The Symbolic Geometry System


 Morgan Casey
 2 years ago
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1 The Symbolic Geometry System This document is intended to grow with the symbolic geometry system, nd provide set of exmples which we cn use s strting point for testing, documenttion nd demonstrtion. 1
2 U S I N G S Y M O L I G E O M E T R Y ontents THE SYMOLI GEOMETRY SYSTEM... 1 ontents... First Exmple...6 Tringles...10 Exmple 1: ltitude of right ngled tringle...11 Exmple : ltitude of the symmetricl ltitude of n isosceles tringle...1 Exmple 3: ltitude of the nonsymmetricl ltitude of n isosceles tringle...13 Exmple 4: ltitude of tringle defined by side lengths...14 Exmple 5: ltitude of n Isosceles Tringle...15 Exmple 6: Tringle res...16 Exmple 7: Intersection of the Medins of Right ngled Tringle...19 Exmple 8: entroid of generl tringle...0 Exmple 9: ngle of Medin... Exmple 10: Orthocenter...3 Exmple 11: Orthocenter oordintes...4 Exmple 1: ircumcircle rdius...5 Exmple 13: ircumcircle rdius in terms of vectors...7 Exmple 14: Incircle Rdius...8 Exmple 15: Vector joining the incircle with the vertex of tringle defined by vectors...9 Exmple 16: Incircle enter in rycentric oordintes...30 Exmple 17: How does the point of contct with the incircle split line...31 Exmple 18: Length of line joining vertex with the point of contct between the incircle nd the opposite side..3 Exmple 19: PythgorsLike igrm...33 Exmple 0: Penequilterl Tringle...34 Exmple 1: nother Penequilterl Tringle...38 Exmple : Folding right ngled tringle...40 Exmple 3: Interior point of n Equilterl Tringle...41 Exmple 4: Right ngled tringle in semicircle...43 Exmple 5: Line Joining the pex to bse of n isosceles tringle...44 Exmple 6: re of Tringle defined by vectors...45 Exmple 7: Similr Tringle...46 Exmple 8: res of tringles bounded by evins...47 Exmple 9: ecomposing vector into components prllel nd perpendiculr to second vector...48 Qudrilterls...49 Exmple 30: igonls of rhombus...50 Exmple 31: igonls of prllelogrm...51 Exmple 3: igonls of Kite...5 Exmple 33: yclic Qudrilterl...55 Exmple 34: Right Trpezoid...56 Exmple 35: res of Qudrilterls...57 Exmple 36: res of tringles in trpezoid...60 Exmple 37: imeter of the circumcircle...6 Exmple 38: Qudrilterl with perpendiculr digonls nd one right ngle...63 Exmple 39: Finding the dimeter of n rc given the perpendiculr offset from the chord...64 Exmple 40: Npoleon s Theorem...65 Exmple 41: n Isosceles Tringle Theorem...66 Exmple 4: Qudrilterl with Perpendiculr igonls...67
3 Exmple 43: Intersection of ommon Tngent with xis of Symmetry of Two ircles...68 Exmple 44: Slope of the ngle isector...70 Exmple 45: Loction of intersection of common tngents...71 Exmple 46: ltitude of yclic Trpezium defined by common tngents of circles...73 Exmple 47: res yclic Trpezi defined by common tngents of circles...74 Exmple 48: Tringle formed by the intersection of the interior common tngents of three circles...75 Exmple 49: istnce between sides of rhombus...76 Exmple 50: ngles of Specific Tringles...77 Exmple 51: Sides of Specific Tringles...79 Exmple 5: ngles in the generl tringle...81 Exmple 53: Some implied right ngles...85 Exmple 54: Tringle defined by ngles nd side...90 Exmple 55: Tringle defined by two sides nd the included ngle...91 Exmple 56: Tringle defined by sides nd the nonincluded ngle...9 Exmple 57: Incenter...93 Exmple 58: Qudrilterl Formed by Joining the Midpoints of the sides of Qudrilterl...94 Exmple 59: Some mesurements on the Pythgors igrm...95 Exmple 60: n unexpected tringle from Pythgorslike digrm...96 Exmple 61: Theorem on Qudrilterls...97 Exmple 6: Rectngle ircumscribing n Equilterl Tringle Polygons Exmple 63: Regulr Pentgons nd more...10 Exmple 64: Regulr Pentgon onstruction Exmple 65: re of Hexgon bounded by Tringle side trisectors ngles nd ircles Exmple 66: Rounding orner Exmple 67: yclic Qudrilterls Exmple 68: ngles subtended by chord Exmple 69: ngle subtended by point outside the circle Exmple 70: ngle t intersection of two circles: Exmple 71: ngle subtended by two tngents ircles Exmple 7: istnce between the incenter nd circumcenter Exmple 73: Rdius of ircle through vertices of tringle tngent to one side Exmple 74: Length of the common tngent to two tngentil circles...10 Exmple 75: Tngents to the Rdicl xis of Pir of ircles...1 Exmple 76: Inverting segment in circle...13 Exmple 77: Inverting circle...14 Exmple 78: The nine point circle...15 Exmple 79: Excircles...16 Exmple 80: ircle tngentil to two sides of n equilterl tringle nd the circle centered t their intersection through the other vertices...18 Exmple 81: ircle tngent to 3 sides of circulr segment...19 Exmple 8: Vrious ircles in n Equilterl Tringle...13 Exmple 83: ircle tngentil to bse of equilterl tringle nd constructing circles Exmple 84: Rdius of the circle through two vertices of tringle nd tngent to one side Exmple 85: ircle Tngent to 3 ircles with the sme rdii Exmple 86: ircles Tngent to 3 ircles of ifferent Rdii: Exmple 87: ircle Tngentil to 3 circles sme rdius Exmple 88: Two circles inside circle twice the rdius, then third Exmple 89: theorem old in Pppus time Exmple 90: Yet nother Fmily of ircles Exmple 91: rchimedes Twins
4 U S I N G S Y M O L I G E O M E T R Y Exmple 9: ircles tngentil to two Touching circles nd their common tngent Exmple 93: The tringle joining the points of tngency of 3 circles Exmple 94: The tringle tngentil to 3 tngentil circles Exmple 95: enter nd Rdius of ircle Given Eqution Exmple 96: limit point Exmple 97: uehler s ircle Exmple 98: ircle to two circles on orthogonl rdii of third...16 Equtions of Lines nd ircles Exmple 99: Intersection of Two Lines Exmple 100: Eqution of Line Through Two Points Exmple 101: Intersection of Line with ircle Exmple 10: Eqution of the Line Joining the Intersection of Two ircles Exmple 103: Projecting Point onto Line Exmple 104: Rdius of the incircle of Tringle formed by 3 Lines Whose Equtions re Known Exmple 105: Eqution of the ltitude of Tringle efined by lines with Given Equtions Exmple 106: Eqution of the Line through Given Point t 45 degrees to Line of Given Eqution Exmple 107: Eqution of Tngent to ircle Rdius r entered