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Barukh
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Equilateral Triangle Construction
« on: Mar 29th, 2004, 4:39am » |
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1. Construct an equilateral triangle so that its vertices lie on 3 given parallel lines.
2. Construct an equilateral triangle so that its vertices lie on 3 given circles.
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Sir Col
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What a beautiful problem, Barukh! Where do you find such gems?
I solved the first question by working backwards...
Consider the diagram below.
Triangle ABC is right angled and AB is perpendicular to the 3rd parallel line.
The hypotenuse length AC is chosen such that when the triangle is rotated 60o about A, length AC' joins the 1st and 2nd parallel lines.
It should be clear that the angle CAC' is 60o, and as AC=AC', we have two sides of an equilateral triangle, ACC'.
The construction is simple...
Begin with a point A on the 2nd parallel line and draw line perpendicular to the 2nd parallel line through A to find B.
Rotate AB through 60o angle to form AB'.
By drawing a perpendicular to the AB' we locate C' on the first parallel line.
It follows that a perpendicular bisector of AC' will locate the point C on the 3rd parallel line.
When I get a little more time I'll have a look at the 2nd problem – I'm already looking forward to it! 
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| « Last Edit: Mar 29th, 2004, 1:18pm by Sir Col » |
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Sameer
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Re: Equilateral Triangle Construction
« Reply #2 on: Mar 29th, 2004, 3:38pm » |
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Hmm I am just wondering if there is any flaw in this one ::
Given point A on the middle parallel line and a length L. Use compass to draw an arc that intersects both the top and bottom parallel lines which give you the other two vertices.
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"Obvious" is the most dangerous word in mathematics.
--Bell, Eric Temple
Proof is an idol before which the mathematician tortures himself.
Sir Arthur Eddington, quoted in Bridges to Infinity
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BNC
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Re: Equilateral Triangle Construction
« Reply #3 on: Mar 29th, 2004, 4:02pm » |
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on Mar 29th, 2004, 3:38pm, Sameer wrote:Hmm I am just wondering if there is any flaw in this one ::
Given point A on the middle parallel line and a length L. Use compass to draw an arc that intersects both the top and bottom parallel lines which give you the other two vertices. |
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I think that for arbitrary L, you'll get an isosceles rather than equilateral triangle. Do you have a method to determine a specific L that would create an equilateral triangle?
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How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
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Sameer
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Re: Equilateral Triangle Construction
« Reply #4 on: Mar 30th, 2004, 2:36pm » |
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Well I am having a hard time figuring out how it would be isosceles unless I am understanding the problem wrong or I am doing something wrong  . Here is some more explanation:
Considering Sir Col's figure let's work backwards. Let's say we can to make a equilateral triangle ACC' of length L=AC.
Now we can plot a random point A on middle parallel line.
Since AC=AC'=L the points C and C' have to lie on a circle with center A. Using compass of length L and center A plot two arcs that intersect the top and bottom parallel lines which would be C' and C respectively.
Now since you have your three points you can construct the triangle ACC'.
THe fact that triangle exists is that L should be greater than or equal to the distance between the middle line to the farthest parallel line.
Am I thinking something wrong here? I am seeing Sir Col's method as what we used to use in Engineering drawing for projection of points and lines on planes but this problem doesn't sound the same. 
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"Obvious" is the most dangerous word in mathematics.
--Bell, Eric Temple
Proof is an idol before which the mathematician tortures himself.
Sir Arthur Eddington, quoted in Bridges to Infinity
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Sir Col
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Re: Equilateral Triangle Construction
« Reply #5 on: Mar 30th, 2004, 3:18pm » |
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As BNC said, your method would certainly form an isosceles triangle with AC=AC', but there are infinitely many such constructions. Unless you do something else in your construction you will not ensure that CC'=L, which is necessary in forming an equilateral triangle.
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Sameer
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Re: Equilateral Triangle Construction
« Reply #6 on: Mar 30th, 2004, 3:56pm » |
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I gotcha now!!! Cool one!
Sir Col How did you get your method? Any proofs?
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"Obvious" is the most dangerous word in mathematics.
--Bell, Eric Temple
Proof is an idol before which the mathematician tortures himself.
Sir Arthur Eddington, quoted in Bridges to Infinity
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Sir Col
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Re: Equilateral Triangle Construction
« Reply #7 on: Mar 30th, 2004, 11:23pm » |
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A puzzle solver's mental wanderings...
I discovered the idea by accident. I started originally by drawing an equilateral triangle with its vertices on three parallel lines (as above), then drawing a box around it (imagine adding a segment perpendicular to the bottom parallel through C up to the top parallel, and extending BA to the top parallel). This produced three right angle triangles. Using the Pythagorean Theorem, I was trying initially to find a relationship between the distances between the parallels and the side of the equilateral triangle, hoping this would present some insight. Then in my mind's eye I imagined the bottom left right angle triangle, ABC, rotating 60o and I could see that the hypotenuse was the other side of the required equilateral triangle.
Thus was born a solution! 
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Barukh
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Re: Equilateral Triangle Construction
« Reply #8 on: Mar 31st, 2004, 5:36am » |
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Well done, Sir Col  !!!
After reading your explanations, I think you've got everything to solve the second problem as well, and maybe even too much
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Sameer
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Re: Equilateral Triangle Construction
« Reply #9 on: Mar 31st, 2004, 11:36am » |
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I think I was looking at it the wrong way but only with a slight modification it can be used to answer the problem in question.
Given point A and length L of equilateral triangle, make a arc intersecting one of the other parallel lines (let's say the bottom; and referring to Sir Col's figure) gives us point C.
