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Explanation

A mother of young twins and a toddler made a rectangular cake for the twins' coming birthday. She put it into a refrigerator and went to bed. At night, the toddler got up, strolled to the refrigerator and helped himself to a rectangular piece of the cake he cut off one of the corners. (The two rectangles are assumed to have parallel sides, notwithstanding the toddler's tender age.) In the morning, the mother was faced with the problem of dividing the remaining cake equally between the twins. Since she was homeschooling her kids, she was always on lookout for an edifying experience. This time, she asked the disappointed twins whether it's possible to divide the cake into two equally worthy parts with a single vertical cut.

Is it?

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

 What if applet does not run?

The answer is yes. And this is how one may proceed.

First, introduce some notations. Let the original rectangle be ABCD, with a small rectangle UZRD removed off the corner D, Z being internal to ABCD. The problem is to draw a line that cuts the L-shape ABCRZDU into to pieces of equal area. The solution is obviously not unique. The one below has been motivated by Euclid I.43.

Had Z lied on the diagonal BD, the problem would have been solved with the cut BZ. We shall consider BZ as the first approximation even where Z is away from BD. In the latter case BZ still solves the problem for a different shape cake. Extend BZ to the intersection with either CD or AD, as the case may be. Assume BZ intersects, say, CD first. Through the point of intersection M draw a line parallel to AD. The latter intersects AB in K and ZU in L. BZ divides the L-shape KBCMZL into equal areas. However, the rectangle AKLU is left over. One solution is to upgrade the cut BZ as to append to the area BCMZ a shape of area half that of AKLU. What comes to mind is the triangle BZQ, with ZQ = LU. The triangle has the same base (LU) and height (KL) as the rectangle AKLU and, therefore, half the area of the latter. Hence BQ provides a solution to our problem.

Another solution with a cut PQ parallel to BZ can be obtained by taking BP = ZQ = LU/2. Finally note that the problem can be easily generalized to a rare case of a cake in the shape of parallelogram. As, the only thing that is important to the above solution is that the toddler cuts off another parallelogram UZRD whose sides will naturally be parallel to those of ABCD.