t the Origin, through given point...17 Exmple 108: Intersection of two tngents to the curve y=x^ Trnsforms Exmple 109: omposition of reflections in prllel lines Exmple 110: ombining Reflections Exmple 111: Prbolic Mirror Exmple 11: illirds Exmple 113: res nd ilttion Exmple 114: Trnslting circle Some onstructions...18 Exmple 115: Eqution of the Perpendiculr isector of Points Exmple 116: Length of the ngle isector of tringle Some Mechnisms Exmple 117: rnk Piston Mechnism Exmple 118: Quick Return Mechnism Exmple 119: Pucellier s Linkge Exmple 10: Hrborth Grph Loci Exmple 11: ircle of pollonius...19 Exmple 1: ircle inside ircle Exmple 13: nother ircle in ircle Exmple 14: Ellipse s locus Exmple 15: rchimedes Trmmel Exmple 16: n lterntive Ellipse onstruction Exmple 17: nother ellipse Exmple 18: ent Strw Ellipse onstruction...00 Exmple 19: Similr construction for Hyperbol...01 Exmple 130: Prbol s locus of points equidistnt between point nd line...0 Exmple 131: Squeezing circle between two circles...03 Exmple 13: Rosce Qutre rnches...04 Exmple 133: Lemniscte...05 Exmple 134: Pscl s Limçon...06 Exmple 135: Kulp Qurtic...07 Exmple 136: The Witch of gnesi
5 Exmple 137: Newton s Strophoid...09 Exmple 138: McLurin s Trisectrix nd other Such Like...10 Exmple 139: Trisectrice de elnge...1 Exmple 140: Foglie del Surdi...13 Exmple 141: onstruction of iocletin...14 Exmple 14: Kpp urve...15 Exmple 143: Kepler s Egg...16 Exmple 144: ruciform urve...17 Exmple 145: Locus of centers of common tngents to two circles...18 Exmple 146: Stedy Rise m urve...0 Exmple 147: Oscillting Flt Plte m...1 Exmple 148: m Str... Exmple 149: Ellipse s Envelope of ircles...4 Exmple 150: Hyperbol s n envelope of circles...5 Exmple 151: Hyperbol s n Envelope of Lines...6 Exmple 15: ustics in cup of coffee...7 Exmple 153: Nephroid by nother route...8 Exmple 154: Tschirnhusen s ubic...9 Exmple 155: ubic Spline...30 Exmple 156: Tringle Spline...31 Exmple 157: nother Tringle Spline
6 U S I N G S Y M O L I G E O M E T R Y First Exmple s first exmple, we ll do Pythgors Theorem. First we drw the tringle: 6
7 s first tsk, let s mke the ngle right. To do this, we select the segments nd, then click the perpendiculr button: Notice tht there is now perpendiculr constrint symbol between the two line segments. Now we need to set the distnce between nd to be x. To do this, we select the segment nd click the distnce constrint button You will get n edit box with the preset vlue of. hnge this to x. Similrly, set the constrint of y between nd : 7
8 U S I N G S Y M O L I G E O M E T R Y 8
9 Now we re redy to disply the vlue of the hypotenuse. Select the segment nd the lculte istnce button The => sign in front of the mesurement shows tht it is n output. 9
10 U S I N G S Y M O L I G E O M E T R Y Tringles The following exmples explore spects of tringles: 10
11 Exmple 1: ltitude of right ngled tringle b +b b 11
12 U S I N G S Y M O L I G E O M E T R Y Exmple : ltitude of the symmetricl ltitude of n isosceles tringle x x b 4 +x b 1
13 Exmple 3: ltitude of the nonsymmetricl ltitude of n isosceles tringle x x b x b  b4 4 x b 13
14 U S I N G S Y M O L I G E O M E T R Y Exmple 4: ltitude of tringle defined by side lengths Heron s formul sys tht the re of tringle is ( + b + c)( + b c)( b + c)( + b + c) 4 You ll need to expnd the expression to see the reltionship to the ltitude below: b + b b  b 4  c 4 c c 4 c 4 c 14
15 Exmple 5: ltitude of n Isosceles Tringle n isosceles tringle is defined by bse length nd one ngle. Wht is its ltitude? sin(θ) cos(θ) θ 15
16 U S I N G S Y M O L I G E O M E T R Y Exmple 6: Tringle res Two sides nd the included ngle: θ b sin(θ) b 16
17 Two sides nd the non included ngle: b b sin(θ) +b cos(θ) sin(θ) b θ 17
18 U S I N G S Y M O L I G E O M E T R Y Three sides: +b+c +bc b+c +b+c 4 c b 18
19 Exmple 7: Intersection of the Medins of Right ngled Tringle Medins will be esy to specify when we hve the Midpoint construction. However, they cn lso be specified simply by setting the distnce to the end of the medin to be hlf the specified length of the side: E F +b 3 b b 19
20 U S I N G S Y M O L I G E O M E T R Y Exmple 8: entroid of generl tringle We look t the centroid (intersection of the medins). We check the length of the medin nd the loction of the incenter on tht medin: 9 + b 9  c 9 E F + b  c 4 b c c We see tht the centroid is locted /3 of the wy down the medin. 0
21 lterntively, if we constrin the tringle by specifying its vertices, we obtin this formul for the loction of the centroid: x 1,y 1 F x 0 +x 1 +x y 0 +y 1 +y, 3 3 x 0,y 0 E x,y 1
22 U S I N G S Y M O L I G E O M E T R Y Exmple 9: ngle of Medin Given tringle with sides length,b nd included ngle θ we derive the ngle the medin mkes with the other side of the tringle: rctn b sin(θ) +b cos(θ) rctn  b sin(θ)  +b θ b
23 Exmple 10: Orthocenter We look t the orthocenter (intersection of the ltitudes) of tringle. gin we derive n expression for its distnce from vertex of the tringle. b F b b +c +b+c (+bc) (b+c) (+b+c) c E 3
24 U S I N G S Y M O L I G E O M E T R Y Exmple 11: Orthocenter oordintes This exmple is tken from the pper Lernen mit einem neuen Tschenrechner Given tht nd re locted t points (,b) nd (c,b) respectively, nd tht point cn move long the line y=d, we find the coordintes of the intersection of the ltitudes. The locus is qudrtic, s shown both by its shpe nd by its eqution. X b  c+b d+x (+c)+y (bd)=0 (t,d) F (,b) E (c,b) 4
25 Exmple 1: ircumcircle rdius The center of the circumcircle of tringle is the intersection of the perpendiculr bisectors. Exmining the expression for the distnce from one vertex to the intersection point shows tht it is symmetric. Hence will be the sme, whichever vertex is chosen (try it). b b F b c +b+c (+bc) (b+c) (bc) E c 5
26 U S I N G S Y M O L I G E O M E T R Y Of course, more direct pproch is just to drw the circumcircle: b c +b+c (+bc) (b+c) (+b+c) b c 6
27 Exmple 13: ircumcircle rdius in terms of vectors If we hve vectors defining circle, wht is the circle s rdius: u 0 u 1 u 0 v + 0 u 0 u 1 v 0 u 0 v 1 u 0 +u 1 + v 0 +v 1 + u 0 v + 0 v +v +v u 1 v 0 u 0 v 1 u 0 +u 1 + v 0 +v 1 u 1 u 0 v 1 v 0 7
