MATH 68090 NonEuclidean Geomety Execise Set : Solutions Pove that the opposite angles in a convex quadilateal inscibed in a cicle sum to 80º Convesely, pove that if the opposite angles in a convex quadilateal sum to 80º, then the quadilateal can be inscibed in a cicle Such a quadilateal is called a cyclic quadilateal The opposite angles of a quadilateal inscibed in a cicle subtend complimentay acs The sum of the cental angles then is 60º The sum of the angles will be half of this, o 80º Now, assume that the opposite angles in a convex quadilateal sum to 80º Any thee points lie on a cicle So choose A, C, and D and let Γ be the cicle that cicumscibes ACD Let BD intesect the cicle in a point X Now, DXC subtends the same ac as does DAC We have CAX CDX, DAC DXC, ACX X, ACD AXD We know that C + ABC 80 C + AXC, so ABC AXC Then in ABC we have that BAC > XAC and BCA > XCA Howeve, ABC + BCA + CAB > AXC + XCA + CAX α + β + γ + δ 80 Thus, thee is a tiangle with angle sum moe than 80º This means that X B, and we ae done Suppose DABC D A B C Show that Ê AB ˆ D ABC Á D ABC Ë AB If the two tiangles ae simila, then by dopping a pependicula fom B to AC and fom B' to A'C', we will ceate simila tiangles so that BD B D k, AB A B whee k is the atio of similaity Anothe way to convey this is BD k BD Thus, ABC BD i AC ( kbd )( kac ) k ABC AB ABC AB Execise Set Solutions Page Sping 004
(The Angle Bisecto Theoem) In an abitay tiangle D ABC, let the inteio angle bisecto at A intesects the side BC at D Show that BD DC AB AC Constuct the line though B paallel to Let E be the point of intesection with AC Then, BEA DAC Now, EAB + B + DAC 80 Since B DAC, we have EAB 80 DAC But, BEA + EAB + ABE 80 Applying the pevious equation gives us that BEA ABE and ABE is isosceles Thus, AE AB E B A D C Now, since BE is paallel to, we have that BEC DAC Thus, EC AC BC DC EA + AC BD + DC AC DC EA AC BD DC AB AC BD DC 4 The pentagon in Figue is egula and each side has length one Show that AF FD + 5 A B Fom the fact that the pentagon is egula, we have that AED 08 Now, D AED is isosceles, since the sides ae equal Thus, the base angles ae equal Since the sum of the angles is 80º, we get that EDA AED 6 Using the same infomation we can ague that CED 6 This makes D FED an isosceles tiangle which is simila to D AED Thus, by simila tiangles we have E F D Figue C Execise Set Solutions Page Sping 004
FD AE FD FD ED Now, it is easy to show that AEF AFE 7 Thus, AF Now, AF + FD + 0 D AEF is isosceles, which makes Using the Quadatic Fomula, we get that Since 5 0 Thus, ± + 4 <, we must have that ( + ) AF FD 5 Ê + 5 ˆ Á Ë + 5 5 (The Tangential Vesion of Powe of the Point) In Figue, suppose PR is tangent to the cicle at R Show that PQ PQ PR Now, D OQQ, D OQR, and D OQ R ae isosceles, so the base angles ae equal Let OQQ OQ Q a, OQR ORQ b, and OQ R ORQ g We intend to show that DPQR D PRQ Clealy, QPR RPQ, since it is the same angle Now, Execise Set Solutions Page Sping 004
Then, QOQ + QOR ROQ ( 80 a) + ( 80 b) ( 80 g ) 80 + g ( a + b) 90 + g a + b ( ) PQR 80 ( a + b) 80 90 + g 90 g PRQ Thus, the thid angles will be equal Theefoe, as we wanted, DPQR D PRQ Thus, PQ PR PR PQ PQ PQ PR R P Q Figue O Q' 6 Use the tangential vesion of the powe of the point to come up with yet anothe poof of the Pythagoean Theoem Daw a cicle centeed at B with adius BC This will make AC tangent to the cicle at C Let Q denote the point whee AB intesects the cicle and let Q' be the othe point of intesection The above theoem gives us that AC AQ AQ, Now, AC b, AQ c c a, and AQ' c + c + a Thus we get b ( c a)( c+ a) a + b c D A C Figue Execise Set Solutions Page 4 Sping 004
7 In Figue, the point C is on the diamete AB of a half cicle The two smalle half cicles have diametes AC and CB The egion bounded by the cuved edges of the half cicles was called an abelos o butche's knife by Achimedes The pependicula at C intesects the lage cicle at D Pove that the cicle with diamete CD has the same aea as the abelos Let AB and AC p Let DC h Then since D lies on the semicicle, the angle B is a ight angle and we have thee ight tiangles The Pythagoean theoem then gives us: p + h x ( p) + h y x + y z whee x and y BD We need h in tems of and p Adding the fist two togethe and equating this in the thid we have x + y ( p + h ) + ( p) + h ( ) h + p p+ 0 h + p p h p( p) The aea of the cicle with diamete CD h is p h 4 This gives The aea of the abelos is given by h p p p( p ) 4 4 Ê Êˆ Ê pˆ Ê pˆ ˆ p p p p p( p ) Á Á Á + Á Ë Ë Ë Ë 4 D C C 4 C 5 O 4 C O 5 A O O P C O B Figue 4 8 Show that the two inscibed cicles in the abelos in Figue 4 have the same adii Execise Set Solutions Page 5 Sping 004
