[Guest]

having produced your equilateral triangle HERE

you now have to put inside three equal size circles touching each other and two sides of the triangle - you have only a compass - none of the circles overlap each other r triangle... Nice snug fit!

Also easy peasy,

Edited December 31, 2008 by Lost in space

[HoustonHokie]

I didn't mean that. Of course the red line is 1/2 of blue line.

But my question is that:

How do you know that centre of the circle need to be at that point? Which calculation gave you the information that the centre is "1/2 of blue line" apart from blue line?. Sorry for wording.

What you're driving at is how I know that the radius point of the circle is exactly 1/2 the base length from the base. Here's the explanation:

The centers of the three circles have to form another equilateral triangle. The lengths of the sides of this smaller triangle are equal to the diameter of the circles. And the new triangle is offset from the old triangle by a distance equal to the radius of the circle. So, the new side length is twice the offset distance. So, if R = radius, S1 = length of original triangle side, and S2 = length of new triangle side, then:

S2 = 2 * R

tan(30) = R / [(S1 - S2) / 2]

By substituting S2/2 for R, you get that S2 = S1 / [1 + 1/tan(30)].

To get the distance from the midpoint of one base to the center of the opposing circle, define 2 distances: X1 is the offset distance between the two triangles, and X2 is the distance from the midpoint of the base of the interior triangle to the opposing vertex. Now:

X1 = R = S2/2

X2 = S2 sin(60)

X1 + X2 = S2 * [1/2 + sin(60)] = S1 / [1 + 1/tan(30)] * [1/2 + sin(60)] = S1 / 2, which is what we were looking for. Thus the need for construction of an arc with radius point at the midpoint of one side and radius equal to half the side length.

By the way, another interesting property forms: the distance from a vertex of the large triangle to the center of one of the two opposing circles is S1 * √2, which means that the radius of a arc from the vertex of the original triangle through the center of one of the small circles is S1 * √2. Not sure that helps much, but I'll just throw it out there for whatever it's worth because that arc might show up in a solution.

Edited January 5, 2009 by HoustonHokie

[Guest]

can be accomplished using only a compass and a series of iterations. Starting with the bottom side of the triangle, put the point of the compass on one end of the line and set the compass to a size just smaller than the length of that line. Strike an arc inside the triangle, and move the point to the other end of the line. Without changing the compass setting, strike another arc intersecting the first arc. Adjust the compass to a slightly smaller setting and repeat. Continue until you have several intersecting arcs inside the triangle. From there you should be able to connect the intersections with a straight line (you'll have to freehand it, since no straightedge was allowed, but if you have enough pairs of arcs, it should be fairly easy). Then repeat the whole process for the other two sides to get your three bisecting perpendiculars, from which you can follow HH's method of creating the 3 circles. I hope my explanation was clear, for I have no good way to illustrate what I'm talking about with this computer. Perhaps tomorrow I'll sketch something at work.

Note that as your arcs get closer to a radius that is half the length of the side, very small changes in your compass setting will result in more evenly spaced intersections to connect. And, theoretically, with infinitesimal changes to the compass setting, you will find the distance that is exactly half the length of the side when the two arcs drawn do not intersect, but are tangent to each other.

And thanks for the explanation, HH. I had that all sketched out as well and used the software to measure the distance from a side to the center of the opposite circle, and it confirmed that it was half the length of one of the sides, but I was still trying to come up with the mathematical proof. Sorry for misinterpreting your question, nobody.

Edited January 5, 2009 by lazboy

[Guest]

What you're driving at is how I know that the radius point of the circle is exactly 1/2 the base length from the base. Here's the explanation:

The centers of the three circles have to form another equilateral triangle. The lengths of the sides of this smaller triangle are equal to the diameter of the circles. And the new triangle is offset from the old triangle by a distance equal to the radius of the circle. So, the new side length is twice the offset distance. So, if R = radius, S1 = length of original triangle side, and S2 = length of new triangle side, then:

S2 = 2 * R

tan(30) = R / [(S1 - S2) / 2]

By substituting S2/2 for R, you get that S2 = S1 / [1 + 1/tan(30)].

To get the distance from the midpoint of one base to the center of the opposing circle, define 2 distances: X1 is the offset distance between the two triangles, and X2 is the distance from the midpoint of the base of the interior triangle to the opposing vertex. Now:

X1 = R = S2/2

X2 = S2 sin(60)

X1 + X2 = S2 * [1/2 + sin(60)] = S1 / [1 + 1/tan(30)] * [1/2 + sin(60)] = S1 / 2, which is what we were looking for. Thus the need for construction of an arc with radius point at the midpoint of one side and radius equal to half the side length.

By the way, another interesting property forms: the distance from a vertex of the large triangle to the center of one of the two opposing circles is S1 * √2, which means that the radius of a arc from the vertex of the original triangle through the center of one of the small circles is S1 * √2. Not sure that helps much, but I'll just throw it out there for whatever it's worth because that arc might show up in a solution.

