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TimMann
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Obtuse triangles
« on: Oct 30th, 2002, 7:09pm » |
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An obtuse triangle has one of its angles greater than 90 degrees. An acute triangle has all of its angles less than 90 degrees. (A right triangle is neither acute nor obtuse.)
The problem: either find a way to cut up any obtuse triangle into pieces, each of which is an acute triangle, or prove that it can't be done.
This one is from an old Martin Gardner book. Gardner claims professional mathematicians have been known to get it wrong. Then again, professional mathematicians get the Monty Hall problem wrong sometimes too.
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| « Last Edit: Nov 6th, 2002, 11:54pm by TimMann » |
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Icarus
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Re: New puzzle: Obtuse triangles
« Reply #1 on: Oct 30th, 2002, 9:29pm » |
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Cute. Almost fooled me -- three times! Then just when I'm sure of myself, I think yeah, but Lobatchevsky can do it! No, I was wrong again!
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TimMann
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Re: New puzzle: Obtuse triangles
« Reply #2 on: Oct 30th, 2002, 10:46pm » |
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Here's Gardner's followup problem: Cut up a square into the smallest possible number of acute triangles.
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TimMann
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Re: New puzzle: Obtuse triangles
« Reply #3 on: Oct 31st, 2002, 9:39pm » |
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on Oct 30th, 2002, 9:29pm, Icarus wrote:| Cute. Almost fooled me -- three times! Then just when I'm sure of myself, I think yeah, but Lobatchevsky can do it! No, I was wrong again! |
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Heh, that's pretty cryptic. Does the Lobatchevsky reference mean that you solved the problem in hyperbolic geometry too? 
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Chronos
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Re: New puzzle: Obtuse triangles
« Reply #4 on: Nov 1st, 2002, 12:30pm » |
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Need we have a finite number of pieces when we're done?
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James Fingas
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Re: New puzzle: Obtuse triangles
« Reply #5 on: Nov 1st, 2002, 2:05pm » |
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Hmm...
Let's make a deal: I'll solve that one for you if you'll solve this one for me:
Chop any right triangle into acute triangles.
And as for the square, I figured out how to do it in 12 acute triangles ... any better?
UPDATE: make that 10 triangles ...
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| « Last Edit: Nov 1st, 2002, 2:26pm by James Fingas » |
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TimMann
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Re: New puzzle: Obtuse triangles
« Reply #6 on: Nov 1st, 2002, 5:50pm » |
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on Nov 1st, 2002, 12:30pm, Chronos wrote:| Need we have a finite number of pieces when we're done? |
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Yes. Very finite indeed. 
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| « Last Edit: Nov 1st, 2002, 5:50pm by TimMann » |
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TimMann
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Re: New puzzle: Obtuse triangles
« Reply #7 on: Nov 1st, 2002, 6:00pm » |
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on Nov 1st, 2002, 2:05pm, James Fingas wrote:Hmm... Let's make a deal: I'll solve that one for you if you'll solve this one for me:
Chop any right triangle into acute triangles. |
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Whom are you wheeling and dealing with here?  The solution for an obtuse triangle works for a right triangle too, of course. Can you do a right triangle with fewer pieces?
Quote:And as for the square, I figured out how to do it in 12 acute triangles ... any better?
UPDATE: make that 10 triangles ... |
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Gardner says he did it in 8. I haven't reproduced that yet, and have managed to resist looking in the answer section. I have a 10-triangle solution too.
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Icarus
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Re: New puzzle: Obtuse triangles
« Reply #8 on: Nov 2nd, 2002, 6:38pm » |
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on Oct 31st, 2002, 9:39pm, TimMann wrote:
Heh, that's pretty cryptic. Does the Lobatchevsky reference mean that you solved the problem in hyperbolic geometry too? |
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I was trying to be a smart aleck and find a hyperbolic solution, but then I remembered that the angle defect of a hyperbolic triangle is proportional to it's area. And since any subtriangle is contained inside the original it must have a smaller area, and thus have an angle sum closer to PI, which spoiled my whole idea (Darn you BOYLAI!  )
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TimMann
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Re: New puzzle: Obtuse triangles
« Reply #9 on: Nov 2nd, 2002, 6:54pm » |
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Ah. It seems like the problem doesn't really make sense if translated into hyperbolic geometry.
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Icarus
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Re: New puzzle: Obtuse triangles
« Reply #10 on: Nov 3rd, 2002, 11:13am » |
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D'oh! You can raise the number of times this one has fooled me to 4 (or 5)! I seem to be in "overlook the obvious" mode, so I will give the solution that has now occured to me, contrary to what I thought before, and maybe someone can point out what I missing this time:
It can be done. For preliminaries, run a bisector out from each angle to the point at which they meet. (Is that the centroid? I don't remember.) Drop a perpendicular from this point to all three sides. This cuts the triangle into 6 small right triangles. To reduce the right angles to acute ones, split the central point into three points, positioned on the 3 angle bisectors a small distance from the original point. By choosing these distances appropriately, the small triangle created by the 3 points is acute. The three perpendiculars are each replaced by two lines eminating from the same point on the side as the perpendicular, but going to each of the two closest middle points. Again, by careful choice of the middle points, the three new triangles formed are acute. Last since the new segments lie on either side of the old perpendicular, they decrease the right angles, making them acute. And if the center triangle is chosen small enough, the other angles of the outside triangles are not enlargened enough to stop being acute. Result: 10 acute triangles.
