MATH 6118-090 Non-Euclidean Geometry Exercise Set 3: Solutions 1. Prove that the opposite angles in a convex quadrilateral inscribed in a circle sum to 180º. Conversely, prove that if the opposite angles in a convex quadrilateral sum to 180º, then the quadrilateral can be inscribed in a circle. Such a quadrilateral is called a cyclic quadrilateral. The opposite angles of a quadrilateral inscribed in a circle subtend complimentary arcs. The sum of the central angles then is 360º. The sum of the angles will be half of this, or 180º. Now, assume that the opposite angles in a convex quadrilateral sum to 180º. Any three points lie on a circle. So choose A, C, and D and let Γ be the circle that circumscribes ∆ACD . Let BD intersect the circle in a point X. Now, ∠DXC subtends the same arc as does ∠DAC . We have CAX CDX ,,, DAC DXC ACX ADX ACD AXD We know that ∠+∠==∠+∠ADC ABC180 ADC AXC , so ∠ABC=∠ AXC . Then in ∆ABC we have that ∠>∠BAC XAC and ∠BCA>∠ XCA . However, ∠+∠+∠>∠+∠+∠=+++=ABC BCA CAB AXC XCA CAX αβγδ180 Thus, there is a triangle with angle sum more than 180º. This means that X = B , and we are done. 2. Suppose DDABC∼ A¢¢¢ B C . Show that 2 ʈAB¢¢ D=ABC D ABC¢¢¢. ˯Á˜AB If the two triangles are similar, then by dropping a perpendicular from B to AC and from B' to A'C', we will create similar triangles so that BD B′ D′ = = k , AB A′′ B where k is the ratio of similarity. Another way to convey this is BDkBD= ′′. Thus, 1 ∆ABC= BDi AC 2 1 = ()kBD′′() kAC ′′ 2 = k2 ∆ ABC′′′ 2 ⎛⎞AB ⎜ ⎟ =⎜ ⎟ ∆ABC′′′ ⎝⎠⎜ AB′′⎟ Exercise Set 3 Solutions Page 1 Spring 2004 3. (The Angle Bisector Theorem) In an arbitrary triangle DABC , let the interior angle bisector at A intersects the side BC at D. Show that BDAB = . DC AC Construct the line through B parallel to AD. Let E be E the point of intersection with AC. Then, ∠≅∠BEA DAC . Now, ∠+∠+∠=EAB BAD DAC 180 . A Since ∠≅∠BAD DAC , we have ∠=−∠EAB180 2 DAC . But, ∠+∠+∠=BEA EAB ABE 180 . Applying the previous equation gives us that ∠BEA=∠ ABE and ∆ABE is isosceles. Thus, AEAB≅ . B D C Now, since BE is parallel to AD, we have that ∆BEC∼ ∆ DAC . Thus, EC BC = AC DC EA++ AC BD DC = AC DC EA BD = AC DC AB BD = AC DC 4. The pentagon in Figure 1 is regular and each side has length one. Show that AF 15+ B = . FD 2 A From the fact that the pentagon is regular, we have that –=AED 108 . Now, DAED is isosceles, since the sides are C equal. Thus, the base angles are equal. Since the sum of the F angles is 180º, we get that –=–=EDA AED 36 . Using the E same information we can argue that –=CED 36 . This makes DFED an isosceles triangle which is similar to DAED . Thus, D by similar triangles we have Figure 1 Exercise Set 3 Solutions Page 2 Spring 2004 FD ED = AE AD FD 1 = 1 AD 1 FD = AD Now, it is easy to show that –=–=AEF AFE 72 . Thus, DAEF is isosceles, which makes |AF| = 1. Now, AD=+ AF FD 1 AD =+1 AD 2 AD--= AD 10 Using the Quadratic Formula, we get that 114±+ AD = 2 Since 150-<, we must have that AD =+(152) Thus, AF 1 = FD AD -1 1 = ʈ-+15 ˯Á˜2 15+ = 2 5. (The Tangential Version of Power of the Point) In Figure 2, suppose PR is tangent to the circle at R. Show that 2 PQ◊= PQ¢ PR . Now, DOQQ¢ , DOQR , and DOQ¢ R are isosceles, so the base angles are equal. Let –=–=OQQ¢¢ OQ Q a , –=–=OQR ORQ b , and –=–=OQ¢¢ R ORQ g . We intend to show that DDPQR∼ PRQ¢ . Clearly, –=–QPR RPQ¢ , since it is the same angle. Now, Exercise Set 3 Solutions Page 3 Spring 2004 R –+–=–QOQ¢¢ QOR ROQ r 180-+ 2abg 180 -= 2 180 - 2 ()()() O 180 += 2gab 2( + ) 90 +=+gab r r Then, P Q –=-+=-+=-=–PQR180 (ab ) 180( 90 g) 90 g PRQ¢ Q' Thus, the third angles will be equal. Therefore, as we wanted, Figure 2 DDPQR∼ PRQ¢ . Thus, PQ PR = PR PQ¢ 2 PQ◊= PQ¢ PR 6. Use the tangential version of the power of the point to come up with yet another proof of the Pythagorean Theorem. Draw a circle centered at B with radius BC. This will make AC tangent to the circle at C. Let Q denote the point where