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Barukh
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3 Triangle Construction Problems
« on: Aug 1st, 2014, 3:42am » |
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This thread is for the lovers of ruler-and-compass constructions.
Construct triangle given:
1. Its 3 medians.
2. Its 3 altitudes.
3. Its 3 angle bisectors.
For every problem, determine conditions under which the construction is possible.
There is at least one other thread with a problem of this kind.
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towr
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1) You can tile the plane with triangle with sides equal to the medians, then draw a few lines to get a triangle that has those medians.
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towr
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2) Make a triangle from the altitudes, get the altitudes from that triangle, use those to make another triangle, then scale that to the right size.
This uses the fact that area ~ height * base, so we can convert the altitudes to the side-lengths in the right proportions.
It doesn't seem to works for obtuse triangles, but I haven't yet pondered on why that might be.
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| « Last Edit: Aug 1st, 2014, 8:42am by towr » |
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Barukh
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Re: 3 Triangle Construction Problems
« Reply #3 on: Aug 1st, 2014, 9:18am » |
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on Aug 1st, 2014, 8:29am, towr wrote:| 2) Make a triangle from the altitudes |
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Is it always possible?
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towr
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Re: 3 Triangle Construction Problems
« Reply #4 on: Aug 1st, 2014, 12:45pm » |
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on Aug 1st, 2014, 9:18am, Barukh wrote:Given sides a <= b <= c, you can make a (proper) triangle if a+b > c
The altitudes are proportional to 1/a >= 1/b >= 1/c, to make a triangle of these you need 1/b+1/c > 1/a
Scale is irrelevant, so let's simplify with a=1, then the question becomes: does 1 <= b <= c and 1+b > c imply 1/b+1/c > 1? There's a trivial set of counter-examples with b>2, so the answer is no.
This set of solutions is a lot more limited than I though.
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dudiobugtron
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Re: 3 Triangle Construction Problems
« Reply #5 on: Aug 1st, 2014, 3:43pm » |
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I assume the scale factor (/side length) isn't important, since we aren't given any lengths to work with?
2) It seems like there might actually be two possible triangles, since you aren't told which side of the (orthocentre?) the vertices are on any of the altitudes.
Rather than prove this, though, I will instead outline a method to construct the triangle in either case:
Choose an altitude, and construct (at an arbitrary point) a line perpendicular to it. Find the points where this line meets the other two altitudes, and voila - you have two vertices and one side of your triangle.
Create another line, perpendicular to one of the other altitudes, and passing through the third altitude at the vertex created above. This is the second side, and gives the final vertex required. Create the last side by making a perpendicular line (to the remaining altitude), which should pass through both remaining vertices. If it does, then you have a triangle.
If it doesn't, then I was wrong about there being two possible triangles. In that case (or, if you just want to construct the other possible triangle), carry out the original construction, but reverse the side of the (orthocentre?) that you make the first perpendicular line through.
I apologise if this is incorrect; I tested it on paper and it seemed correct, and it also seems correct just by examination, but I don't have a proper program to test it in. Does anyone know a good ruler/compass construction program that they can link to?
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towr
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Re: 3 Triangle Construction Problems
« Reply #7 on: Aug 2nd, 2014, 12:04am » |
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on Aug 1st, 2014, 3:43pm, dudiobugtron wrote:| I assume the scale factor (/side length) isn't important, since we aren't given any lengths to work with? |
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I'd assume the lengths are the only thing we're given.
If you're given the position of the start and end point, all three problems are just a matter of connecting the start points.
If you're given the lines, then the bisector one is not solvable[edit](based on wrong presumption; so I don't know)[/edit].
NB I also use Cinderella for my constructions.
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| « Last Edit: Aug 2nd, 2014, 12:35pm by towr » |
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Barukh
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Re: 3 Triangle Construction Problems
« Reply #8 on: Aug 2nd, 2014, 6:07am » |
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on Aug 1st, 2014, 12:45pm, towr wrote:
[hide]This set of solutions is a lot more limited than I though. |
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So, your solution can't be considered complete?