28 U S I N G S Y M O L I G E O M E T R Y Exmple 14: Incircle Rdius b E +bc b+c +b+c +b+c c 8
29 Exmple 15: Vector joining the incircle with the vertex of tringle defined by vectors u 1 u 0 +v0 +u0 u 1 +v1 u 0 v 0 u 0 +v0 + u1 +v1 + u0 u 1 + v0 +v 1 v 1 u 0 +v0 +v0 u 1 +v1 F u 0 +v0 + u1 +v1 + u0 u 1 + v0 +v 1 E u 1 v 1 Exmine the expression the denomintor of ech coefficient is the perimeter of the tringle, so the vector cn be written: u v 0 0 u u u u P 1 P + v u
30 U S I N G S Y M O L I G E O M E T R Y Exmple 16: Incircle enter in rycentric oordintes x 1,y 1 E x,y x 0,y 0 x x 0 +x 1 + y 0 y 1 +x 1 x 0 +x + y 0 y +x 0 x 1 +x + y 1 y x 0 +x 1 + y 0 y 1 + x 0 +x + y 0 y + x 1 +x + y 1 y y x 0 +x 1 + y 0 y 1 +y 1 x 0 +x + y 0 y +y 0 x 1 +x + y 1 y, x 0 +x 1 + y 0 y 1 + x 0 +x + y 0 y + x 1 +x + y 1 y If we let =, b= nd c= nd let, nd be the position vectors of the points,, then the incircle center is: + b + c + b + c 30
31 Exmple 17: How does the point of contct with the incircle split line +bc G b H bc F c E 31
32 U S I N G S Y M O L I G E O M E T R Y Exmple 18: Length of line joining vertex with the point of contct between the incircle nd the opposite side E b  b 3  b c+ b c+3 c + b c  c 3 H b c 3
33 Exmple 19: PythgorsLike igrm rw tringle (not necessrily right ngled), nd set the side lengths to be, b nd c. Now drw squres on ech side, constrining them with right ngles nd lengths of sides. Now look t the distnce between neighboring corners of the squres: b + c c c c b  + b + c + b c b b 33
34 U S I N G S Y M O L I G E O M E T R Y Exmple 0: Penequilterl Tringle Strting with tringle whose sides re length,b,c, we construct squres on ech side, join the corners of the squres, then join the midpoints of these lines to crete tringle: F J G E c L b K I H 34
35 The tringle looks to the nked eye s if it is equilterl. Try drgging the originl points nd observe tht the new tringle still looks equilterl: I c H G K b L E J F 35
36 U S I N G S Y M O L I G E O M E T R Y Is it in fct? + b + 3 c 4 + +b+c +bc b+c +b+c F J b + c + +b+c +bc b+c +b+c I G E c L b K K I H 4 + b 4 + c b+c +bc b+c +b+c 16 Not quite, we observe the sides re gurnteed to be close in size, but not identicl unless the originl tringle is isosceles. In fct, the difference in squres of the sides of the new c tringle is. 4 4 Notice we cn repet the process drwing squres on JK, KL nd JL. The difference in c squres of the ensuing sides will be
37 y repeting this process, we cn crete tringle s close s we like to n equilterl tringle, but still not exctly one. O V M F J P G E N L b K U I H T Q R 37
38 U S I N G S Y M O L I G E O M E T R Y Exmple 1: nother Penequilterl Tringle We cn do similr construction bsed on equilterl tringles drwn on the sides of n originl tringle: E Side Length b c b+c +bc b+c +b+c 8 G b H re c b + 3 c +5 +b+c +bc b+c +b+c 3 I E We see tht the difference in squres of the sides of the new tringle corresponding to c nd b is. If we repet the process, this tringle will eventully become equilterl, but not s quickly s the previous tringle. In fct, if S(n) is the sum of the sides t the nth stge in the itertion, nd (n) is the re of the tringle on the nth stge of the itertion, the bove equtions for side length nd re show tht the following re true: 38
39 ) ( 8 5 1) ( 3 3 ) ( 1) ( 6 3 1) ( 8 5 ) ( + = + = n n S n n n S n S Which reltion we cn solve to give: ( ) = (0) (0) ) ( ) ( S n n S n
40 U S I N G S Y M O L I G E O M E T R Y Exmple : Folding right ngled tringle If you crete second right ngled tringle by cutting perpendiculr to the hypotenuse, then tht tringle is similr to the originl one. Here re its side lengths: b x +b b x E x 40
41 Exmple 3: Interior point of n Equilterl Tringle If point is interior to n equilterl tringle, the sum of the perpendiculr distnces to the sides of the tringle is independent of the loction of the point. The following construction shows this (try dding the three distnces). F x y 6 G y 3 ( x+y) 3 E 3 x y 3 x 41
42 U S I N G S Y M O L I G E O M E T R Y Here is simpler wy to define this one, using perpendiculr distnce constrints. In this cse we specify x to be the perpendiculr distnce between nd, nd y to be the perpendiculr distnce from to. We then sk for the perpendiculr distnce from to. x y  3 +x+y Notice tht the sum of these 3 distnces is 3, which is independent of x nd y. 4
43 Exmple 4: Right ngled tringle in semicircle This digrm shows tht the center of the hypotenuse of right ngled tringle is hlf hypotenuse wy from the opposite vertex: d d d 43
44 U S I N G S Y M O L I G E O M E T R Y Exmple 5: Line Joining the pex to bse of n isosceles tringle Tke n isosceles tringle whose equl sides hve length nd divide its bse in lengths x, y. Join the dividing point to the opposite vertex, nd exmine its length. The expression is surprisingly simple: x y x y From this result, you cn deduce, for exmple, tht for ny two chords through prticulr point interior to circle, the product of the lengths of the chord segments on either side of the point is the sme. 44
45 Exmple 6: re of the tringle is: re of Tringle defined by vectors u 1 v 0 u 0 v  1 u 0 v 0 u 1 v 1 45
46 U S I N G S Y M O L I G E O M E T R Y Exmple 7: Similr Tringle Lengths on similr tringle go by rtio: b E c b c d c d b E c 46
47 Exmple 8: res of tringles bounded by evins The rtios of res in this digrm re simple qudrtics in t. n you derive these qudrtics from the sine rule? re of z 4 +b+c +bc b+c +b+c 4 re of F z 3 t +b+c +bc b+c +b+c 4  t +b+c +bc b+c +b+c 4 t<1 z 3 tt re of F/ = z 4 t<1 b t b E t F c c t re of EF z 0 +b+c +bc b+c +b+c re of EF/ = 13 t+3 t z 0 z +b+c +bc b+c +b+c b+c +bc b+c +b+c 4 3 t +b+c +bc b+c +b+c t +b+c +bc b+c +b+c t +b+c +bc b+c +b+c t +b+c +bc b+c +b+c > t +b+c +bc b+c +b+c t +b+c +bc b+c +b+c >
48 U S I N G S Y M O L I G E O M E T R Y Exmple 9: ecomposing vector into components prllel nd perpendiculr to second vector v 0 u 1 v 0 +u 0 v 1 u 1 v 1 u 0 v0 u 0 u 1 v 0 u 0 v 1 u 0 v0 u 0 u 0 u 1 v 0 v 1 u 0 v 0 u 0 v0 v 0 u 0 u 1 v 0 v 1 u 0 v0 48
49 Qudrilterls Here re some exmples using qudrilterls 49
50 U S I N G S Y M O L I G E O M E T R Y Exmple 30: igonls of rhombus rhombus hs sides length nd one digonl length b, wht is the length of the other digonl? 4 b b 50
