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The following example illustrates the same point.

2. To find at what point in a given straight line the angle subtended by the line joining two given points, which are on the same side of the given straight line, is a maximum.

Let CD be the given st. line, and A, B the given points on the same side of CD. It is required to find at what point in CD the angle subtended by the st. line AB is a maximum.

First determine at what point in CD, the st. line AB subtends a given angle.

This is done as follows:

On AB describe a segment of a circle containing an angle equal to the given angle.

III. 33. If the arc of this segment intersects CD, two points in CD are found at which AB subtends the given angle: but if the arc does not meet CD, no solution is given.

In accordance with the principles explained above, we expect that a maximum angle is determined at the limiting position; that is, when the arc touches CD, or meets it at two coincident points.

[See page 231.] This we may prove to be the case.

Describe a circle to pass through A and B, and to touch the st. line CD.

[Ex. 21, p. 253.] Let P be the point of contact.

Then shall the L APB be greater than any other angle subtended by AB at a point in CD on the same side of AB as P.

For take Q, any other point in CD, on the same side of AB as ; and join AQ, QB.

B Since Q is a point in the tangent other than the point of contact, it must be without the circle ; :: either BQ or AQ must meet the arc of the segment APB.

Let BQ meet the arc at K: join AK. Then the L APB=the L AKB, in the same segment : but the ext. L AKB is greater than the int. opp. L AQB.

:: the L APB is greater than AQB. Similarly the L APB may be shewn to be greater than any other angle subtended by AB at a point in CD on the same side of AB:

that is, the 2 APB is the greatest of all such angles. Q. E. D. NOTE. Two circles may be described to pass through A and B, and to touch CD, the points of contact being on opposite sides of AB; Join AP,

hence two points in CD may be found such that the angle subtended by AB at each of them is greater than the angle subtended at any other point in CD on the same side of AB.

We add two more examples of considerable importance.

3. In a straight line of indefinite length find a point such that the sum of its distances from two given points, on the same side of the given line, shall be a minimum. Let CD be the given st. line of

B
indefinite length, and A, B the
given points on the same side of
CD.
It is required to find a point

P in
CD, such that the sum of AP, PB is
a minimum.
Draw AF

perp:

to CD;
and produce AF to E, making FE
equal to AF.
Join EB, cutting CD at P.

PB.
Then of all lines drawn from A and B to a point in CD,

the sum of AP, PB shall be the least.
For, let Q be any other point in CD.

Join AQ, BQ, EQ.
Now in the A8 AFP, EFP,
AF =EF,

Constr.
Because {and FP is common;

and the L AFP=the _ EFP, being rt. angles.

:: AP=EP.
Similarly it may be shewn that

AQ=EQ. Now in the A EQB, the two sides EQ, QB are together greater than EB;

hence, AQ, QB are together greater than EB,

that is, greater than AP, PB. Similarly the sum of the st. lines drawn from A and B to any other point in CD may be shewn to be greater than AP, PB. .. the sum of AP, PB is a minimum.

Q. E, D. NOTE. It follows from the above proof that the L APF=the L EPF

I. 4. =the L BPD. Thus the sum of AP, PB is a minimum, when these lines are equally inclined to CD.

I, 4.

I. 15.

4. Given two intersecting straight lines AB, AC, and a point P between them ; shew that of all straight lines which pass through P and are terminated by AB, AC, that which is bisected at P cuts off the triangle of minimum area.

Let EF be the st. line, terminated by AB, AC, which is bisected at P.

Then the A FAE shall be of mini

с

mum area.

For let HK be any other st. line passing through P.

IM
Through E draw EM parto AC.
Then in the As HPF, MPE, A

E KB the L HPF =the L MPE,

I. 15. Because and the L HFP=the _ MEP,

I. 29. and FP=EP;

Hyp. the A HPF=the AMPE. 1. 26, Cor. But the A MPE is less than the A KPE; in the A HPF is less than the A KPE:

to each add the fig. AHPE;

then the A FAE is less than the À HAK. Similarly it may be shewn that the A FAE is less than any other triangle formed by drawing a st. line through P:

that is, the A FAE is a minimum.

EXAMPLES.

