QUESTIONS AND PROBLEMS 313
A
FIGURE 25. Archimedes' trisection of an angle: ZATA = ^ZAAE.
10.15. Show that Archimedes' result on the relative volumes of the sphere, cylinder,
and cone can be obtained by considering the cylinder, sphere and double-napped
cone formed by revolving a circle inscribed in a square about a midline of the
square, the cone being generated by the diagonals of the square. In this case the
area of a circular section of the cone plus the area of the same section of the sphere
equals the area of the section of the cylinder since the three radii form the sides
of a right triangle. The radius of a section of the sphere cuts off a segment of the
axis of rotation from the center equal to the radius of the section of the cone, since
the vertex angle of the cone is a right angle. These two segments form the legs of
a right triangle whose hypotenuse is a radius of the sphere, which is equal to the
radius of the section of the cylinder.
10.16. A minor work attributed to Archimedes called the Book oj Lemmas contains
an angle trisection. In Fig. 25 we are given an acute angle ZAAE, whose trisection
is required. We draw a circle of any radius r about Ä, the vertex of the angle.
Then, using a straightedge, we mark off on it two points Ñ and Q separated by the
distance r. Setting the straightedge down so that Ñ is at point à on the extension
of the diameter EAZ, Q is at point Β on the circle, and the point A is also on
the edge of the straightedge, we draw the line ΑÃ. By drawing EH parallel to AT,
we get ZATE = ZYEH. By joining AB, we obtain the isosceles triangle ÃΒΑ.
Now since ZBAZ is a central angle on the arc BZ and is equal to ÆÂÃÄ, which
is equal to ZZEH, which is inscribed in the arc ZH, it follows that ZH — 2 BZ.
Since the arcs AE and Β Ç are equal (being cut off by parallel chords), we now get
AE — BH= 3 BZ. Therefore, Ä AT Ε = ZBAZ = ±ZAAE.
Why is this construction not a straightedge-and-compass trisection of the angle,
which is known to be impossible? How does it compare with the neilsis trisection
shown above? Show how to obtain this same result more simply by erasing every-
thing in the figure below the diameter of the circle.
10.17. Show that the problem of increasing the size of a sphere by half is equivalent
to the problem of two mean proportionals (doubling the cube).
10.18. A circle can be regarded as a special case of an ellipse. What is the latus
rectum of a circle?
