Stumped by the simplest geometry problem

[dextercioby]

Yes, one more reason to be humble, I know. This is the simplest problem I couldn't solve so far.

Assume we have a circle of center O, a ruler of arbitrary size and a pencil.
We use the ruler and the pencil to choose 4 points on the circle - the extremities of two diametral/diagonal segments.
They build a rectangle. So far so good.

Using the ruler only and the pencil, how can I draw the segment ED (with E the intersection of a line passing through D and the prolongation of AB) which is parallel to the diagonal segment BC?

Thank you for illuminating me.

 

[scottdave]

Is the distance AB something that you can measure with the ruler, such that you can measure off BE to be the same distance as AB? In order for it to be a parallelogram, you will need for BE to be the same distance as DC, which also is the same as AB.

 

[mfb]

With just an unmarked (?) ruler, all you can do is extending existing lines or drawing new lines through existing points. ABCD has all possible lines drawn already, and extending them outwards does not lead to new intersections that could give new points of relevance. I don't see how you could do anything without a compass or something similar. With a compass it is trivial.

 

[jedishrfu]

Besides extending the lines, you can also draw tangents to the circle. Maybe the solution lies with tangents to the four circle points.

 

[mfb]

Can you do that? It looks a bit problematic to do that with a ruler.

 

[jedishrfu]

Can you do that? It looks a bit problematic to do that with a ruler.

Well its a geometry problem so you can say some line is a tangent line at point X and intersects a line at some other point Y implying you have a right angle and from there you might be able to make an inference that will help you solve it.

 

[TeethWhitener]

Kind of a deep dive into Euclid, but I'm not seeing how it's done. Elements Prop 1.31 gives the construction you want, but it relies on Prop 1.23, which itself requires the construction of several circles.

Well its a geometry problem so you can say some line is a tangent line at point X and intersects a line at some other point Y implying you have a right angle and from there you might be able to make an inference that will help you solve it.

I don't think constructing a tangent to a circle is trivial. Euclid constructs the tangent to a circle by drawing another circle: Prop 3.17

Whether the constructions have to be done like this, I'm not sure. My own tongue-in-cheek solution is a cheat:

Spoiler

Assume the ruler has a finite width less than the radius of the circle and parallel sides. Unfortunately, I think this violates the rules: the ruler is of arbitrary length, but not arbitrary width. For completeness, here would be the rest of the solution:

Spoiler

Put one side of the ruler on the center of the circle and choose points B and D from where the ruler intersects the circle. Construct C and A by passing the straight edge of the ruler through the center of the circle and B and D, respectively. Now place one edge of the ruler on BC and draw a straight line on the other edge. This will intersect CD at its midpoint by nature of the construction of the rectangle ABCD. Repeat this step, placing one edge of the ruler on your new line and drawing another line using the opposite edge of the ruler. Now, this line intersects point D and is parallel to BC.

 

[symbolipoint]

Yes, one more reason to be humble, I know. This is the simplest problem I couldn't solve so far.

Assume we have a circle of center O, a ruler of arbitrary size and a pencil.
We use the ruler and the pencil to choose 4 points on the circle - the extremities of two diametral/diagonal segments.
They build a rectangle. So far so good.

Using the ruler only and the pencil, how can I draw the segment ED (with E the intersection of a line passing through D and the prolongation of AB) which is parallel to the diagonal segment BC?

Thank you for illuminating me.

View attachment 197189

I did not read any of the responses...

I'd say, use a compass. Match the length AB, and mark a small curve distance AB from point B. Now, draw your segment to intersect the small curve. The intersection of the small curve and the "line" you just drew is your point E. NOW connect point E and D using your straight-edge.

 

[jedishrfu]

It says though to use a ruler only

 

[symbolipoint]

Well its a geometry problem so you can say some line is a tangent line at point X and intersects a line at some other point Y implying you have a right angle and from there you might be able to make an inference that will help you solve it.

Clever. Good luck trying to draw the tangent from point D, but some or much of my Geometry is lacking right now.

 

[jedishrfu]

I wasn''t suggesting drawing it but rather defining a tangent line to see where you can go from there.

One idea I had was to fold the paper and use an origami technique to locate the E ie fold along BD and then A and E will coincide and you can draw a line to it.

 

[symbolipoint]

Well its a geometry problem so you can say some line is a tangent line at point X and intersects a line at some other point Y implying you have a right angle and from there you might be able to make an inference that will help you solve it.

In previous post, I misunderstood what you said about the tangent.

My current limitation still makes me want to use a compass.

 

[UW-PurpleHusky]

Is the distance AB something that you can measure with the ruler, such that you can measure off BE to be the same distance as AB? In order for it to be a parallelogram, you will need for BE to be the same distance as DC, which also is the same as AB.

This is correct...
1) "Measure" segment AB (or CD).
2) Align the ruler with the segment AB and continue the line through point B until the line you have drawn is equal to your "measurement".
3) Draw a line from D to the end of the line you just drew in step 2.