1. Two sides of a triangle are given in length; how must they be placed in order that the area of the triangle may be a maximum ?

2. Of all triangles of given base and area, the isosceles is that which has the least perimeter.

3. Given the base and vertical angle of a triangle; construct it so that its area may be a maximum.

4. Find a point in a given straight line such that the tangents drawn from it to a given circle contain the greatest angle possible.

5. A straight rod slips between two straight rulers placed at right angles to one another ; in what position is the triangle intercepted between the rulers and rod a maximum ?

6. Divide a given straight line into two parts, so that the sum of the squares on the segments

(i) may be equal to a given square ;

(ii) may be a minimum. 7. Through a point of intersection of two circles draw a straight line terminated by the circumferences,

(i) so that it may be of given length;

(ii) so that it may be a maximum. 8. Two tangents to a circle cut one another at right angles ; find the point on the intercepted arc such that the sum of the perpendiculars drawn from it to the tangents may be a minimum.

9. Straight lines are drawn from two given points to meet one another on the convex circumference of a given circle: prove that their sum is a minimum when they make equal angles with the tangent at the point of intersection.

10. Of all triangles of given vertical angle and altitude, that which is isosceles has the least area.

11. Two straight lines CA, CB of indefinite length are drawn from the centre of a circle to meet the circumference at A and B; then of all tangents that may be drawn to the circle at points on the arc AB, that whose intercept is bisected at the point of contact cuts off the triangle of minimum area.

12. Given two intersecting tangents to a circle, draw a tangent to the convex arc so that the triangle formed by it and the given tangents may be of maximum area.

13. Of all triangles of given base and area, that which is isosceles has the greatest vertical angle.

14. Find a point on the circumference of a circle at which the straight line joining two given points (of which both are within, or both without the circle) subtends the greatest angle.

15. A bridge consists of three arches, whose spans are 49 ft., 32 ft. and 49 ft. respectively : shew that the point on either bank of the river at which the middle arch subtends the greatest angle is 63 feet distant from the bridge.

16. From a given point P without a circle whose centre is C, draw a straight line to cut the circumference at A and B, so that the triangle ACB may be of maximum area.

17. Shew that the greatest rectangle which can be inscribed in a circle is a square.

18. A and B are two fixed points without a circle : find a point Pon the circumference, such that the sum of the squares on AP, PB may be a minimum. [See p. 161, Ex. 24.]

19. A segment of a circle is described on the chord AB : find a point C on its arc so that the sum of AC, BC may be a maximum.

20. Of all triangles that can be inscribed in a circle that which has the greatest perimeter is equilateral.

21. Of all triangles that can be inscribed in a given circle that which has the greatest area is equilateral.

22. Of all triangles that can be inscribed in a given triangle that which has the least perimeter is the triangle formed by joining the feet of the perpendiculars drawn from the vertices on opposite sides.

23. Of all rectangles of given area, the square has the least perimeter.

24. Describe the triangle of maximum area, having its angles equal to those of a given triangle, and its sides passing through three given points.

VI.

HARDER MISCELLANEOUS EXAMPLES.

1. AB is a diameter of a given circle ; and AC, BD, two chords on the same side of AB, intersect at E: shew that the circle which passes through D, E, C cuts the given circle orthogonally.

2. Two circles whose centres are C and D intersect at A and B; and a straight line PAQ is drawn through A and terminated by the circumferences : prove that

(i) the angle PBQ=the angle CAD

(ii) the angle BPC=the angle BQD. 3. Two chords AB, CD of a circle whose centre is O intersect at right angles at P: shew that

(i) PAP+PB2+ PC2+ PD2=4 (radius)2.

(ii) AB2+CD2 + 40P2 =8 (radius)? 4. Two parallel tangents to a circle intercept on any third tangent a portion which is so divided at its point of contact that the rectangle contained by its two parts is equal to the square on the radius.

5. Two equal circles move between two straight lines placed at right angles, so that each straight line is touched by one circle, and the two circles touch one another : find the locus of the point of contact.

6. AB is a given diameter of a circle, and CD is any parallel chord : if any point X in AB is joined to the extremities of CD, shew that

XC2+ XD2=XA2+ XB2.

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