Now taking center as C and length L make a arc intersecting the remaining parallel line (in our case the topmost) to intersect in C' giving you the three points of equilateral triangle.
All I had to do was change the center while getting the points.
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"Obvious" is the most dangerous word in mathematics.
--Bell, Eric Temple
Proof is an idol before which the mathematician tortures himself.
Sir Arthur Eddington, quoted in Bridges to Infinity
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Barukh
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Re: Equilateral Triangle Construction
« Reply #10 on: Mar 31st, 2004, 11:47pm » |
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[on Mar 31st, 2004, 11:36am, Sameer wrote:I think I was looking at it the wrong way but only with a slight modification it can be used to answer the problem in question.
Given point A and length L of equilateral triangle… |
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Sameer, the problem with your solution is that you are not given the length L! And there is just one specific value for L (for a given configuration of parallel lines) that will result in an equilateral triangle.
So, you cannot accomplish your construction before you have a way to “construct” the length L.
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Sameer
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Re: Equilateral Triangle Construction
« Reply #11 on: Apr 1st, 2004, 7:14am » |
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Ahaaa... then you need Sir Col's method to find out L. *sigh* hard skull needs lot of pounding to get things in... 
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"Obvious" is the most dangerous word in mathematics.
--Bell, Eric Temple
Proof is an idol before which the mathematician tortures himself.
Sir Arthur Eddington, quoted in Bridges to Infinity
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Aryabhatta
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Re: Equilateral Triangle Construction
« Reply #12 on: Oct 15th, 2004, 1:19pm » |
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Here is a nice solution to the 3 parallel lines problem.
Lines are L1, L2 and L3. L2 between L1 and L3.
Pick an arbitrary point Q on L2. Draw AQ and BQ so that AQ and BQ make 60o with L2 and A lies on L1 and B lies on L3 and A and B are on the same side of the perpendicular line (to L2) through Q. Pick a point C on L2 such that angle ACB is 60o. ABC is the required triangle!
This method can be modified to construct a triangle on 3 parallel lines such that the triangle has given angles [alpha], [beta] and [gamma].
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Barukh
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Re: Equilateral Triangle Construction
« Reply #13 on: Oct 16th, 2004, 10:30am » |
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Aryabhatta, that's a really nice solution! I haven't seen it before. Thanks for posting it here.
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Grimbal
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Re: Equilateral Triangle Construction
« Reply #14 on: Oct 16th, 2004, 3:37pm » |
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For 3 concentric circles:
Black are the initial circles, gray the intermediate lines and blue and red 2 solutions.
This one is for concentric circles (I misunderstood the problem). But in fact, they don't have to be concentric. The trick is to choose a point on one circle, and then, to move one circle 60° left and the other 60°right around that point. The matching intersections and the choosen point form a 60° angle. And the distance to the choosen point is equal.
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rloginunix
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Re: Equilateral Triangle Construction
« Reply #15 on: Oct 19th, 2013, 9:17pm » |
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Guys, I've found a different invariant that solves this problem: two circles bisecting a perpendicular between any two parallels cut each other at points that are exact middles of the side opposite to the vertex on the third parallel.
If you want more explanations and sketches please go to (remove the spaces for a valid address - for some reason this site won't let me post the link):
this link
Sorry for providing the link - it's a small part of a bigger thing. In my articles I'm interested not in the answers themselves but rather in the ways one can find them. You may find other things of interest there.
This is an eleven-step construction, but I think that number can be reduced.
Outline: random point on any parallel, perpendicular through it, two circles bisecting any segment between any two lines, line through their intersection point and a point on the remaining parallel, perpendicular to that last line locates two remaining vertexes.
// Made it a proper link --towr
[edit]
Made the link to my site more inconspicuous.
[/edit]
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| « Last Edit: May 21st, 2014, 9:06am by rloginunix » |
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towr
wu::riddles Moderator
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Re: Equilateral Triangle Construction
« Reply #16 on: Oct 20th, 2013, 6:59am » |
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on Oct 19th, 2013, 9:17pm, rloginunix wrote:| (remove the spaces for a valid address - for some reason this site won't let me post the link) |
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Yeah, sorry about that, spammers make us make things harder for everyone. It's a proper link now.
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Wikipedia, Google, Mathworld, Integer sequence DB
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rloginunix
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Re: Equilateral Triangle Construction
« Reply #17 on: Oct 20th, 2013, 5:48pm » |
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Thank you.
I'll try to insert the image from my site so that folks don't have to switch to a different tab in the browser:
Point A is picked at random. Perpendicular to the left parallel through A produces B and C. Circles bisecting segment BC produce two points. One of them is D. D is the middle of the side opposite to vertex A.
Line AD. Perpendicular to AD locates E and F - remaining two vertexes...
[edit]
Moved the drawing file to this forum.
[/edit]
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| « Last Edit: Apr 5th, 2014, 4:17pm by rloginunix » |
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rloginunix
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Re: Equilateral Triangle Construction
« Reply #18 on: Oct 25th, 2013, 6:30pm » |
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My previous construction took 11 steps to accomplish. I came up with a shorter one - 8 steps.
Anyone interested in my reasoning please follow this link.
Invariant this time - an equilateral triangle built on the outer parallels locates two vertexes one of which is on the inner parallel:
In the sketch above the triangle AGH locates A and C. That observation lead me to the following 8-step construction (the lines are numbered):
[edit]
Moved the drawing files to this forum.
[/edit]
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| « Last Edit: Apr 5th, 2014, 4:22pm by rloginunix » |
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