 

[symbolipoint]

This is correct...
1) "Measure" segment AB (or CD).
2) Align the ruler with the segment AB and continue the line through point B until the line you have drawn is equal to your "measurement".
3) Draw a line from D to the end of the line you just drew in step 2.

Once the task of "measure" segment AB or CD is done, the tool most indicated would be the compass. I wish someone would make clearer how you use just the straight-edge to make a measurement. I can only understand this: Match the two pointers of the compass to each endpoint length to be taken. One puts the pin pointer on one endpoint and adjust the pencil pointer to the other endpoint, and this then holds the distance between the two endpoints. One might do something 'like' this using a straight edge, but the proper tool for this should be the compass.

 

[dextercioby]

Ok, thank you for your responses. Let me restate my original problem as follows. Using just a circle of arbitrary radius, arbitrary placed center O in the 2D plane, a ruler with no markings on and arbitrary length and a pencil, I must convert a segment AB (i.e. am given A,B and the line between them, again of arbitrary length with respect to the circle's radius and arbitrary placement with respect to the circle's center) into a segment AM 18 times longer than AB and, moreover, the 17 points equally spaced inbetween A and M should be determined (am given only B, the rest of 16, N_1, N_2,..., N_16 should be determined).
Therefore, I need to able to draw parallel lines, 18 of them, using just the given circle and the ruler. See attached.

 

[arydberg]

Yes, one more reason to be humble, I know. This is the simplest problem I couldn't solve so far.

Assume we have a circle of center O, a ruler of arbitrary size and a pencil.
We use the ruler and the pencil to choose 4 points on the circle - the extremities of two diametral/diagonal segments.
They build a rectangle. So far so good.

Using the ruler only and the pencil, how can I draw the segment ED (with E the intersection of a line passing through D and the prolongation of AB) which is parallel to the diagonal segment BC?

Thank you for illuminating me.

View attachment 197189

If your ruler is really of arbitary size all you need is to make the ruler of infinite length. Then extend line BC to the left to infinity. Now place the ruler on the end of BC and point D and draw the line.

( the real answer of course is you cannot do any of these problems without a compass)

 

[Nidum]

@jedishrfu : Besides extending the lines, you can also draw tangents to the circle. Maybe the solution lies with tangents to the four circle points .

 

[Nidum]

@dextercioby: Do these problems that you are posting have any rational origin or purpose ?

 

[bahamagreen]

For those with continuing interest in the first problem, here is my solution assuming only unmarked straight edge and pencil.

Draw a new diameter through circle center O such that the diameter projects out of the circle and
- intersects only the extended line projections of sides AB and CD
- does not intersect either rectangle side AB or CD
(So to be clear, for example, the new diameter might project "left" from O down to below A, and "right" from O up to above D.)

Call these new intersections on the line projection of sides AB and CD as L and R.

Draw the lines B to R and C to L (these lines are parallel).

Those lines also intersect the circle (call the intersections N and S)

The line through N O S is parallel to AB and CD.

The line N O S intersects the rectangle side BD at its midpoint, call it M.

A line through C and M intersects line projection of AB at the point E.

Draw line DE.

 

[mfb]

The line through N O S is parallel to AB and CD.

Why?
Draw the first new line very close to A and D. Then S will be very close to A and N will be very close to D.

There is also a problem earlier, but that can be solved by asking to draw a better line:

Those lines also intersect the circle (call the intersections N and S)

They don't have to. Draw the initial line nearly parallel to AB, and the new lines won't intersect the circle.The approach with the new line looks interesting, but so far I don't see how it would help.

 

[bahamagreen]

Thanks for checking, looks like you are correct. If the midpoint of BD (or any side) can be found then the solution follows; I had tried some different hand drawn tests with rectangles of differing aspect ratio... but not rotated. Noticing angles BNC and BSC will always be right angles... so projecting a new diameter can be used to construct rotated interior rectangles... if only could identify which construction puts NOS parallel to AB and CD.

 

[mfb]

Noticing angles BNC and BSC will always be right angles

Sure - Thales. We don't need the rest of the construction for that.

 

[Aufbauwerk 2045]

Today I learned not to get involved in this sort of thread. I'm not going to even attempt to solve anything here. Free at last!

:)

P.S.

When you say ruler and pencil, it sounds quite odd. We normally use a straight edge, not a ruler. A pencil or some other writing implement is normally assumed.

 

[Nidum]

With just an unmarked (?) ruler, all you can do is extending existing lines or drawing new lines through existing points. ABCD has all possible lines drawn already, and extending them outwards does not lead to new intersections that could give new points of relevance. I don't see how you could do anything without a compass or something similar.

This has to be correct . Solutions involving tangents and auxiliary lines require 'personal judgement' to locate the lines and there is no certainty of an accurate result .