Class Note for MATH 409 with Professor Martin at KU (2)
Class Note for MATH 409 with Professor Martin at KU (2)
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CHAPTER 4 Circles and Regular Polygons Circles and regular polygons are the subject of Books 111 and IV of The Elements Euclid39s abstract exposition of the interrelation of chords arcs and tangents lines is augmented with the computation of the circle s circumference and area 1 The Neutral Geometry of the Circle Equal circles are circles that have equal radii A chord of a circle is a line segment that joins two of its points A diameter is a chord that contains the center of the circle An arc of a circle is a portion of the circle that joins two of its points Every chord determines two arcs of the circle Consequently it takes at least three letters to denote an arc unambiguously and the two arcs of the circle of Figure 41 with endpoints A and B should be denoted properly speaking by arcAEB and arcAFB Nevertheless it is customary to label both of these arcs arcAB and to rely on the context for clarification A segment of a circle is the portion between a chord and either of its arcs A sector of a circle is the portion between two radii The arcs determined by a diameter are each called a semicircle That the two semicircles determined by a diameter are equal in length is a 41 proposition that Euclid mentions in Definition 17 Chapter 2 This observation is proved as part of Proposition 411 below Figure 4 1 A central angle of a circle is one both of whose sides are radii Every arc subtends a central angle that is either greater or less than 1800 according as the arc is greater or less than a semicircle Every chord subtends a central angle that is at most 0 180 The following four propositions of Euclid39s are established here with a single unified proof PROPOSITION 411III26 In equal circles equal central angles stand on equal arcs PROPOSITION 411III27 In equal circles central angles standing on equal arcs are equal to one another PROPOSITION 411III28 In equal circles equal chords cut off equal arcs the greater equal to the greater and the less to the less PROPOSITION 411III29 In equal circles equal arcs are subtended by equal chords 42 41 THE NEUTRAL GEOMETRY OF THE CIRCLE GIVEN Equal circles centered at E and E39 respectively Points A B on the first circle and points A B39 on the second Fig 42 T0 PROVE The following are equivalent 1 arcAB arcA B 2 AB 2 AB 3 LAEB LA E B B A V A E Figure 42 PROOF 1 gt 2 Since the given circles are equal it is possible to apply the circle centered at E to that centered at E39 so that E and A fall on E39 and A respectively and arc AB falls along arc A39B39 The two arcs having equal lengths B falls on B39 It follows from PT 1 that the chord AB falls on the chord AB and hence by CN 4 AB A39B39 2 gt 3 AAEB s AA39E B by SSS because AB A B39 Given AE A E39 Given BE 2 B E39 Given 4 AEB A A39E39B 43 41 THE NEUTRAL GEOMETRY OF THE CIRCLE 3 gt 1 Since A AEB A A39E B it is possible to apply the first circle to the second so that E falls on E 39 and these angles coincide Since the circles have equal radii it follows that arc AB falls on arc A39B39 Consequently these arcs have equal lengths QED COROLLARY 412 In a circle all the semicircles are equal to each other See Exercise 1 PROPOSITION 4131113 In a circle a radius bisects a chord not through the center if and only if the radius and the chord are perpendicular to each other See Exercise 2 EXERCISES 41A PW Fquot Prove Corollary 412 Prove Proposition 413 Prove that in a circle a diameter is greater than any chord which is not a diameter Prove that two chords of a circle are equal if and only if they are at equal distances from its center Exercise 23N2 can be used to produce a neutral proof Prove that a circle cannot contain three collinear points 1112 Prove that in a circle the radius perpendicular to a chord bisects that chord39s central angle and arc Prove that in a circle two equal intersecting chords cut each other into respectively equal segments Prove that of two unequal chords in a circle the greater one is closer to the center This is Proposition 11115 It can be easily proved on the basis of the Theorem of Pythagoras but such a proof is would not be neutral Euclid39s neutral proof is based on Proposition 124 Exercise 23Q5 44 10 11 12 13 14 15 16 17 41 THE NEUTRAL GEOMETRY OF THE CIRCLE Let A be a given point and p a circle centered at C If the point P moves along the circle 17 prove that the midpoint of AP describes a circle centered at the midpoint of CA See Exercises 31D778 It is necessary to consider three cases depending on the relative positions of A and p Construct the midpoint of a given arc on a given circle Given an arc of a circle construct the center of the circle Given points A B C D construct a circle through A and B Whose center is equidistant from C and D Given a point A inside a circle construct a chord that is bisected by A Prove that this chord is the shortest of all the chords through A Given an angle a and a line segment a construct a circle Whose center is on one side of a and which cuts a segment equal to a on the other side Given a circle 17 and a point A outside it construct a straight line through A which cuts the circle so that the segment from A to the circle equals the segment in the circle See Exercises 31D778 Comment on Proposition 411 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi Comment on Proposition 413 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi An infinitely extended straight line is said to be tangent to a circle if they have exactly one point in common and that point is called their point of contact PROPOSITION 414III16 18 If a straight line intersects a circle then they are tangent if and only if the straight line is perpendicular to the radius through the point of contact GIVEN Circle C CP straight line PT Fig 43 T0 PROVE PT is tangent to C CP if and only if CPJ PT 45 41 THE NEUTRAL GEOMETRY OF THE CIRCLE P Q Figure 43 PROOF Suppose first that PT is tangent to C CP Hence if Q is any point of PT that is distinct from P it must lie outside the circle so that CP lt CQ Consequently CP J PT PN 2324 Conversely suppose that CPJ PT Then by Proposition 2324 for any point Q of PT that is distinct from P CP lt CQ Consequently no such point Q can lie on the circle C CP and hence PT is tangent to it QED EXERCISES 41B 1 Suppose S and T are the contact points of the tangents to a circle from a point P outside it Prove that PS 2 PT 2 Prove that in a circle the contact points of two parallel tangents are the endpoints of a diameter 3 Prove that the straight line that joins the center of a circle to the intersection of two of its tangents bisects the angle between these tangents 4 Prove that for each side of the triangle there is a circle that is tangent to that side at one of its interior points and tangent to the other two sides at points on their extensions Construct these circles Use Exercise 3 above Two circles are said to be tangent if they intersect in exactly one point If one circle lies inside the other the tangency is said to be internal otherwise it is external 5 Prove that if two circles are externally tangent then the line segment joining their centers contains the