Math 300 Great Moments in Mathematics

Spring 2002

Hints for some of the exercises

 
1.7 The example worked out for 2 ÷ 13 begins by halving; but in both cases here, the answers begin with unit fractions that are multiples of 3.  Therefore, start your computations by taking thirds.
1.42 Solving a problem in "standard form" is discussed at the top of p. 36.
2.11 Check out Fig 2.20 on p. 66.  Choose a point I between C and H so that the triangle given by the statement of the theorem is GHI.
2.15 In Fig 2.29(a) an inscribed angle is pictured; call it ABC, with B on the circle.  There is also and the corresponding central angle AOC (O the center of the circle).  Both angles cut off the same arc AC on the circle.  In both triangles ABO and OBC, two of the sides are radii of the circle, so...(use Props. i.5 and i.32).
2.36 Suppose a number n had two different prime factorizations p1 p2 ... pr and q1 q2 ... qs ; order both sets of factors from smallest to largest.  Now divide out common factors from the front until you get a situation that you can't do this anymore.  Why is this situation guaranteed?  If p is the first prime on the left side and q the first prime on the right which are not equal to each other, then you can apply VII-30 to get a contradiction.
2.39 Let A, B, C be consecutive vertices of the regular decagon inscribed in a circle of radius 1 centered at O.  Then d = AB = AC is the length of the side of the decagon, AO = BO = CO = 1 and p = AC is the length of the side of the regular pentagon.  Let D be the intersection of AC with the radial line that bisects angle BOC.  Then show that triangle AOC is a 54°-54°-72° triangle and that triangle BDC is an 18°-18°-144° triangle.  It will follow that triangles AOC and ADO are similar, as are triangles ABC and BDC.  Set up proportions from parts of these triangles.  Combine with the fact that the side h of the regular hexagon inscribed in this circle is h = 1 (why?).
3.6 Archimedes used estimates of the perimeters of regular circumscribed polygons to help determine pi, starting with a circumscribed hexagon: refer to Fig 3.5a (p. 109) and suppose that t1 = AC represents half of one side of this hexagon and u1 = OC is the distance from the center of the circle to one of the vertices of the polygon.  Then angle COA = 30° (why?).  Bisecting this angle determines the point D, and t2 = AD represents half of one side of the 12-sided circumscribed polygon (why?) and u2 = OD the distance from the center of the circle to one of this polygon's vertices.  Archimedes studied the geometry of this figure and discovered how to measure t2 and u2 from knowledge of t1 and u1; this relationship is expressed in the recursive formulas at the bottom of page 109.  They can be reused to determine the lengths of the corresponding elements in the 24-gon, 48-gon, etc. circumscribed about the circle, and each new polygon has a perimeter that more closely approximates the circumference of the circle, hence more closely computes pi.  This problem asks you to use geometry of the 30°-60°-90° triangle COA (assume throughout that the circle has unit radius: r = 1) to compute the values of t1 and u1 exactly, and then, setting the circumference of the circle equal to the perimeter of the polygon, determine a (crude) estimate (call it, say, p1) for pi.  (Further hints: OC is twice AC (why?), so the Pythagorean theorem can be used to determine the lengths of the sides of the triangle COA.)  Then, using your calculator (or a computer), apply the formulas to compute the same three values, t2, u2, and p2, for the 12-gon (watch out for round-off errors!), and so on for the higher order polygons, until you have a value of pi that approximates pi to 5 decimal-place accuracy.  How many steps of iteration are necessary for this?
3.10(a) Refer to Fig 3.26 (p. 130).  Quadrilateral LACE is a square (why?) so MS = AC = AS + SC.  Also angle CAE = 45° so SQ = AS.  Thus, (MS)(SQ) = (AS)2 + (SC)(AS) (explain).  Finally, AS, SC, and OS satisfy the relation (SC)(AS) = (OS)2 by Euclid vi.13 (why?).
