Which Leg Do You Put An Anklet On

shaded area in a decagon

Statement

I'll start by posting a solution that has provided me Peiró Ricard, which, playing with cabri and Ptolemy's Theorem , has found a solution.

regular decagon

Obviously, we begin by drawing a shape (like a) a regular decagon. If we draw with ruler and compass, first drawing a regular pentagon and divide into two parts from the center of their faces. Drawing a circumcircle, easily find the cutoffs of the decagon.

In any case, there are many possible segments that can play the role of A 1 4 A and A ₁ A 2. In the drawing Ricard, has found a ring formed with sides and diagonals of a decagon, using the vertices A 1, A 2, A and A ₄ 6.

is easy to see that A ₁ A ₆ is a diameter of the circle, that A ₁ A 2 is a side of decagon, that A ₂ 4 A and A ₄ A ₆ are corresponding sides of the pentagon, A ₁ A ₄ is one of segments appear in the problem, and A 2 6 A is a diagonal of the pentagon.

If you have studied some geometry of a regular pentagon, you know that the ratio between a diagonal and the side is called the golden ratio, represented by the Greek letter φ = (1 + √ 5) / 2. In our case, In a decagon there is also another very special relationship as the central angle of each side is 360/10 = 36 degrees, drawing two radios will form an isosceles triangle with unequal angle 36 degrees, which is similar to that formed between one side Pentagon and two diagonals. This means that a radius divided by the side of the pentagon is also equal to φ. Among the points we have chosen, this means that the product of the diagonals equals the sum of the product from its opposite sides. In fact, a registered arc derived result and is very useful when working with circles and segments. In our ring, applying this theorem means that ie, Romans and Carthaginians a couple of the other side, then a trip to take two Greeks who complete the group. However, at a given time to do go back a group of people for the further implementation of the rules. Obviously, the boat can not go no one to row, and you should always be busy. One way, as suggested by a comment from Claudio Meller our facebook group would be as follows, where, for short, is named for the initials soldiers (and a number representing the amount, if more than one) there on one side in the boat or the other side. RCG Cruzan in the boat, leaving 3R3C3G on the shore. Dejan on the other side and returns RC G in the boat, gathering 3R3C4G. Cruzan

3C in the boat, leaving on one side 3R4G and the other (when they) r4c. Back R in the boat, leaving the other side 4C and the starting 4R4G. Now cross 3R 1R4G leaving and joining the other side, where there are now 3R4C. Now comes the difficult step, because they must return with the boat full. Any other move us back to the same place we were, or breach of any rule. R2C are back in the boat, leaving the other side 2R2C. When coupled to the edge of departure, are 2R2C4G. Now cross

2G, leaving 2R2C2G and joining an identical amount on the other side. Again

must return the boat filled with RCG, leaving the other side of RCG and the initial side 3R3C3G. back across the boat with 3G, reaching the final edge RC4G and leaving the starting 3R3C. The boat returns with a C. Now there is in the starting 3R4C and the arrival R4G. Cruzan R2C, leaving the starting 2R2C and the arrival 2R2C4G. back the boat to RC, leaving the RC4G arrival and the departure 3R3C. Cruzan 3C now, leaving the starting 3R and the arrival R4C4G. Back _R now, leaving the finishing 4C4G, and the starting 4R. _Cruzan 3R, leaving the R and conducted an in 3R4C4G arrival. Back R in the boat, leaving the 2R4C4G arrival and the departure 2R. Cruzan the last two in the boat.