Oscars Chocolate Mould

Sweeping the park

Statement

This problem starts out simple but gets complicated later.

points forming an equilateral

The first property is well known to everyone. The technique to find points that are equidistant from the two points that you have is to draw two circles, centered on two points, and whose radius is the distance that separates them. Like all points of the circle are equidistant from the center, which will cut the minimum distance of two points. Clearly, there are two different solutions, one for "up" and one "below" the line that joins them.

points forming an isosceles

^{The second property is stranger. What we seek is the third vertex of an isosceles triangle. There are three possibilities, but most people see only the first of them. If the two points that we make up the other side, then the third point will be equidistant from both, which will be on the bisector, ie the vertical line through the center of the segment. Therefore, all points of the bisector (except the center of the segment, and points that we encountered in the previous section) could be the point that we lack. However, many forget that the segment that we may be one that is equal to another. The other point is the distance that marks the segment of one of the two ends of the segment, for what will be on one of the two circles that can be drawn on one of the center and the segment as radius. As in the line, we remove two points from the previous year and those who are directly in line with our initial segment.}

Points are parallelogram

The third paragraph also often difficult, because we forget some point. The idea is, since they are parallelograms, draw parallels to the sides we have for the item that does not happen that way. However, as we have three points, we consider three segments. One of them is diagonal and the other two sides. According to what we consider to be diagonal, the lines will be different and the fourth point falls into one of three different positions. It is curious that the three points form a triangle similar solution to the initial one, but two-scale, ie, containing four triangles identical to the original.

points forming three isosceles

Finally, for the fourth section must return to reason as in paragraph b), because you have to do the same design for each of the pairs three vertices of the triangle. The cutoff points could be points we seek. But the picture becomes so complex that there are up to 10 points of intersection (one inside the triangle, three very close, three forming an equilateral triangle double scale, and three further away). It is easy to see that for each of these four families of points is satisfied that the three triangles which we are isosceles.