. There are two possibilities for the line,
either between the two legs, or from a leg to the hypotenuse.
The latter leads to a cut length of AB, while the former leads to
a cut length of AB*Sort(Sqrt(2) - 1) ~= AB*0.6436-.
The lengths of the equal sides of the isosceles triangle formed
by it are AB/FourthRoot(2).