point of contact Hint Proceed by contradiction and examine the triangle formed by the centers and the contact point 46 10 11 12 13 14 15 16 17 18 19 20 21 22 41 THE NEUTRAL GEOMETRY OF THE CIRCLE Prove that if two circles are internally tangent then the line joining their centers contains the point of contact Prove that if two circles are tangent to each other then they have a common tangent line at their point of contact Prove that if two circles lie outside each other then they have four different common tangent lines Let m and n be common tangents to unequal circles such that both circles lie inside one of the angles formed by these tangents Prove that the line joining the centers of the circles bisects this angle Let m and n be common tangents to unequal circles such that the circles lie in vertically opposite angles formed by these tangents Prove that the line joining the centers of the circles bisects these angles Given two circles with the same center and unequal radii prove that all the chords of the larger circle that are tangent to the smaller circle have the same length Construct a circle with a given radius tangent to a given line Construct a circle with a given radius tangent to a given line and containing a given point Construct a circle containing a given point and tangent to a given straight line at a given point on the line Construct a circle that is tangent to two given parallel straight lines Construct a circle that is tangent to two given parallel straight lines and contains a given point between them How many solutions are there Construct a circle that is tangent to two intersecting straight lines Construct a circle that is tangent to two given intersecting straight lines and contains a given point Construct a circle that is tangent to two given parallel straight lines as well as to a given third line that intersects them Given a circle 17 and a point A construct a straight line containing A such that its segment inside 17 has a given length Hint See Exercise 11 Construct a point such that the lengths of the tangents from it to two given circles are given Comment on Proposition 414 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi The following proposition was proved by Euclid in its entirety The proof offered in this text is incomplete in two ways In the first place the argument is restricted to rational values of the ratios in question Moreover given an angle 4 ABC and a positive 47 41 THE NEUTRAL GEOMETRY OF THE CIRCLE integer m this argument makes use of the angle LABC n even though it has not been demonstrated that such an angle can be constructed within Euclid39s system PROPOSITION 415VI33 In equal circles central angles are proportional to the arcs on which they stand GIVEN Equal circles with centers G and H respectively Fig 44 LBGL arcBL TO PROVE LEHN arcEN H4 E E2 Figure 44 SUPPORTING ARGUMENT The argument is limited to the case where the given ratios are rational In other words it is assumed that there exist positive integers m and n such that LBGL a i e LBGL LEHN LEHN n I I m n Let a be an angle such that LBGL LEHN m n I 48 41 THE NEUTRAL GEOMETRY OF THE CIRCLE It follows that there exist points B1 BZ Bmi1 on arcBL and points E1 E2 En1 on arcENsuch that LBGB1 ABIGB2 LerlGL LEHE1 A ElHE2 A EnilHN 1 Hence by Proposition 411 arcBBl arcB 132 arcBm1L arcEE 1 arcE 1E2 arcEm1L If the common length of these arcs is denoted by it then arcBL mi LBGL arcEN n n LEHN QED Proposition 415 was used by Eratosthenes ca 275 194 BC director of the Alexandrian library to obtain a remarkably accurate estimate of the circumference of the earth He knew that on the summer solstice the sun shone down at mid day directly into a well in the city of Syene whereas in Alexandria 5000 stadia to the north the shadows indicated that the sun formed an angle of 150 of 3600 720 with the vertical Assuming that the sun is so far away that its rays can be considered to be parallel when they reach the earth Fig 45 he then used Proposition 415 to obtain the equation 49 41 THE NEUTRAL GEOMETRY OF THE CIRCLE circumference of earth 360 distance from Alexandria to Syene 720 SSW Syene Figure 45 from which he concluded that the circumference is 505000 2 250000 stadia In order to make his answer divisible by 60 probably because of the in uence of the Babylonian sexagesimal number system he adjusted this result to 252000 stadia The standard stade of the time had a length of 1786 meters which converts his rounded estimate to 45007 km an overestimate of 123 since the circumference of the earth is actually 40075 km EXERCISES 41C 1 A circle has circumference 10 ft Find the lengths of the arcs that subtend the following angles at the center of the circle 0 o 0 a 10 b 30 c 90 0 0 0 d 110 e 120 f 180 2 A location on earth has latitude 250 N Find its distance from the equator and from the North Pole 3 A location on earth has latitude 700 N Find its distance from the equator and from the North Pole 410 5000 0 10 41 THE NEUTRAL GEOMETRY OF THE CIRCLE A location on earth has latitude 700 S Find its distance from the equator and from the North Pole A location on earth has latitude 700 S Find its distance from the equator and from the North Pole A location on earth lies 2000 km north of the equator Find its latitude A location on earth lies 1234 km north of the equator Find its latitude A location on earth lies 1000 km south of the equator Find its latitude A location on earth lies 617 km south of the equator Find its latitude Comment on Proposition 415 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi 2 The NonNeutral Geometry of the Circle The next proposition is one of the most surprising in The Elements Unlike those appearing the previous section its implications are quite unexpected PROPOSITION 421III20 In a circle the angle at the center is double of the angle at the circumference when the angles have the same arc as base GIVEN Points A B C on the circumference of a circle centered at E Fig 46 T0 PROVE A BEC 24 BAC A Case 1 411 42 THE NONNEUTRAL GEOMETRY OF THE CIRCLE PROOF It is necessary to distinguish three cases Case 1 One of the sides of A BAC contains the center E A BEC A BAC A ECA Exterior PN 316 A BAC A ECA AE 2 EC PN 235 A BEC 2 A BAC Case 2 The center E lies in the interior of A BAC Let F be the other intersection of XE with the circumference of the given circle Then A 2 2 A 1 Case 1 A 4 2 A 3 Case 1 A BEC 2 A BAC CN 2 Case 3 The center E lies outside of A BAC Let F be the other intersection of XE with the circumference of the given circle Then 42 241 Case 1 24 223 Case 1 LBEC 2LBAC CN 3 QED Proposition 421 has several corollaries whose proofs are relegated to the exercises PROPOSITION 422III21 In a circle the angles in the same segment are equal to one another GIVEN Points A B C D on the circumference of a circle such that A and D lie on the same side of BC Fig47 TO PROVE A BAC A BDC 412 42 THE NONNEUTRAL GEOMETRY OF THE CIRCLE Figure 47 PROOF See Exercise 1 This proposition is somewhat counterintuitive Suppose the points A and B in Figure 48 are fixed whereas P slides clockwise around the circle occupying positions P1 P2 P5 successively Proposition 422 implies that as long as the point P remains in the interior of the upper or longer arcAB the angle APB