r (1+cos(θ)) sin(θ) C θ 2 r cos θ 2


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1 icles
2 xmple 66: Rounding one ssume we hve cone of ngle θ, nd we ound it off with cuve of dius, how f wy fom the cone does the ound stt? nd wht is the chod length? (1+cos(θ)) sin(θ) θ 2 cos θ 2
3 xmple 67: yclic Qudiltels Opposite ngles of yclic qudiltels dd up to 180, so the exteio ngle equls the opposite inteio ngle:
4 xmple 68: ngles subtended by chod The ngle t the cente subtended by chod is twice the ngle t the cicumfeence subtended by tht chod: 2
5 Why? b π2 π2 b 2 (+b) b
6 xmple 69: ngle subtended by point outside the cicle We genelize the bove esult fo the ngle subtended by point outside the cicle: b 2 +b
7 xmple 70: ngle t intesection of two cicles: Hee is fmili method of mking the 120 degee ngle needed to dw hexgon: 2 π 3
8 xmple 71: ngle subtended by two tngents π 22
9 xmple 72: istnce between the incente nd cicumcente ist, n isosceles tingle b 2 +b 2 b b Then, moe genelly:
10 b c4 +c 3 (b)+c bb 2 +c 32 b b 2 +b 3 +b+c (+bc) (b+c) (+b+c) b c
11 xmple 73: Rdius of icle though 2 vetices of tingle tngent to one side b c 2 b +b+c (+bc) (b+c) (+b+c)
12 xmple 74: cicles succinct fomul: Length of the common tngent to two tngentil 2 b b lso the intenl common tngent bisects this:
13 2 b b b b
14 xmple 75: Tngents to the Rdicl xis of Pi of icles The dicl xis of pi of cicles is the line joining the points of intesection. The lengths of tngents fom given point on this xis to the two cicles e the sme b 2 +b c 22 b 2 c 2 +c 4 +4 c 2 d 2 2 c c G b d b 2 +b c 22 b 2 c 2 +c 4 +4 c 2 d 2 2 c
15 xmple 76: Inveting segment in cicle To invet point in cicle dius, you find point on the line such tht = 2 If = nd = b nd = c, then we find the length of the inveted segment c 2 b c 1 b 2 b
16 xmple 77: Inveting cicle We cn invet cicle, by inveting its points of tngency: H R 2 R2 2 R I R G
17 xmple 78: The nine point cicle The nine point cicle is the cicle though the midpoint of ech side of tingle. Tking tingle with one vetex t the oigin, nothe t (x,0) nd nothe t (x1,y1), we look t the cente nd dius of the nine point cicle: x 1,y 1 x 4 + x 1 2, x x 1 x 2 2 +y y 1 G (0,0) (x,0) x22 x x 1 +x 1 2 +y 12 x 1 2 +y 12 x 4 x y 1
18 xmple 79: xcicles The thee excicles of tingle e tngent to the thee sides but exteio to the cicle: b G +b+c +bc +b+c 2 b+c c H I
19 We exmine the tingle joining the centes of the excicles: b G +b+c +bc +b+c 2 b+c c 2 b c (+bc) (b+c) H I
20 xmple 80: icle tngentil to two sides of n equiltel tingle nd the cicle centeed t thei intesection though the othe vetices This cicle hs dius one thid the side of the tingle 3
21 xmple 81: icle tngent to 3 sides of cicul segment Hee e the dii of the cicle tngent to two sides nd the cicul c of cicul segment, fo one o two popul ngles: π 2
22 π 3
23 π 4
24 xmple 82: Vious icles in n quiltel Tingle We look t the dii of vious cicles in n equiltel tingle: G
25 I H 3 54 G Wht would the next length in the sequence be?
26 xmple 83: icle tngentil to bse of equiltel tingle nd constucting cicles Now we cete cicle centeed on though nd, nd mke cicle tngentil to it the thn to line. We see the cicle hs dius 3/8 3 8
27 xmple 84: Rdius of the cicle though two vetices of tingle nd tngent to one side b 2 c +b+c (+bc) (b+c) (bc) b c
28 xmple 85: icle Tngent to 3 icles with the sme dii G H nd the extenl one:
29 I H G 3+2 3
30 xmple 86: icles Tngent to 3 icles of iffeent Rdii: s H G s + 1 t +2 1 s + 1 t + 1 s t I t J s  1 t +2 1 s + 1 t + 1 s t
31 xmple 87: icle Tngentil to 3 cicles sme dius 3 cicles of the sme dius, two e tngentil nd the thid is distnce b nd c fom those b H G c 2 +b+c 2 +b+c 2 +bc 2 b+c 8 b c+42 +b+c 2 +b+c 2 +bc 2 b+c b 2 +b 4 +c 4 +c b 2
32 xmple 88: thid Two cicles inside cicle twice the dius, then H 2 3 G 2
33 nd if we keep on going: H 2 3 G J 3 I L K 2 11 N M 9 2
34 The genel cse looks like this: 1 x H G x x 2 We cn copy this expession into Mple to genete the bove sequence: > 1/(1/2*1/+1/x+2*sqt(1/2*1/(^2)+1/2/x/)); x x > subs(=1,%); x x > f:=x>1/(1/2+1/x+sqt(2+2/x)); 1 f := x x x > f(1);
35 2 3 > f(2/3); > f(1/3); > f(2/11); > f(1/9); > f(2/27); > little nlysis of the seies cn led us to postulte the fomul 2/(n^2+2) fo the n th cicle: Let s feed the n1th tem into Mple: > f(2/((n1)^2+2)); ( n 1 ) 2 ( n 1 ) 2 In ode to get the expession to simplify, we mke the ssumption tht n>1: > ssume(n>1); > simplify(f(2/((n1)^2+2))); n~ 2 We see tht this is the next tem in the seies. y induction, we hve shown tht the n th cicle 2 hs dius n
36 xmple 89: theoem old in Pppus time theoem which ws old in Pppus dys (3 d centuy ) eltes the dii to height of the cicles in figues like the bove: 2 3 H G 3 J I 2 L 11 K
37 pplying the genel model, we get fomul: x  2 x 22 x2 x x +x 3 2 gin, we cn copy this into Mple fo nlysis: > (2*sqt(x)*^34*x^(3/2)*^22*x^(5/2)*)*sqt(2* 2*x)/(+x)^2/; ( 2 x 3 4 x ( 3/ 2 ) 2 2 x ( 5/ 2 ) ) 2 2 x ( + x ) 2 > simplify(%); 2 x 2 2 x > subs(x=2/(n^2+2),=1,%); n~ n~ 2 > simplify(%); n~ n~ 2 We see tht the height bove the centeline fo these cicles is the dius multiplied by 2n.