3.12 Choose a coordinate system so that (1) one corner O of the parabolic segment is at the origin, (2) the parabola is oriented so that its axis of symmetry is vertical, (3) it opens down, and (4) the entire segment lies in the first quadrant.  (Draw a figure.)  The scale on the coordinate axes are chosen so that the equation of the parabola has the form y = x(a - x) (this way the x-coordinates are at x = 0 and x = a).  The base of the parabolic segment is a line of nonnegative slope  y = bx.  Show that the curves intersect at O and the point P on the curve with x-coordinate x = a - b, so that the area of the segment is the integral from 0 to a - b of the difference between these two functions, which evaluates to (a - b)3/6.  Now the midpoint M of the base OP of the parabolic segment is the point on the line with x-coordinate x = (a - b)/2, and the vertex V of the triangle inscribed in the segment is the point on the parabola with this same x-coordinate.  To find the area of this triangle, view it as decomposed into two halves, one to the left of MV (triangle OMV) and the other to the right (triangle PMV).  Treating MV as the base of each half, the (horizontal) heights of the two subtriangles have the same value, so OMV and PMV have the same area.  Adding, we find that the area of the entire triangle OVP is (a - b)3/8 (why?).  Combine this with the computation of the area of the parabolic segment to finish the problem.
3.23 Choose a coordinate system in which the equation of the parabola has the form y = ax2, with a > 0 (so the parabola opens up).  Comparison with Fig 3.20 (p. 124) means that E is the origin, the y-axis is the line AB (!) with B on the positive side, and the x-axis is vertical.  Pick a point C(x0, y0) on the curve; then y0 = ax02.  Determine the coordinates of the point A (see the text for an explanation of where A lies), then show that the line AC is tangent to the curve at C by comparing its slope with the derivative of the curve at C.
3.33 Refer to p. 110 for a description of the quadratrix curve.  There we see the proportion ZL : BA = arc ZG : arc BD (keyed to Fig 3.6).  Consequently, if Y is the intersection point with BA of the horizontal through Z, then angle BAZ is proportional to the length of the segment BY (why?).  To trisect angle BAZ, then, first trisect BA.  That is, construct the point W on BA so that BW : BA = 1 : 3; then the horizontal through W which cuts the curve in a point X will allow angle BAX : angle BAZ = 1 : 3.
4.1 The formula for the chord of the supplement of an angle must be used first (middle of p. 144) to find  crd 120°.  Then use the half-angle formula (bottom of p. 144) to find  crd 30°.  Repeat to find  crd 150°, crd 15°, and so forth.  Express all answers with two sexagesimal fractional places.
4.3 Modify Fig 4.16 so that  AB = crd a  and  CD = crd b.
4.4 Recall from #4.1 that you will need to combine use of the formula for the chord of the supplement of an angle with the half-angle formula.  Use the values of  crd 72°, crd 120°, and  crd 108°  in the table on p. 149 to get started.  Express answers with two sexagesimal fractional places.
4.15 From the middle of p. 154, we find that Ptolemy has calculated the declination of the sun at longitude  l = 60°  to be  d = 20°30'9".  Then, as in the example at the top of p. 155, one can determine the coordinate s = EC (from Fig 4.24).  This is used to compute  L(l,f)  which formula is in the middle of p. 155.  Convert this angle to hours and minutes, using the fact that one complete revolution is 360° = 24hr.  For times of sunrise and sunset, remember that on the day of the vernal equinox, noon is precisely at mid-day.
4.24 Recall from #2.39 (Elements xiii.10) that if a is the side of the pentagon and r is the radius of the circumscribed circle, then  a2 = r2 + d2  where d is the side of the inscribed decagon.  The geometry of the decagon allows us to decompose it into 10 congruent isosceles 36°-72°-72° triangles, each with base d and sides r; bisecting one of the base angles creates another similar 36°-72°-72° triangle and a 36°-36°-108° triangle.  Show from this that d = [(sqrt(5) - 1)/2] r, then that  a = [sqrt((10 - 2*sqrt(5))/4)] r.  Now decompose the pentagon into five congruent 54°-54°-72° triangles and use the information above to compute the area of the pentagon.
5.12 NOTE THE TYPO: the number 2120 should be 2170.  To solve the problem, put  x = z + a/2.
5.19 The result says that: given a line segment AB, if C is the point  on the segment so that AB/AC = AC/BC, then (AB)2 + (BC)2 = 3(AC)2.  So assume the result and derive the hypothesis.  (Is the analysis valid?  That is, can the argument be reversed to prove the desired result?)