retains a constant acute value When P passes through A or B APB is no longer an angle Finally when P is in the interior of the shorter or lower arcAB the angle APB Figure 48 A discontinuous function assumes a different obtuse value In other words even though the point P moves in a continuous manner A APB varies as a discontinuous function of the position of P 413 42 THE NONNEUTRAL GEOMETRY OF THE CIRCLE PROPOSITION 423III31 In a circle the angle subtended by a diameter from any point on the circumference is a right angle See Exercise 2 PROPOSITION 424III32 Let AB be a chord ofa circle and let ltXT be any straight line at A Then the line ET is tangent to the circle ifand only if A TAB is equal to the angle at the circumference subtended by the intercepted arc GIVEN Circle p with chord AB straight line KT arcAB Fig 49 T0 PROVE AT is tangent to p if and only if A TAB equals the angle at the circumference of p subtended by arcAB PROOF Let AD be the diameter of the circle containing A and join BD Figure 4 9 By Proposition 423 A ABD 900 Hence the following statements are all equivalent to each other lt gt AT is tangent to the circle LDAT900 41 900 42 41 43 QED 414 42 THE NONNEUTRAL GEOMETRY OF THE CIRCLE PROPOSITION 425III36 37 Let P be apoint outside a given circle p let T be a point on p and let PAB be a secant line with chord AB Then PT is tangent to p if and only if PA PB 2 PT2 Figure 410 GIVEN Point P outside circle p straight lines PT and PAB that intersect p in T A B Fig 410 TO PROVE PTis tangent to p ifand only if PA PB 2 PT2 PROOF The following statements are all equivalent to each other The line PT is tangent to p A 1 A 2 PN 424 A TPA and A BPT are similar to each other PN 357 3 PT PB PA PB 2 PT2 QED A polygon is said to be cyclic if all of its vertices lie on a circle 415 PROPOSITION 426III22 42 THE NONNEUTRAL GEOMETRY OF THE CIRCLE The opposite angles of a cyclic quadrilateral are equal to two right angles See Exercise 3 EXERCISES 42A MP9 10 11 12 13 14 15 16 Prove Proposition 422 Prove Proposition 423 Prove Proposition 426 Prove that in a circle parallel chords enclose equal arcs Prove that if the quadrilateral ABCD is cyclic then the exterior angle at A equals the interior angle at C In a circle the extensions of the chords AB and KL intersect in a point P outside the circle Prove that A AKP A LBP and A BKP A LAP In a circle the extensions of the chords AB and CD intersect in a point P outside the circle Prove that PAPB PCPD Proposition 11135 In a circle the chords AB and CD intersect in a point P inside the circle Prove that PAPB PCPD Proposition 11136 Prove that two equal and parallel chords in a circle constitute the opposite sides of a rectangle Prove that if the hexagon ABCDEF is cyclic and the interior angles at A and D are equal then BC H EF 1n the cyclic quadrilateral ABCD AD 2 BC Prove that the interior angles at A and B are equal to each other as are those at C and D Prove that the sum of the interior angles at A C and E in the cyclic hexagon ABCDEF is four right angles Prove that if the perpendicular chords AB and CD of a circle intersect at the point M inside the circle then the straight line through M that is perpendicular to AD bisects the chord BC Prove that every cyclic rhornbus is a square Prove that if A and B are two distinct points and D is any other point on AB then the locus of AP AD all the points P in the plane such that E is a circle This is the circle of Apollonius State and prove the converse of Proposition 426 416 42 THE NONNEUTRAL GEOMETRY OF THE CIRCLE 17 Given a line segment AB construct the circle which consists of all the points from which AB subtends an angle of 900 18 Given a line segment AB construct the arc which consists of all the points from which AB subtends an angle of 600 19 Given a line segment AB construct the arc which consists of all the points from which AB subtends an angle of 1200 20 Given a line segment AB construct the arc which consists of all the points from which AB subtends an angle equal to a given angle a 21 Construct a triangle given the data a ahbhc b ahaa c amaa d abchaa 22 Construct a parallelogram given its two diagonals and one of its angles 23 Given line segment AB and CD and angles a and constructapoint P such that A APB a and 4 CPD 2 24 In a given AABC construct a point P such that A APB A BPC A CPA 25 Comment on Proposition 421 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi 26 Comment on Proposition 422 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi 27 Comment on Proposition 423 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi 28 Comment on Proposition 426 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi 29C Use a computer application to verify the following propositions a 421 b 422 c 423b d 424 Three or more straight lines are said to be concurrent if they all contain the same point PROPOSITION 427 The three perpendicular bisectors 0f the sides of a triangle are concurrent 417 42 THE NONNEUTRAL GEOMETRY OF THE CIRCLE GIVEN A ABC DD 39 EE39 FF 39 are the perpendicular bisectors of AB AC and BC respectively Fig 411 TO PROVE DD 39 EE39 F F 39 are concurrent B F C B C Figure 411 PROOF Exercise 31A4 guarantees that DD and EE39 intersect in some point M Draw AM BM CM Then AM 2 BM PN 2312 AM 2 CM PN 2312 BM 2 CM CN 1 M is on the perpendicular bisector to BC PN 2313 QED A circle is said to circumscribe a triangle if all of the triangle s vertices are on the circle Its center and radius are respectively the triangle s circumcenter and circumradius PROPOSITION 428IV5 About a given triangle to circumscribe a circle See Exercise 1 418 42 THE NONNEUTRAL GEOMETRY OF THE CIRCLE PROPOSITION 429 The bisectors 0f the three interior angles of a triangle are concurrent GIVEN A ABC AA BB CC are the bisectors of A BAC A ACB 4 ABC respectively Fig 412 TO PROVE AA 3339 CC are concurrent A A G E AMA B C B F C Figure 412 PROOF Since 1 1 0 4143 24 lt 5180 PN2321 it follows from Postulate 5 that 3339 and CC intersect in some point D Let EF and G be those points on AB BC CA respectively such that DE J AB DF J BC and DG J AC Then DE DF DG PN 2331 DE DG CN 1 DA bisects 4 BAC PN 2332 QED 419 42 THE NONNEUTRAL GEOMETRY OF THE CIRCLE A circle that lies in the interior of a triangle and is tangent to all of its sides is said to be inscribed in the triangle Its center and radius are respectively the triangle s incenter and inradius PROPOSITION 4210IV4 In a given triangle to inscribe a circle See Exercise 2 EXERCISES 42B MPWP 10 11 12 13 14 15 Prove Proposition 428 Prove that similar triangles have circumradii that are proportional to their sides Prove Proposition 4210 Prove that similar triangles have inradii that are proportional to their sides Prove that the circumcenter of a right triangle is the midpoint of its hypotenuse Prove that if the circumcenter of A ABC lies inside the triangle then the triangle is acute if the center is on a side the triangle is right and if the center is outside the triangle then the triangle is obtuse Prove that the circumcenter of an acute triangle lies inside the triangle Prove that the circumcenter of an obtuse triangle lies outside it Prove that the area of the triangle equals the product of half its perimeter with the inradius Hint Examine the