[bonanova]

OK this is a little off beat, but it points out an interesting geometrical result.

The problem is to [describe how to] construct an equilateral triangle.

Your tools are

1. Pencil
   
2. Straight edge
   
3. A device that trisects angles
   

I can think of two ways - one is pretty simple.

Post it if you think of it, but there's another way that's perhaps unexpected. [same tools]

Using the pencil as a pencil, the straight edge as a straight edge...

i.e. no arcs, etc.

[Guest]

1-Will we construct the triangle on a paper by using the pencil? or we don't have a paper, and construct a real eq. triangle, not an imaginary one on a paper?

2- The angle trisector can divide 180^o to 60^o

I don't suppose it to be such easy, I've missed something???

On a straight horizontal line, making two 60 degree angles (one is towards upper left, other is towards upper right) will intersect while making an eq. triangle???

[bonanova]

1-Will we construct the triangle on a paper by using the pencil? or we don't have a paper, and construct a real eq. triangle, not an imaginary one on a paper?

2- The angle trisector can divide 180^o to 60^o

I don't suppose it to be such easy, I've missed something???

On a straight horizontal line, making two 60 degree angles (one is towards upper left, other is towards upper right) will intersect while making an eq. triangle???

Yeah, that's the "pretty simple" way.

Draw a straight line

Mark a point on the line, which becomes the apex of a straight angle.

Trisect that angle, and draw a line along one of the 60^o angles.

Mark a point somewhere on that line, and do another trisection.

Extend one of the new 60^o lines back to the first line.

That creates a 60-60-60 [equilateral] triangle.

And, yes, assume you have some paper. Sorry.

Now what's the other way?

[Guest]

Now what's the other way?

my method was slightly different and since it's not harder, it can't be what you look for:

Draw a straight line. As you did, draw a line at 60 degree to upper left. Then go some otherwhere towards left on the first line, take another point on it. Make a 60 degree line towards upper right. These upper left and upper right lines will intersect at 60 degree, making a eq. triangle??

[bonanova]

my method was slightly different and since it's not harder, it can't be what you look for:

Right - it's not a variant of that approach at all.

[Guest]

Right - it's not a variant of that approach at all.

Another try, even more simpler?

If the paper we got is a regular paper, that is square or rectangle, it should have a right angle corner:

Fold the paper obliquely around corner c, define two points that are a,b. As you folded the paper, ac and bc lengths will be equal.

Place the medial arms of the trisector to a and b. Lean the straight edge to back of trisector. Total angle of trisector=180, each angle of it is 60, the angle around "a" is 120/60 (look at the medial quadrangle). Thus the shaded area is an equlateral triangle.

Though I don't think this to be the answer, only a variation.

[Guest]

I would start the same way, but create my 60 degree angles in the same direction creating parallel lines. Now fold the paper in half such that the 180 degree line doubles itself somewhere between the start points of the other angles. When you look through the paper, you should see a nice equilateral triangle formed from the overlapping lines.

Something like this:

// folded to become X if that makes any sense.

[Guest]

Draw a horizontal line with the straight edge. Put the straight edge at the midpoint of the horizontal line and draw another line perpendicular to the first. Put the straight edge on one end of the horizontal line and draw a slanted line that connects to the top of the perpendicular line. Repeat on the other end of the horizontal line. You wind up with a triangle with all three sides the same.

Also since we're not allowed a ruler here, perhaps it is cheating to blithely find the midpoint. OTOH, without a ruler, we'd never know for sure the triangles had lines the right length.

[Guest]

Draw a horizontal line with the straight edge. Put the straight edge at the midpoint of the horizontal line and draw another line perpendicular to the first. Put the straight edge on one end of the horizontal line and draw a slanted line that connects to the top of the perpendicular line. Repeat on the other end of the horizontal line. You wind up with a triangle with all three sides the same.

Also since we're not allowed a ruler here, perhaps it is cheating to blithely find the midpoint. OTOH, without a ruler, we'd never know for sure the triangles had lines the right length.

First off, Welcome to the Den! Please keep in mind the use of spoilers is recommended when posting solutions. Information on the topic is found here.

You can ensure that the angles are the same with the given tools, so the sides must be equal as well. Math is cool.

edit: clarification

Edited December 24, 2008 by Grayven

[bonanova]

Draw a horizontal line with the straight edge. Put the straight edge at the midpoint of the horizontal line and draw another line perpendicular to the first. Put the straight edge on one end of the horizontal line and draw a slanted line that connects to the top of the perpendicular line. Repeat on the other end of the horizontal line. You wind up with a triangle with all three sides the same.