Let C denote the oute cicle of the abelos, C, the lefthand inne cicle, and C the ighthand inne cicle Let C 4 denote the cicle that is tangent to C, C, and CD Assume that the cente of cicle C i is O i Let C i have adius i Thus, AO BO and Let OO p Note then that + p and + Fom these two equations it follows that p We want to solve fo the adius of C 4 in tems of,, and This means that when we do the same computation fo the cicle C 5, we want to get the same answe, meaning that the two cicles have the same adius, which is what we want to show Let P be the foot of the pependicula dopped fom O to AB Note that since the cicle C 4 is tangent to CD, then PC 4 Also, note that C 4 is tangent to C and O 4 lies on a adius to C though O Theefoe, OO 4 4 Likewise, O 4 lies on a adius to C though O, so OO 4 + 4 The tiangles D OOP 4 and D OOP 4 ae ight tiangles with common leg PO 4 Thus, OO 4 OP OO 4 OP + 4 4 4 4 ( ) ( ) ( ) ( ) ÈÎ È Î ( + 4) + ( 4) ( + 4) ( 4) [( 4) + ( 4)][( 4) ( 4)] ( )( 4) ( + 4)( + ) 4 ( ) ( ) 4 4 + 4 4 4 + Thus, the adius of this cicle tangent to the oute cicle, the inne cicle and the pependicula segment depends only on the adii of the oute cicle and the two inne cicles This means that the computation fo the adius of cicle C 5 will yield the same esult The adical axis of two cicles G is the set of point P with the popety that the powes of P with espect to both G ae equal 9 Let two cicles G have distinct centes O and O' Pove that the adical axis of G and G is a line pependicula to OO' Let me fist do a calculation Let D ABC be a tiangle, let M be the midpoint of AB and let CH be the altitude with H Œ AB Label the sides BC, AC, and AB as a, b, and c, espectively Then fom the Pythagoean Theoem we have Execise Set Solutions Page 6 Sping 004
a b BH AH ( BH AH )( BH AH ) + c MH Now, let P be a point on the adical axis of G with centes O and O and adii and Then, because P lies on the adical axis PO PO PO PO OO MH whee M is the midpoint of the segment O O and H is the foot of P in the line OO Thus, MH OO H O M O is a constant This means that Figue 5 H neve moves any futhe fom M Thus, evey point on the adical axis has the same foot in O O, so the adical axis is pependicula to O O 0 Suppose two cicles intesect at A and B Pove that the adical axis of the two cicles is the line AB Since D OAO @ D OBO, A and B have equal powes with espect to the two cicles Since the adical axis is a line and it must pass though A and B, it must be the line AB Let the cicle G intesect the cicle G at A and B, and the cicle G at C and D Let P be the point of intesection of AB and CD Pove that P lies on the adical axis fo G Since P lies on AB it lies on the adical axis of G Since it lies on CD, it lies on the adical axis of G Both of these follow fom the pevious poblem (The Radical Cente) Let G, G, be thee cicles Show that the thee adical axes of these cicles intesect at a common point This point is called the adical cente of the thee cicles Execise Set Solutions Page 7 Sping 004
Let the adical axis fo G and the adical axis fo G intesect in a point P We want to show that P is on the adical axis fo G Now, PO PO PO The fist equality is tue because P lies on the adical axis fo G and the second equality is tue because P lies on the adical axis fo G Thus, P lies on the adical axis fo G Let P be the adical cente of thee cicles G, G, Suppose P is outside cicle G Pove that P is also outside cicles G Because P lies outside G, PO 0 0 > Since all thee quantities ae equal we have PO PO PO > This puts them all outside thei espective cicles 4 Let P be the adical cente of thee cicles G, G, and suppose P is outside cicle G Show that P is the cente of a cicle which intesects G, G, pependiculaly Futhemoe, show that the adius of this cicle is P ( P) Take the cicle centeed at P with adius P ( P) Note that P ( P) POi i At the point at which the cicle intesects cicle G i we have a tiangle with sides P ( P), i, and PO i Since i + P ( P) POi, we have a ight tiangle, which means that the adius of the cicle G i and ou cicle ae pependicula, and we ae done Note that these could be paallel Take thee cicles all centeed on the same line, the middle cicle equidistant fom the othe two Then the adical axes ae all pependicula to the line containing the centes, and hence paallel Execise Set Solutions Page 8 Sping 004