PROBLEMS,

WHICH REFER TO THE THIRD AND FOURTH BOOKS.

CONSTRUCTION OF PROPORTIONAL LINES.

PROBLEM I.

To divide a given line into any number of equal parts.

To make a definite case, suppose

we wish to divide AB into five

pq, qr, rs, sb. Join the last extremity 6 with B, and through the other points of division draw parallel to Bb, lines cutting AB in the points P, Q, R, and S (B. II., P. VI.).

The straight line will thus be divided into five equal parts (B. III., T. I.).

PROBLEM II.

To divide a given line into parts proportional to given lines. As a definite case, suppose we wish

to divide AB into three parts, which shall be to each other as the three lines P, Q, R.

P.

R

P

As in the last Problem, draw any indefinite line AX, making an angle with

P

B

AB. Make Ap, pq, qb respectively equal to P, Q, and R. Draw Bb, and parallel to it draw pP', qQ', and the line AB will be divided at P' and Q' as required (B. III., T. III.)

Scholium. As a particular case, suppose we wish to divide a straight line AB into two parts proportional to the lines M and N.

The straight line AB will thus be divided at D in the ratio required.

For, the two triangles DAC, DBC', are evidently similar, and give

AD: DB:: AC: BC':: M: N.

If, instead of taking BC', equal to N, in the direction of BY, opposite to AX, we had taken BC", equal to N, in the same direction with AX, then the line CC" would have met AB produced at D', which is called the conjugate point of the point D (B. III., T. XII., S.).

PROBLEM III.

To find a fourth proportional to three given lines M, N, P.

Form any angle, as XAY, and take on AX, AB=M, BC= N, and on AY, AD=P; then draw BD, and through the point C draw CE parallel to BD. The line DE will be the fourth proportional required.

For, we have (B. III., T. III.),

Α

Y

E

D

X

B

C

M

N

P

AB: BC:: AD: DE, or M:N:: P: DE.

Cor. I. If P were equal to N, the above proportion would become

M:N::N:DE.

That is, DE would in this case be a third proportional to the two lines M and N.

Cor. II. We deduce immediately the following problem:

A point O being given within an angle YAX, to draw through O a straight line DOE, such that the segments DO, OE may be to each other in the ratio of M to N.

and draw EOD, which will be the line required.

For, since OB is parallel to AD, we have

OD: OE:: AB: BE:: M: N.

C

X

B

E

When M = N, it is sufficient to take on AX, BE = AB, and then to draw EOD.

PROBLEM IV.

To find a mean proportional between two given lines M and N.

On the indefinite line AX, take AB=M, BC= N, and on AC, as a diameter, describe the semi-circumference ADC. Through B draw BD perpendicular to AC, and BD will be the mean proportional sought. For, we have (T. I., S. I.),

M

N

D

X

B

C

AB: BD:: BD: BC, or M: BD:: BD: N.

PROBLEM V.

To divide a given line into mean and extreme ratio.

We have already noticed this kind of division (T. III., S. I.). And we have also actually solved this problem algebraically (T. XI., S. I.).

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, AL 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.