three triangles formed by the center of the circle with the triangle39s three sides In a given circle inscribe a triangle similar to a given triangle Prove that the altitudes of the triangle are concurrent Hint Through each vertex of the triangle draw a line parallel to the opposite side Then show that the altitudes in question are the perpendicular bisectors of the triangle formed by these parallels Comment on Proposition 427 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi Comment on Proposition 428 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi Comment on Proposition 429 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi Comment on Proposition 4210 in the context of the following geometiies a spherical b hyperbolic c taxicab d maxi 4 20 42 THE NONNEUTRAL GEOMETRY OF THE CIRCLE 16C Use a computer application to verify the following propositions a 425 b 427 3 Regular Polygons A polygon is regular if all of its sides and all of its interior angles are equal The equilateral triangles are the regular triangles and they are the subject matter of Proposition 1 of Book I Squares are the regular quadrilaterals and they are constructed in Proposition 331 Book IV of The Elements is mostly concerned with the constructibility of other regular polygons and their inscription in circles Regular hexagons are also easily constructed PROPOSITION 431IV 15 In a given circle to inscribe a regular hexagon GIVEN Circle p 0 r Fig 413 TO CONSTRUCT Points A B C D E F on p such that ABCDEF is a regular hexagon Figure 413 CONSTRUCTION Let A be an arbitrary point on the circle p Let B be the intersection of an arc of radius r and center A with p Let C be the intersection of an arc of radius r and center B with p and let D E F be constructed in a similar manner Then ABCDEF is a regular hexagon 421 43 REGULAR POLYGONS PROOF By construction A AOB A BOC A COD A DOE and A EOF are all equilateral so that A AOB A BOC A COD A DOE A EOF 600 It follows that A F 0A 2 3600 5600 2 600 and hence the isosceles A F 0A is also equilateral Thus FA 2 0A and so each of the sides of ABCDEF has length r It also follows that each of the interior angles of ABCDEF equals 1200 Thus ABCDEF is a regular hexagon QED A slight variation on the construction of the regular hexagon yields the flower like configuration of Figure 414 Figure 414 The construction of the regular pentagon is a considerably more difficult matter Some of the technically demanding details are isolated in the following lemma Others were listed as exercises above PROPOSITION 432IV10 To construct an isosceles triangle having each of the angles at the base equal to double of the remaining one TO CONSTRUCT A ABC such that 4 ABC 2 A ACB 2L BAC Figure 415 CONSTRUCTION Let AB be an arbitrary line segment and let D be a point such that AB BD 2 AD2 PN 341 Then the required A ABC is that triangle such that AC 2 AB and BC 2 AD PN 2327 4 22 43 REGULAR POLYGONS Figure 415 PROOF By construction BC2 2 AD2 ABBD It therefore follows that from Proposition 425 that BC is tangent to the circle p that circurnscribes AACD Hence 41 42 PN424 43 44 4244 45 46 DC BC AD 1 l l 41 442243 246 245 QED If the smallest of the angles of the triangle of Proposition 432 is denoted by x then the other two angles are each 226 and so by Proposition 316 from which it follows that x 360 Hence the following corollaries hold PROPOSITION 433 To construct angles of 360 and 720 4 23 43 REGULAR POLYGONS We are now ready to construct the regular pentagon PROPOSITION 434IV11 In a given circle to inscribe a regular pentagon GIVEN Circle p 0 r Fig 416 TO CONSTRUCT Points A B C D E on p such that ABCDE is a regular pentagon CONSTRUCTION At the center 0 of the circle construct five non overlapping central angles of 720 PN 433 Label the successive intersections of their sides with the circle A B C D E Then ABCDE is the required pentagon Figure 416 PROOF The five constructed isosceles triangles are all congruent by SAS It follows that the five sides AB BC CD DE and EA are all equal Moreover the base angles of 1 these triangles are 5 1800 720 540 each and hence all of the pentagon39s interior angles are equal to 1080 each QED Euclid took the trouble to prove that the regular 15 sided polygon is constructible Exercise 3 It is reasonable to suppose that this was his way of pointing out that there is 4 24 43 REGULAR POLYGONS an interesting question to be pondered here Namely for which integers n can the regular n sided polygon be constructed It has already been shown above that this is possible for n 3 4 5 6 Some more such 11 can be easily produced by simply doubling the number of sides of any constructible regular polygon see Exercises 1 2 4 However this does not answer the question for such numbers as 7 9 11 13 14 17 The surprising intricacy of the construction of the pentagon indicates that such polygons might pose an even greater challenge This problem continued to excite the interest of mathematicians after Euclid but no progress was made for over 2000 years until the young Gauss demonstrated the constructibility of the 17 sided polygon in 1796 Actually he did much more Using the newly emergent theory of complex numbers Gauss proved that a regular p sided polygon can be constructed for every prime integer p 2quot that has the form 2 1 for some nonnegat1ve integer n These include the values 0 1 221 211 3 221 221 5 22 4 23 8 2 1 2 21 17 2 1 2 21 257 4 22 1 2161 65537 25 32 Curiously the next number in this sequence namely 2 1 2 1 4294967297 fails to be a prime since it can be factored as 6416700417 a fact that had already been noted by Euler over fifty years earlier The same is true for all the numbers of this form for n 6 7 16 and several other values including 11 1945 In fact it is not known whether there are any more primes p that can be expressed in this form above and beyond the five listed above Gauss completely resolved the issue of the constructibility of regular polygons as follows 4 25 43 REGULAR POLYGONS PROPOSITION 435 It is possible to construct in the sense de ned by Euclid a regular ggon g 2 3 if and only if the factorization of g into primes has the form k g 2 p1p2pm Yl where m 2 0 and p1p2 pm are distinctprimes each of which has theform 22 1 Thus the regular 2040 gon is constructible because 2040 233517 whereas the regular 28 sided and 100 sided polygons are not constructible because 28 227 and 100 2252 EXERCISES 43 PW 10 11 12 Prove that the regular octagon is constructible Prove that the regular decagon is constructible Prove that the regular 15 sided polygon is constructible Proposition IV16 Let g be a positive integer Prove that if the regular gisided polygon is constructible so is the regular 2gisided polygon Let p and q be two prime integers Prove that if the regular pisided and qisided polygons are constructible so is the regular pqisided polygon Let g and h be relatively prime integers such that the regular gisided and hisided polygons can be constructed Prove that the regular ghisided polygon can be constructed Let g h gt 1 be integers such that h is an integer multiple of g Prove that if the regular hisided polygon is constructible so is the regular gisided polygon 26 Use a calculating device to prove that 2 1 is not a prime integer For which n 3 