Thank you HH. I didn't think to use trigonometry, as it would mess up everything. Your simple explanation is very nice.

[Guest]

And thanks for the explanation, HH. I had that all sketched out as well and used the software to measure the distance from a side to the center of the opposite circle, and it confirmed that it was half the length of one of the sides, but I was still trying to come up with the mathematical proof. Sorry for misinterpreting your question, nobody.

Note that as your arcs get closer to a radius that is half the length of the side, very small changes in your compass setting will result in more evenly spaced intersections to connect. And, theoretically, with infinitesimal changes to the compass setting, you will find the distance that is exactly half the length of the side when the two arcs drawn do not intersect, but are tangent to each other.

An interesting approach - normally the bisection is done by drawing arcs either side using both end points of the line and connecting the line - normally requiring a straight edge - you would be winging it with a best guess and free hand use of a pencil on paper or scribe on steel....

How to get that pinpoint accuracy?

[HoustonHokie]

An interesting approach - normally the bisection is done by drawing arcs either side using both end points of the line and connecting the line - normally requiring a straight edge - you would be winging it with a best guess and free hand use of a pencil on paper or scribe on steel....

How to get that pinpoint accuracy?

I think we can finally get there... It takes a while, but we can get there. Check out the answer to this topic: Going around in circles, but getting where you want to go. The elusive bisection of a line using only a compass is solved, and I think that's all we were missing here

[Guest]

Well HH, do you want the honor of putting a tidy answer in there and call it a wrap? - I thought one of two young ladies may have put this t'bed before you came back to it.

[HoustonHokie]

Well HH, do you want the honor of putting a tidy answer in there and call it a wrap? - I thought one of two young ladies may have put this t'bed before you came back to it.

I thought they might, too - but I guess they've left this one alone. Actually, I've still got a little snag, or I would finish it up.

Obviously, one of the missing pieces was a method to bisect one side of the triangle, which we've got a procedure for now.

But I still need a means for cleanly getting the center of the circles. After bisecting a triangle edge, I have a method for putting one arc through the center of the circles, but it would be really good to have another arc to go through them so I can define the centers based on that intersection and not have to rely on any eyeballed tangents. And to be really precise, I'd like to have a means for getting the exact radius of the small circles by intersections without trying to be tangent to the triangle edge.

I know a radius of S1 * √2 or S1 * (√3 - 1) / 2 will do to find the center of the circles if I try to use one of the triangle corners as the center of my construction arc. But I don't have one of those distances yet (nor can I find a power of 2 multiple of them, which I could eventually subdivide). I'd prefer to use the S1 * √2 distance, because then I'd have a real intersection and not the common point of two tangent arcs. There may be other distances from other points of intersection from the circles I've constructed so far, but none that are working for me yet.

The radius of the small circles is S1 * (√3 - 1) / 4, which is half of S1 * (√3 - 1) / 2 noted above, so I could get that, assuming I could find the center of the circles (with precision). So, I guess it's not quite done yet - at least, not as done as I'd like...

[Guest]

A tidy answer? Maybe...

OK, based on some info from previous posts and some math, we can say (given a triangle with unit length sides):

The height of the triangle is √3/2.

The distance along the perpendicular bisector from one edge of the triangle to the center of the opposite internal circle is 1/2.

The distance from one vertex of the triangle to the center of either opposite internal circle is √2/2 (the hypotenuse of a triangle whose legs are each 1/2).

The radius of the internal circles is half the distance from the center of the circle to the nearest vertex of the triangle = (√3-1)/4 or (√3/2 - 1/2)/2.

OK now to the construction:

Call the vertices of the triangle A, B, and C.

1. Create a circle about C radius CB. Create a circle about B with same radius. Mark the point of intersection as D.

2. Create a circle about D radius DC. Mark the intersection with circle B as point E.

3. Create a circle about E radius EC (this radius is double the height of the triangle = √3). Create a circle about A radius AD (also length √3). Mark the intersection as point F.

4. Notice that the length of BF is √2 (one leg of a right triangle with hypotenuse = √3 and other leg = 1). Construct the midpoint (as above) from B to F, and mark it as G.

5. Now, length of BG is √2/2. Create a circle about B radius BG. This circle will pass through the center of both internal circles opposite from B.

6. Create a circle about A and C also radius BG. This marks the centers of all three internal circles.

7. Construct the midpoint from any interior circle's center to the nearest vertex. This midpoint is on the selected interior circle. You can use this radius to create all three circles.

All in all, I don't think I'd call it easy peasy...

[HoustonHokie]

A tidy answer? Maybe...

OK, based on some info from previous posts and some math, we can say (given a triangle with unit length sides):

The height of the triangle is √3/2.