51 Exmple 31: igonls of prllelogrm Given prllelogrm whose sides mesure nd b, nd one digonl is c, wht is the length of the other digonl? b c + b c b simple enough result, but cn you derive it? I used the cosine rule, but cn you do it by Pythgors lone? 51
52 U S I N G S Y M O L I G E O M E T R Y Exmple 3: igonls of Kite ontinuing in this theme, if we hve kite whose nonxis digonl is length c, nd whose sides re length nd b, wht is the length of the other digonl? c +b  c + +c c b+c bc b b n you see the similrity to the prllelogrm? 5
53 Wht bout the nonconvex kite with the sme side lengths nd digonl? c b b +c c  b+c bc 53
54 U S I N G S Y M O L I G E O M E T R Y How bout if we re specified the xis, wht is the nonxis digonl? + b  4 d + b d  b 4 d d d b b This is very similr to the eqution of the ltitude of tringle. Why? 54
55 Exmple 33: yclic Qudrilterl qudrilterl is inscribed in circle of rdius. 3 sides of the qudrilterl hve length b. Wht is the length of the fourth side? b b E b b 96 b + b 4 4 Wht is the reltionship between nd b when the fourth side hs length b? 55
56 U S I N G S Y M O L I G E O M E T R Y Exmple 34: Right Trpezoid h +(+b) b h 56
57 Exmple 35: Trpezium: res of Qudrilterls h b h (+b) 57
58 U S I N G S Y M O L I G E O M E T R Y Kite: b b 58
59 Prllelogrm: θ b b sin(θ) 59
60 U S I N G S Y M O L I G E O M E T R Y Exmple 36: res of tringles in trpezoid One pir of tringles formed by the digonls of trpezoid re equl in re. b h (+b) b h (+b) h b 60
61 The other pir re not: h (+b) h b h (+b) b 61
62 U S I N G S Y M O L I G E O M E T R Y Exmple 37: imeter of the circumcircle Here is digrm which llows us to find the dimeter of the circumcircle of the tringle whose sides hve lengths, b nd c. Why? b c b c +b+c (+bc) (b+c) (bc) 6
63 Exmple 38: Qudrilterl with perpendiculr digonls nd one right ngle Qudrilterl hs 3 sides length,b, c nd right ngle. It s digonls re perpendiculr. Wht is the length of the remining side? b  +b +c c 63
64 U S I N G S Y M O L I G E O M E T R Y Exmple 39: Finding the dimeter of n rc given the perpendiculr offset from the chord. Here is the digrm. Verify tht it is correct: h E h+ 4 h 64
65 Exmple 40: Npoleon s Theorem Npoleon s Theorem sttes tht if you tke generl tringle nd drw n equilterl tringle on ech side, then the tringle formed by joining the incenters of these new tringles is equilterl. You cn see tht the length is symmetricl in,b,c nd hence identicl for the three sides of the tringle. P O b Q P E b b c S c c T 6 + b 6 + c b+c +bc b+c +b+c 6 F 65
66 U S I N G S Y M O L I G E O M E T R Y Exmple 41: n Isosceles Tringle Theorem is isosceles. E=. We show tht EF=F. c E b 4 +c  b c 4 b F c b 4 +c  b c 4 66
67 Exmple 4: Qudrilterl with Perpendiculr igonls Given two sides, the lengths of the digonls nd the fct tht they re perpendiculr, wht re the lengths of the other two sides of qudrilterl? d +d b+c+d b+cd bc+d b+c+d  b c b +c b+c+d b+cd bc+d b+c+d  b 67
68 U S I N G S Y M O L I G E O M E T R Y Exmple 43: Intersection of ommon Tngent with xis of Symmetry of Two ircles Line hs length. is perpendiculr distnce x from line, nd is perpendiculr distnce y from. Find the distnce of the intersection point between nd from nd : y x+y y E x x x+y 68
69 Wht if we chnge the digrm slightly so tht E is externl to : y xy x xy E y x 69
70 U S I N G S Y M O L I G E O M E T R Y Exmple 44: Slope of the ngle isector Wht is the slope of the ngle bisector of line with slope 0 nd line with slope m? m m m 0 70
71 Exmple 45: Loction of intersection of common tngents ircles nd hve rdii r nd s respectively. If the centers of the circles re prt, nd E is the intersection of the interior common tngent with the line joining the two centers, wht re the lengths E nd E? r r r+s E s r+s s 71
72 U S I N G S Y M O L I G E O M E T R Y How bout the exterior common tngent? r r r+s s  s r+s E 7
73 Exmple 46: ltitude of yclic Trpezium defined by common tngents of circles Given circles rdii r nd s nd distnce prt, wht is the ltitude of the trpezium formed by joining the intersections of the 4 common tngents with one of the circles? r J r s H s F G I E Notice tht this is symmetricl in r nd s, nd hence the trpezium in circle hs the sme ltitude. 73
74 U S I N G S Y M O L I G E O M E T R Y Exmple 47: res yclic Trpezi defined by common tngents of circles Look t the rtio of the res of the trpezi in the previous exmple: z 0 r s r  r ss + r s r + r ss r J H s F G I E z 0 z 1 r s z 1 r s r  r ss + r s r + r ss 74
75 Exmple 48: Tringle formed by the intersection of the interior common tngents of three circles Notice tht if is the re of the tringle formed by the centers of the circles, then re STU is: rst ( r + s)( s + t)( r + t) r s t +b+c +bc b+c +b+c (r+s) (r+t) (s+t) I s N S K H G c T O r Q F t L J b U M E P R 75
76 U S I N G S Y M O L I G E O M E T R Y Exmple 49: istnce between sides of rhombus Given rhombus with side length nd digonl b, wht is the perpendiculr distnce between opposite sides? b b +b b 76
77 Exmple 50: ngles of Specific Tringles Here re some tringles, with their ngles displyed π 3 π 3 π 3 77
78 U S I N G S Y M O L I G E O M E T R Y π 4 π 4 rctn b rctn b b 78
79 Exmple 51: Sides of Specific Tringles Here re some specific ngledefined tringles: π 4 3 π 6 79
80 U S I N G S Y M O L I G E O M E T R Y π π
81 Exmple 5: 3 S I E S E F I N E ngles in the generl tringle c rccos  +b +c b c b 81
82 U S I N G S Y M O L I G E O M E T R Y T R I N G L E W I T H S I E S N I N L U E N G L E c π+rctn b sin() c+b cos() b 8
83 T R I N G L E W I T H S I E S N N O N  I N L U E N G L E c πrcsin b sin() c b 83
84 U S I N G S Y M O L I G E O M E T R Y T R I N G L E W I T H T W O N G L E S Here s the exterior ngle: s+t s t 84
85 Exmple 53: Some implied right ngles The medin of n isosceles tringle is lso its perpendiculr bisector, nd ltitude: π b b 85
86 U S I N G S Y M O L I G E O M E T R Y tringle whose medin is the sme length s hlf its bse is right ngled: π 86
87 The digonls of kite re perpendiculr π b b 87
88 U S I N G S Y M O L I G E O M E T R Y The line joining the intersection points of two circles is perpendiculr to the line joining their centers: π 88
