COR. 3. If we consider the portion of the circle in Cor. 2, which is convex to C, the intercept of a variable tangent made by the sides of the angle subtends a constant angle a at the centre of the circle (2α = π- -C).

1

1

1

Hence the variable triangle AB10 has a constant vertical angle (a) and altitude (y), and therefore its base and area are minima when 4,0=B,0. In this case the point of contact P, is the middle point of A,B1. Therefore, having given a circle and two fixed tangents, the portion of a variable tangent intercepted by the fixed tangents becomes a minimum in two positions, viz., when its point of contact bisects the arc XY internally or externally.

In the latter case the area cut off (ABC) is a minimum but in the former a maximum;

For

therefore, since CXOY is constant, when A,B,0 is a minimum, A,B,C is a maximum.

EXAMPLES.

1. The triangle of least area and perimeter escribed to a circle is equilateral.

[For the point of contact of each side bisects the arc between the other; cf. Art. 8, Ex. 1.]

2. The polygon of least area and perimeter escribed to a circle is regular. (By Ex. 1.)

3. Having given the vertical angle C of a triangle in position and magnitude, and the in- or corresponding ex-circle, to prove that the line LM joining the middle points of the sides forms with the centre of the circle a triangle of constant area.

[For the ex-circle: if p be the perpendicular of ABC drawn from C to the base, and r, the radius, we have 20LM = c(+p+r3) =ABC+AOB=OCXY=const., etc.]

13. Problem.-Given an angle O of a triangle and a point P on the base, construct it such that AP. BP is a minimum.

Through P draw AB so that the triangle ABO is isosceles. Describe a circle touching the sides of the angle at A and B, and draw any other line A'PB'.

It is evident that AP. PB < A'P. B'P, and is therefore a minimum.

EXAMPLE.

1. Through the point of intersection P of two circles draw a line APB such that PA. PB is a minimum.

[This reduces to describe a circle touching the two given ones at A and B such that A, B and P are in a line.

It will be afterwards seen that this line passes through a point Q, on the line of centres 0,02 of the circles where 201/Q02=the ratio of the radii.]

14. Theorem. If a right line be divided into any two parts a and b, their rectangle is a maximum when the line is bisected.

hence ab is a maximum when a-b=0 or when a

=

COR. The continued product of the segments of a line is a maximum when the parts are equal.

EXAMPLES.

1. Through any point P on the base of a triangle parallels PX and PY are drawn to the opposite sides; the area of the parallelogram PXCY is a maximum when the base AB is bisected at P.

[For the triangles APX and BPY are constant in species, hence PX.PY¤AP.BP. But the area of the parallelogram =PX. PY sin CxPX. PY; therefore, etc.]*

2. The maximum rectangle inscribed in a given segment of a circle is such that if tangents BC and AC be drawn at its vertices X and Y, then BX=CX and CY=AY.

[For NX is the maximum rectangle that can be inscribed in the triangle BCN, and therefore greater than any other X'N. Hence from the symmetry of the figure the rectangle on the side XY is greater than that on X'Y', and a fortiori greater than that on X"Y".

*Hence, the maximum parallelogram inscribed in a triangle is half the area of the triangle.

The construction of the maximum rectangle is as follows:-Let BL be drawn perpendicular to OL, the diameter of the circle parallel to AB. Join OX and let it meet BL in P. Since the triangles OCX and BPX are equal in every respect (Euc. I. 26)

but PL-PB is given; hence the segment PB is known, and since it is equal to OC, C' is determined.

In the general case the line AB does not meet the circle, the segment is therefore imaginary, and the proposition may be thus stated-given a line AB and a circle; construct the maximum rectangle, having two of its vertices X and Y on the circle and the remaining two on the line.]

3. Draw a chord XY of a circle in a given direction such that the area of the quadrilateral ABXY, where AB is a given diameter, is a maximum.

[Draw a diameter YX', and AYBX' is a rectangle, hence AX' is equal and parallel to BY. Join BX and XX', and draw BC parallel to XX'.

Then since

triangle AXX + triangle BXY=triangle ABX',

reject the common part AMX' and let BMX be added to each,

The quadrilateral is a maximum therefore when BXX' is a

maximum. It is easy to see that the latter is half the rectangle inscribed in a given segment BC.

For since BC is parallel to XX', AC is perpendicular to XX' and therefore parallel to PX, hence BAC=a.

The problem is thus reducible to Ex. 2.]

4. If a given finite line be divided into any number of parts a, b, c .; to find when a b c ... is a maximum, where a, ß, y are given quantities.

[This expression is a maximum when

(9)*(-3)*(-)*

α Ᏸ γ

...

is a maximum........

...... (1) but a/a is one of the a equal parts into which the segment a may be divided; hence (a/a)a is the product of the equal subdivisions. Similarly (b/B) is the product of the ẞ equal subdivisions of b, and so on. Therefore (1) attains its greatest value when the subdivisions of a, b, c are all equal; i.e., when

...

5. Find a point 0 with respect to a triangle such that the product of the areas (BOC)COAXAOB) is a maximum.

[Since BOC+COA+AOB is constant, when BOC=COA=AOB, by Ex. 4, or when O is the centroid of the triangle.]

6. The maximum triangle of given perimeter is equilateral. [From the formula A2=s(s-a)(s-b)(s-c); since the sum of the factors on the right hand side is constant, A is a maximum when s-a-s-b=s-c; therefore, etc.]