4 100 is the regular nigon constructible For which n 101 102 200 is the regular nigon constructible Comment on Proposition 431 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi Comment on Proposition 432 in the context of the following geometries 4 26 43 REGULAR POLYGONS a spherical b hyperbolic c taxicab d maxi 13 Comment on Proposition 434 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi 14 Show that in taxicab geometry equilateral triangles are not necessarily equiangular Can all three of the angles of an equilateral taxicab triangle be distinct 15C Perform the construction of Proposition 431 using a computer application 4 Circle Circumference and Area The fundamental observation that the circumference of a circle is proportional to its diameter and hence also its radius goes back several millennia BC Surprisingly Euclid says nothing on this topic in The Elements PROPOSITION 441 Circumferences of circles are proportional to their radii GIVEN Circles p1 01 r1 and p2 02 r2 of circumferences c1 and 02 respectively Fig 417 01 71 TO PROVE 02 72 Figure 417 SUPPORTING ARGUMENT It follows from Proposition 431 that it is possible to inscribe a regular hexagon in each of the given circles By repeatedly bisecting the 4 27 44 CIRCLE CIRCUMFERENCE AND AREA central angles subtended by the sides of the polygons it is possible to inscribe in each circle pi i 1 2 a regular n sided polygon of side say am where n is an integer of the form 32 It is clear that 4 1310101 4 320202 3600n Since A 013101 and A 0232C2 are isosceles they must be equiangular PN 316 and hence they are similar PN 357 In other words the sides of the inscribed polygons are proportional to the radii Making the reasonable assumption that for large n the difference between the circumferences of each circle and that of its inscribed polygon is negligible it follows that g quot91 91 62 quot92 2 r2 QED An alternate supporting argument that makes use of calculus is described in Exercise 1 It follows from the above proposition that if c and r denote the circumference and radius of an arbitrary circle then the ratio has a constant value say a This constant number can be used to restate the above proposition in the following form PROPOSITION 442 There is a number 0L such thatif c and r are respectively the circumference and radius ofany circle then c 2 ar 4 28 44 CIRCLE CIRCUMFERENCE AND AREA The numerical value of a is of course of interest and will be estimated at the end of this section Next the area of the circle is examined The following proposition was proved by Euclid using the Method of Exhaustion which was the Greeks version of the integral calculus This method was developed by Euclid39s predecessor Eudoxus whom Archimedes 287 212 BC credits with this and other similar propositions PROPOSITION 443XII2 The areas of circles are proportional to the squares of their radii GIVEN Circles p1 01 r1 and p2 02 r2 of areas A and A2 respectively Fig417 A1 r3 TO PROVE A 7 2 r 2 SUPPORTING ARGUMENT It follows from Proposition V119 Exercise 35E10 that the areas of A OlBlC1 and A 0232C2 are proportional to the squares of the radii r1 and r2 Making the reasonable assumption that for large n the difference between the areas of each circle and that of its inscribed polygon is negligible it follows that 2 nA 013101 A 01310 r1 A2 nA 023202 A 023202 r QED An alternate supporting argument makes use of calculus see Exercise 2 It follows from the above proposition that if A and r denote the area and radius of an arbitrary circle then the ratio NIDgt 4 29 44 CIRCLE CIRCUMFERENCE AND AREA has a constant value say 7 This number can be used to restate the above proposition in the following form PROPOSITION 444 There is a number if such that if A and r are the area and radius respectively of any circle then A 2 m2 The numerical value of n is of course of interest but the relationship between a and 7139 needs to be addressed first The discovery of this relationship is attributed by Proclus to Archimedes PROPOSITION 445 The proportionality constants of the circumference and area of a circle are related by the equation a 2n SUPPORTING ARGUMENT Suppose the circle p is divided into n equal sectors Figure 418 each of which has a central angle of 3600n and let OBD be a typical sector Fig 418 If n is large it may be assumed that OBD is a triangle with altitude 0C 2 r Applying Proposition 325 it follows that this triangle has area rarcBD2 Hence the circle p has area 430 44 CIRCLE CIRCUMFERENCE AND AREA r a 2 7ir r r Aniarc BD ci ar 2 2 2 2 a It now follows from Proposition 444 that n 3 or a 2n QED COROLLARY 446 The circumference of a circle of radius r is 2m H 2 O Figure 419 An alternate supporting argument for Proposition 445 can be based on Figure 419 Imagine that the circle of radius r on the left is filled with circular strands Cut the circle along the vertical dashed radius and straighten out all the strands as indicated until they form an isosceles triangle that its sides are straight follows from Proposition 442 It follows from Proposition 325 that the area of this triangle and hence also the area of the circle is A circumferencer2 cr2 arr2 a2r2 and the rest of the argument proceeds as before Appealing as this argument is it is fraught with logical perils which are discussed in Exercise 21 431 44 CIRCLE CIRCUMFERENCE AND AREA Yet another alternative argument in support Proposition 445 calls for slicing the circle into an even number of equal sectors and rearranging these to form the near parallelogram at the top of Figure 420 As the number of sectors increases to infinity the Halfcircumference Halfcircumference Half circumference Half circumference Figure 420 near parallelogram converges to the rectangle below it whose area is clearly A 2 half circumferencer cr2 arr2 a2r2 The procedure of successive approximations used in Proposition 441 also yields a method for obtaining numerical estimates of the constant of proportion 71 This was first carried out by Archimedes and constitutes the first of a long and still ongoing series of scientific estimations of 71 Let an denote the length of the chord AB which is one side of the regular n gon inscribed in a circle of radius 1 Fig 421 If C is the midpoint of the corresponding arcAB then the chord AC has length a2 Two 432 44 CIRCLE CIRCUMFERENCE AND AREA Figure 421 applications of the Theorem of Pythagoras then yield 2 a a2 AC2 AD2 DCZ 7 2 0C 0D2 a 2 7 2 1 OAZADZ a 2 7 2 1 V1 an2f anz anz anz 7 12 1 7 1 7 2 4 ai Hence PROPOSITION 447 If an denotes the length of the regular polygon with n sides that is inscribed in a circle of radius 1 then 433 44 CIRCLE CIRCUMFERENCE AND AREA 612 2 4 an Since the length of the side of the inscribed regular hexagon equals the radius it follows that a6 1 and so 2 l4 1 2 E 5176380902 24 2 H 2610523844 6148 2 4 2 2 2 2 2 B 1308062585 6196 2 2 2 2 B 0654381656 Since the length of any arc exceeds that of its chord another one of those reasonable assumptions the circumference of the circle exceeds that of any inscribed polygon As the circle of radius 1 has circumference 271 it follows that for each positive integer n and in particular 71 gt 48a96 314103195089 434 44 CIRCLE CIRCUMFERENCE AND AREA To obtain upper bounds for the value of n Archimedes examined