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Theorem 4. If the ends of a line of given length are joined by a straight line, and the area of the figure enclosed is a maximum, it takes the form of a semicircle.

Given a line APB (the curve

in the figure), of given length,

and AB joining its end-points.

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S

a

S2

Proof. 1. Let P be any point on the line; then joining A and P, B and P, let the segments cut off by AP, BP be called s1, S2, and ▲ ABP called t, as in the figure. 2. Then P is a right angle; for if not, without changing S1, S2, the area of t could be increased by making P right.

Th. 1

3. But this is impossible if ABP is a maximum.
4. Similarly for any other point on APB... APB is a
semicircumference.

Why?

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2. Then S1, S2 are maxima when they are semicircles,

and AB is a diameter.

Why?

EXERCISE. 513. If the diagonals of a parallelogram are given, its area is a maximum when it is a rhombus.

Theorem 6. Of all equal plane figures the circle has the minimum perimeter.

P

X

Given

To prove

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that circumference C<perimeter P.

Proof. 1. Suppose X a circle of circumference equal to

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Theorem 7. A polygon with given sides is a maximum

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suppose congruent segments constructed on a, b,..... (opposite P').

2. Then P+ΣA > P' + A.

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Th. 5 Why?

Theorem 8. Of all isoperimetric polygons of a given number of sides, the maximum is regular.

Given

P the maximum polygon of a given perimeter and a given number of sides.

To prove

that P is regular.

B X

P

Proof. 1. Any two adjacent sides, AB, BC, must be equal. For if unequal, as AX, XC, then ▲ AXC could be replaced by ABC, thus enlarging P without changing the perimeter. But this is impossible because P is a

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Theorem 9. Of two isoperimetric regular polygons, that having the greater number of sides is the greater. Proof. 1. Let ABCD be a square, P a point

on DA, A PCX isoperimetric with

A PCD and having CX PX.

2. Then A PCX > PCD. Th. 2

3. .. area ABCXP> ABCD.

Ax. 4

4. But pentagon ABCXP would,
with the same perimeter, be

greater if it were regular.

X

Th. 8

5. .. a regular pentagon is greater than an isoperimetric square. Similarly, a regular hexagon would be greater than a regular isoperimetric pentagon, and so on.

EXERCISE. 514. Considering only the relation of space inclosed to amount of wall, what would be the most economical form for the ground plan of a house?

EXERCISES.

515. Why is the most economical form for piping that

with a circular cross-section ?

516. Of all triangles in a given circle, that which has the greatest perimeter is equilateral.

517. Of all triangles in a given circle, what is the shape of the one having the greatest area? Prove it.

518. A cross-section of a bee's cell is a regular hexagon. Show that this is the best form for securing the greatest capacity with a given amount of wax (perimeter).

519. Find a point in a given straight line such that the tangents drawn from it to a given circle contain the maximum angle.

520. A straight ruler, 1 foot long, slips between the two edges of the floor (the edges making a right angle). Find the position of the ruler when the triangle formed by the edges and ruler is a maximum; also the area of that triangle.

521. Through a point of intersection of two circumferences draw the maximum line terminated by the two circumferences.

522. Of all triangles of a given base and area, the isosceles has the greatest vertical angle.

523. Draw the minimum straight line between two non-intersecting circumferences.

524. Find the maximum rectangle that can be inscribed in a given semicircle.

525. The minimum square that can be inscribed in a given square has half its area.

526. Draw the minimum tangent from a variable point in a given line to a given circle.

527. Given the point P within the XOY, to draw through P a line cutting OX, OY and making with them the minimum triangle. (See the note under th. 3; symmetry suggests either that the line be bisected at P, or that an isosceles ▲ be formed.)

528. What is the area of the largest triangle that can be inscribed in a circle of radius 5 ?

529. Given a square of area 1. Find the area of an isoperimetric (1) equilateral triangle, (2) regular hexagon, (3) circle.

530. The sum of the lines from the center of an equilateral triangle to the three vertices is less than the sum of the lines from any other point to those vertices.

531. In a given triangle, each of whose angles is less than 2 π/3, to find the point the sum of whose distances from the vertices is a minimum. (Proposed by Fermat to Torricelli.)

Section 2.-Concurrence and Collinearity.

Theorem 10. If X, Y, Z are three points on the sides a, b, c, respectively, of a triangle ABC, such that the perpendiculars to the sides at these points are concurrent, then

(BX2XC2) + (CY2YA2) + (AZ2 - ZB2) = 0;

and conversely.

Proof. 1. Let P be the point of con

currence, and draw PA,

PB, PC.

2. Then (BX2-XC2)

+ (CY2 - YA2)

+(AZ2 - ZB3)

PB2 PC2+PC2- PA2

+PA2PB20.

3. To prove the converse, suppose the

meet at P; and suppose PZ' c.

4. Then by step 2

12

Why?

from X, Y, to

(BX2 — XC2) + (CY2 — YA2) + (AZ12 — Z'B2) = 0. 5. But (BX2-XC2) + (CY2- YA2) + (AZ2— ZB3)=0. 6... AZ2 Z'B2 AZ2 - ZB2.

=

Why? 7... AZ12 AZ2 Z'B2- ZB2; but these differences have opposite signs and cannot be equal unless each is zero.

8.

EXERCISES.

.. Z = Z'.

532. Show that the following is a special case of th. 10: The perpendicular bisectors of the sides of a triangle are concurrent. 533. Also, the perpendiculars from the vertices of a triangle to the opposite sides are concurrent.

534. If three circumferences intersect in pairs, the common chords are concurrent.

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