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towr
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Nope, it can't. But I found a better one, with the same basic idea. The trouble was finding the inverse for the altitudes, which you can't do in one triangle for all valid combinations of altitudes. But there's simpler ways of finding an inverse.
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| « Last Edit: Aug 2nd, 2014, 11:40am by towr » |
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rloginunix
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Re: 3 Triangle Construction Problems
« Reply #10 on: Aug 2nd, 2014, 3:05pm » |
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Didn't look at any of the previous posts, so if mine is a duplicate I'll remove it.
2). Problem number two - the heights (altitudes).
The method of square areas works again (that's three in a row)! Expressing the square area of the whole through the square areas of its parts (you know the drill, yada, yada, yada ...) we can say that:
a*h(a) = b*h(b) = c*h(c) and hence
a/h(b) = b/h(a)
b = c*(h(c)/h(b)) and hence b/h(a) = c* h(c)/(h(a)*h(b)) which means:
a/h(b) = b/h(a) = c/v, where v = (h(a)*h(b)/h(c))
In other words the {a, b, c} triangle sought after is similar to the {h(b), h(a), v} triangle. From that observation it follows that we can construct the latter triangle - AB'C' (see the drawing below).
Then. Construct the perpendicular to B'C' through A. Draw the Cir(A, h(a) = AC') till it intersects the perpendicular at E locking in h(a) = AE. Construct the line through E parallel to B'C' which will intersect the extensions AB' and AC' at B and C correspondingly:
Three heights (altitudes) triangle construction drawing.
To construct the fourth proportional "v" all we have to do is use B6.P12.
From the above construction analysis follows easily: this construction is possible iff the "triangle's inequalities" stand:
1). h(b) + h(a) > v
2). h(b) + v > h(a)
3). h(a) + v > h(b)
I'd imagine some possibly more convenient rearrangements of the above terms are possible.
[edit]
Hid the chunks of my solution as per Barukh's notice.
[/edit]
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| « Last Edit: Aug 3rd, 2014, 11:57am by rloginunix » |
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rloginunix
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Re: 3 Triangle Construction Problems
« Reply #11 on: Aug 2nd, 2014, 3:21pm » |
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Clarification: I've carried out the above constructions in an honest way.
First, I constructed an upper left triangle ABC to obtain some realistic heights h(a), h(b), h(c).
I then carried these distances with circles to obtain the fourth proportional "v".
I then constructed a new triangle AB'C' carrying the distances with circles, hiding them to avoid the clutter.
Lastly, I've showed all the important lines.
[edit]
ditto
[/edit]
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| « Last Edit: Aug 3rd, 2014, 11:58am by rloginunix » |
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Barukh
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Nice solutions, towr and rloginunix. Attached is another solution, based on the same idea.
on Aug 2nd, 2014, 3:05pm, rloginunix wrote:| Didn't look at any of the previous posts, so if mine is a duplicate I'll remove it. |
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This is fine, but it would be also nice if others don't look at your solution.
Now, when we answered the question "How?", let's answer the question "When?"
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towr
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Re: 3 Triangle Construction Problems
« Reply #13 on: Aug 3rd, 2014, 10:17am » |
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You mean, when can three numbers be the altitudes of a triangle?
Maybe I'm missing something, but isn't it enough for altitudes a<=b<=c if 1/b+1/c > 1/a ? i.e. the reciprocals can form a triangle.
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Barukh
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Re: 3 Triangle Construction Problems
« Reply #14 on: Aug 3rd, 2014, 10:56am » |
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on Aug 3rd, 2014, 10:17am, towr wrote:You mean, when can three numbers be the altitudes of a triangle?
Maybe I'm missing something... |
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No, you don't miss anything on this one. But there are 2 other problems...
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rloginunix
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Re: 3 Triangle Construction Problems
« Reply #15 on: Aug 3rd, 2014, 12:02pm » |
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Though I didn't look at towr's "medians" problem solution I doubt I'll add anything new and still sound half way intelligent. So may be instead someone will find it more interesting not the solution itself but the search for one.
So let's say you are stuck. Drawing on the "Gun with infinite bullets" experience we observe that if it's the current problem you can't solve - there exists a simpler problem you can't solve. Find it. Repeat.