regular n gons whose sides were tangent to the circle Suppose a regular n gon is inscribed in a circle of radius 1 and at each of the vertices a straight line tangent to the circle is constructed It is easily verified that the resulting polygon surrounding the circle is also a regular n gon of bu 7 Figure 422 Comparing a circumscribed polygon with an inscribed one side say I Fig 422 The side I can be estimated by showing that see Exercise 9 2 i 2 4 b 2 b6 2 E and b2 2 bn 1 Alternately the figure above can be used to show that b 2a I 2 n 4 an This gives us 1796 0654732208 and hence 961796 71 lt 2 31427145996 435 44 CIRCLE CIRCUMFERENCE AND AREA Archimedes did not have the decimal number system at his disposal and he had to work within the much more cumbersome systems that were then current Using some of the quite complicated methods for the estimation of square roots by means of fractions that were then known he was able to show that 6336 m 48d96 gt i gt 371 31408 20174 and 14688 1 481796 lt 1 lt 37 31428 4673 Thus Archimedes proved that PROPOSITION 448 3 lt n lt 3 This section and chapter conclude with a discussion of some paradoxes and problems regarding the areas of circles Consider Figure 423 where A ABC is both right and A a B a C Figure 423 isosceles with legs of length a and hypotenuse of length The outside arc is a semicircle with diameter AB and radius m5 2 whereas the inside arc is a quarter 436 44 CIRCLE CIRCUMFERENCE AND AREA circle centered at C with radius 61 Regions bounded between two such arcs are called lunes but these are different from the spherical lunes Note that the area of this lune is the difference between the entire figure and the quartercircle centered at C Thus area of lune 2 area of semicircle area of AABC area of quartercircle 1 m5 2 i L642 L642 i Luz i 2 22394423942 2 area of AABC This surprising equation is due to Hippocrates of Chios A similar equation appears in Exercise 10 There are two unexpected aspects to this equation First the area of the lune whose boundary consists of circular arcs turns out to have an expression that is free of 71 Second this curvilinear figure has the same area as a triangle This leads naturally to the question of whether it is possible to construct a triangle or a square for that matter whose area equals that of a given circle the simplest of all the curvilinear regions The operative word here is construct It is clear that any circle of radius r has the same area as a square of side The difficulty lies in constructing r7 I within the framework of Euclidean geometry This problem drew the attention of many mathematicians and non mathematicians both in classical times and during the subsequent two and a half millennia Although many individuals dedicated their lives to the solution of this problem and some even deluded themselves into believing they had discovered the construction all their efforts were in fact wasted In 1882 the German mathematician C L Ferdinand von Lindemann 1852 1939 proved a theorem which had the following corollary amongst many others 437 44 CIRCLE CIRCUMFERENCE AND AREA Given a line segment a it is impossible to construct in the sense of The Elements a square whose area equals that of the circle of radius a EXERCISES 44A 10 b 2 The length of the graph of the function y fx for a s x s b is l f39 dx Use this a formula to prove Proposition 441 Use calculus to prove Proposition 443 Using 314 for the value of n in a circle of radius 10quot find the length of the arc and the area of a sector determined by a central angle of a 60 b 20 c 90 d 100 e 180 f 230 Compute the radical expressions for 1192 and 12192 and use them and a calculator to obtain decimal bounds of the value of J17 Compute the radical expressions for 1384 and 22384 and use them and a calculator to obtain decimal bounds of the value of J17 Compute the radical expressions for 0768 and 12768 and use them and a calculator to obtain decimal bounds of the value of J17 Compute the radical expressions for 11536 and 121536 and use them and a calculator to obtain decimal bounds of the value of J17 Prove Equations 1 Prove that of two circular arcs joining two given points the one with the longer radius has shorter length Semicircles are constructed on the sides of a right A ABC Fig 424 Prove that the sum of the areas of the two lunes l and II equals the area of AABC Figure 424 438 11 12 13 14 15 16 17 18 19 20 21 44 CIRCLE CIRCUMFERENCE AND AREA Show that the circumference of a circle of spherical radius r on a sphere of radius R is 2nR sin rR 2 Show that the area of a spherical circle of spherical radius r on a sphere of radius R is 21R 1 7 cos rR Comment on Proposition 441 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi Comment on Proposition 442 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi Comment on Proposition 443 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi Comment on Proposition 444 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi Comment on Proposition 445 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi Comment on Proposition 446 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi Comment on Proposition 447 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi Comment on Proposition 448 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi Explain the following paradox Suppose the method that was used to convert a circle into a triangle see paragraph following Proposition 446 is applied to the same square ABCD in two different manners 7 first by slicing from a corner to the center and second by slicing from the middle of a side to the center Fig 425 The two triangles so obtained have their bases equal to the perimeter of the square but their altitudes are clearly different Why are two triangles of different areas obtained 439 44 CIRCLE CIRCUMFERENCE AND AREA Figure 425 45 Impossible Constructions Mostly Part of the legacy that the Greek mathematicians passed on to their successors was a collection of construction problems they could not resolve by ruler and compass alone While most of these problems have already been discussed Sections 23 and 43 it might be a good idea to reexamine this topic here in order to provide a better perspective on its outcome We begin by listing the specific construction problems in question I To divide a given angle into three equal parts 2 To construct a regular n gon for each integer n 2 3 3 To construct a square whose area equals that of a given circle 4 To construct a cube whose volume is double that of a given cube The reader will recall that Cartesian coordinates were invented for the purpose of expressing geometrical problems in the language of algebra Since construction problems are geometrical this applies to them as well Some of this relation between geometry and algebra has already been pointed out If a and b are the lengths of two given line segments then it is possible to construct line segments of lengths a 17 Exercise 23A3 and a 17 Proposition 233 Assuming a to be a unit length Exercise 35B13 shows how given segments of lengths b and c it is possible to construct a segment of length bc Assuming c to be a unit length the same exercise can be used to 440 IMPOSSIBLE CONSTRUCTIONS construct for any given segments of lengths a