AB intersects the circle and let Q' be the other point of intersection. The above theorem gives us that 2 AC=◊ AQ AQ¢ , Now, |AC| = b, |AQ|= c – r = c - a, and |AQ'|= c + r = c + a. Thus we get bcaca2 =-()() + ab22+= c 2 D A C Figure 3 Exercise Set 3 Solutions Page 4 Spring 2004 7. In Figure 3, the point C is on the diameter AB of a half circle. The two smaller half circles have diameters AC and CB. The region bounded by the curved edges of the half circles was called an arbelos or butcher's knife by Archimedes. The perpendicular at C intersects the larger circle at D. Prove that the circle with diameter CD has the same area as the arbelos. Let |AB|=r and |AC|=p. Let |DC|=h. Then since D lies on the semicircle, the angle –ADB is a right angle and we have three right triangles. The Pythagorean theorem then gives us: p22+=hx 2 ()rp-+=22 h y 2 x222+=yz where x = |AD| and y = |BD|. We need h in terms of r and p. Adding the first two together and equating this in the third we have rxy222=+ =++-+()()p22hrph() 22 =+-+22hpprr22 2 2 02=+-hppr22 2 2 hppr=-() The area of the circle with diameter |CD| = h is p h2 4 . This gives h2 p p =-p()pr 44 The area of the arbelos is given by 22 2 1 ʈʈrprp Ê ˆ Ê- ˆ p Á˜ppÁ˜-+ Á ˜ p Á ˜ =-p()pr 22˯˯ Ë 2 ¯ Ë 2 ¯ 4 D C C1 4 C5 O4 O5 C2 A O O P C O B 2 1 3 Figure 4 8. Show that the two inscribed circles in the arbelos in Figure 4 have the same radii. Exercise Set 3 Solutions Page 5 Spring 2004 Let C1 denote the outer circle of the arbelos, C2, the left-hand inner circle, and C3 the right- hand inner circle. Let C4 denote the circle that is tangent to C1, C2, and CD. Assume that the center of circle Ci is Oi. Let Ci have radius ri. Thus, rAOBO11== 1 and. Let OO12= p. Note then that rrp12=+ and rrr231+=. From these two equations it follows that p =-=rr12 r 3 We want to solve for the radius of C4 in terms of r1, r2, and r3. This means that when we do the same computation for the circle C5, we want to get the same answer, meaning that the two circles have the same radius, which is what we want to show. Let P be the foot of the perpendicular dropped from O3 to AB. Note that since the circle C4 is tangent to CD, then |PC| = r4. Also, note that C4 is tangent to C1 and O4 lies on a radius to C1 through O1. Therefore, OO14=- r 1 r 4. Likewise, O4 lies on a radius to C2 through O2, so OO24=+ r 2 r 4. The triangles DOOP42 and DOOP41 are right triangles with common leg PO4. Thus, 22 22 OO42-= OP 2 OO 41 - OP 1 222 2 ()()()(rr24+--=---- rr 24 rr 14 rrr 234 ) È˘È˘(rr24++-) ( rr 24) ( rr 24 +--) ( rr 24) =[][]()( rr 14 -+-- rrr 234 )()( rr 14 ---- rrr 234 ) Î˚Î˚ (22r2)( r 4) =+-- ( rrr 123 2)( rrrr 4123 -+ ) 4(22)2rr24=- r 2 r 4( r 3) rr24+= rr 34 rr 23 rr r = 23 4 rr+ 23 rr = 23 r1 Thus, the radius of this circle tangent to the outer circle, the inner circle and the perpendicular segment depends only on the radii of the outer circle and the two inner circles. This means that the computation for the radius of circle C5 will yield the same result. The radical axis of two circles G and G¢ is the set of point P with the property that the powers of P with respect to both G and G¢ are equal. 9. Let two circles G and G¢ have distinct centers O and O'. Prove that the radical axis of G and G¢ is a line perpendicular to OO'. Let me first do a calculation. Let DABC be a triangle, let M be the midpoint of AB and let CH be the altitude with HABŒ . Label the sides BC, AC, and AB as a, b, and c, respectively. Then from the Pythagorean Theorem we have Exercise Set 3 Solutions Page 6 Spring 2004 22 a22-= b BH - AH =-()BHAHBHAH() + =◊2cMH Now, let P be a point on the radical axis of G and G¢ with centers O1 and O2 and radii r1 and r2. Then, because P lies on the radical axis 2222 PO11-= r PO 2 - r 2 22 22 r12-= r PO 2 - PO 1 =◊2 OO12 MH where M is the midpoint of the segment O1O2 and H is the H O M O foot of P in the line OO12. 1 2 Thus, rr22- MH = 12 2 OO12 is a constant.