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PROPOSITION V. THEOREM.

753. Of all polygons formed with given sides, that which can be inscribed in a circle is the maximum.

F

E

B

Given: Two mutually equilateral polygons, ABCDE and A'B'C'D'E', of which only ABCDE can be inscribed in a circle; To Prove: ABCDE is greater than A'B'C'D'E'.

Draw the diameter AF, and join FC, FD.

Upon c'D' (= CD) construct ▲ F'C'D' = ▲ FC D, and join A'F'. Since polygon ABCF is inscribed in a semicircle,

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(Ax. 5)

.. ABCFDE-FCD > A'B'C'F'D'E'-F'C'D';

i.e., ABCDE> A'B'C'D'E'.

Q.E.D.

754. SCHOLIUM. The area of the inscribed polygon will be the same in whatever order the sides are arranged. For these sides are chords which cut off equal segments, in whatever order they occur; and the polygon is the difference between the circle and these segments.

EXERCISE 960. Of all rectangles that can be inscribed in a given circle, the greatest is a square.

PROPOSITION VI. THEOREM.

755. Of isoperimetric polygons of the same number of sides, the maximum is a regular polygon.

B B'

Given: ABCD, the maximum of isoperimetric polygons of n sides;

To Prove: ABCD is a regular polygon.

The polygon ABCD must be equilateral.

For if any two of its sides, BA, BC, were unequal, then upon AC as base we could construct an isosceles triangle B'A C, having the sum of its sides, B'A, B'C, equal to BA+B C. The triangle B'AC would be greater than BAC (748), and therefore the polygon AB'CD would be greater than the maximum polygon ABCD. But this is impossible; hence ABCD must be equilateral. It can also be inscribed in a circle (753); hence it is a regular polygon (380).

EXERCISE 961. Of all triangles having the same vertical angle, and whose bases pass through a given point, that whose base is bisected by the given point is least.

962. The rectangle contained by the segments of a line is a maximum when the segments are equal.

963. Through a given point within a given circle, draw the maximum and minimum chords that pass through that point.

964. From a given point without a given circle, draw the secant whose outer segment is a minimum. What about its inner segment ?

PROPOSITION VII. THEOREM.

756. Of isoperimetric regular polygons, that having the greatest number of sides is the maximum.

B

Given: A regular polygon A B C, of n sides;

To Prove: ABC is less than an isoperimetric regular polygon of n + 1 sides.

In one of the sides, as AB, take any point P.

We may now regard the given polygon as an irregular polygon of n+1 sides, in which the sides AP, PB, make with each other an angle equal to two right angles.

This irregular polygon is less than the regular polygon of the same perimeter and having n + 1 sides (755); that is, a regular polygon of n sides is less than the isoperimetric regular polygon of n + 1 sides.

Q.E.D.

757. COR. The circle contains a maximum area within a given perimeter.

EXERCISE 965. On the circumference of a given circle find the point such that the sum of the squares of its distances from two given points without the circle shall be a minimum.

966. Of two given circles, one lies wholly within the other. Find the maximum and minimum chords of the outer that are tangent to the inner circle.

967. A line A B C is perpendicular to an indefinite line CM. Find the point Pin CM at which the angle APB is a maximum.

968. Find a point within a quadrilateral such that the sum of the lines drawn from that point to the vertices shall be a minimum.

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758. Of two regular polygons having equal areas, that having the greater number of sides has the less perimeter.

P

R

Given: P and Q, regular polygons of the same area, but with the greater number of sides;

To Prove: Perimeter of Q is less than perimeter of P.

Let R be a regular polygon having the same perimeter as Q and the same number of sides as P.

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759. COR. The circumference of a circle is less than the perimeter of any polygon of equal area.

EXERCISE 969. Through a given point within a given angle draw the intercept that cuts off the triangle of maximum area.

970. Through a point of intersection of two circles draw that intercept between the two circumferences which is a maximum.

971. In a given line find a point such that the sum of its distances from two given points without the line, and on the same side of it, shall be a minimum.

972. In a given line find the point such that the tangents drawn from it to a given circle contain the maximum angle.

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