We will now proceed with its geometrical solution.

Let AB be the line which we are

required to divide.

A

D

B

Draw BO perpendicular to AB and equal to one half of AB; then draw AO, with O as a centre; with OB for a radius, describe the circumference meeting AO in C. Finally, make AD equal to AC, and the line AB will be divided at the point D as required. In effect, from the construction, AB is a tangent to the circle OB; and if AO is produced until it meets the circumference in C', we shall have (T. III.),

Consequently,

AC: AB::AB: AC.

AC-AB: AB:: AB-AC: AC.

=

Now, AC' AC+CC' AC+ AB; consequently, AC' AB=AC=AD, and AB-AC=AB-AD=BD hence the proportion evidently becomes

AD: AB:: BD: AD,

or interchanging means and extremes,

AB: AD:: AD: BD.

We will suppose the problem solved, and that MN, mn are the two tangents, the one exterior and the other interior, meeting the line of the centres OO' in two points C, C'.

It is evident that, if these points were known, it would be sufficient to draw through each of them a tangent to one of the circles (B. II., P. XII.), and it would of necessity be tangent to the other; the problem would thus become resolved.

Now, in drawing the radii OM and O'N, Om and O'n, we shall evidently obtain two couple of similar triangles OMC and O NC, OmC and O'nC', which give the proportions

OC O'COM: O'N,

OC': O'C'::OM: O'N.

But the radii OM, O'N, are given lines. We see then that the points C, C' are the conjugate points (B. III., T. XII., S.; also P. II., S.), which divide the distance OO' in the ratio of OM to O'N. From this results the following construction:

Draw any diameter KOk of the circle O, and through the point O' draw the radius O'L parallel to OK; join the points K and k with the point L.

The points C, C', where the straight lines KL, kL meet the line of the centres, are the points sought; since we have (B. III., T. IV.),

OC: O'COM: O'N,

OC': O'C':: OM: O'N.

Now draw through each of these points a tangent to either of 'these circles, and it will be tangent to the other.

Scholium. This problem is evidently susceptible of four, three, two, or only one solution, or it may not admit of any, according to the five relative positions of the two circumferences (B. II., T. XVIII., S.).

PROBLEMS OF AREAS.

PROBLEM VII.

To transform a polygon into another having one side less than the first, and finally into a triangle.

Let ABCDE, etc., be any polygon whatever, which we here represent by a broken line, in order that the generality of the construction may be the better shown.

Through the point A draw the diagonal AC cutting off the triangle ABC. Draw afterwards, through the point B, the line BI parallel to AC, meeting DC, produced, in I, then draw AI.

E

B I

L

The polygon AIDE, etc., is equivalent to the given polygon ABCDE, etc., and it has one side less.

For, since BI is parallel to AC, the two triangles AIC, ABC are equivalent (B. III., T. XIX., C.), and if to these two triangles we add the portion of the surface ACDE, etc., we shall have, in the first case, the polygon AIDE, etc., and in the second case, the polygon ABCDE, etc.; consequently these two polygons are equivalent.

It is evident, moreover, since the side CI of the triangle AIC is the prolongation of the side DC, that the two sides AB, BC of the first polygon have been replaced by the single side AI; hence the second figure has one side less than the first.

Actually operate upon the polygon AIDE, etc., as upon the first; that is to say, draw the diagonal AD cutting off the triangle AID; afterwards draw IL parallel to AD meeting ED. produced in L; we shall thus obtain a new polygon ALE, etc., equivalent to the second and having one side less.

In continuing, in this way, we shall necessarily reach a polygon of three sides, and the problem will be resolved.

PROBLEM VIII.

To transform a polygon into a square.

If the given polygon is a triangle, let b represent its base, and h its height, and a the side of the square sought.

We have this condition,

x2=b×1h,

from which we obtain the following proportion:

b:x::x:h.

Thus, the side of the square sought is a mean proportional between the base and half the height of the triangle.

This line being constructed (P. IV.), we easily obtain the construction of the square.

Whatever may be the polygon, we commence by transforming it into a triangle (P. VII.); afterwards we transform, as above, the triangle into a square.

When the given polygon is a parallelogram, or a rectangle, or in short, any figure whose area depends immediately upon the product of two lines, all that is necessary in order to obtain the side of an equivalent square is to construct a mean proportional between these two lines.

Thus, in the case of a regular polygon, it is sufficient, after having developed on an indefinite straight line the perimeter of the polygon, to seek a mean proportional between the half perimeter and the radius of its inscribed circle.

To transform a circle into a square, it is necessary first to rectify the circumference, and afterwards to determine a mean proportional between the radius and the rectified semi-circumference.

Scholium. The quadrature of the circle is wholly dependent upon the rectification of a circumference; and thus, up to the present, we have not been able, by the assistance of the Ruler and the Compass, to construct rigorously a square equivalent to a circle, as we can do for rectilinear figures, since all known methods give only approximate values for the ratio of the circumference to the diameter.

CONSTRUCTION OF SIMILAR POLYGONS UNDER CERTAIN

CONDITIONS.

PROBLEM IX.

Upon a given line, to construct a polygon similar to a given polygon.

Let ab be the given line, and ABCDEF the given polygon. After decomposing the given polygon into triangles by drawing diagonals from the corner A to the other corners, form at the

points a and b two angles cab, abc, respectively equal to the angles CAB, ABC. We shall thus obtain a triangle abc similar

to the triangle ABC. Construct, in the same manner, on ac a triangle acd similar to the triangle ACD; and so on for the other triangles.

The polygon abcdef thus obtained will be similar to the given polygon ABCDEF.

PROBLEM X.

Two similar polygons being given, it is required to construct a third polygon similar to the two first, and equivalent to their sum or to their difference.

The solution of this problem is an immediate consequence of C. IV., T. XXIX., B. III.

Let a, a' be two homologous sides of the given polygons A, A'. Upon these sides, regarded as the sides about the right angle, or as the hypotenuse and one of the sides of the right angle, construct a right-angled triangle; afterwards, on the third side a' of the triangle thus obtained, construct (P. IX.) a polygon A" similar to one of the given polygons.

The polygon thus constructed will be the polygon required. For, by construction, a" a2+a", or a'a2-a2; hence, (B. III., T. XXIX., C. IV.),

=

A"=A+A', or A"= A — A'.

PROBLEM XI.

To find a square which shall be to a given square in the ratio

of two given lines.