<p>slurpz you are just full of pick-up lines tonight :D</p>
<p>that last one was the best slurpz.</p>
<p>[holding a wash rag] "Does this smell like chloroform to you"</p>
<p>"holding a wash rag] "Does this smell like chloroform to you" "</p>
<p>HAHAHAH thats classic Ksanders...Kudos man</p>
<p>hey, how are you?</p>
<p>ksanders you're the mna</p>
<p>i love the insurance one. :)</p>
<p>ding</p>
<p>i love this...but idk anything cuz i have enough fun w/ girls i already know... and i rarely say hi to anyone... so yea i would never use pickuplines</p>
<p>"I got into Yale."</p>
<p>"I know how you make you come with just one finger..."</p>
<p>::motions her over with index finger::</p>
<p>slurpz, when i read the upper portion, i thought that it was going to be a really dirty pickup line.</p>
<p>"Cates, you sinner!"</p>
<p>hotpiece101, it IS dirty.</p>
<p>
[quote]
"I got into Yale."
[/quote]
</p>
<p>i'm sure that'll work on any CC girl</p>
<p>guy to a girl</p>
<p>your last name must be gilet, because your the best a man can get.</p>
<p>Say this:</p>
<p>Napoleon's Theorem</p>
<p>Napoleon's theorem states that if we construct equilateral triangles on the sides of any triangle (all outward or all inward), the centers of those equilateral triangles themselves form an equilateral triangle, as illustrated below.</p>
<p>This is said to be one of the most-often rediscovered results in mathematics. The earliest definite appearance of this theorem is an 1825 article by Dr. W. Rutherford in "The Ladies Diary". Although Rutherford was probably not the first discoverer, there seems to be no direct evidence supporting any connection with Napoleon Bonaparte, although we know that he did well in mathematics as a school boy. According to Markham's biography,</p>
<p>Regarding the idea that Napoleon might actually have discovered what we now call Napoleon's Theorem, Coxeter and Greitzer have said that</p>
<p>The possibility of [Napoleon] knowing enough geometry for this feat is as questionable as the possibility of his knowing enough English to compose the famous palindrome, ABLE WAS I ERE I SAW ELBA.
(I wonder how many times that palindrome has been independently re-discovered... but I digress.) It's actually not very difficult to prove this theorem, nor to discover it, so I don't think it would have been impossible for Napoleon to have found it. Consider the drawing below, where the original triangle is shaded in gray, and we have added a 4th equilateral triangle to the left hand vertex.</p>
<p>Notice that if we rotate the figure counter-clockwise through an angle of p/3 about the point "c", the triangle originally centered at "b" moves to the position originally occupied by the triangle centered at "d". This proves the line segments cb and cd are of equal length and make an angle of p/3. Likewise if we rotate the figure clockwise through an angle p/3 about the point "a", the triangle centered at "b" again moves to the position of the triangle at "d", so the line segments ab and ad are of equal length and make an angle of p/3. Consequently the line ac bisects the angles at "a" and "c", so the triangle abc has an angle of p/6 at each vertex, so it is equilateral (as is acd). This is a simple enough observation that it could easily be made by a bright school boy. On the other hand, since Rutherford's article first appeared in 1825, just four years after Bonaparte died on St Helena, it's also conceivable that Rutherford just decided to name his theorem after the famous fallen Emperor.</p>
<p>In any case, the theorem really just expresses the nice tiling pattern, as shown below:</p>
<p>This makes the theorem visually obvious, and is also suggestive of various generalizations, but it doesn't provide much quantitative information. Given the edge lengths A,B,C of the original triangle, what is the edge length of "Napoleon's equilateral triangle"? It's not too difficult to give a proof of Napoleon's theorem using coordinate geometry and a little algebra that also yields the answer to this question. It turns out to be closely related to Heron's formula for the area of a triangle in terms of the edge lengths.</p>
<p>If we construct equilateral triangles on any two of the sides of the given triangle, the distance from their centers is the same, regardless of which two sides we choose. Hence, if we express this distance in terms of the edge lengths, it must be a symmetrical function of those lengths, i.e., any permutation of the edges should leave the result unchanged. The same is true for the area of a triangle, so we shouldn't be surprised if there turns out to be a relation between the sides of Napoleon's equilateral triangle and the area of the original triangle. On the other hand, these two quantities can't be identical, nor even proportional, because we can have a degenerate triangle with zero area but still a non-zero edge length for Napoleon's equilateral triangle.</p>
<p>The center of the equilateral triangle constructed on the side a is to be found by moving from the midpoint of that side perpendicularly outward a distance of a/(12)1/2. The components m,n of this segment satisfy the requirements</p>
<p>where x,y are the coordinates of the vertex opposite the side c. Thus, noting that x2 + y2 = a2, we find</p>
<p>Likewise we can find the components for the segment leading from the midpoint of the side b to the center of the equilateral triangle constructed on b. Making use of x = (a2 - b2 + c2)/(2c) we have</p>