89 Here is prticulr tringle (from the book The urious Incident of the og in the Night ) 1+n π n 1+n 89
90 U S I N G S Y M O L I G E O M E T R Y Exmple 54: Tringle defined by ngles nd side sin(t) sin(s+t) πst sin(s) sin(s+t) s t 90
91 Exmple 55: included ngle Tringle defined by two sides nd the b +c  b c cos() c rctn b sin() c+b cos() b 91
92 U S I N G S Y M O L I G E O M E T R Y Tringle defined by sides nd the non Exmple 56: included ngle rctn   c sin() c cos() sin() c+ c sin() cos()+c cos() c πrcsin c sin() 9
93 Exmple 57: Incenter The incenter is the intersection of the ngle bisectors. We hve tringle with ngles nd b nd bse d: d cos(+b) sin(+b)  d cos(+b) sin(+b) d (cos(+b)cos(+b)) sin(+b) d b b We see tht the perpendiculr distnce to the other sides is the sme. This shows tht is the center of circle tngent to nd. 93
94 U S I N G S Y M O L I G E O M E T R Y Exmple 58: Qudrilterl Formed by Joining the Midpoints of the sides of Qudrilterl x x y 0 y + x 1,y 1 F x,y x x y 1 y + 3 E x x y 1 y  3 G x 0,y 0 H x x y 0 y  x 3,y 3 Quick inspection of the side lengths shows tht the new figure is prllelogrm 94
95 Exmple 59: Some mesurements on the Pythgors igrm rw right tringle nd subtend squre on ech side. The two red lines in the digrm re equl in length. G H +(b) F I b +(b) E 95
96 U S I N G S Y M O L I G E O M E T R Y Exmple 60: n unexpected tringle from Pythgorslike digrm Regrdless of the originl tringle the resulting tringle from this digrm is right ngledisosceles: +c + +b+c +bc b+c +b+c J F E G b π c K H I Exmintion of the length J shows tht it is symmetric in nd b, nd hence identicl to K. 96
97 Exmple 61: Theorem on Qudrilterls This theorem sttes tht if you drw squre on ech side of qudrilterl, then connect the center of opposite sides, the resulting lines hve the sme length, nd re perpendiculr. Here is the result in Geometry Expressions  b + c +b d c d + e +b e +c e +d e +e 4 + e +d+e +de d+e +d+e + e b+c+e b+ce bc+e b+c+e + +d+e +de d+e +d+e b+c+e b+ce bc+e b+c+e e J b b I H E e G c c K F d d L  b + c +b d c d + e +b e +c e +d e e 4 + +d+e +de d+e +d+e b+c+e b+ce bc+e b+c+e e If we crete the length of the other side we cn by creful exmintion see tht the lengths re identicl. lterntively, we cn do some simplifiction. Our constrints re necessrily symetric Geometry Expressions will not let you overconstrin the digrm, nd one digonl is sufficient to define the qudrlterl. However, we might expect the formul to be simpler if expressed in terms of both digonls. 97
98 U S I N G S Y M O L I G E O M E T R Y lose inspection of the formul for the length shows tht it incorportes the squre of the other digonl of the figure, s well s Heron s formul for the res of the tringles nd. The following Mthemtic worksheet contins the formuls from Geometry Expressions for L the length of the desired line, nd f the length of the other digonl. little mnipultion gives simple formul for L^f^/. This cn be simplified further by noting tht the remining terms re e^/ nd twice the re of the qudrlterl: JJ b + c + b d  c d + e + b e + c e + d e + e 4 + e " + d + e " + d  e "  d + e " #  + d + e + e " b+ c+ e " b+ c e " # b  c + e " #  b + c + e +" # + d + e " # + d  e " #  d + e " #  + d + e " # b + c+ e " # b+ c  e " # b c + e " #  b+ c + en 1 e I,I b + c +b d  c d + e +b e +c e +d e +!! 4 b +c e e b  c +e! b +e! b+c! +!  +d +e! +d +! b +c  e! +d  e! b  c +e!  b +c+e! b +c+e!  d +e!  +d +e! +d +em J"# J b + c + b d  c d + e + b e + c e + d e  e +" # 4 + d + e " # + d  e " #  d + e " #  + d + e " # b+ c+ e " # b+ c  e " # b c+ e " #  b+ c+ en 1! ei,i b + c +b d  c d + e +b e +c e + d e  e +! 4 b +c  e! +d  e! b  c +e!  b +c+e! b +c+e!  d +e!  +d +e! +d +em 1 4e I b + c +b d  c d + e +b e +c e +d e +e +!! 4 b +c  e e b  c +e!  b +c +e! b +c +!! +d  e e  d +e!  +d +e! +d +! b +c  e! +d  e! b  c +e!  b +c +e! b +c +e!  d +e! 1 e I b + c +b d  c d + e +b e +c e +d e  e +! 4 b+c  e! +d e! b  c +e!  b +c +e! b +c +e!  d +e!  +d +e! +d+em +! 1 Ie b +c e! b  c +e!  b +c +e! b +c +! +d  e!  d +e!  +d +e! +d +em In[6]:= L = 1 e Out[6]= In[4]:= f = 1 e Out[4]= In[7]:= L^ Out[7]= In[8]:= f^ Out[8]= In[11]:=  f^ Out[11]= e f L = + + From which we cn derive tht:! +d  e e  d +e  +d +e! +d +em 98
99 res re simpler when expressed in terms of ngles. Here is revision of the digrm with ngles inserted. This gives us more of clue of how to prove the result: b +c + d +b c sin(θ)+c d sin(φ)b d sin(θ+φ)b c cos(θ)c d cos(φ) J b + c +b c sin(θ) c +d  c d cos(φ) b π 4 H θ π 4 G I E b +c  b c cos(θ) F L d π φ 4 c c + d +c d sin(φ) K 99
100 U S I N G S Y M O L I G E O M E T R Y Exmple 6: Tringle Rectngle ircumscribing n Equilterl b 4  b 4 b b b F b E b 4 + b 4 b irectly from the digrm we hve the following theorem: The re of the lrger right tringle is the sum of the res of the smller two. This ppers in pge 191 of Mthemticl Gems, by Ross Honsberger (nd vrious other plces). 100
101 Polygons Some exmples with pentgons, hexgons 101
102 U S I N G S Y M O L I G E O M E T R Y Exmple 63: Regulr Pentgons nd more Wht length should the side of regulr pentgon be? We cn ddress this problem by mking pentgon with ll but one side length, then solving such tht the finl side is lso length : E r r 6 r 4 r 4 F > solve(sqrt(16*^^8/r^6+8*^6/r^40*^4/r^)=,); æ5 ç + 1 è 5 ö r, ø æ5 ç  1 è 5 ö r, 3 r, 0 ø 10
103 Ignoring the 0 solution, there re 3 solutions. Lets pick the middle one nd pste it in s : r 55 E r r 55 F We see this indeed gives us the regulr pentgon. Wht do the other solutions give us? 103
104 U S I N G S Y M O L I G E O M E T R Y r r F r 5+ 5 E The first solution gives the regulr five pointed str. How bout the third solution? 104
105 r 3 r 3 r E F t first sight we seem to hve lost of the sides. ut they re in fct there, sitting on top of. 105
106 U S I N G S Y M O L I G E O M E T R Y Exmple 64: Regulr Pentgon onstruction Here is construction for regulr pentgon inside circle of rdius r. is the midpoint of, E is congruent to nd F is congruent to E. r 55 r F I E G H We see this yields side of the requisite length. 106
107 Exmple 65: trisectors re of Hexgon bounded by Tringle side 3 b 3 b H K F I J L G 3 O N M b 3 c E c 3 3 c +b+c +bc b+c +b+c 40 The re cn be seen to be 1/10 the re of the originl tringle 107
