mersenneforum.org Axioms and Proof
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 2007-08-29, 16:17 #1 cherrycherry   Aug 2007 2 Posts Axioms and Proof Success can be achieved only by hard work. (write down what the given and the proven part of statement is, then write the converse and determine if true or false) Every equilateral triangle is equiangular. (same directions as above.)
 2007-10-05, 04:56 #2 nibble4bits     Nov 2005 18210 Posts 1) By converse, you mean that if the three angles are not equal, that it is impossible for the triangle to be equilateral? Before we can go further, it would be nice to know. I probably shouldn't assume that you mean that if the angles are the same, then it must be equilateral. There is a subtle difference in these statements although it seems moot in this context. ;) 2) By definition a triangle has to be 'connected' in a loop consisting of three lines whos ends meet exactly once each. In other words: Three lines x, y, and z are looped as "xy -> yz -> zx -> xy" where each possible pair of letters represent the intersections. They must not cross more or less than 3 times and only at the end points of the lines. 3) I'll assume Euclidian geometry, since this depends on the axium A+B+C=Pi where A,B, and C are the angles of the three corners. It would be interesting to make an argument for a curved surface. Note that not all surfaces are like planes or even spheres... Beware counterexamples! 4) Do you want a geometric solution? Algebraic proof? Other? You know that this is also called a regular polygon, right? This is a nice classical problem. The solution is in many books but it's fun to try other approaches. Good luck, all! If two corners (or more) of a triangle have the same relative angles, then the two (or more) sides must be equal. This is because the lines have the same absolute slope. (Try graphing y=ax+c and y=c-ax. Notice how they meet at y=c no matter what 'a' is? Of course c<>0.) Trigonometric proofs aren't too hard but I like the linear equation example above in addition to the trigonometric theorems. There is another equivilent theorem that states that if any two sides have equal length, then the angles of the opposite corners must also be equal. Good counterexample could made from the attached? (what happens if the cube is distorted or the lengths of the triangle are fiddled with?) ;) Attached Thumbnails   Last fiddled with by nibble4bits on 2007-10-05 at 05:08
 2007-10-05, 06:28 #3 Mr. P-1     Jun 2003 22218 Posts I think you missed the point of the question. It does not ask that the solver prove either of the two statements, or their converses, merely to state which part is proven. (The first statement, and it's converse are, in any case, not amenable to mathematical proof, though they can still be analysed for their logical content, which is the point of the question.) The second statement can be restated as Code: IF a triangle is equilateral THEN it is equiangular. The 'given' part is that the triangle is equilateral. The 'proven' part is that it is equi-angular. The converse is the same statement written with 'given' and 'proven' parts reversed: Code: IF a triangle is equiangular THEN it is equilateral. . or, using the form of the original statement: Code: Every equiangular triangle is equilateral. The solver is asked to determine whether this is true of false. He is not asked to prove it, so I think a bare assertion is sufficient: It's true.
 2007-10-05, 12:38 #4 nibble4bits     Nov 2005 2·7·13 Posts So, what really should have been asked is to break it into logical conditions and results and to state if they're true? After that, do it for the converse? Heh, OK that should be a lot simpler than what I did. Yes, because of the properties of a triangle with two or more equal angles, the first statement is true. The converse is also true because of the properties of triangles with two or more equal lengths. Darn, I made it too complicated. :) Statement 1: Equilateral leads to Equilangular Statement 2: Equilangular leads to Equilateral So A -> B, and B -> A, and thus the two statements are interchangable. I seem to remember that if two (or more) statements can be converted into each other and still always be true, then the loop is complete and they're all just different ways of saying the same thing. Like: A -> B -> C -> D -> E, but if E doesn't always lead to A, then you can't assume if E is true, that A is true, but the converse is always usable. Wow, I actually remember some of this stuff!

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