and b a segment of length ab Finally assuming 17 to have unit length Exercise 35B 14 can be used to construct for any given segment of length a a line segment of length a Thus the four arithmetical operations as well as the taking of square roots can be mimicked by ruler and compass constructions The Cartesian coordinate system can be used to argue that the power of ruler and compass constructions cannot be extended beyond these five algebraic operations To do this it is first necessary to formalize some notions A configuration is a set of points straight lines and circles An elementary ruler and compass construction is any of the following five operations 1 Draw the line joining two given points 2 Draw a circle with a given center and radius 3 Find the intersection of two given straight lines 4 Find the intersection of a given circle and a given straight line 5 Find the intersection of two given circles A configuration T is said to be constructible from configuration S provided every element of T can be obtained from the elements of S by a succession of elementary ruler and compass constructions In particular note that every construction problem stipulates a given configuration and aims at the derivation of a desired configuration Assume now that a Cartesian coordinate system has been chosen to which all the configurations below are referred The numerical aspects of the point x y are x and 441 IMPOSSIBLE CONSTRUCTIONS y The numerical aspects of the straight line with equation ax by c 0 are a b and c The numerical aspects of the circle with equation x2 y2 ax by c 0 are a b and c A real number r is said to be a Hippasian function of the set S provided it is obtainable from the elements of S and the rational numbers by rational operations and extractions of real square roots this terminology honors Hippasus of Metapontum the discoverer of the irrationality of V2 For example the numbers below are all Hippasian functions of the set S 71 3V2 e g n3e 3VE WE 5 e ZW whoaE 1 The Hippasian numbers are those that are obtainable from the rational numbers alone by the rational operations and the extractions of real square roots In fact this is tantamount to saying that they are obtainable from the number 1 by the said operations The numbers below are all Hippasian numbers 3 11ME Lg linHm Jim The following theorem formalizes the intuitively plausible connection between constructibility and Hippasian functions 4 42 IMPOSSIBLE CONSTRUCTIONS THEOREM 451 If configuration T is constructible from configuration S then the numerical aspects of T are Hippasianjunctions of the numerical aspects of S OUTLINE OF PROOF Suppose configuration T is obtained from configuration S by the elementary construction i where i 1 2 3 4 5 If i 5 for example let the two given circles have equations x2y2axbyc0 and x2y2a cb yc 0 By Exercise 5 the intersection point of these two circles if it exists has coordinates which are Hippasian functions of a b c a b c The proof of the cases i 1 2 3 4 is similar see Exercises 1 4 and we conclude that if T is constructible from S by any ruler and compass operations then the numerical aspects of T are obtainable from those of S in the desired manner QED We note in passing that the converse of Theorem 451 is also valid albeit somewhat harder to prove As this converse is not needed for the proof of the impossibilities below it is relegated to Exercise 6 The strategy for demonstrating the non feasibility of a ruler and compass construction calls for demonstrating that the numerical aspects of the desired configuration are not Hippasian functions of those of the given configuration Matters can and will be simplified below by setting things up so that the numerical aspects of the given data are either integers or Hippasian numbers and hence it will suffice to show that the desired configuration has a non Hippasian number as one of its numerical aspects 4 43 IMPOSSIBLE CONSTRUCTIONS The following unproven proposition provides an easily applied criterion for recognizing non Hippasian numbers It is found in a more general form in many undergraduate modern algebra texts PROPOSITION 452 Let x be a real solution of the equation ax3bx2cxd0 1 where a b c d are integers Then x is Hippasian if and only if this equation has a rational solution u Unlike the above proposition the next one is found in many precalculus texts and is easily proven Exercise 7 PROPOSITION 453 The Rational Zeros Theorem Let Px anx an1x 1 alx a0 where all the ai s are integers If pq is a rational number in lowest terms such that Ppq 0 then p is afactor of a0 and q is afactor of an u 444 IMPOSSIBLE CONSTRUCTIONS The four construction problems of antiquity are now reexamined one at a time The simplest of these turns out to be the doubling of the cube DOUBLING THE CUBE If it were possible to double a given cube by ruler and compass constructions then it would certainly be possible to construct a cube of volume 2 The length of the side of such a cube would be 3V2 and it would have to be a Hippasian number However 3V 2 is clearly a solution of the equation x3 220 2 and hence by Proposition 452 this equation would have to have a rational solution say pq in lowest terms By Proposition 453 p must be a factor of 2 and q a factor of 1 Hence pq must be one of the numbers 11 t 21 none of which by Exercise 9 is a solution of the Equation 2 Thus the supposed feasibility of a ruler and compass doubling of the cube has lead to a contradiction and we conclude that The cube cannot be doubled by ruler and compass alone ANGLE TRISECTION We next argue that there is no method for trisecting angles by ruler and compass alone Suppose to the contrary that such a method exists and is used to trisect the 60 angle of an equilateral triangle whose side has unit length Here it may be supposed that the given configuration consists of the three points 00 0 A0 1 445 IMPOSSIBLE CONSTRUCTIONS B12 32 Fig 426 all of whose numerical aspects are Hippasian numbers The hypothetical construction yields an angle of 20 that may be placed at the origin with one side on the x axis Fig 426 The constructible intersection P of this angle s other side with the circle 0 1 has coordinates cos 20 sin 20 and hence it follows from Theorem 451 that cos 20 is a Hippasian function of Hippasian numbers Consequently cos 20 is a Hippasian number We now go on to obtain an analog of Equation 2 If x cos 20 then by Exercise 8 12 cos 60 2 4 cos3 20 3 cos 20 2 4x3 73x and hence 8x3 6x 120 3 By Proposition 452 this equation has a rational solution say pq where p and q are integers By Proposition 453 pq must be one of the fractions 1 12 14 18 none of which by Exercise 10 is a root of Equation 3 Thus the assumption of the feasibility of a ruler and compass method for trisecting angles has resulted in a contradiction and hence 446 IMPOSSIBLE CONSTRUCTIONS There is no method for trisecting angles by ruler and compass alone y B P 600 20 0 A x Figure 426 REGULAR nGONS Whether or not a regular ngon is constructible by ruler and compass turns out to depend on the value of n In general such a polygon can be constructed if and only if an angle of 360011 can be constructed When this angle is placed at the origin with one side on the x axis the other side intersects the circle 