A Math Problem

Let ABCE be a square. Let D be a point, so that CED is right angle and ED has the same length as CD. Use 2 straight lines to cut the pentagon into three pieces and arrange these, so that they form a square.

Screen Shot 2014-05-28 at 7.32.03 PM

Answer:

Let the side EA have the length 1. The area of the figure is 1 and 1/4, since EDC is 1/4 of AECB. The square that we want to construct has the area 1 and 1/4 and the side length (5/4)^1/2

We can construct this length by connecting A to the middle of CB. Now we can construct a line that connects the middle of BC with D. Now we can guese a solution.Screen Shot 2014-05-28 at 8.31.00 PM

First we show that the lines that are together have the same length. We know that the 2 and 3 fit together since the sides that are touching each other  were created by halving the line BC. We know that 2 and 1 fit together because the sides that are touching each other from 2 and 1 are both sides from the square ABCE. We know that 3 and 1 fit together since DC is equal to DE. The angle where 3 2 and 1 come together is 360 degrees, since the angle at 3 is 135 degrees, the angle at 1 is 135 degrees and the angle at 2 is 90 degrees since it was a corner of the square. The angle where only 1 is is 90 degrees. To show this, lets draw this picture:Screen Shot 2014-05-28 at 8.47.42 PM

 

We will show that ABC is congruent to CDE. BA has the same length as CD per definition. BC has the same length as DE per definition. ABC and CDE are both 90 degrees. It follows that ABC is congruent to CDE. Since the triangle has a 90 degree angle, the other angles added together are also 90 degrees. Since BCD is a straight line, ACE has 90 degrees. It also follows that the angle at 2 and 3 is 90 degrees. Since the angle at 2 and 1 were 90 degrees before we arranged the square, they are 90 degrees afterwards too. It follows that our arrangement is a square. This completes the proof.

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