Are tangent points allowed in ruler and compass constructions?

[Genady]

In classical constructions with ruler and compass, a construction proceeds from point to point where the points are intersections of lines and circles, i.e., where the lines and the circles cross each other. Are the points where the lines and the circles touch each other rather than cross, allowed as well?

More specifically, if I have constructed two circles and I know that they touch at some point, can I proceed with the construction using this point?

[Genady]

Even more specifically... With ruler and compass I could construct the point where the circles touch as a point where the line connecting their centers cross them. However, what I deal with is a construction using compass only, no ruler. I don't know if constructing that point by crossing only some circles is possible.

[studiot]

Well yes the tangent is perpendicular to the line joining the centres.

So it can be constructed as a perpendicular line anywhere along the common centreline.

Two things.

Firstly constructed using a ruler and compass has two variations.

The otiginal strict meaning was that the ruler could only be used as a straight edge, not for linear measurement.

These days many relax that and allow the use of the ruler as a scale device.

 

Secondly you haven't mentioned if you know the diameters of at least one of the circles ?

[Genady]

4 minutes ago, studiot said:

it can be constructed as a perpendicular line anywhere along the common centreline.

Yes, but as I've clarified in the post before yours, I don't have a centerline, or any line, because I don't have a ruler.

 

5 minutes ago, studiot said:

The original strict meaning was that the ruler could only be used as a straight edge, not for linear measurement.

These days many relax that and allow the use of the ruler as a scale device.

If I had ruler, I would use it only as a straightedge, i.e., in the classical meaning.

 

6 minutes ago, studiot said:

you haven't mentioned if you know the diameters of at least one of the circles

The circle is drawn, and its center is known, so it is not a problem to take its radius with compass. Then, the diameter can be constructed by doubling the radius, which is doable with compass only, i.e., no ruler is needed for this.

[studiot]

I'm not being awkward, but what you are attempting is still not clear to me.

Classical geometric construction is not coordinate geometry.

It starts with a blank sheet.

So what do you mean by the centre is known ?

And what do you mean by I don't have a ruler ?

I'm only trying to help after all.

[Genady]

10 minutes ago, studiot said:

I'm not being awkward, but what you are attempting is still not clear to me.

Classical geometric construction is not coordinate geometry.

It starts with a blank sheet.

So what do you mean by the centre is known ?

And what do you mean by I don't have a ruler ?

I'm only trying to help after all.

I should just formulate the original problem, and all will become clear:

Given two points, A and B, construct a point C, using only a compass, such that B is a midpoint between A and C.

[studiot]

Just now, Genady said:

I should just formulate the original problem, and all will become clear:

Given two points, A and B, construct a point C, using only a compass, such that B is a midpoint between A and C.

Ah I see, thank you I will give it some thought.

[Genady]

Good, but this is not the problem that led to the question in the thread. Here is a solution of this problem:

Spoiler

1. Draw a circle of radius AB with the center at A.

2. Draw a circle of radius AB with the center at B.

3. Mark one of their intersections, D.

4. Draw a circle of radius AB with the center at D.

5. Mark its intersection with the circle of step 2, E.

6. Draw a circle of radius AB with the center at E.

7. Its intersection with the circle of step 2 is the point C.

Now, the problem in question is this:

Given points A and B construct point C in the middle between them using only compass.

I've solved this problem, but my solution relies on being able to mark a point where two circles touch rather than cross each other. I don't know if this is legitimate. That was why I've asked.

[Genady]

I've found that the last problem can be solved also by using intersections of circles, i.e., circles crossing rather than touching at points. It can be done by following constructions in the proof of Mohr–Mascheroni theorem - Wikipedia. Although, these constructions are very long and convoluted. 

[Genady]

And another AI fail. A meaningless "solution":

### Steps to Construct Point C (Midpoint of A and B) Using Only a Compass:

 

1. **Draw a circle centered at A with radius AB:**

   - Place the compass at point A and adjust its width to the distance AB.

   - Draw a circle centered at A with radius AB.

 

2. **Draw a circle centered at B with radius AB:**

   - Without changing the compass width, place the compass at point B.

   - Draw a circle centered at B with radius AB.

 

3. **Find the intersection points of the two circles:**

   - The two circles will intersect at two points. Let’s call these points P and Q.

 

4. **Draw a circle centered at P with radius PA:**

   - Place the compass at point P and adjust its width to the distance PA (which is equal to AB).

   - Draw a circle centered at P with radius PA.

 

5. **Draw a circle centered at Q with radius QA:**

   - Similarly, place the compass at point Q and adjust its width to the distance QA (which is also equal to AB).

   - Draw a circle centered at Q with radius QA.

 

6. **Find the intersection points of the new circles:**

   - The circles centered at P and Q will intersect at two points: one of these points is A, and the other is the midpoint C.

 

7. **Identify the midpoint C:**

   - The intersection point that is not A is the midpoint C between A and B.

 

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[Genady]

Solved. The answer to the OP question is yes, because I've found a simple way to convert a "touching case" into a "crossing case."