0 1 in the point cos 360011 sin 360011 By Theorem 451 and Exercise 18 this general ruler and compass construction is feasible if and only if cos 360011 is a Hippasian number We now show that there is no ruler and compass construction for the regular 7 gon Set A 2 36007 and x cos 36007 If such a method existed then 6 would be a Hippasian number However by Exercises 8 and 12 cos 3A 4cos3A 3cosA 4x3 3x cos4A 8cos4A 8cos2A1 8a4 8x21 4 47 IMPOSSIBLE CONSTRUCTIONS Since 3A 4A 3600 it follows that cos 3A 2 cos 4A and hence 4263 326 2 8 4 8x21 S 4x3 8x23x1 0 x 18x34x2 4x 1 0 However cos 36007 1 and hence 8x34x2 4x 10 It follows from Proposition 453 that the only possible rational solutions of this equation are again 1 12 14 18 Since by Exercise 10 none of these is a solution it follows from Proposition 452 that cos 36007 is not a Hippasian number and hence no such method for constructing regular 7 gons can exist According to Exercise 11 cos 36005 is a Hippasian number and so the regular pentagon is indeed constructible as demonstrated in Proposition 434 As was mentioned above the regular 17 gon is also constructible and hence cos 360017 must also be a Hippasian number In fact it is known to equal 1161 16116134 21 448 IMPOSSIBLE CONSTRUCTIONS 18117 31E 134 24E 2x34 245 SQUARING THE CIRCLE To square a circle of unit radius is tantamount to constructing a line segment of length a V75 As it happens neither 7 nor J39l3 are the solutions of any equation of form 1 In fact it was proven by Lindemann that there exists no polynomial Px with integer coefficients of any degree such that either 71 or 171 are solutions of Px 0 Since every Hippasian number is known to be the solution of such an equation it follows that V7 is not a Hippasian number and hence It is impossible to square a circle by ruler and compass alone EXERCISES 45 1 Show that the straight line joining the points a b and a b has equation b 7 bx a7a y a b7ab 0 2 Show that the circle with center a b and radius r has equation x2 y2 7211M 72by a2 b2 7 r2 0 3 Show that if the two lines with equations ax by c 0 and 1 by c 0 intersect then their point of intersection has coordinates bc 7 b cab 7 a b a c 7 ac ab 7 ab 4 Show that if the line with equation ax by c 0 and the circle with equation x2 y2 1 by c 0 intersect then their points of intersection have coordinates x 7B 1 Bz 7 4AC2A and y 7ax7cb where A a2 b2 B 2m a b27abb and C 027 bb c c bz 5 Show that if the two circles with equations x2 y2 ax by c 0 andx2 y2 a c by c 0 intersect then their points of intersection have coordinates x 7B 1 Bz 7 4AC2A and 4 49 IMPOSSIBLE CONSTRUCTIONS y ia cic wl Where Aa 2b Z B 2a c a b 21z1 b b Czc aib bb kc b Z and a aia b bib c cic 6 State and prove the converse of Theorem 451 7 Prove Proposition 453 8 Prove that cos 3A 4 cos3A 7 3 cosA 9 Verify that none of the numbers 111 121 is a solution of the equation x3 1 2 0 10 Verify that none of the numbers 111 112 114 118 is a solution of the equation 813 1 6x 1 1 0 11 Prove thatif x cos 72quot then 4x22x11 0 12 Prove that cos 4A 8 cos4A 7 8 coszA 1 13 Prove that it is impossible to construct a regular 97gon by ruler and compass alone 14 Prove that it is impossible to triple a cube by ruler and compass alone 15 Prove that it is impossible to halve a cube by ruler and compass alone 16 Prove that the following numbers are not Hippasian a 35 b 2 35 c 12 7 3W 17 Is it possible to construct an angle of 1 18 Let A be any number Explain Why sinA is a Hippasian number if and only if cos A is a Hippasian number 4 50 IMPOSSIBLE CONSTRUCTIONS CHAPTER REVIEW EXERCISES 9 10 11 12 13 14 15 Circle 17 intersects two concentric circles Prove that the arcs of 17 cut off by the two circles are equal Prove that if each of the sides of a square is extended in both directions by the length of the radius of the circle that circumscribes the square one obtains the vertices of a regular octagon Two circles centered at C and D respectively intersect at apoint A Prove that if PAQ is the lldouble chordll that is parallel to CD then PQ 2CD The area of the annular region bounded by two concentric circles equals that of a circle whose diameter is a chord of the greater circle that is tangent to the smalller one In a circle a diameter bisects the angle formed by two intersecting chords Prove that the chords are equal Prove that every equiangular polygon all of whose sides are tangent to the same circle is regular Through the center of a circle passes a second circle of greater radius and their common tangents are drawn Prove that the chord joining the contact points of the greater circle is tangent to the smaller circle Prove that every cyclic equilateral pentagon is regular Three circles through the point 0 and of radius r intersect pairwise in the additional points A B C Prove that the circle circumscribed about A ABC also has radius r Prove that in the regular hexagon ABCDEF the diagonals AC and AE cut the diagonal BF into three equal segments Each of the sides of a cyclic quadrilateral is the chord of a new circle Prove that the other four intersection points of these new circles also form a cyclic quadrilateral A circle of radius r is inscribed in A ABC in which A ACB is a right angle Prove that a b c 2r The chord AB of a circle of radius 1 has the property that if the circle is folded along AB so as to bring AB39s arc into the circle then the arc passes through the center of the circle Compute the lengths of the chord AB and its arc In a given AABC construct a point whose distances from the sides of the triangle are proportional to three given line segments Through the midpoint M of a chord PQ of a circle any other chords AB and CD are drawn chords AD and BC meet PQ at points X and Y Prove that M is the midpoint of XY Thisis the notorious Butter y Problem 451 16 17 IMPOSSIBLE CONSTRUCTIONS The points of intersection of the adjacent trisectors of any triangle are the vertices of an equilateral triangle This is known as Morley39s Theorem Are the following statements true or false Justify your answers a b c d e f g h i j k 1 m n 0 The Greeks believed that the world is flat The area of a plane Euclidean figure whose perimeter is composed of circular arcs must involve J17 in its eXpression The Greeks knew that J17 314 If two sectors have equal angles then their arcs are proportional to their radii Given a line segment a it is impossible to construct in the sense of The Elements a square whose area equals that of the circle of radius 1 Given a line segment a it is impossible to construct in the sense of The Elements an equilateral triangle whose area equals that of the circle of diameter 1 Of two equal chords in unequal circles the one in the larger circle lies further from the center The diameter is the circle s longest chord It is possible to construct a regular 3407sided polygon in the sense of The Elements It is possible to construct a regular 1407sided polygon in the sense of The Elements In a circle all the angles subtended by a chord are equal to each other In a circle arcs are proportional to their chords Every circle has only one center Every circle has only one tangent line If two chords of a circle bisect each other then they are both diameters 4 52
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