Cyclic quadrilaterals Andrew Tung March 2021 1 Introduction Definition A cyclic quadrilateral is a quadrilateral that can be inscribed in a circle. B A C D Four points A, B, C, and D are called concylic if they all lie on the same circle (i.e. quadrilateral ABCD is cyclic). The main theorem regarding cyclic quadrilaterals (with regard to angle chasing) is the following. Proposition 1 If ABCD is a convex quadrilateral, the following are equivalent. 1. ABCD is cyclic 2. \DCA = \DBA 3. \CDA = 180 − \CBA The proof of this is by the inscribed angle theorem, and is not hard1. Note that Propo- sition1 can be used to prove a quadrilateral is cyclic, or to get some angle information from 1Left as an exercise to the reader. 1 a known cyclic quadrilateral. Both directs are important. Cyclic quadrilaterals also give similar triangles: Proposition 2 If ABCD is a cyclic quadrilateral whose diagonals intersect at P , then 4P BA ∼ 4PCD and 4P BC ∼ 4P AD. B A P C D Proof We prove 4P BC ∼ 4P AD; the other case is symmetric. Using Proposition1, \DBC = \DAC. So \P BC = \DAP . Also \BP C = \AP D because they are vertical angles. Hence by AA similarity, 4P BC ∼ 4P AD. Note that Proposition2 does not imply ABjjCD, since that similarity would go the \other way" (i.e. 4P BA ∼ 4P DC). Also note that the side ratios resulting from Proposi- tion2 imply Power of a Point: BP · PD = AP · PC. Some other computational tools relating to cyclic quadrilaterals include: Proposition 3 (Ptolemy's Theorem) In a cyclic quadrilateral ABCD, we have (AB)(CD) + (BC)(DA) = (AC)(BD) The proof of Ptolemy's Theorem is either by trig (which is slightly messy), or by con- structing an isogonal conjugate of a diagonal, which leads to some similar triangles. Proposition 4 (Brahmagupta's Formula) If a cyclic quadrilateral has side lengths a, b, c, d, and if a + b + c + d s = 2 then the area of the quadrilateral is p(s − a)(s − b)(s − c)(s − d) The proof of Brahmagupta's Formula is by trig or by repeated applications of Heron's formula. Both are messy (and are omitted). 2 2 Examples 2.1 Using cyclic quads for angle chasing These first two problems illustrate one of the basic uses of cyclic quadrilaterals: angle chasing. Problem (2006 AIME II 12) Equilateral 4ABC is inscribed in a circle of radius 2. Extend AB through B to point D so that AD = 13; and extend AC through C to point E so that AE = 11: Through D; draw a line l1 parallel to AE; and through E; draw a line l2 parallel to AD: Let F be the intersection of l1 and l2: Let G be the point on the circle that is collinear with A andp F and distinct from A: Given that the area of 4CBG p q can be expressed in the form r ; where p; q; and r are positive integers, p and r are relatively prime, and q is not divisible by the square of any prime, find p + q + r: A C B G E D F The key claim is that 4BGC ∼ 4F DA To prove this, we use the cyclic quadrilateral ABGC. Note \BCG = \BAG = \F AD since ABGC is cyclic. Similarly, \CBG = \CAG = \EAF = \AF D, using the fact that 3 ADF E is a parallelogram. So to find the area of 4CBG, we first find the area of 4F DA, and then multiplyp by BC2 143 3 AF 2 . Note \F DA = \BGC = 180 − \BAC = 120, so the area of 4CBG is 4 . Using Law of Cosines, AF 2 = 433. Hence our answer is p p 143 3 4 143 3 = 4 433 433 Problem (2018 AMC 12A 20) Triangle ABC is an isosceles right triangle with AB = AC = 3. Let M be the midpoint of hypotenuse BC. Points I and E lie on sides AC and AB, respectively, so that AI > AE and AIME is a cyclicp quadrilateral. Given that a− b triangle EMI has area 2, the length CI can be written as c , where a, b, and c are positive integers and b is not divisible by the square of any prime. What is the value of a + b + c? B M E A X I C In this case the cyclic quadrilateral is given: AIME. We note that \MAI = \MAB = 45, since M is the midpoint and ABC is isosceles. So \MEI = \MAI = 45, and 4EMI is isosceles. Hence MI = 2. Let the foot of the perpendicular from M to AC be X. Then MX = 3 , and since p p 2 7 3− 7 MI = 2, XI = 2 by the Pythagorean Theorem. So CI = 2 . 