108 U S I N G S Y M O L I G E O M E T R Y ngles nd ircles 108
109 Exmple 66: Rounding orner ssume we hve corner of ngle θ, nd we round it off with curve of rdius r, how fr wy from the corner does the round strt? nd wht is the chord length? r (1+cos(θ)) sin(θ) θ r r cos θ 109
110 U S I N G S Y M O L I G E O M E T R Y Exmple 67: yclic Qudrilterls Opposite ngles of yclic qudrilterls dd up to 180, so the exterior ngle equls the opposite interior ngle: E 110
111 Exmple 68: ngles subtended by chord The ngle t the center subtended by chord is twice the ngle t the circumference subtended by tht chord: 111
112 U S I N G S Y M O L I G E O M E T R Y Why? b π E π b (+b) b 11
113 Exmple 69: ngle subtended by point outside the circle We generlize the bove result for the ngle subtended by point outside the circle: F E b +b 113
114 U S I N G S Y M O L I G E O M E T R Y Exmple 70: ngle t intersection of two circles: Here is fmilir method of mking the 10 degree ngle needed to drw hexgon: π 3 114
115 Exmple 71: ngle subtended by two tngents E π  115
116 U S I N G S Y M O L I G E O M E T R Y ircles 116
117 Exmple 7: istnce between the incenter nd circumcenter First, n isosceles tringle F E b +b b b Then, more generlly: 117
118 U S I N G S Y M O L I G E O M E T R Y b c4 +c 3 (b)+c  +3 bb +c 3  b b +b 3 +b+c (+bc) (b+c) (+b+c) E F b c 118
119 Exmple 73: Rdius of ircle through vertices of tringle tngent to one side b c b +b+c (+bc) (b+c) (+b+c) 119
120 U S I N G S Y M O L I G E O M E T R Y Exmple 74: Length of the common tngent to two tngentil circles succinct formul: b b lso the internl common tngent bisects this: 10
121 b b F b b E 11
122 U S I N G S Y M O L I G E O M E T R Y Exmple 75: Tngents to the Rdicl xis of Pir of ircles The rdicl xis of pir of circles is the line joining the points of intersection. The lengths of tngents from given point on this xis to the two circles re the sme. 4  b +b 4  c  b c +c 4 +4 c d c c G b F d E 4  b +b 4  c  b c +c 4 +4 c d c 1
123 Exmple 76: Inverting segment in circle To invert point in circle rdius r, you find point E on the line such tht E = r If = nd = b nd = c, then we find the length of the inverted segment 11 1 r c r b c r 1 b r b 13
124 U S I N G S Y M O L I G E O M E T R Y Exmple 77: Inverting circle We cn invert circle, by inverting its points of tngency: F H R R r R I r E +r R +r G 14
125 Exmple 78: The nine point circle The nine point circle is the circle through the midpoint of ech side of tringle. Tking tringle with one vertex t the origin, nother t (x,0) nd nother t (x1,y1), we look t the center nd rdius of the nine point circle: x 1,y 1 x 4 + x 1, x x 1 x +y y 1 G E F (0,0) (x,0) x  x x 1 +x 1 +y 1 x 1 +y 1 x 4 x y 1 15
126 U S I N G S Y M O L I G E O M E T R Y Exmple 79: Excircles The three excircles of tringle re tngent to the three sides but exterior to the circle: E b F G +b+c +bc +b+c b+c c H I 16
127 We exmine the tringle joining the centers of the excircles: E b F G +b+c +bc +b+c b+c c b c (+bc) (b+c) H I 17
128 U S I N G S Y M O L I G E O M E T R Y Exmple 80: ircle tngentil to two sides of n equilterl tringle nd the circle centered t their intersection through the other vertices This circle hs rdius one third the side of the tringle E 3 18
129 Exmple 81: ircle tngent to 3 sides of circulr segment Here re the rdii of the circle tngent to two sides nd the circulr rc of circulr segment, for one or two populr ngles: 1+ E π 19
130 U S I N G S Y M O L I G E O M E T R Y E π 3 130
131 E π 4 131
132 U S I N G S Y M O L I G E O M E T R Y Exmple 8: Vrious ircles in n Equilterl Tringle We look t the rdii of vrious circles in n equilterl tringle: G F 3 18 E
133 I H 3 54 G F E Wht would the next length in the sequence be? 133
134 U S I N G S Y M O L I G E O M E T R Y Exmple 83: ircle tngentil to bse of equilterl tringle nd constructing circles Now we crete circle centered on through nd, nd mke circle E tngentil to it rther thn to line. We see the circle hs rdius 3/8 E
135 Exmple 84: Rdius of the circle through two vertices of tringle nd tngent to one side b c +b+c (+bc) (b+c) (bc) b c 135
136 U S I N G S Y M O L I G E O M E T R Y Exmple 85: rdii ircle Tngent to 3 ircles with the sme r r G H r 3+ 3 F r E nd the externl one: 136
137 I 3 r+ 3 r+r r H G F r r r 3+ 3 E 137
138 U S I N G S Y M O L I G E O M E T R Y Exmple 86: Rdii: ircles Tngent to 3 ircles of ifferent s r H G 1 1 r + 1 s + 1 t + 1 r s + 1 r t + 1 s t I t E F J 11 r  1 s  1 t + 1 r s + 1 r t + 1 s t 138
139 Exmple 87: ircle Tngentil to 3 circles sme rdius 3 circles of the sme rdius, two re tngentil nd the third is distnce b nd c from those F E b H G c  +b+c +b+c +bc b+c 8 b c+4  +b+c +b+c +bc b+c b +b 4 +c 4 +c 8  b 139
140 U S I N G S Y M O L I G E O M E T R Y Exmple 88: then third Two circles inside circle twice the rdius, H r 3 G r r E r F 140
141 nd if we keep on going: H r 3 G J r 3 I L K r 11 N M r 9 r r E r F 141
142 U S I N G S Y M O L I G E O M E T R Y The generl cse looks like this: 1 F x H G 1 r + 1 x r r x r r E We cn copy this expression into Mple to generte the bove sequence: > 1/(1/*1/r+1/x+*sqrt(1/*1/(r^)+1//x/r)); r x 1 + r x r > subs(r=1,%); x 1 + x 14
143 > f:=x>1/(1/+1/x+sqrt(+/x)); 1 f := x x > f(1); > f(/3); > f(1/3); > f(/11); > f(1/9); > f(/7); > little nlysis of the series cn led us to postulte the formul /(n^+) for the n th circle: Let s feed the n1th term into Mple: > f(/((n1)^+)); ( n 1 ) ( n 1 ) In order to get the expression to simplify, we mke the ssumption tht n>1: x 143
144 U S I N G S Y M O L I G E O M E T R Y > ssume(n>1); > simplify(f(/((n1)^+))); 1 + n~ We see tht this is the next term in the series. y induction, we hve shown tht the n th circle hs rdius + n 144
145 Exmple 89: theorem old in Pppus time theorem which ws old in Pppus dys (3 rd century ) reltes the rdii to height of the circles in figures like the bove: r 3 H G r 3 J I r L 11 K 4 r 3 1 r 11 r 4 r 3 r E r F 145
146 U S I N G S Y M O L I G E O M E T R Y pplying the generl model, we get formul: F x   r x r x x r x r+x r 3 r E gin, we cn copy this into Mple for nlysis: > (*sqrt(x)*r^34*x^(3/)*r^*x^(5/)*r)*sqrt(*r *x)/(r+x)^/r; ( x r3 4 x ( 3/ ) r x ( 5/ ) r ) r x ( r + x ) r > simplify(%); > subs(x=/(n^+),r=1,%); x r x 146
