Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

The transversal PMN such that the parallels OM and ON to the sides of the angle intersect on PQ is the required line.

For draw any other line PM'N. Join M'N. Then the triangles MON and M'ON are equal (Euc. I. 37), but M'ON> M'ON'; therefore MON > M'ON'.

[blocks in formation]

Prop. II.-On the sides BC and CA of a triangle, to find points M and N such that if the lines AM and BN meet in O the triangle MON may be a maximum.

[graphic][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed]

Regarding A as a point on the base produced of BCN and AOM a transversal to the sides, MON is maximum when ON' and MN' parallels to these sides respectively meet on AC. Similarly since B is on the base produced of ACM and BON a transversal to the sides, OM' and NM' parallels to the sides meet on the base.

Then we have ANM'O and CN'OM' equal parallelograms (Euc. I. 36), therefore AN- CN', also BM = CM'. But by Prop. I. AN. AC-AN'2, therefore AN. AC = CN2; similarly BM. BC= CM2, or the sides of the triangle ABC are divided in extreme and mean ratio, the greater segments being measured from the vertex.

Prop. III. Through one extremity A of the diameter APB of a semicircle draw a chord AMN to meet a perpendicular through P to the diameter AB in M and the circle in N, such that the triangle MBN may be a maximum.

[graphic][subsumed][subsumed][subsumed][subsumed][subsumed]

Suppose a tangent is drawn at the required point N. Let it meet PM in S. Join AS. From the centre C let fall CX perpendicular on AS. Join CN.

By Prop. I. the parallels MQ to the tangent and NQ to PS meet on AB, for then with respect to the angle PSN the triangle MBN is maximum; therefore a fortiori it is the maximum triangle whose vertex N lies on the circle ANB.

LANS=ABN=ANQ. Hence since MN, the diagonal of a parallelogram MSNQ, bisects the angle N, the figure is a rhombus, and NQNS. Then the triangles ANS and ANQ are equal in every respect (Euc. I. 4), therefore ASN is a right angle; hence CNSX is a rectangle, and SX is equal to the radius of the circle.

Also CPSX is a cyclic quadrilateral, therefore

[blocks in formation]

which is known. Therefore we have the rectangle and difference of AS and AX, from which data these lines are at once determined. Then we can construct the rightangled triangle ACX, which fixes the point X; therefore, etc.

[graphic][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed]

COR. In the particular case when PMS is a vertical radius, if SN meet the tangent AT in T and AB in T', we have AS. AX=2, therefore by parallels AT'. AC=CT".

=

Similarly TT'. TS-T'S, but TT'. TS- AT2 = TN2; therefore TNT'S, and TS-T'N.

But when a line TT' is divided in extreme and mean ratio in S and from the greater segment a part T'N is taken equal to the less TS, T'S is divided into extreme and mean ratio.

Ex. Draw the transversal AMN such that the quadrilateral MNBP may be a maximum.

Prop. IV.* Through a given point O in the tangent at C to a circle draw a secant AB such that the triangle ABC may be of maximum area.

[graphic][subsumed][subsumed][subsumed][subsumed]

Draw tangents at A and B to meet in T. The required triangle is such that the parallels through A and B to the tangents at these points meet on OC in P.

For since 0 and T, O and C, are pairs of conjugate points with respect to the circle, CT is the polar of 0.

*This Proposition may be omitted on the first reading.

Let OC', the second tangent from 0, meet PT in C′, Since PBTA is a rhombus, AB is at right angles to PT; also since TCMP is a harmonic row, we have

therefore

TC/CM=TP/PM=2;

TM or PM=3MC'.

Then a given angle COC is divided by the required line AB, such that the ratio of the tangents of its segments is known; therefore, etc.

Ex. If a, b, c denote the sides of the maximum triangle ABC, prove that

OA c2-a2

(1)

OB b2-c2'

[blocks in formation]
« ΠροηγούμενηΣυνέχεια »