The Elements of Plane and Solid Geometry ...

Proposition 23. Theorem.

472. Of regular polygons with a given perimeter, that which has the greatest number of sides has the greatest area.

Hyp. Let P be a regular poly

gon of three sides, and Q a regular

polygon of four sides, with the same given perimeter.

Proof. In any side AB of P take any pt. C.

The polygon P may be regarded as an irregular polygon of four sides, in which the sides AC, CB make with each other a st..

Then, since the polygons P and Q are isoperimetric, and have the same number of sides,

(Нур.)

.*. the irregular polygon P < the regular polygon Q. (471)

In the same way it may be shown that Q < the regular isoperimetric polygon of five sides, and so on.

Q.E.D.

473. COR. The circle has a greater area than any polygon of the same perimeter.

EXERCISES.

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

2. The shortest chord which can be drawn through a given point within a

circle is the perpendicular to the diameter which passes through that point.

Proposition 24. Theorem.

474. Of regular polygons having the same given area, the greater the number of sides the less will be the perimeter. Hyp. Let P and Q be regular polygons having the same area, and let Q have the greater number of sides.

To prove the perimeter of P > that of Q.

Proof. Let R be a regular poly

gon having the same perimeter as

Q and the same number of sides as P.

Then, since Q has the same perimeter as R and a greater number of sides,

(Cons.)

(472)

(Hyp.)

(379)

.. the perimeter of P > the perimeter of R.

.. the perimeter of P > the perimeter of Q. Q.E.D. 475. COR. The circumference of a circle is less than the perimeter of any polygon of the same area.

Proposition 25. Theorem.

476. Given two intersecting straight lines AB, AC, and a point P between them; then 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.

Hyp. Let EF be the st. line, termi

nated by AB, AC, which is bisected. at P.

To prove ▲ AEF is a minimum. Proof. Let HK be any other st. line passing through P.

In the same way it may be shown that ▲ AEF < any other A formed by a st. line through P.

... ▲ AEF is a minimum.

Q. E.D.

Proposition 26. Problem.

477. 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.

Given, the st. line CD, and the pts. A, B, on the same side of CD.

Required, to find at what pt. in CD the subtended by the st. line AB is a maximum.

Cons. Describe a to pass through

A, B, and to touch the st. line CD. (339) and Ex. 40 in (354)

Let P be the pt. of contact.

Then the APB is the required max. Z.

Proof. Take any other pt. in CD as Q, and join AQ, BQ.

Then,

ZAQB <ZAPB.

(246)

... APB > any other subtended by AB at a pt. in CD.

Q.E.F.

Proposition 27. Problem.

478. 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.

Given, the st. line CD, and the pts. A, B, on the same side

of CD.

Required, to find a pt. P in CF

CD, so that the sum of AP, PB

is a minimum.

Cons. Draw AF 1 to CD;

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

Join AP, PB.

Then, P is the required pt., and of all lines drawn from A and B to a pt. in CD, the sum of AP, PB is a minimum. Proof. Let Q be any other pt. in CD.

... APPB < AQ + QB.

... the sum of AP and PB is a minimum.

Q.E.F.

479. COR. The sum of AP and PB is a minimum, when these lines are equally inclined to CD;

for

ZAPC

= ZEPC =

BPD.

NOTE. In order that a ray of light from A may be reflected to a point B, it must fall upon a mirror CD at a point P where APC ZBPD; i.e., by (478) the ray pursues the shortest path between A and B and touching CD.

EXERCISES.

THEOREMS.

1. If AB be a side of an equilateral triangle inscribed in a circle, and AD a side of the inscribed square, prove that three times the square on AD is equal to twice the square on AB.

2. Show that the sum of the perpendiculars from any point inside a regular hexagon to the six sides is equal to three times the diameter of the inscribed circle.

3. The area of the regular inscribed hexagon is twice the area of the inscribed equilateral triangle.

4. The area of the regular inscribed hexagon is threefourths of that of the regular circumscribed hexagon.

5. The area of the regular inscribed hexagon is a mean proportional between the areas of the inscribed and circumscribed equilateral triangles.

6. If the perpendicular from A to the side BC of the equilateral triangle ABC meet BC in D, and the inscribed. circle in G; prove that GD = 2AG.

See figure of (269).

7. If three circles touch each other externally, and a triangle be formed by joining their centres, and another triangle by joining their points of contact, the inscribed circle of the former triangle will be the circumscribed circle of the latter.

8. If ABCD be a square described about a given circle, and P any point on the circumference, of the circle; prove that the sum of the squares on PA, PB, PC, PD, is three times the square on the diameter of the circle.

Use (333).

9. ABC is a triangle having each of the angles B, C double the angle A; the bisectors of the angles B, C meet AC and the circle circumscribing the triangle ABC respectively in D, E: prove that ADBE is a rhombus.

10. ABCDE is a regular pentagon inscribed in a circle, P is the middle point of the arc AB: show that the difference of the straight lines AP, CP is equal to the radius of the circle.

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