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TimMann
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Re: New puzzle: Obtuse triangles
« Reply #11 on: Nov 3rd, 2002, 7:01pm » |
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Good try, Icarus! You're still some distance from the minimal number of pieces, though.
I won't give a hint on that one yet, but here's a small hint on the minimal dissection of a square: Icarus's method for dissecting the obtuse triangle may provide inspiration.
By the way, full disclosure here: although I solved the obtuse triangle on my own, I gave up on the square and looked at the answer. 
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Icarus
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Re: New puzzle: Obtuse triangles
« Reply #12 on: Nov 3rd, 2002, 7:41pm » |
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Hey, I'm just happy at this point to get it right at all! Until this morning, I thought it was impossible.  I keep fooling myself in believing I've got it figured out, only to realize later "You Idiot, you missed ...". This problem has proven really hard on my ego!
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| « Last Edit: Nov 3rd, 2002, 7:43pm by Icarus » |
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TimMann
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Re: New puzzle: Obtuse triangles
« Reply #13 on: Nov 4th, 2002, 12:30am » |
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If anyone gets desperate and wants to throw in the towel, see my drawing of the minimal solution for an obtuse triangle. Gardner gives a short proof of minimality, which you can probably reproduce.
I've also drawn the minimal solution for a square. Gardner asserts without giving the proof that if the dissection is to have bilateral symmetry, the two interior points must lie within the region bounded by four red arcs, as shown. He also says that asymmetric solutions are possible with one of the points outside that area as long as it remains outside the two large semicircles. I haven't tried to verify all that.
I used Juno-2 to make the drawings.
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| « Last Edit: Nov 4th, 2002, 12:41am by TimMann » |
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TimMann
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Re: New puzzle: Obtuse triangles
« Reply #14 on: Nov 4th, 2002, 12:39am » |
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p.s. Gardner also gives minimal dissections of the pentagram (6 acute triangles) and Greek cross (20 acute triangles). A Greek cross is a cross with four equal arms, made from five squares assembled in this pattern:
O
OOO
O
I didn't get a chance to work on those before seeing the answers. I haven't made drawings either.
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| « Last Edit: Nov 9th, 2002, 6:55pm by TimMann » |
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Garzahd
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Re: New puzzle: Obtuse triangles
« Reply #15 on: Nov 4th, 2002, 1:43am » |
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Regarding the red arcs, there's a geometry theorem saying that if two points of a triangle form the diameter of a circle, and the third point is on that circle, then the triangle is a right triangle with the hypotenuse as the diameter.
Given this, it's pretty easy to see why the intersections need to lie outside the arcs. If the points are inside, the triangles are obtuse; if the points are on the circles, the triangles are right.
Also... pentagram = 5-pointed star, correct? Not a pentagon with a star inside it?
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James Fingas
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Re: New puzzle: Obtuse triangles
« Reply #16 on: Nov 4th, 2002, 9:32am » |
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I have solved it! It turns out that every obtuse triangle (and every right triangle) can be divided into 7 acute triangles, and that this is the minimum possible number as far as I can tell. I'm down to 9 triangles for the square, but 8 still eludes me--I thought I had it, but I didn't ...
Hint:
If you create a node inside the triangle, it must be the meeting point of at least 5 acute triangles. A node on the edge of the triangle must be the meeting point of at least 3 acute triangles.
Full Solution:
Label the obtuse triangle ABC, with angle ABC being the obtuse angle. Locate a point D and a point E on the line AC, and locate points F and G on top of point B. Draw a line DF, which is perpendicular to AC and intersects point B. Now draw a line EG, which is the same as DB.
Now we are going to move the points D and E apart, keeping DF and EG perpendicular to AC, and keeping points F and G on lines BA and BC respectively. Keep the lengths of DF and EG the same as they are moved apart. When the length of DE is equal to the lengths of DF and EG, then stop.
Notice that DEGF forms a square inside the obtuse triangle, with one side parallel to AC, and the other side touching BA and BC. Side question: is this the largest square that can be inscribed in this obtuse triangle?
Keeping D and E in place, move G and F infinitesmally farther from B, keeping them on the sides BA and BC respectively. Now the triangles ADF and CEG are acute. Cut them off.
You are left with an irregular pentagon. Locate the point H so that it is on the perpendicular bisector of DE, infinitesmally farther from the line DE than half the distance DE.
Now the triangles DHE, EHG, GHB, BHF, and FHD are all acute. Cut them apart.
Now it's one thing to say that it works, but it's quite another thing to prove that it works. Of course, I would never do such a thing
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| « Last Edit: Nov 4th, 2002, 10:44am by James Fingas » |
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James Fingas
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Re: New puzzle: Obtuse triangles
« Reply #17 on: Nov 4th, 2002, 11:48am » |
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Phew, I found the 8-triangle solution for the square. Looking at it now, I can't believe it took me so long.
<insert witty comment using 'obtuse' or 'acute'>
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TimMann
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Re: New puzzle: Obtuse triangles
« Reply #18 on: Nov 4th, 2002, 2:16pm » |
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on Nov 4th, 2002, 1:43am, Garzahd wrote:| Also... pentagram = 5-pointed star, correct? Not a pentagon with a star inside it? |
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Yes, a 5-pointed star.
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