147 > simplify(%); n~ + n~ n~ 4 + n~ We see tht the height bove the centerline for these circles is the rdius multiplied by n. 147
148 U S I N G S Y M O L I G E O M E T R Y Exmple 90: Yet nother Fmily of ircles We generlize the sitution from couple of exmples go. We look t the fmily generted by two circles of rdius nd b inside circle of rdius +b: b (+b) + b+9 b b (+b) + b+4 b I b (+b) + b+b G b (+b) + b+16 b K M N L J H b E +b F The pttern is pretty obvious this time: the rdius of the nth circle is: b( + b) n + b + b 148
149 To prove this, we derive the formul for the generl circle rdius x, nd nlyze in Mple: 1 1 b + 1 x  1 +b + 1 b x  1 b (+b)  1 x (+b) H G E x F +b b Now we try feeding in one of the circle rdii into this formul in mple (first mking the ssumption tht the rdii re positive (long with n>1 for lter use): > ssume(>0,b>0,n>1); > f:=x>1/(1/b+1/x1/(+b)+*sqrt(1/(x*b)1/((+b)*b) 1/((+b)*x))); f := x 1 b x + b x b ( + b ) b > f((+b)*b*/(9*^+b*+b^)); 1 ( + b ) x 149
150 U S I N G S Y M O L I G E O M E T R Y 1 1/ b~ 9 ~ + b~ ~ + b~ + ( ~ + b~ ) b~ ~ 1 ~ + b~ 9 ~ + b~ ~ + b~ 1 + ( ~ + b~ ) b~ ~ ( ~ + b~ ) b~ > simplify(%); ( ~ + b~ ) b~ ~ 16 ~ + b~ ~ + b~ Let s try the generl cse, feeding in the formul for the n1 st rdius: 9 ~ + b~ ~ + b~ ( ~ + b~ ) b~ ~ > simplify(f((+b)*b*/((n1)^*^+b*+b^))); b~ ~ ( ~ + b~ ) b~ ~ + ~ n~ + b~ y induction, we hve proved the generl result. 150
151 1 1 b + 1 x  1 +b + 1 b x  1 b (+b)  1 x (+b) x b +b 151
152 U S I N G S Y M O L I G E O M E T R Y Exmple 91: rchimedes Twins The given circles re mutully tngentil with rdius, b nd +b. rchimedes twins re the circles tngentil to the common tngent of the inner circles. We see from the symmetry of the rdius expression tht they re congruent. L b +b K G J I b +b E b +b F H 15
153 Exmple 9: ircles tngentil to two Touching circles nd their common tngent Here is nother fmily of circles., bsed on originl circles of rdius r r r 153
154 U S I N G S Y M O L I G E O M E T R Y Notice tht the two red ones hve the sme rdii: r 5 r 16 r 9 r 5 r 4 154
155 Exmple 93: The tringle joining the points of tngency of 3 circles Given tngentil circles rdii,b,c, we find the side length of the tringle joining the points of tngency: F H c E G b c b+ c+b c+c (+c) (b+c) I b 155
156 U S I N G S Y M O L I G E O M E T R Y Exmple 94: The tringle tngentil to 3 tngentil circles We look t the tringle formed by the common tngents to these 3 circles in the cse where the circles ll hve rdius, this is n equilterl tringle, with length s shown below: I E F H + 3 G 156
157 Exmple 95: enter nd Rdius of ircle Given Eqution Find the center nd rdius of the circle whose eqution is x + y + x + by + c = 0 +b 4 c , b X +Y +X +Y b+c=0 157
158 U S I N G S Y M O L I G E O M E T R Y Exmple 96: limit point In the exmple, E is the intersection of with the x xis. We exmine the x coordinte of E s r tends to 0 r (0,0) (1,0) E r + +r r,0 This cn be done by mens of picture: r (0,0) (1,0) E (4,0) Or nlyticlly in Mple: 158
159 > limit(r^/(sqrt(r+)*sqrt(r+)),r=0); 4 159
160 U S I N G S Y M O L I G E O M E T R Y Exmple 97: uehler s ircle Given circle inside nd tngent to nother circle, we crete n isosceles tringle whose bse connects the intersection between the xis of symmetry nd the two circles, nd whose pex lies on the outer circle. We get the height of this tringle: H s r rs+s rs r s>r rs (0,0) F 0 G E 160
161 Using this s input, we construct the circle tngent to both circles, whose center is on the line through G perpendiculr to. We exmine its rdius nd the distnce from its center to the line. We infer tht it is tngentil to this line.  r (rs) r+s M r (r+s) r+s s >r H s L r r+s r+s (0,0) F 0 G E 161
162 U S I N G S Y M O L I G E O M E T R Y Exmple 98: ircle to two circles on orthogonl rdii of third In the digrm, ircles E nd FG re centered on perpendiculr rdius lines nd. ircle HI is tngentil to ll three. We observe tht FH is rectngle. r F 0, r s r+s I H rs, r s r+s G s E (0,0) 0 16
163 Equtions of Lines nd ircles 163
164 U S I N G S Y M O L I G E O M E T R Y Exmple 99: Intersection of Two Lines Wht is the intersection of the lines Y=c+mX nd Y=d+nX? Y=c+X m Y=d+X n E cd m+n, d m+c n m+n 164
165 Exmple 100: Eqution of Line Through Two Points x 1,y 1 x 1 y 0 +x 0 y 1 +Y x 0 +x 1 +X y 0 y 1 =0 x 0,y 0 165
166 U S I N G S Y M O L I G E O M E T R Y Exmple 101: Intersection of Line with ircle x+ r  x+ y x+ y+ 0 +,y r  x+ y x+ y+ 0 + r E (x,y) 0 +X +Y =0 166
167 Exmple 10: Eqution of the Line Joining the Intersection of Two ircles Given two circles, wht is the eqution of the line joining their intersections? Y= X 1 +X c 1 +c b 1 b F E X +Y +X 1 +Y b 1 +c 1 =0 X +Y +X +Y b +c =0 167
168 U S I N G S Y M O L I G E O M E T R Y Exmple 103: Projecting Point onto Line Project the point (x,y) onto the line c+*x+b*y = 0: (x,y) c+ x+b y +b c+b (b x+ y), b c+ b x y  b  b c+x +Y b=0 168
169 Exmple 104: Rdius of the incircle of Tringle formed by 3 Lines Whose Equtions re Known c b 0 c 0 b  b 1 c b c 01 b 0 c + 0 b 1 c b 1 0 +b01 b 0 +b0  b 0 1 +b1 +0 b 1 +b1 +1 b 0 +b 0 b 1 +b c 0 +X 0 +Y b 0 =0 E c+x 1 +Y b 1 =0 c +X +Y b =0 This eqution is simpler if we ssume tht the line equtions re normlized, nd hence tht the squre roots in the denomintor re equl to
170 U S I N G S Y M O L I G E O M E T R Y Exmple 105: Eqution of the ltitude of Tringle efined by lines with Given Equtions X 0 Y=  b 0 c 0 b 0 Y= X 1 b 1  c 1 b 1 Y= X b  c b X b 0 b 1 +X 0 b 1 b + 1 c 0 +b 1 b c c b 0 b 1 c Y= 1  b b 170
171 Exmple 106: Eqution of the Line through Given Point t 45 degrees to Line of Given Eqution Y= X +X b xb x+ yb y b (x,y) π 4 c+x +Y b=0 171
172 U S I N G S Y M O L I G E O M E T R Y Exmple 107: Eqution of Tngent to ircle Rdius r entered t the Origin, through given point Y= X r x+r x +r y X y r +x +y r yx r +x +y r (x,y) (0,0) 17
173 Exmple 108: Intersection of two tngents to the curve y=x^ We crete the point (x,x^) nd drw its locus s x goes from 3 to 3. Now we crete two tngents point proportionl long the curve, nd exmine their intersection x 1 x 0 x,x x 0 +x 1,x 0 x The x coordinte of the intersection is the verge nd the y coordinte the product of the x coordintes of the tngent points. 173