Also since we're not allowed a ruler here, perhaps it is cheating to blithely find the midpoint. OTOH, without a ruler, we'd never know for sure the triangles had lines the right length.

Your tools don't include a means of constructing a perpendicular.

Your slanted line must be at 60^o.

How will you ensure this?

[Guest]

Your tools don't include a means of constructing a perpendicular.

Your slanted line must be at 60^o.

How will you ensure this?

As long as I have a straight edge of finite length I can construct a square with all sides equal; then I can connect the corners of the square with diagonals and the point where they cross is the center of the squre. A line drawn from that center to any side is perpendicular.

[bonanova]

As long as I have a straight edge of finite length I can construct a square with all sides equal; then I can connect the corners of the square with diagonals and the point where they cross is the center of the squre. A line drawn from that center to any side is perpendicular.

A square is a figure with four equal sides.

How do you draw equal length lines using the available tools?

A straight edge is only that - don't assume it's a ruler.

But OK, I'll give you that - use the entire length of a finite-length straight edge.

If you can use a straight edge in any meaningful way, it must have finite length.

A square has another property: four right angles.

How do your construct right angles using the available tools?

A line drawn from the center to any side is perpendicular to the side iff it's parallel to two sides.

How do you construct a parallel line with the available tools?

But more to the point, your post would only construct an isosceles triangle.

We need an equilateral triangle.

[Guest]

How do you construct a parallel line with the available tools?

Can we assume that the legs on the trisector tool are the same length, as would seem prudent from a design standpoint?

If so, all one needs to do is use the straightedge to draw a line. Place the trisect tool on the line and describe your 60degree angles. Leaving the tool in place, place the straightedge across the ends of the legs. You can then draw a paralell line across the straightedge. Depending on the design of the trisecting tool, you could stop right there, as the legs and the straighedge create an equilateral triangle. Again, this is all based on an assumption.

[bonanova]

Can we assume that the legs on the trisector tool are the same length, as would seem prudent from a design standpoint?

If so, all one needs to do is use the straightedge to draw a line. Place the trisect tool on the line and describe your 60degree angles. Leaving the tool in place, place the straightedge across the ends of the legs. You can then draw a paralell line across the straightedge. Depending on the design of the trisecting tool, you could stop right there, as the legs and the straighedge create an equilateral triangle. Again, this is all based on an assumption.

I have no idea what an angle trisector looks like.

The OP says use it [only] to trisect angles.

But, given a line, put the trisector on it, trisect it for a 60^o line, move down the 60^o line and trisect it.

One of those 60^o lines will be parallel to the first line.

We're pretty close to the first method, disclosed in one of the early posts.

But we need to find the second method.

Or, given there might be a third method, any method that's fundamentally different from method 1.

[Guest]

I have no idea what an angle trisector looks like.

The OP says use it [only] to trisect angles.

But, given a line, put the trisector on it, trisect it for a 60^o line, move down the 60^o line and trisect it.

One of those 60^o lines will be parallel to the first line.

We're pretty close to the first method, disclosed in one of the early posts.

But we need to find the second method.

Or, given there might be a third method, any method that's fundamentally different from method 1.

The trisector could trisect a 360 degree angle (or a slightly smaller angle within a reasonable margin of error). Trisecting the 180 degree angle from the same base would produce the second side. The length wouldn't matter as long as the two intersect both the base and each other.

[Guest]

I have no idea what an angle trisector looks like.

The OP says use it [only] to trisect angles.

But, given a line, put the trisector on it, trisect it for a 60^o line, move down the 60^o line and trisect it.

One of those 60^o lines will be parallel to the first line.

We're pretty close to the first method, disclosed in one of the early posts.

But we need to find the second method.

Or, given there might be a third method, any method that's fundamentally different from method 1.

I suppose my paper folding solution wasn't fundamentally different. The same effective solution could be achieved without folding. I'm still trying to imagine what you might be after.

using 120degree angles to create a hexagon, then connect all the opposite corners through the middle point to make triangles, but how to ensure sides the same length?

It seems you must start with a line trisected into 60's. From there, my mind gravitates toward the simple solution. The next step eludes me thus far.

Is there another way to begin that we need to look for?

[HoustonHokie]

using 120degree angles to create a hexagon, then connect all the opposite corners through the middle point to make triangles, but how to ensure sides the same length?

