PROBLEM 15.

To make a pentagon on the line A B.

From B draw B F perpendicular to A B, and equal to half of it, join A F and produce it to G, making F G equal to B F. With centres A and B and radius E A G, describe arcs cutting each other at D; with centres A and D, and radius A B describe arcs cutting each other at E; and with centres B and D, describe arcs cutting at C. Join A E, DE, B C, D C, which forms the pentagon.

A

B

The line B G is the radius of the circle circumscribing the pentagon.

PROBLEM 16.

To make a hexagon on a line A B.

With the radius A B and centres A and B, describe arcs cutting in O. Then with the centre O and radius A B, describe the circle A B C D E F, and apply F A B the necessary number of times round the circle, and the figure will be the hexagon.

0

D

B

PROBLEM 17.

To make an octagon in a line A B.

Produce A B to I and K, and draw A F, B E perpendicular to AB. Bisect the angles IA F, K BE, and make A Hand B C each equal to AB; draw HG, CD parallel to A F, and each equal to A B. With centres G and D and radius A B describe arcs cutting each other in F and E; join G F, F E, E D, and the octagon is complete.

F E

G

H

PROBLEM 18.

From a point A draw a line touching the circle E D.

Find the centre O of the circle, join A 0, upon which describe the semicircle, cutting the given circle in E, join A E which is the line touching the circle.

If the point A be on the circumference, this construction fails. In this case, join O A, to which draw a perpendicular from this will be the touching line required.

A;

PROBLEM 19.

To describe an ellipse upon the diameters A B, C D, and centre O. The diameters being at right angles.

With centre C and radius A O describe arcs cutting A B in F and f; these points are called the foci of the ellipse. Take a thread equal in length to AB and fasten it with two pins at Fand f; keep the thread tight, and it will reach the ellipse in P; and by moving a pencil round with the thread, observing

C

T

B

D

always to keep it tight, the pencil will trace out the ellipse.

Otherwise. Assume any point Ton the transverse diameter A B, and with a radius TA and centre F, describe arcs at D and E, and with a radius TB and centre f, describe arcs at D and E; then the points D and E are in the curve of the ellipse.

To describe an ellipse on the diameters A B and C D.

in the curve of the ellipse. Proceed in the same manner for ach of the remaining three quadrants.

PROBLEM 20.

To describe a parabola with the abscissa A B, and ordinate B P, at right angles to it.

Join A P, and draw P N perpendicular to it, meeting A B K produced in N; then B N is the parameter, one-fourth of which is the focal distance Af.

Make A LA ƒ, and draw LK and AM perpendicular to AB; in L K, the directrix, take

any point H, and join ƒ H cutting

H

M

ᏞᎪf

B

AM in I. Then, draw H p parallel to A B, and I p perpendicular to Hf, and p is a point in the parabola.

PROBLEM 21.

To describe an hyperbola on the diameters A B and C D at
right angles to each other.

Make O F, Of each equal to B C, then F and ƒ are the foci of the hyperbola whose diameters are A B and CD. Produce A B indefinitely, and assume in it any point T; with T A as radius and centre f, describe an arc at P; with T B as radius, and centre F, describe an arc at P; then P is a point in the hyperbola.

If the diameters be equal, the hyperbola is called equilateral. And if a parallelogram C O BN be drawn, the line ON produced is called an asymptote, which has the singular property of constantly approaching the curve without ever meeting it.

PROBLEM 22.

To describe a cycloid on the line A B.

Bisect A B in C, draw CD perpendicular to A B and equal to the diameter of a circle whose circumference is A B. In C D assume any point E, and draw E F P at right angles to CD, and make P F equal to the arc D F of the circle; then P is a point in the cycloid.

D

A

A

This curve was incidentally referred to by Aristotle, in alluding to the paradox, that a nail on the rim of a wheel describes a longer line than the axle. Roberval, Fermat, and Descartes shewed that a tangent at P is parallel to the chord F D, and the area A D B is equal to & of ABX CD.

Huygens showed that the arc PD of the cycloid was double the chord D F; and also that a body would fall along it from any point, P to D, in the same time.

Newton and the Bernoullis showed it to be the curve of quickest descent, or that a body would fall by gravity from A to D in less time than by any other route.

PROBLEM 23.

To describe a catenary.

A catenary is the curve which a regular and flexible chain or rope A C B will assume, if suspended from the ends A and B. Galileo confounded this curve with the parabola. Its correct nature was first detected by

James Bernoulli. The equation

to this curve is very difficult, and there is no easy method of describing it. Its application, however, to suspension bridges has given a considerable degree of importance to its various properties.

For further information, see Leslie's "Geometry of Curved Lines," page 383.

PROBLEM 24.

To describe an involute of a circle A B C.

Draw the diameter A B, and in the circumference A CB take any point C. Join O C, and draw P C perpendicular to O C, or a tangent to the circle, making P C equal to the circumference A C, then P is a point in the involute.

In toothed wheels there is less waste of power in passing from one tooth to another A when they are of the form of the involute of a circle than in any other case.

PROBLEM 25.

P

To describe the equable spiral of Archimedes.

With C B as a radius, describe the circle A B D. Divide the circumference into any number of equal parts 1, 2, 3, 4, &c., and the radius C B into the same number of equal parts 1, 2, 3, 4, &c., Join 1 C, 2 C, C, 4 C, &c., on which take the distances 1 C, 2 C, 3 C, 4 C, &c., as given on C B, and these points will form the equable spiral of Archimedes.

This curve is much used in architectural ornaments, and also for the volutes of the capitals of columns.

PROBLEM 26.

To find any number of points in the circle passing through A B and C, where A D is equal to B D and C D perpendicular to A B.

Make the angle E C B equal to DC B, and in CD take any point F, through which draw A P. In CE take C G equal to CF, and draw B G, cutting A B in P; then P is a point in the circle.

E