 Most of what we think about the geometry and mathematics of circles, spheres, cylinders and cones comes to us from Archimedes. Archimedes lived between 287 and 212 BC in Syracuse. Syracuse was actually a Greek colony in Italy. It was apparently related to King Heron of Syracuse. Heron had allied with Rome, but after Heron's death his nephew, Heronimus, became king and decided he would break with the Alliance. Big mistake, the result was a Roman invasion of Syracuse and Archimedes apparently died during the course of the Roman conquest. And there's a lot of stories about Archimedes' actions during the battle, most of which are probably mythological. Our focus is on the mathematics of Archimedes. So let's talk about circles. What can we say about the area of a circle? Both the Mesopotamians and the Egyptians gave two different area formulas for a circle. The Egyptians gave a formula that we can read as 8 ninths the diameter squared, and the Mesopotamians gave a different formula, one-twelfth of the square of three times the diameter. And as we noted earlier, the fact that they don't give the same area might have led to the creation of deductive geometry. Now Euclid proved the proposition that circles are to each other as the squares on their diameters, and this corresponds to A equals KD squared. And this leads us to the work of Archimedes. First, note that all previous formulas for giving the area of a circle expressed in terms of the diameter. Archimedes talked about circles in a revolutionary new way. The area of a circle is equal to the triangle with base equal to the circumference and height equal to the radius. And this is the first time that the radius shows up as an important part of the computation of the area of a circle. So Archimedes proves this by a double proof by contradiction. Let K be a triangle with base equal to the circumference and height equal to the radius. If the area of the circle is not equal to K, it's either greater or less. Suppose the area of the circle is greater, inscribe in the circle a regular polygon with area greater than K. The area of this polygon is equal to that of a triangle with base equal to the polygon's perimeter and the triangle's height will equal the opothem, the perpendicular to the side drawn to the center. But the perimeter is less than the circumference, so the triangle's base will be less than that of K. Meanwhile, the opothem is less than the radius, and so the triangle's height is less than that of K. And so the polygon has the area of a triangle that is shorter and narrower. Remember, the polygon was supposed to have an area greater than K, so this is impossible. Similarly, if the area of the circle is less than K, we can use a circumscribed polygon to arrive at a contradiction. Since the area can be neither greater nor lesser, it must be equal. Archimedes then approximates the perimeter of a 96-sided polygon circumscribed about the circle and a similar polygon inscribed in the circle. The work is more tedious than interesting, and the truly remarkable feature is that Archimedes found rational bounds on the perimeters by beginning with the inequality, 265-153 is less than the square root of 3 is less than 1351-700-80. We don't know where these approximations came from, but as we saw earlier, it's possible that a variation of the side and diameter numbers may have produced them. At the end of it, Archimedes concludes that the ratio of the circumference to the diameter is less than 3-7, but greater than 3 and 10-71. And you'll notice that this upper bound gives us our approximation 22-7. Archimedes also went on to prove the volume and area formulas for spheres. They're too complicated to get into here. Somewhat more comprehensible is Archimedes' results on the quadrature of a parabola, finding the area of a parabolic segment. And what Archimedes had to say about that, the segment of a parabola is four-thirds the triangle with the same base and vertex. Now in Archimedes' time, a vertex was where the diameter of a curve met the curve, and the diameter is any line that bisects parallel lines cutting the curve. So Archimedes proves this result as follows. Let QQ be the base of a segment, and bisect QQ at V, then bisect QV at M. And since we have a line that's been bisected, we can draw a diameter going through the vertex P, and we'll draw another line, RM, parallel to PV. So remember by the property of the parabola, the ratio of the squares of the lines drawn ordinate-wise is equal to the ratio of the ordnance. And so QV squared is to RW squared as PV is to PW. Now by construction, QV is twice MV, which is the same as twice RW. And so this ratio over on the left, well that's really a four-to-one ratio. So RM is three-fourths PV, and four-thirds RM is equal to PV. Consequently, this triangle PRQ is one-eighth of the triangle PQQ. You should prove that. Here's a hint. These two triangles have the same base and the same height. So all you need to do is figure out what the height of one of them is and what the base is. Now what this means is that if we take our triangle PQQ, we can describe additional triangles with total area one-fourth of this triangle. And these form these little projections here. But now, lather, rinse, repeat. And because all of these triangles are going to be inscribed in the segment, then the area of the segment has to be greater than PQQ plus a quarter of PQQ plus a sixteenth and so on. And that's the geometric series that sums the four-thirds the original triangle. And since the segment is greater than any partial sum, it's either greater or equal to four-thirds PQQ. But it can't be greater since each additional triangle reduces the difference between the segment and the triangles by more than half, so it must be equal. And so that gives us the area of the parabolic segment as four-thirds the area of the triangle with the same base and vertex.