PROBLEM XXXI.

In any given triangle to inscribe a circle.

Bisect any two of the angles with the lines AO, BO, and O will be the centre of the circle. Then, with the centre O, and radius the nearest distance to any one of the sides, describe the circle.*

PROBLEM XXXII.

About any given triangle to circumscribe a circle.

Bisect any two of the sides A B, BC, with the perpendiculars mO, nO. With the centre O, and distance to any one of the angles, describe the circle.

A

c

B

PROBLEM XXXIII.

In, or about, a given square to describe a circle.

Draw the two diagonals of the square, and their intersection O will be the centre of both the circles. Then, with that centre, and the nearest distance to one side for radius, describe the inner circle; and with the distance to one angle for radius, describe the outer circle.t

*For if perpendiculars be let fall from O on each of the sides, it may easily be shown, that these perpendiculars are equal. And consequently O is the centre of the required circle.

The diagonals of a square mutually bisect each other.

PROBLEM XXXIV.

In, or about a given circle to describe a square, or an octagon.

Draw two diameters A B, C D, perpendicular to each other. Then connect their extremities, and they will give the inscribed square A C B D. Also through their extremities draw tangents, each parallel to the other diameter, and they will form the outer square m n o p.

m

D

NOTE. If any quadrant, as A C, be bisected in q, it will give one eighth of the circumference, or the side of the octagon.

PROBLEM XXXV.

In a given circle to inscribe a trigon, a hexagon, or a dode

cagon.

Then

B

F

E

The radius is the side of the hexagon. Therefore from any point A in the circumference, with the distance of the radius, describe the arc BO F. is A B the side of the hexagon; and therefore, being carried round six times, it will form the hexagon, or divide the circumference into six equal parts, each containing 60 degrees. The second of these, C, will give A C, the side of the trigon, or equilateral triangle, and the arc A C one third of the circumference, or 120 degrees. Also the half of A B, or A n, is one twelfth of the circumference, or 30 degrees, and gives the side of the dodecagon.

Y y

NOTE. If tangents to the circle be drawn through all the angular points of any inscribed figure, they will form the sides of a like circumscribing figure.

PROBLEM XXXVI.

In a given circle to inscribe a pentagon, or a decagon..

Draw the two diameters A P, mn, perpendicular to each other, and bisect the radius On in q. With the centre 7, and radius q A, describe the arc Ar; and with the centre A, and radius Ar, describe the

arc rB.

Then is A B one

fifth of the circumference;

B

נוב

S

વ્

P

and A B, carried round five times, will form the pentagon. Also the arc A B, bisected in S, will give A S, the tenth part of the circumference, or the side, of the decagon.*

square of

* If a regular pentagon be inscribed in a circle, the the radius is to the square of its side, as 2 to 5—✓5. Suppose a right line drawn from A to r, and A to g. Then Ar2 Aq2+rq2 -2 rq Xo q = 2 A q2 — 2 Aq xo q = 2 Aq2 -Aqx Ao;

but Aq2=Ao2+oq2={Ao2,

hence Aq=Ao√5,

and consequently, Ar= A B is the side of the pentagon.

As the square of the side of a regular pentagon, inscribed in a circle, is equal to the sum of the squares of the radius and of the side of a regular decagon, inscribed in the same circle, and Ar2=Ao2+or2,

ro➡ the side of the decagon.

ANOTHER METHOD.

Inscribe the isosceles triangle ABC, having each of the angles ABC, ACB, double the angle BAC. Then bisect the two arcs ADB, AEC, in the points D, E; and draw the chords AD, DE, AE, EC; so shall ADBCE be the inscribed pentagon required.* And the decagon is thence obtained as before.

D

B

E

NOTE. Tangents, being drawn through the angular points, will form the circumscribing pentagon or decagon.

PROBLEM XXXVII.

To divide the circumference of a given circle into twelve equal parts, each being 30 degrees.

Or to inscribe a dodecagon by another method.

Draw two diameters 17 and 4 10 perpendicular to each other. Then, with the radius of the circle, and the four extremities 1, 4, 7, 10, as centres, describe arcs through the centre of the circle; and they will cut the circumference in the points required, dividing it into 12 equal parts at the points marked with the numbers.t

10

The angle ACB at the circumference, standing on the arc ADB, is double the angle BAC; consequently the arc ADB = double the arc BC. For the same reason the arc AEC = dou. ble the arc BC. Therefore the chords AD, DB, BC, CE, EA are equal to each other; and ADBCE is the required pentagon.

†The radius being equal to the chord of 60°, the arc 13, in the quadrant 1 4,2 4 = 60°. Therefore the arc 1 2 = 2 3

PROBLEM XXXVIII.

To divide a given circle into any proposed number of parts by equal lines, so that those parts shall be mutually equal, both in area and perimeter.

Divide the diameter A B into the proposed number of equal parts at the points a, b, c, &c. Then on A a, A b, A c, &c. as diameters, describe semicircles on one side of the diameter AB; and on B d, B c, B b, &c. describe semicircles on the other

b

A

B

side of the diameter. So shall the corresponding joining se micircles divide the given circle in the manner proposed. And in like manner we may proceed, when the spaces are to be in any given proportion. As to the perimeters, they are always equal, whatever may be the proportion of the spaces.*

34; and the chords of these equal arcs are equal. The same may be said of each of the other quadrants. Therefore the problem is truly solved.

The several diameters being in arithmetical progression, the common difference being equal to the least of them, and the diameters of circles being as their circumferences, the cir cumferences are also in arithmetical progression. But in such a progression the sum of the extremes is equal to the sum of each two terms, equally distant from them; therefore the sum of the circumferences on AC and CB is equal to the sum of those on AD and DB, and of those on AE and EB, &c. and each sum equal to the semicircumference of the the given circle on the diameter AB. Therefore all the parts have equal perimeters, and each is equal to the circumference of the given circle.

Again the same diameters being as the members 1, 2, 3, 4, &c. and the areas of circles being as the squares of their diameters, the semicircles will be as the numbers 1, 4, 9, 16, &c. and conse¬