6.16 Let Bhaskara's formula (see the top of p. 215) be called B(x).  (Take R = 1.)  To determine the percentage error in approximation with the sine function, consider a graph of  B(x)/sin x  on the interval  0 < x < 90.
6.20 Brahmagupta's method (the kuttaka method) is described and illustrated on pp. 218-220.  For this problem, the array we use, analgous to the one at the bottom of p. 219, has first column only 4 numbers tall.
6.26 Here the model is the problem illustrated on pp. 221-222.  The given solution (u0, v0) = (1, 9) for 'subtractive' c0 = -2  means that 83(1)2 - 2 = (9)2.  Compose this solution with itself to get a solution for 'additive' c = 4.  Divide this equation through by 4 to get a solution to the original problem.
7.13 Begin with equation (7.2) on p. 256; put k = 4.  Rather than set n = 4 as al-Haytham does, leave the variable n in your equations.  The sum of the fifth powers is the first summation on the right in (7.2), and this is ultimately what you want to solve for.  The inner summation of fourth powers (the one that runs from 1 to p) in the second term on the right can be expressed as a polynomial in p (use the result that appears at the top of p. 258, with p instead of n, of course).  Now replace the counting variable p with i and collect like terms.  You can express the desired sum of fifth powers now in terms of sums of fourth, third, and first powers.  Substitute the appropriate resulting formulas and simplify to finish.
7.16 To show that the method works, substitute the equation for the hyperbola into that for the parabola.  When you sketch the curves (and only the branch of the hyperbola in the first quadrant is being considered here), you will see how it is possible to geometrically determine whether the cubic has zero, one, or two solutions.  The important case is the one with one solution.  Here, the hyperbola and parabola intersect in just one point and the curves are tangent to each other: notice that their tangent lines at this point must coincide.  This will point to the necessary conditions on b and d that the problem calls for.  Of course, they must be the same as those found by Sharaf al-Din (pp. 262-263).
7.29 The solution is by the same method as the one worked out for the qibla for Jerusalem in the text (pp. 279-280).  I have worked out the details for you in a page I have linked to the course calendar for March 14.  Model your solution on this.
8.6 While the given circle has diameter 33, Abraham bar Hiyya's table (p. 294) is keyed to a circle of diameter 28, so solve the similar problem in the circle of diameter 28.  You will need to use linear interpolation (that is, assume that lengths of arcs that are not tabulated are linear between the tabulated values.
8.12 This is handled very much like the previous problem: Leonardo's table (p. 298) is keyed to a circle of diameter 42.  The first two columns give arcs and the last four give chords in units of rods, feet, unciae and points.  (1 rod = 6 feet, 1 foot = 18 unciae, 1 uncia = 20 points.)
8.38 Write out the areas of the first seven or eight quadrilaterals, and look for a pattern.  Use the geometric series formula for summing powers of 1/2.
8.39 If the integer n lies between two consecutive powers of 2: 2k < n < 2k+1, then  1/n < 1/2k.
9.14 It requires 17 successive computations of intermediates (see p. 349) to arrive at the desired approximation.
9.31 Solutions to the given equation correspond to x-intercepts of the graph of the function  y = x3 - cx - d.  To have three real roots, this cubic curve must have three intercepts, hence it must also have two extreme values--a local maximum and a local minimum.  Use calculus to identify the coordinates of these two points; the desired set of three real roots is guaranteed if we can show that the maximum has a positive y-coordinate and the minimum has a negative y-coordinate.  Explain how the stated inequality allows us to prove that the maximum has a positive y-coordinate.
9.33 The resolvent cubic has a small integer root, so can be factored without much trouble into a linear and quadratic term.  Unfortunately, this quadratic factor does not have rational roots.  In the case of the integer value of b, it should be easy to determine the four values of x that solve the original equation, but this is more complicated in the case of the irrational values of b.  However, if you are careful, you will find that in each case, you obtain the same four values of x.