174 U S I N G S Y M O L I G E O M E T R Y Trnsforms Here re some exmples using trnsforms: 174
175 Exmple 109: omposition of reflections in prllel lines If we hve line with eqution cos(t)x + sin(t)y + = 0, nd second line prllel nd distnce wy, nd if we reflect point in the first line, then reflect its imge in the second, the result is equivlent to trnsltion of twice : x 0 + cos(t),y 0 + sin(t) x 0,y 0 ' '' 0 +X cos(t)+y sin(t)=0 175
176 U S I N G S Y M O L I G E O M E T R Y Exmple 110: ombining Reflections Reflection in two lines through the origin is equivlent to rottion bout the origin of twice the ngle between the lines: ' (x,y) θ '' (y sin( θ)+x cos( θ),x sin( θ)+y cos( θ)) (0,0) 176
177 Reflection in two prllel lines is equivlent to trnsltion of twice the distnce between the lines: (x,y) ' u v '' ( u+x, v+y) 177
178 U S I N G S Y M O L I G E O M E T R Y Exmple 111: Prbolic Mirror prbolic mirror focuses prllel rys into single point. Where is tht point? We crete the prbol y = x nd reflect ry prllel to the y xis in the tngent to the curve. We exmine the y intercept of the imge: X=0 1 0, 4 x, x 178
179 Exmple 11: illirds In mericn billirds, you need to bounce your cue bll off cushions before hitting the other bll. Wht is the length of this pth? We cn crete the correct pth by reflecting the trget bll in the two cushions, drwing line between cue bll nd twicereflected trget, then reflecting this line bck. The pth length is the sme s the distnce between cue bll nd twice reflected trget. X=0 x 0,y 0 x 0 x 1 + y0 y 1  x 1 y 0 x 0 y 1 0, x 0 x 1 x 1,y 1 Y=0 '' ' Note the y coordinte of point. Of course, this is the criticl thing to compute when plying the gme. 179
180 U S I N G S Y M O L I G E O M E T R Y Exmple 113: res nd ilttion rete tringle with sides length, b nd ngle θ. Look t its re. Now ilte by fctor of k. Look t the re of the trnsformed tringle. k b sin(θ) θ b b k sin(θ) ' ' Notice it is k² times the originl re. 180
181 Exmple 114: Trnslting circle circle rdius r is trnslted by the vector (u,v). Wht is the length of the line joining the intersection points between the circle nd its imge? E ' r 4 r u v ' F u v 181
182 U S I N G S Y M O L I G E O M E T R Y Some onstructions We hve the constructions Midpoint, Perpendiculr isector, nd ngle isector. Here re some exmples using these: 18
183 Exmple 115: Eqution of the Perpendiculr isector of Points x 1,y 1 x 0,y 0 Y=  X x 0 +x + X x 0 1 x 1 +y 0 y 1 y 0 y 1 183
184 U S I N G S Y M O L I G E O M E T R Y Exmple 116: Length of the ngle isector of tringle Given tringle side lengths,b,c, wht is the length of perpendiculr bisector? b b+c c b+c b b b c +c 3 +c  +b b+c c 184
185 Some Mechnisms Some exmples involving mechnisms: 185
186 U S I N G S Y M O L I G E O M E T R Y Exmple 117: rnk Piston Mechnism For crnk length c nd connecting rod length L, we compute piston displcement: b t b  sin(t) + cos(t) 186
187 Exmple 118: Quick Return Mechnism The crnk G opertes quick return mechnism whose endeffector is t point F. The formul shows the horizontl displcement of F in terms of t nd the vrious prmeters of the geometry:,b,u,v: F b H E v G 1 t u b  uv+ u  1+u  u cos(t) cos(t) 1+u  u cos(t)  sin(t) 1+u  u cos(t) 187
188 U S I N G S Y M O L I G E O M E T R Y Exmple 119: Pucellier s Linkge In Pucellier s linkge, we look t the height of the endeffector: E c b  +b c < b +t <b b t c G 1 F We see this is invrint in t. The mechnism ws the first to convert purely rotry motion (of the crnk ) to exct liner motion of the end effector. 188
189 Exmple 10: Hrborth Grph The Hrborth grph looks like this: It is the miniml known 4regulr mtchstick grph. Tht is plnr grph, ech of whose vertices re djcent to 4 edges, nd ech edge being the sme length (if you look closely t the bove picture the lengths re not ll the sme). Here is the ccurte drwing from Geometry Expressions 189
190 U S I N G S Y M O L I G E O M E T R Y Here is how this quite complicted digrm is put together. It is bsed on mechnism tht looks like this (which is then replicted by reflecting in I nd then in GF: F G 1 E π 3 I π θ 180 π 3 1 π 3 1 π 3 3 The trpezoid is 3 units on the long bse nd units on the short bse. The ngle E is tweked until GF is pproximtely perpendiculr to I. This mkes the reflected mechnisms join up nicely. The pproprite vlue is degrees. 190
191 Loci Here re some locus exmples 191
192 U S I N G S Y M O L I G E O M E T R Y Exmple 11: ircle of pollonius The ircle of pollonius is the locus of points the rtio of whose distnce from pir of fixed points is constnt: X  +X 1+k +Y 1+k =0 t k t (0,0) (,0) Wht is the center nd rdius? 19
193 Exmple 1: ircle inside ircle Points nd E re proportion t long the rdii nd of the circle centered t the origin nd rdius r. The intersection of nd E trces circle E t F θ (0,0) t (r,0) X r t+x (1+t)+Y (1+t)= Show tht it goes through the origin. Wht is the center of the circle? Wht is its rdius? 193
194 U S I N G S Y M O L I G E O M E T R Y Exmple 13: nother ircle in ircle More generlly if is proportion s long, we hve the following circle: r s  r s tr t + r s t +X 1 s t+s t +Y 1 s t+s t +X  r s+ r s t+ r s t r s t =0 E t F θ (0,0) s (r,0) Wht is the center of this circle? n we find the rdius of this perhps by copying the expression into n lgebr system nd working on it there? 194
195 Here is one pproch, in Mple. First we substitute Y=0., then solve for X to determine the x intercepts of the circle. The rdius cn be found by subtrcting these nd dividing by. > subs(y=0,s^*r^+*t*s^*r^+t^*r^*t^*s*r^+( *t*s+1+t^*s^)*x^+(*t*s+1+t^*s^)*y^+( *t*r+*t*s*r+*t^*s*r*t^*s^*r)*x = 0); s r + t s r + t r  t s r + ( t s t s ) X + ( t r + t s r + t s r  t s r) X = 0 > solve(%,x); r (t + s) t s  1, r (t  s + t s) t s  1 > (r*(t+s)/(t*s1) r*(ts+*t*s)/(t*s1))/; r (t + s) ( t s  1)  r (t  s + t s) ( t s  1) > simplify(%);  r s 1 + t ( ) t s
196 U S I N G S Y M O L I G E O M E T R Y Exmple 14: Ellipse s locus Here is the usul string bsed construction of n ellipse foci (,0) (,0): L 4 +4 L Y +4 L +X 4 L 16 =0 t Lt (,0) (,0) 196
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