Here we first want to solve Problem Zero (P0): construct a triangle given three linear measures (sides). That we can do. And that gives us the condition of when this construction is possible.
Problem One (P1): construct a triangle given two sides and one median. Two cases are possible: 1). all three given object share a common vertex or 2). not. You basically construct a {c, b, 2*m(a)} ABE triangle. Condition of when follow.
Problem Two (P2): construct a triangle given one side and two medians. Two cases are possible: 1). the side b is included between the m(a) and m(c) or 2). not. I show only 1): you basically construct a {(4/3)m(c), (4/3)m(a), 2*b} FMG triangle. Condition of when follow.
Problem Three (P3): construct a triangle given three medians. You solve this problem in terms of the previously solved one: P1, by constructing a {(4/3)m(c), (4/3)m(a), (4/3)m(b)} triangle using the HMG triangle as your reference: two sides are (4/3)m(c) and (4/3)m(a) and its median is (2/3)m(b). Condition of when follow:
Since 4/3 cancel each other out you get the triangle inequality applied to m(a), m(b), m(c):
a unique solution exists iff |m(a) - m(c)| < m(b) < m(a) + m(c).
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rloginunix
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Re: 3 Triangle Construction Problems
« Reply #16 on: Aug 3rd, 2014, 12:34pm » |
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As far as 3). goes I remember that the situation is very slippery - like a good politician.
On the one hand, using a ruler and a compass alone your success rate of constructing a triangle given its three angle bisectors is equal to the success rate of trisecting an angle (with the same tools). The situation is aggravated even more - you can't do it even if you had a theoretic "angle trisector" tool.
On the other, I vaguely recall a theorem (or some such) that for any three positive numbers there exists a triangle whose angle bisectors' lengths are equal to those numbers. But I'm not so sure. Someone with a heavy duty math knowledge must step in.
I think at this point Barukh excavated my cranium cavity pretty well.
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towr
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Re: 3 Triangle Construction Problems
« Reply #17 on: Aug 3rd, 2014, 12:41pm » |
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on Aug 3rd, 2014, 10:56am, Barukh wrote:| No, you don't miss anything on this one. But there are 2 other problems... |
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I figured for the first one the construction spoke for itself, since it goes both ways.
I still can't get a fix on the last one, neither how nor when.
[edit]According to this it can't be done in general. (Like rloginunix also said.) But, it can certainly be done in some cases, e.g. when they are the same length.[/edit]
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| « Last Edit: Aug 3rd, 2014, 12:57pm by towr » |
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rloginunix
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Re: 3 Triangle Construction Problems
« Reply #18 on: Aug 3rd, 2014, 1:21pm » |
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towr, by Steiner–Lehmus theorem if two angle bisectors are equal in length then the triangle is isosceles.
May be that can help.
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dudiobugtron
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Re: 3 Triangle Construction Problems
« Reply #19 on: Aug 3rd, 2014, 6:04pm » |
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Thanks for the links rloginunix! 
on Aug 2nd, 2014, 12:04am, towr wrote:| I'd assume the lengths are the only thing we're given. |
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Oooh, right. I've never heard those terms refer specifically to a length before. So, for each question, the information you are given is just the lengths of three line segments?
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rloginunix
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Re: 3 Triangle Construction Problems
« Reply #20 on: Aug 3rd, 2014, 6:25pm » |
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Yes, the lengths.
In ruler and compass problems the word "length" is usually omitted but always assumed.
"Construct ... given ... sides/medians/altitudes/bisectors/radii" means given the line segments of certain lengths that represent the sides/medians/altitudes/bisectors/radii.
At first you assume that these lengths are "good" - the object at hand is constructable. Then you find out the how - a finite series of steps. That search should provide enough information to analyze when the construction is possible and how many solutions it can yield.
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Barukh
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Re: 3 Triangle Construction Problems
« Reply #21 on: Aug 4th, 2014, 2:19am » |
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Here
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Littleton
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Re: 3 Triangle Construction Problems
« Reply #22 on: Nov 25th, 2014, 2:40am » |
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23
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