How to construct a regular Pentagon using only circles and lines

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We first need to practice doing easy things such as making a perpendicular to a given line at a given point.  This is not too hard:

A perpendicular to a line at a point

Suppose our line is l  and our point is P. Here we go.

1. Obtain any point on the line other than P. [This is a more common request than you might think.] Call this point A.
2. Make a circle centered at P, radius PA. This circle will cross the line l again at another point, call it B.
3. Make the perpendicular bisector of AB! This will go through P, as desired, and be perpendicular to l.

Next, let's create a line of length √5.  For this purpose, we first note that 2² + 1² = 5.   So if we make a right triangle with its two shorter sides of length 1 and 2 respectively, its longest side (its hypotenuse, that is) will be the length we want.  This is not too hard.  In what follows, let's denote the length of a line segment AB as |AB|.  The notation AB will continue to represent the line segment itself.

Segment of length √5

[Actually, given a segment AB, we're going to construct a segment of length √5|AB|.]

1. Given: segment AB.
2. Construct the line AB.  In other words, extend AB in both directions.
3. Make a circle center at B, radius BA, and let it cross the line again at C.  At this point, |AC| = 2|AB|.
4. Draw another circle, center C, radius CB.
5. Draw a perpendicular at C.  This will cross the last circle at two points; just pick one of them, and call it D.  CD, of course, is the same length as AB.

Let's give you a picture to look at here.  If all has gone well, the triangle ACD is a right triangle, one side has length 2|AB|, and the perpendicular side has length |AB|.  The hypotenuse (the third and longest side) will have length √5|AB|, so the construction will be complete.

The next step is to construct a segment whose length is ½(1+√5).  If you think about it, this is just the average of the numbers 1 and √5.  There's an easy way of doing this generally.  If you want to find a segment whose length is the average of the lengths of PQ and RS, it's easy:

1. Base the construction right at PQ.  Extend the line PQ past Q.
2. At Q, construct a segment QT whose length is equal to RS.
3. Now PT has length equal to the two given segments combined.  But you want the average.  So, find the midpoint of PT; call it M.
4. PM has the length you want.

Construct a Pentagon!

We're going to build a regular Pentagon on a given side AB.

1. Construct AZ along the line BA so that |ZA| = √5|AB|.  (All we have to do is, working with the previous figure, draw a (large) circle, center A, radius AD.  This circle should cut the line AB on the far left of A at some point Z.  That's the point we want.
2. Construct a point Y on ZB so that YB is ½(1+√5)|AB|.  To do this, let Y be the midpoint of ZB.  (Why does this work?  ZA has length √5|AB|, AB has length |AB|, so put them together, they have length (√5+1)|AB|.  When you find the midpoint, the right half should have length as desired.  We show you a picture up to this point, and from here on, we will make these early construction lines very light and grey.)
3. Draw a circle of radius YB centered at B.  Why?  Because the third point from A (oh dear; I could have called it D, but I've used it up...), call it E, has to be on this circle, because the radius of this circle is the length of the diagonal!
4. Draw the perpendicular bisector of AB.  Why?  Because the third point from A, which we have called E, must lie on this perpendicular bisector, just from common sense.
5. The next steps might be a little surprising at first: construct the circle passing through A, B and E, and using circles in the obvious way, find the two last points of the pentagon, which must both lie on this circle.

Well, there you have it.  It is sometimes useful to look at a process in large stages:
First, construct a line whose length is √5|AB|.
Secondly, use this segment to construct another segment whose length is the average of the one above, and AB itself.  This is the length of the diagonals of the Pentagon.
Thirdly, draw a circle with radius equal to the diagonals at B (or at A; doesn't really matter), and construct the "highest" point on the Pentagon where this circle hits the perpendicular bisector of AB.
Finally, construct the circumcircle of the three points of the Pentagon you've got so far, and using circles of radius AB, mark on the circumcircle the remaining two points of the Pentagon.

This is a quite complex construction, even if it is short.  It is algebraic in inspiration, since we base it on the theoretical length of the diagonals.  The various alternative constructions of Pentagons are all extremely informative about the properties of pentagons, and the concepts of geometric rule-and-compass constructions generally.

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