Show, how given a distance r, one can construct a regular pentagon, whose circumradius is r, using only ruler and compass...

Note 1: The ruler doesn't have any marks, so it's no good for measuring. It's only good for joining 2 points by a line.

Note 2: The circle which circumscribes the polygon, such that the polygon lies entirely within the circle and all of whose vertices lie on the circumference of the circle is the circumcircle of the polygon. The radius of this circle is the circumradius of the circle.

With the distance r, we open the compass with a distance r, then we mark two distances r in the X axis and a distance r to the Y axis, to form a rectangular triangle of sides r and 2r. It's hypotenusa must be √5*r, so we find the midpoint of the hypotenusa (√5*r/2) and open the compass to get the distance of the middle of such hypothenusa. To that lenght, we add the middle of r (r/2) so in the compass we have an opening equal to (.5*r+.5√5*r) or ((1+√5)/2)*r that is the golden ratio. So, with the golden ratio founded, we simply find the intersection of such distance just by puting the compass origin un the endpoint of the initial distance r. We get an isoceles triangle of sides r, and two sides with lenght of the golden ratio. We open the compas to a lenght r and draw a light circular reference where the next vertex could be, we do this from one vertex of sides golden ratio and r. We open the compass with the lenght of golden ratio and put the origin in the vertex thas has tow sides of the golden ratio, from there we intersect the golden ratio distance with the previously marked circle. We make the same operation to find the vertex next to the vertex of twice the golden ratio. We could find all the vertex makeing the same operation, and voila! we have a perfect pentagon. I assume that all of you know how to draw a perpendicular line and get the midpoint of a given line using a compass, because I didn't mentioned in order to write less.

Posted by Antonio on 2003-08-31 03:43:05