2(d) ("If two rational lines commensurable in square only be added together, the whole is irrational.") The meaning of this hinges on what "the whole is irrational" means. As De-Vonna wrote, "Euclid's use of the words `rational' and `irrational' are not equivalent to the modern uses." Referring to Euclid Book X we see that two lines are "rational" if either they, or their squares, are commensurable. So "the whole" can be irrational only in comparison with some other line, and the wording is not terribly clear as to what that other "magnitude" is. I think the statement means that the sum line is not rational with respect to either of the original lines (which, we recall, are rationally related to each other).
Let's translate this situation into modern numerical language. (I'm following and extending De-Vonna's paper.) Suppose that a2 = cm and b2=cn, where m and n are integers and c is a real number, but that a and b themselves are not commensurable in this sense (a = dp and b = dq is false). Now we are to consider sqrt{cm} + sqrt{cn}, and compare its square with cm and cn. Well, the square is cS, where S = m + n + 2 sqrt{mn}, so the theorem would be false if S is an integer. We can assume without loss of generality that m and n are relatively prime (otherwise dump the common factor into c), so the only way that S can be integral is if both are perfect squares. For example, suppose m = 25 and n = 4. Then a = 5 sqrt{c} and b = 2 sqrt{c}, contradicting the assumption that a and b are commensurable in square only. Conversely, if S is not an integer, then sqrt{mn} is not even a rational number (remember last week's problem about sqrt{5}), and therefore the assertion is valid.
3 (perpendicular to hypotenuse) I'll invite two instructive and different responses to be posted.
6 (parabola and triangle) Let A, B, C be three points on a parabola, as on p. 9 of Allen, "Archimedes". The assertion can't be true for any three points, because the area of the triangle approaches 0 as B approaches A or C. We therefore assume that the slope condition was intended as part of the problem statement: The tangent line at B has same slope as the secant line AC. On the other hand, we do not want to assume that A is the vertex of the parabola.
With apologies to Archimedes, we shall take advantage of what we've learned from Descartes and Leibniz. By displacing and scaling the coordinates (x' = ex + f, y' = gy + h) we can assume that the parabola is in the standard form, y = x2. (The areas will change by the determinant of the scaling transformation, but only the ratios of areas matter in the problem.) Let A = (a,a2), etc. The slope of the secant is
(c2 - a2)/(c - a) = a + c,
and the slope of the tangent is 2b, so b = (a + c)/2 -- i.e., B is midway between A and C in abscissa! (This step may not be strictly necessary, but the new condition is certainly easier for my post-Cartesian mind to embrace.)
We can now reformulate the problem as integral calculus. Let y = f(x) be the equation of the secant line, y = g(x) be the equation of the broken line forming the other two sides of the triangle, and y = h(x) be the equation of the parabola. Let I be definite integration between a and c. Then the area of the triangle is If - Ig, and the area of the parabolic segment is If - Ih. What makes this problem so interesting to me is that it now reduces to a famous fact of numerical analysis:
If the graph of a function k passes through A, B, and C (equally spaced in abscissa), then If is the crudest approximation to Ik by the trapezoidal rule, Ig is the next crudest trapezoidal approximation, and Ih is the crudest approximation to Ik by Simpson's rule. (In particular, Simpson's rule is exact if k is quadratic -- i.e., k = h.) Romberg's formula states
Ih = (4/3)Ig - (1/3)If.
Thus you can evaluate Simpson's rule by doing the trapezoidal rule with more and more points until the numbers look good, then applying Romberg's formula to get an answer much better than either of your last two trapezoidal results! Intuitively, the formula says that the coarser trapezoid result is in error in the same direction as the finer one, but even more so; so you improve your estimates by combining them in a weighted average where the coarser one is penalized by a negative coefficient. (If you have Stewart's Calculus (the textbook we use at TAMU), you will find a similar discussion in the section on numerical integration, but using the midpoint approximation instead of the finer trapezoidal one.) Romberg's formula can be proved by simple algebra if you evaluate the three integrals, or it can be derived more abstractly by seeking a linear combination of the two trapezoidal approximations that makes the error term of order (step size)2 cancel out. In any case, trivial algebra shows that Romberg's formula is equivalent to Archimedes's.
11 (cubes as sums) Some of you still have trouble with mathematical induction, so I'll try to find a good solution to post.
16 (two numbers in mean proportion between two others) There was some confusion about what this phrase means. It means to find numbers x and y such that a/x = x/y = y/b. Now if b = 2a, then x/a = cube root of 2; so if you can construct x by straightedge and compass, then you have duplicated the cube! (Or halved the cube, at least.) The Greeks knew this reduction rather early, hence the great interest in the problem.
If un+1 = un + un-1 and un = rn, then we should be able to find r. Substitution and cancellation yields r2 = r + 1, whence
r = (1/2)(1 +/- sqrt{5}).
One of these numbers is the famous "golden section" and the other is the negative of its reciprocal. (Why? Because the product of the two roots of r2 +br + c = 0 is c, which is -1 in our case.)
But this is not the end of the story. We also would like to impose the "initial conditions" that u1 = 1 = u2. In fact, any linear combination
un = c+ r+n + c- r-n
of the two simple power solutions is also a solution. The two starting values give two equations to solve for c+/-. This technique is all but identical to the solution of constant-coefficient homogeneous linear differential equations by the substitution y(x) = erx, if you remember that.
Rich found on the Web another solution using "generating functions", which is very similar to solving differential equations by Laplace transforms, if you remember that.