It seems you must start with a line trisected into 60's. From there, my mind gravitates toward the simple solution. The next step eludes me thus far.

Is there another way to begin that we need to look for?

Can we assume "normal" paper was supplied (i.e., rectangular)? If so, we might start by trisecting one of the corners into 3 15-degree angles and go from there. Actually, you could trisect two of the corners on the "short edge" of the paper into 15-degree angles and extend the 30-degree lines until they meet. Then you'd have an equilateral triangle consisting of one edge of the paper and two constructed lines. Is this approach legal, and if so, is it "fundamentally" different?

[Guest]

Can we assume "normal" paper was supplied (i.e., rectangular)? If so, we might start by trisecting one of the corners into 3 15-degree angles and go from there. Actually, you could trisect two of the corners on the "short edge" of the paper into 15-degree angles and extend the 30-degree lines until they meet. Then you'd have an equilateral triangle consisting of one edge of the paper and two constructed lines. Is this approach legal, and if so, is it "fundamentally" different?

I think you intended 30⁰ when saying 15-degree and 60⁰ when saying 30-degree. This is nearly the same with first method. Extending two 60_⁰ angles where they meet.

Bonanova wants a ingenious or magical method

I propose him to make some restrictions. So that we will not be able to use first method, othervise we will all stroll around first method.

My paper folding method seemed me different but not admitted, so some restrictions will help us.

[bonanova]

I think you intended 30⁰ when saying 15-degree and 60⁰ when saying 30-degree. This is nearly the same with first method. Extending two 60_⁰ angles where they meet.

Bonanova wants a ingenious or magical method

I propose him to make some restrictions. So that we will not be able to use first method, othervise we will all stroll around first method.

My paper folding method seemed me different but not admitted, so some restrictions will help us.

Hmmmmm .... restrictions.

Well, I can start you off on the solution, but restrictions are just

use the pencil as a pencil, the straight edge to make sure a line

you want to be straight is straight, and the angle trisector to give

you the two lines that trisect any angle you may have drawn.

Now the method ... if this helps:

1. Draw at least one straight line.
   
2. Make some angles.
   
3. Trisect some or all of the angle(s)
   
4. Make an equilateral triangle.
   

That adds some restrictions,I guess.

Like, don't fold the paper.

Don't try to draw arcs or circles.

Don't use the [assumed square] corners of the paper.

Don't give up.

Does that help?

[Guest]

it will be scalene, trisect and use the meeting points to give your equialteral - I'll try a pic if needs be - assuming the trisector is usable

I think this is basic theory ???

[bonanova]

it will be scalene, trisect and use the meeting points to give your equialteral - I'll try a pic if needs be - assuming the trisector is usable

I think this is basic theory ???

Yeah, that's it.

Morley's theorem. Pretty cool.

[Guest]

Yeah, that's it.

Morley's theorem. Pretty cool.

Finally...

Without being aware of this theorem, I don't think that anybody could solve this, unless being a genious.

Bonanova supposes that we are genious. Maybe LIS is so, but I suggest he was aware of theorem, or got some help.

If OP had told that the trisector could work for only in acute angles (<90⁰), our first solutions wouldn't work, and maybe would lead us to get closer to solution. Though nothing can halp me to discover it.

[Guest]

That's an approach I obviously hadn't considered LIS. That is a very cool geometric result which makes total sense in retrospect.

the angles in a scalene triangle, you will find center (the point equidistant from all three sides). It makes sense that as you create angles that move away from those lines proportionately, the resultant intersections should move away from center at an equal rate, if my visualization is accurate.

[Guest]

I am guessing no compasses allowed?

[Guest]

I am guessing no compasses allowed?

The OP allows a pencil, straightedge, and a "tool that trisects angles".

edit: and paper!

edit edit: And welcome to the DEN! Enjoy the ride.

Edited December 31, 2008 by Grayven

[Guest]

Finally...

Without being aware of this theorem, I don't think that anybody could solve this, unless being a genious.

Bonanova supposes that we are genious. Maybe LIS is so, but I suggest he was aware of theorem, or got some help.

If OP had told that the trisector could work for only in acute angles (<90⁰), our first solutions wouldn't work, and maybe would lead us to get closer to solution. Though nothing can halp me to discover it.

I'd accept geometrically gifted at best - it wa a puzzle, along with others when I was 15.. Practice for exams the following year. Thankfully got this right, but we started with a scalene triangle.