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0|0|0|||||Isoperimetric inequality|Jonathan|1|13:15:49|01/11/2011|I very much enjoyed your page on the Isoperimetric inequality %28http%3A%2F%2Fwww.cut-the-knot.org%2Fdo_you_know%2Fisoperimetric.shtml%29.  I had a lot of trouble understanding it%2C as some of the phrases seemed ambiguous to me.  %28E.g.%2C the statement of the lemma makes it seem as if points S and T must be fixed.  Also I got lost by the phrase %22If it%27s not.%22%29%0D%0A%0D%0AHowever%2C I finally rewrote it for myself so that I could make sense of it.  It%27s not perfect%2C but I figured I%27d share what I wrote in case it%27s helpful%3A%0D%0A%0D%0ALemma 2%0D%0AConsider all the arcs with a given length and whose endpoints are called S and T. The curve that encloses the maximum area between it and the straight line between S and T is a semicircle.%0D%0A%0D%0A%0D%0AProof of Lemma 2%0D%0AThe proof relies on two results %28whose proofs are left as an exercise for the reader%29%3A%0D%0A%28a%29 For every arc with endpoints S and T%2C if all angles formed on a point of the arc by lines from that point through S and T are right angles%2C then the arc is a semicircle.%0D%0A%28b%29 Among all triangles with two given sides%2C the one where those two sides form a right angle has the largest area.%0D%0A%0D%0ANow consider an arc with endpoints S and T for which there exists a triangle whose vertices are S%2C T%2C and a point on the arc%2C where the angle formed at that point is not a right angle %28Figure 4%29.  For convenience%2C we will call such an arc an %22arc with a non-right angle.%22 In that case%2C we could slide S %28either to the left or right%2C depending on the whether the angle is acute or obtuse%29%2C so as to create a triangle where the angle is right. Let pieces of the arc move along %28Figure 5%29. By result %28b%29 above%2C the area of the triangle grew.  Since the area of the two red regions did not change but the area of the triangle grew%2C the whole area between the new arc and the line ST has increased.  Therefore%2C for any arc with a non-right angle%2C another arc of the same length can be formed enclosing a larger area between it and the line connecting its endpoints.  Hence%2C all such arcs enclose less area than ones where the angle is always right %5Bnot sure if that%27s phrased right%3B may need to state another logical step%5D.  But by result %28a%29%2C any arc where the angle is always right is a semi-circle.%0D%0A%0D%0ABy the way%2C I wonder if there%27s a way%2C instead of doing lemma 2%2C to just skip it altogether and extend lemma 1%3F  Instead of dividing the closed curve into two%2C divide it into N segments%2C each with the same perimeter length%2C and two line segments going to the centroid. %28e.g.%2C like cutting out pieces of a pie%29.  Then one could use similar logic as lemma 1 and show that as N approaches infinity%2C the pie pieces approach line segments%2C and if they are all equal%2C then one has a circle by definition.%0D%0A%0D%0AI realize though that this involves the added complexity of proving that if the pie pieces are cut to the centroid%2C the reasoning paralleling in lemma 1 still holds %28obviously it wouldn%27t if one just picked an arbitrary point%29.
1|1|0|||||RE%3A Isoperimetric inequality|alexb||08:24:42|01/14/2011|Thank you for pointing out stylistically obtuse pieces. I put in some modifications that I hope make the text more transparent.%0D%0A%0D%0AI do not believe that you can split the shape into N pieces as you suggest - at least not easily. The requirement of them having the same perimeter is rather a stringent constraint%2C even if you choose as the nexus of the construction the centroid of the shape.