triangle cake measures exactly 8 inches. What is the length of the shortest possible cut that will divide the cake into two pieces with equal area?

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For a straight cut, I believe the shortest is parallel to one of the sides and has a length that is 1/sqrt(2) times the base length. In this case the length is 8/sqrt(2) = 4*sqrt(2) = 5.65685. This divides the cake into an equaliateral triangle and trapezoid that have 1/2 the area of the cake.

This cut is shorter than an altitude cut, which has a length of 4*sqrt(3) = 6.9282.

For a curved cut, I'm not sure. If there was a curve with a shorter length than the straight cut, it would probably be a circular arc.

EDIT: Here's how to make a curved cut that has a shorter length than the straight cut. Make a 60 degree circular arc centered at one of the triangle's vertices such that the radius of the arc is 4*sqrt(3*sqrt(3)/pi). The area of the wedge cut by the arc will be 8*sqrt(3) which is exactly half the area of the cake. The length of the arc will be 4*sqrt(pi/sqrt(3)) = 5.38709, which is shorter than the first answer I gave. I don't know if this is THE shortest cut.

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That's cool, but why do you think the optimal solution is a circular arc?

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I found the longest cut..........................................

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Haha Nice!