Auto Mechanic License Ontario

Statement

This problem I found fascinating. I had not found a problem with a very interesting challenge.

The idea I continued to explore all the possibilities is to draw circles. Imagine that over the point from which more equal lines go. At least, you have two equal, since they start three of them. The idea that occurred to me to explore all distributions of points from there.

distribution

First

The first possibility is that the three points are at equal distance from it (in a circle). If we consider one of the points, think you also have the other two points the same distance (on another circle). This situation determined by the intersection of two circles of same radius) the position of the points. The form is that of a diamond, and only a distance will be different (and longer).

Second and third distribution

The second possibility is that the three points are on the circumference, but only two of them are within radio distance (and thus make him an equilateral triangle) . The third point, which can not be the same distance from the center of one of the other points, it would repeat the same arrangement as before, will be equidistant from the two, so they form an isosceles triangle is say, will be on the bisector. There are two possibilities, obviously. The first has greater distances than the original radio, and the other children. The picture one after the other.

Fourth

distribution

Once we have exhausted this possibility, we can only assume that none of the three points is the same distance from another central point. but that means that the three are at the same distance, forming an equilateral triangle. This is the fourth distribution points.

Therefore, we now assume that only two points are the same distance from one of them (because we have already explored all such distributions.) We got to the section more difficult. Imagine

central point of the circle. On the circle can only have two points. And the distance between them and can not be equal to the radius, as if it were, form an equilateral triangle, and the third point would be the same distance from the three (which we have seen) or again be at a distance of radius of one of the points, so that point would be the same distance from all three.

Now, the distance between those two points will be different. If the fourth item is out of the circle, its distance from the center matches the distance you are the other two points. Again, can not be that the missing point is at the same distance of the two new girth, then again form an equilateral triangle can not be. Thus, we have two possibilities: either the last point is a radius distance of the circumference of the two is either a radio distance and other distance of the other.

From here in the first case is easy to understand that form a square, as all four sides "outside" are equal, and both interiors are equally between them. This is the figure that came as an example in the statement.

sixth and final distribution

The latest figure is more difficult to visualize. Three sides are equal, and the fourth is like the distance to the cross (diagonal). Form in this case no fewer than four isosceles triangles between points, the same pairs. If we study the relations that exist between the angles, we find that this ring is neither more nor less than a trapeze, and not just any, if not an isosceles trapezoid whose largest angle is 108 degrees.