The distance along the perpendicular bisector from one edge of the triangle to the center of the opposite internal circle is 1/2.

The distance from one vertex of the triangle to the center of either opposite internal circle is √2/2 (the hypotenuse of a triangle whose legs are each 1/2).

The radius of the internal circles is half the distance from the center of the circle to the nearest vertex of the triangle = (√3-1)/4 or (√3/2 - 1/2)/2.

OK now to the construction:

Call the vertices of the triangle A, B, and C.

1. Create a circle about C radius CB. Create a circle about B with same radius. Mark the point of intersection as D.

2. Create a circle about D radius DC. Mark the intersection with circle B as point E.

3. Create a circle about E radius EC (this radius is double the height of the triangle = √3). Create a circle about A radius AD (also length √3). Mark the intersection as point F.

4. Notice that the length of BF is √2 (one leg of a right triangle with hypotenuse = √3 and other leg = 1). Construct the midpoint (as above) from B to F, and mark it as G.

5. Now, length of BG is √2/2. Create a circle about B radius BG. This circle will pass through the center of both internal circles opposite from B.

6. Create a circle about A and C also radius BG. This marks the centers of all three internal circles.

7. Construct the midpoint from any interior circle's center to the nearest vertex. This midpoint is on the selected interior circle. You can use this radius to create all three circles.

All in all, I don't think I'd call it easy peasy...

Nice... Finally!

I pulled out the old engineering compass and tried to make it work. The proof is in the doing, right? And the OP says you have a compass (on steel, as modified, to prevent folding), not a computerized drafting tool that will only draw circles.

After 3 attempts, I gave up. When you combine the 2 methods and get the full procedure, it takes a dozen circle constructions to simply identify the centers of the small circles, and that's only because several of the points required to find point G as the midpoint of BF have already been defined and you can skip several steps in the first bisection procedure. By that point, I was far enough off that the centers were in the wrong place and I didn't even try the second bisection procedure (requiring 7 more constructions) to get the radius of the small circles. And I started with the largest triangle that I could (2" sides) and still make all the points that I required with a 6" compass (largest theoretical radius is 5.66").

So it seems we may have 3 equally valid procedures, given the difficulty of using a real-life compass to construct these tangent circles:

1. the procedure I suggested in post 19

2. the procedure I suggested in post 4, after using sihyunie's bisection procedure (instead of folding) to find the midpoint of one side of the triangle

3. jb_riddler's procedure in post 34

If I had the worthless computerized drafting tool which could only draw circles, I think all three are valid, because the tool would pick out intersections or tangents with exact precision. With a compass, I doubt either of the three will actually get you the three small circles, unless you maybe have a very large bar compass and the means to keep the point and the pencil perpendicular to the plane of the triangle.

It would be interesting to introduce an assumed error into the puzzle and figure out which is the best method to use in real life. For example, assume:

1. the pencil thickness and compass point are both about 1/50 of an inch thick (based on 0.5 mm pencil lead, meaning the distance plotted could be 1/25 of an inch away from true distance)

2. the location of a marked intersection is also about 1/50 of an inch from it's true location (true center of shape defined by two intersecting 1/50 inch-thick arcs)

3. the location of a marked tangent is about 1/20 of an inch from it's true location (true point of tangency, if one exists, or point of minimum distance between non-intersecting circles, or maximum distance between intersecting circles)

For each method, figure out how close the circle centers are likely to be to their theoretical locations, how close the circle radius is likely to be to the theoretical value, and how close the circles are likely to being tangent with each other and with the sides of the triangle.

It should be possible to see how error could propogate through each step of the construction and somehow weigh the relative benefits of each construction method. But that would not be an easy peasy problem, either.

All in all, a very good puzzle... but not at all easy peasy!

[Guest]

Uhmm easy peasy was a dig - thought you guys could take that - may have thrown you off, but very good job - and maximum kudos goes to.... me for setting it - sorry, being tight with the kudos.

[Guest]

I have now read through all of the replies to this intriguing puzzle, and can assure you that the solution is indeed easy peasy, but only if one knows anything about so-called Mascheroni Constructions, which indeed was referred to in one reply, if not by name.

Mascheroni, in 1797, correctly stated that "Any construction which can be made using compasses and a straight rule only, can be made with compasses only." He was not in fact the first person to come to this conclusion, but Mascheroni was not aware of this, and his is the eponym for the construction. The first person who discovered the technique was a Dane by the name of Georg Mohr over a century earler in 1672.

I intend, over the next couple of hours, to provide a simple set of procedures which will arrive at the correct result. What I cannot guarantee is that the procedure I am about to transmit is the simplest, but I think that it is either the simplest or almost so.

[Guest]

I have attached my procedure.

3 Circle Mascheroni.doc

Edited August 18, 2009 by jerbil