38
39 xmple 90: Yet nothe mily of icles We genelize the sitution fom couple of exmples go. We look t the fmily geneted by two cicles of dius nd b inside cicle of dius +b: b (+b) 2 + b+9 b 2 b (+b) 2 + b+4 b 2 I b (+b) 2 + b+b 2 G b (+b) 2 + b+16 b 2 K M N L J H b +b The ptten is petty obvious this time: the dius of the nth cicle is: b( + b) 2 2 n + b + b 2
40 To pove this, we deive the fomul fo the genel cicle dius x, nd nlyze in Mple: 1 1 b + 1 x  1 +b +2 1 b x  1 b (+b)  1 x (+b) H G x +b b Now we ty feeding in one of the cicle dii into this fomul in mple (fist mking the ssumption tht the dii e positive (long with n>1 fo lte use): > ssume(>0,b>0,n>1); > f:=x>1/(1/b+1/x1/(+b)+2*sqt(1/(x*b)1/((+b)*b) 1/((+b)*x))); f := x 1 b x + b x b ( + b ) b > f((+b)*b*/(9*^2+b*+b^2)); 1 9 ~ 2 + b~ ~ + b~ 2 1 1/ + b~ ( ~ + b~ ) b~ ~ ~ + b~ > simplify(%); 9 ~ 2 + b~ ~ + b~ ( ~ + b~ ) b~ 2 ~ ( ~ + b~ ) b~ 1 ( + b ) x 9 ~ 2 + b~ ~ + b~ 2 ( ~ + b~ ) 2 b~ ~
41 ( ~ + b~ ) b~ ~ 16 ~ 2 + b~ ~ + b~ 2 Let s ty the genel cse, feeding in the fomul fo the n1 st dius: > simplify(f((+b)*b*/((n1)^2*^2+b*+b^2))); b~ ~ ( ~ + b~ ) b~ ~ + ~ 2 n~ 2 + b~ 2 y induction, we hve poved the genel esult. 1 1 b + 1 x  1 +b +2 1 b x  1 b (+b)  1 x (+b) x b +b
42 xmple 91: chimedes Twins The given cicles e mutully tngentil with dius, b nd +b. chimedes twins e the cicles tngentil to the common tngent of the inne cicles. We see fom the symmety of the dius expession tht they e conguent. L b +b K G J I b +b b +b H
43 xmple 92: icles tngentil to two Touching cicles nd thei common tngent Hee is nothe fmily of cicles., bsed on oiginl cicles of dius
44 Notice tht the two ed ones hve the sme dii:
45 xmple 93: The tingle joining the points of tngency of 3 cicles Given tngentil cicles dii,b,c, we find the side length of the tingle joining the points of tngency: H c G 2 b c b+ c+b c+c2 (+c) (b+c) b I
46 xmple 94: The tingle tngentil to 3 tngentil cicles We look t the tingle fomed by the common tngents to these 3 cicles in the cse whee the cicles ll hve dius, this is n equiltel tingle, with length s shown below: I H G
47 xmple 95: ente nd Rdius of icle Given qution ind the cente nd dius of the cicle whose eqution is x y + x + by + c = 0 2 +b 24 c 22, b 2 X 2 +Y 2 +X +Y b+c=0
48 xmple 96: limit point In the exmple, is the intesection of with the x xis. We exmine the x coodinte of s tends to 0 (0,0) (1,0) ,0 This cn be done by mens of pictue: (0,0) (1,0) (4,0) O nlyticlly in Mple: > limit(^2/(sqt(+2)*sqt(+2)2),=0); 4
49 xmple 97: uehle s icle Given cicle inside nd tngent to nothe cicle, we cete n isosceles tingle whose bse connects the intesection between the xis of symmety nd the two cicles, nd whose pex lies on the oute cicle. We get the height of this tingle: H s  s+s s s> s (0,0) 0 G
50 Using this s input, we constuct the cicle tngent to both cicles, whose cente is on the line though G pependicul to. We exmine its dius nd the distnce fom its cente to the line. We infe tht it is tngentil to this line. 2 (s) +s M 2 (+s) +s s 2 > 2 H s L +s +s (0,0) 0 G
51 xmple 98: icle to two cicles on othogonl dii of thid In the digm, icles nd G e centeed on pependicul dius lines nd. icle HI is tngentil to ll thee. We obseve tht H is ectngle. 0, 2 s +s I H s, 2 s +s G s (0,0) 0
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