Re: [math-fun] Triangle puzzle
Michael writes: << Dan Asimov wrote:
Determine which shape(s) of triangles, if any, can be dissected into 3 congruent triangular pieces.
Then do the same for 3 *similar* triangles.
Huh? You can cut any right triangle into 2 similar triangles, then do this again to one of these, resulting in the original triangle cut into 3 similar ones. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele
I strongly suspect that the only solution in the congruent case is a 30-60-90 triangle dissected by the perp bisector of the hypotenuse, and the angle-bisector of the 60 angle. /| / | / | / | / | /_____| / \ | / \ | / \ | - \ | - \| Rough heuristic: two congruent triangles fit together to form a parallelogram or kite [since if two edges are to fit against one, rriangle wd be degenerate]. Only in the latter case can the quadrilateral be completed to a triangle. R. On Mon, 21 Jan 2008, Dan Asimov wrote:
Michael writes:
<< Dan Asimov wrote:
Determine which shape(s) of triangles, if any, can be dissected into 3 congruent triangular pieces.
Then do the same for 3 *similar* triangles.
Huh? You can cut any right triangle into 2 similar triangles, then do this again to one of these, resulting in the original triangle cut into 3 similar ones.
--Dan
_____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele
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Huh? Equilateral -> 120-30-30s.
Huh? Equilateral -> 120-30-30s.
That's what I said at first too--it's just the scope of the word "congruent" in the statement is ambiguous. Anyway the presumably-intended harder puzzle is the pieces must also be congruent to the original triangle.
Huh? Equilateral -> 120-30-30s.
That's what I said at first too--it's just the scope of the word "congruent" in the statement is ambiguous.
Anyway the presumably-intended harder puzzle is the pieces must also be congruent to the original triangle.
They might be similar, but they can't possibly be congruent. I think both equilateral -> 120-30-30 and 30-60-90 -> 30-60-90 are solutions to the original puzzle. Only Dan knows for sure, but it's certainly legitimate to ask whether there are any other examples, even allowing the smaller congruent pieces to be of a shape unrelated to the original big triangle. --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.
Determine which shape(s) of triangles, if any, can be dissected into 3 congruent triangular pieces.
Then do the same for 3 *similar* triangles.
Huh? You can cut any right triangle into 2 similar triangles, then do this again to one of these, resulting in the original triangle cut into 3 similar ones.
Yes indeed, but is this the only way to do it? (The proof I came up with for the "congruent" case gave the answer to the "similar" as an intermediate step, which is why I mentioned it.) --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.
participants (5)
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Dan Asimov -
Marc LeBrun -
Michael Kleber -
R. William Gosper -
Richard Guy