9.37 Following Bombelli, you should be able to identify numbers a and b that satisfy the conditions  a2 = b - 1  and  a2 + 3ab = 18.  This last result shows that  3ab < 18, so b < 6/a; thus, a2 = b -1 < 6/a - 1  and  a3 < 6 - a < 6, so  a < 2.  The only integer possibility for a (namely a = 1) fails to satisfy these conditions, so a must be nonintegral.  If you try some other simple possibilities, you should be able to find the necessary values of a and b by trial and error.
10.5 Build the requested table of values that measure the distance (in cm) from your equator to the latitude lines at 0, 5, ... , 30 degrees.  To draw the chart, you will need a sheet of legal-size (8" x 14") paper.  The scale is  1° = 1 cm, so your chart is 10cm wide (from 75°W to 85°W) and  D(30) cm tall.  Mark your lines of latitude and longitude.
10.8 Use the law of sines (see p. 400).
10.22 You can express  Nlog(xy)  and  Nlog(x/y)  very easily in terms of  Nlog x, Nlog y, and Nlog 1.  See p. 418  to find the connection between the Naperian and the natural logarithm.
10.27 Suppose the projectile is being fired from the origin into the first quadrant.  If we denote its position at time  t  as  (x(t), y(t)), then since its firing speed  v0  breaks up into horizontal and vertical components and gravity acts on it (i.e., it is uniformly accelerated) in only the vertical direction, we have  x(t) = (v0cosa)t  and  y(t) = (v0sina)t - (g/2)t2.  The given data is for a firing angle of  a = 45°.  Use the first equation to solve for  t, then the second to solve for v0.  Now use this value of the firing speed to determine the range of the projectile for the other two angles: use the second equation to determine the time t  of flight then the first equation to find its range.  You will find that Galileo's value of 17,318 is slightly too small.
11.2 You will obtain the equation of a circle in an oblique coordinate system.  So this curve will not be a circle in the standard (x, y)-coordinate system.  What else could it be?
11.7 My edition has a misprint here: the equation should read  z2 = az - b2.  Add to Figure 11.6 the perpendicular NS from the cener of the circle to the line RM.  Then in the one case, QM = SM - SQ, and in the other, RM = SM + RS.  In each case, square both sides, then use the Pythagorean Theorem on triangles NSQ and NSR; collecting terms ought to lead to the expressions a·QM - b2  and a·RM - b2.  This proves that QM and RM satisfy the required equation.
12.6 The method of Fermat is outlined at the bottom of p. 471.
12.11 Descartes' method of normals is explained on pp. 472-473.  The idea is to equate coefficients of like terms in the equation 
[f(x)]2 + (v - x)2 = (x - x0)2q(x).
In this instance, f(x) = sqrt(x) so the left hand side is already a quadratic polynomial, so q(x) = q must be constant.
12.17 Use the fact that the surface area of a sphere satisfies the formula S = 4pr2.
12.25 See p. 488.  There it is explained how Wallis finds  a1/2,3/2 = (4/3)[] .  Now use the symmetric relation  ap,n = an,p  and his formula ap,n = ap,n-1·(p+n)/n  to find  a3/2,n  for  n = 0, 1/2, 1, 3/2, 2, 5/2, ... .
12.36 First, work out the details for the determination of  p = (1/4)x.  To solve for q, we set  p = (1/4)x + q  in the equation, then remove all terms which are of degree higher than 2 in x (realizing that q is itself an expression in x whose lowest order term is of degree 2).
12.40 See the method illustrated in the middle of p. 510; this is the same equation as in the example, but you will be using a different arithmetic progression.  Work out the algebra to get a rational expression for x´/y´  (as in the example).  Then use the original equation to substitute for both expressions  x3  and  y3  to simplify the numerator and denominator expressions.  You should obtain the same answer as for the example on p. 510.  Can you see how using any arithmetic progression would yield the same answer?
12.46 Near the bottom of p. 527 is the similar proof of the product rule.  For the quotient rule, take this hint: multiply by  (y - dy)/(y - dy).
12.47 Write  y = a + bx + cx2 + dx3 + ex4 + ...  as a power series.  Since  y = log(x + 1), you can immediately determine  a.  Next, use the differential equation in the form (x + 1)dy = dx, together with the expression for  dy  that you obtain from the series to determine values for  b, c, d, e, ...  by equating like terms on both sides.
12.49 This is just a nasty Calc I problem!

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