PROPOSITION XIX. PROBLEM.

399. To inscribe in a given circle a regular polygon similar to a given regular polygon.

_ DI

CA

Let A B C D, etc., be the given regular polygon, and

C'D' E' the given circle.

It is required to inscribe in C'D' E a regular polygon similar to A B C D, etc. From 0, the centre of the polygon A B C D, etc.

draw 0 D and 0 C.
From O' the centre of the O C' D'E',

draw O'C' and O'D',
making the 20= 2 0.

Draw C" D'. Then C" D' will be a side of the regular polygon required.

For each polygon will have as many sides as the 20 =20') is contained times in 4 rt. As.

.. the polygon C' D'E', etc. is similar to the polygon C D E, etc,

§ 372 (two regular polygons of the same number of sides are similar).

Q. E. F.

PROPOSITION XX. PROBLEM. 400. To circumscribe about a circle a regular polygon similar to a given inscribed regular polygon.

Let H MRS, etc., be a given inscribed regular polygon.

It is required to circumscribe a regular polygon similar to HMRS, etc.

At the vertices H, M, R, etc., draw tangents to the O, intersecting each other at A, B, C, etc..

Then the polygon A B C D, etc. will be the regular polygon required.

Since the polygon A B C D, etc.

has the same number of sides as the polygon H MRS, etc.,

it is only necessary to prove that A B C D, etc. is a regular polygon.

§ 372

the A BHM, B MH, CMR, and CRM are equal, $ 209

(being measured by halves of equal arcs) ;

.. the A BHM and C M R are equal, § 107 (having a side and two adjacent 1 of the one equal respectively to a side and

two adjacent of the other).

..ZB=LC,

(being homologous 5 of equal A ).
In like manner we may prove 2C=2 D, etc.

.. the polygon A B C D, etc., is equiangular.
Since the A BHM, CMR, etc. are isosceles, § 241
(two tangents drawn from the same point to a O are equal),
the sides B H, BM, CM, C R, etc. are equal,

(being homologous sides of equal isosceles S). .. the sides A B, BC, C D, etc. are equal, Ax. 6

and the polygon A B C D, etc. is equilateral. Therefore the circumscribed polygon is regular and similar to the given inscribed polygon.

§ 372 Q. E F.

Ex. Let R denote the radius of a regular inscribed polygon, g the apothegm, a one side, A one angle, and C the angle at the centre; show that

1. In a regular inscribed triangle a = R V3, r = { R, A = 60°, C= 120°.

2. In an inscribed square a = RV2, n=1 RV2, A = 90°, C = 90°.

3. In a regular inscribed hexagon a = R, p = \ R V3, A= 120°, C = 60°.

4. In a regular inscribed decagon a = (V3_ ) go= IR V10 + 2 V5, A = 144°, C=36o.

PROPOSITION XXI. PROBLEM.

401. To find the value of the chord of one-half an arc, in terms of the chord of the whole arc and the radius of the circle.

D

B

Let A B be the chord of arc A B and A D the chord

of one-half the arc A B.

It is required to find the value of A D in terms of A B and R (radius). From D draw D H through the centre 0,

and draw 0 A. HD is I to the chord A B at its middle point C, $ 60 (two points, 0 and D, equolly distant from the extremities, A and B, determine the position of a I to the middle point of A B). The 2 HA D is a rt. Z,

§ 204 (being inscribed in a sernicircle), .. A D = DH X DC,

$ 289 (the square on one side of a rt. A is equal to the product of the hypoten use b?

the adjacent segment made by the I let fall from the vertex of the rt. 2).

Now

and

DH= 2 R,
DC=DO – C0=R-CO;
.:AD' = 2 R (R – CO).

substitute for C 0 its value VAR? - ", then ADP = 2 R (r – V4 RAB),

= 2 B – R (V4 #2 – AZ"). ..AD=v2 r – R (V+ P – ĀB).

2

Q. E. F.

402. COROLLARY. If we take the radius equal to unity,

omes

the equation A D=V2 P2 – R (V4* – AB") becomes

AD=V2-V4-AB.