PEAUCELLIER'S CELL.

This consists of a framework of six rods, OQ, OR, PQ, PR, P'Q, P'R hinged at their extremities, and such PQ=PR=P'Q=P'R,

and OQ OR.

R
A

This arrangement is called a linkage, the rods being linked or hinged together.

(1) Since O, P, P' are equidistant from Q and R they lie on the r

bisector of QR.

... O, P, P' are in a straight line, however the rods may move. difference of sqs. on OQ, PQ,

(2) Also rect. OP, OP'

which is constant. (See Ex. 195.) .. if O be fixed and P be made to move along any curve, P' will move along the inverse of that curve with respect to O.

If, therefore, we introduce another link DP fixed at a point D, and such that DPDO, the point P will move along a circle which passes through O, and the point P' will therefore describe a straight line perpendicular to OD. (See p. 251.)

Ex. 491.-A framework is formed of four rods, CD, DE, EF, FC in one plane hinged at their extremities, CD being equal to EF and DE equal to CF; CD, EF being opposite sides, and CF, ED diagonals of

an axe.

Show that the mid-points O, P, P', O' of CD, CF, DE, EF are in a straight line, and that the rectangle OP, OP' is constant.

The above linkage is called Hart's Contraparallelogram.

If Q and R are the mid-points of DF, CE it can easily be shown (Ex. 65) that

so that the relative position of the five points O, Q, R, P, P is the same as in Peaucellier's cell.

For further information on 'Linkages' the student is referred to an interesting little treatise by Mr. A. B. Kempe, How to Draw a Straight Line.

CONTACT PROBLEMS.

I. To describe a circle passing through two given points to touch

(a) a given straight line,

(6) a given circle.

(a) Let A and B be the two given pts., XY the given st. line, and suppose that AB is not to XY; let them meet in C.

X

DY

Along XY take CD, CD' such that sq. on CD=rect. AC, CB=sq. through A, B, D or A, B, D'.

on CD' and describe a

It will touch XY by III. 37.

The case in which AB is parallel to XY is left as an exercise to the student.

Ex. 492.-Find a point in a given straight line XY of unlimited length at which a given finite straight line AB subtends the greatest angle.

If AB meets XY in C, XY is divided into two parts, CX, CY, in each of which there exists a point at which the angle subtended by AB is a maximum for that part.

Ex. 493.-D is any point in the side AB of a triangle ABC; show how to draw a straight line through D, cutting AC in E, so that DE may be equal to the sum of the perpendiculars let fall from D and E upon BC.

(b) Let A and B be the two given pts., DEF the given

.

Take any

If this does

pt. E on C DEF and describe a through A, B, E.
not touch DEF let it cut it in F; and suppose that EF is not || to
AB; let them meet in C.

Ε

E

From C draw the tangents CD, CD' to D, E, F, and describe a

through A, B, D or A, B, D'.

It will touch C DEF.

Sq. on CD=rect. CE.CF

=rect. CB. CA;

... CD touches through A, B, D.

Similarly CD' touches through A, B, D'.

[III. 36.

[III. 36, Cor.

The case in which EF is parallel to AB is left as an exercise to the student.

Note that one of the tangents, CD', from C might be in a straight line with AB, in which case it would be impossible to describe a circle through A, B, D'.

Ex. 494.-Find the points on a given circle (DEF) at which a given finite straight line (AB) subtends the greatest and least angles.

If AB when produced cuts circle DEF it will divide it into two parts, in each of which there exists a point at which the angle subtended by AB is a maximum for that part.

If AB when produced does not meet circle DEF, there exist two points, at one of which the angle subtended by AB is a maximum, at the other a minimum. (Compare Ex. 492.)

U

II. To describe a circle to pass through a given point and to touch, (c) two given straight lines,

(d) a given straight line and a given circle,

(e) two given circles.

(c) Let A be the given pt., OC, OD the two given st. lines. Draw AN Ir to the intl. bisector of the COD in which A lies, and produce it to B, making NB=AN.

Describe a to pass through A and B, and touch OC at C:

(a) it shall also touch OD.

For its centre E lies on the line ON. And if we draw ED Lr to
OD, it = EC,

and.. the touches OD at D.

The problem is thus reduced to (a).
There will therefore be two solutions.

P

Α

(d) Let B be the given pt., A'P' the given st. line, and OAP the given Let the diamr. OA of the which is r to A'P' meet it in A'. OB and produce it to C, such that OB. OC=OA.OA'.

Join

[I. 45.