<p>In terms of these lengths the distance s between the centers of the equilateral triangles constructed on the edges a and b can be expressed as</p>
<p>Expanding the squares gives</p>
<p>Noting that the last term is x2/3, we can combine this with y2/3 to give a2/3. Also, we can combine c2/4 and c2/12 to give c2/3. Furthermore, recalling that 2cx = a2 - b2 + c2, we can make this substitution and collect terms to give</p>
<p>Now, since y2 = a2 - x2, the quantity cy can be written as</p>
<p>This is just twice the area of the triangle with sides a,b,c, as discussed in the note Heron's Formula For Triangle Area, and it has the symmetrical factorization</p>
<p>Hence the distance between the centers of the equilateral triangles constructed on the sides a and b is</p>
<p>Since this is perfectly symmetrical in the three sides, it's clear that the distances between the centers of the equilateral triangles constructed on any two sides of the triangle are the same, and so the triangle formed by connecting those centers is equilateral, which proves Napoleon's theorem. The theorem can be generalized to say that the centers of regular n-gons constructed on the sides of a regular n-gon form a regular n-gon, and naturally it also applies to n-gons subjected to conformal and affine transformations. </p>
<p>Recall that the edge lengths of a triangle are proportional to the sines of the opposite angles, which implies that the lengths of the segments from the vertex to the centroid of the equilateral triangles centered on points a,b,c are proportional to sin(a), sin(b), sin(g) respectively. Therefore, noting that the angle between the segments Pb and Pc is a + p/3, the normalized squared length of segment can be expressed using the law of cosines as</p>
<p>Adding and subtracting sin(a)2 and expanding the cosine of a + p/3 in the above expression gives</p>
<p>Napoleon's Theorem is true if and only if the corresponding expressions for the squares of the other two normalized lengths sab and sca equal this same value. These expressions are given by simply permuting the angles a, b, and g. Since the overall expression is obviously symmetrical under transpositions of b and g, and the first four terms are symmetrical in all three of the angles, it only remains to consider the effect of transposing a and b on the quantity in the square brackets. Hence Napoleon's Theorem is equivalent to the equality</p>
<p>Solving for sin(g) gives</p>
<p>We can write this in a more familiar form by making the substitutions sin(x)2 = (1-cos(2x))/2 on the right hand side, and defining u = a - b and v = a + b. This gives</p>
<p>This identity is easily verified by writing the sines and cosines in exponential form, and carrying out the multiplications.</p>
<p>Yet another approach to Napoleon's Theorem is by representing points in the plane as complex numbers. Both the premise and the conclusion of the theorem involve equilateral triangles, so it will be useful to know the necessary and sufficient on three complex numbers z1, z2, z3 to be the vertices of an equilateral triangle. The centroid of these points is simply z0 = (z1 + z2 + z3)/3, and the rays from the centroid to the vertices are (z1 - z0), (z2 - z0), and (z3 - z0). The vertices form an equilateral triangle if and only if these rays are of equal length L and separated by the angle 2p/3. Hence the necessary and sufficient condition is</p>
<p>for some arbitrary angle q. It follows that</p>
<p>Substituting for z0 and expanding the products, we arrive at the condition</p>
<p>This condition on the vertices of an equilateral triangle can be expressed in several alternate (but algebraically equivalent) forms, the most fundamental of which is</p>
<p>Another equivalent form is</p>
<p>Also, since the quantity in the left parentheses is 3z0, this can be written as</p>
<p>which represents an alternate expression for the centroid of an equilateral triangle.</p>
<p>Given any two points, these expressions enable us to compute a third points that forms an equilateral triangle with the first two. Of course, the relation is quadratic in each of the three points, so there are two solutions, corresponding to the two possible directions that the third point can lie in with respect to the other two. Given z1 and z2, an equilateral triangle is formed by setting z3 to either of the values</p>
<p>Therefore the centroid of this triangle is at</p>
<p>The root given by the + sign places the centroid such that the loop 1,2,0 proceeds in the counter-clockwise direction. Now, if we are given three arbitrary points p1, p2, p3 in the complex plane, Napoleon's Theorem asserts that the centroids of the outer equilateral triangles on the bases (p1,p2), (p1,p3), and (p2,p3) form an equilateral triangle. Letting c12, c13, and c23 denote these centroids, we can use the preceding formula to express these in terms of the points p1, p2, p3. For example, we have</p>
<p>It is then straightforward to verify that</p>
<p>so the centroids do indeed satisfy the condition for being the vertices of an equilateral triangle. Furthermore, letting p0 denote the centroid of the original triangle (not necessarily equilateral), the right-hand quantity is 3p02, but we know that the middle quantity equals 3c02 where c0 is the centroid of the equilateral triangle formed by the three original centroids. Consequently, the centroid of Napoleon's triangle coincides with the centroid of the original three points.</p>
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