4 2.2 Using cyclic quads for length ratios These next two problems illustrate the use of cyclic quadrilaterals for finding ratios of lengths, using the similar triangles in Proposition2. Problem (2002 AIME I 13) In triangle ABC the medians AD and CE have lengths 18 and 27, respectively, and AB = 24.p Extend CE to intersect the circumcircle of ABC at F . The area of triangle AF B is m n, where m and n are positive integers and n is not divisible by the square of any prime. Find m + n. C D G B A E F The midpoints in this problem suggest area ratios. Note that [4AGE] is known, since AG = 12, AE = 12, and GE = 9. Hence we relate [4AF B] to [4AGE]. 1 GE 1 Note that [4AGE] = 3 [4ACE] since CE = 3 . Also note [4AF B] = 2[4EF B] since E is the midpoint of AB. But 4ACE ∼ 4F EB since ACBF is cyclic. The ratio of similarity CE 27 is just EB = 12 . 5 p 27 55 A simple computation reveals [4AGE] = 4 . So as outlined above [4AF B] = 2[4EF B] 122 = 2 [4ACE] 27 42 = 2 (3)[4AEG] 9 p ! 32 27 55 = 27 4 p = 8 55 Problem (2021 AIME I 11) Let ABCD be a cyclic quadrilateral with AB = 4; BC = 5;CD = 6; and DA = 7. Let A1 and C1 be the feet of the perpendiculars from A and C, respectively, to line BD; and let B1 and D1 be the feet of the perpendiculars from B m and D; respectively, to line AC. The perimeter of A1B1C1D1 is n , where m and n are relatively prime positive integers. Find m + n. B B1 X C A C1 D We focus on B and C for now, since the other vertices are symmetric. Note that \BB1C = \BC1C = 90, so quadrilateral BB1C1C is cyclic. So 4B1XC1 ∼ BXC, so B X B C 1 = 1 1 BX BC 6 B1X But note BX = cos \AXB. Extending this to the other sides, we note that each side of A1B1C1D1 is simply cos \AXB times the corresponding side of ABCD. So it suffices to find cos \AXB. p This is not hard. By Brahmagupta's, [ABCD] = 2 210, and by Ptolemy's Theorem, the product of the diagonals is 59. So p 1 [ABCD] = 2 210 = (59)(sin AXB) 2 \ p 4 210 11 11 and sin \AXB = 59 . Thus cos \AXB = 59 , so the perimeter of A1B1C1D1 is 59 (4 + 242 5 + 6 + 7) = 59 . 2.3 Finding cyclic quads The previous problem shows that sometimes, important cyclic quadrilaterals are not imme- diately obvious. The following two problems continue this theme. Problem (1991 AIME 12) Rhombus P QRS is inscribed in rectangle ABCD so that vertices P , Q, R, and S are interior points on sides AB, BC, CD, and DA, respectively. It is given that PB = 15, BQ = 20, PR = 30, and QS = 40. Let m=n, in lowest terms, denote the perimeter of ABCD. Find m + n. AB P S X Q R CD First note that \PXQ = 90, since P QRS is a rhombus. Since \P BQ = 90, P BQX is cyclic. Also note that X is the center of the rectangle ABCD, since the height of X is half of BC (using segment PR) and similarly for the horizontal direction. Using the cyclic quadrilateral P BQX, \XPQ = \XBQ. Since 4PXQ and 4BCD are both right and share an angle, 4PXQ ∼ 4BCD. 7 Using Ptolemy's on P BQX, we find BX = 24, so BD = 48 and the similarity ratio is 25 28 672 48 . So the perimeter of ABCD is 2(BC + CD) = 2 25 (PX + XQ) = 5 . Problem (2011 AIME II 10) A circle with center O has radius 25. Chord AB of length 30 and chord CD of length 14 intersect at point P . The distance between the midpoints 2 m of the two chords is 12. The quantity OP can be represented as n , where m and n are relatively prime positive integers. Find the remainder when m + n is divided by 1000. C AB E P F D O Let E and F be the midpoints of AB and CD, respectively. The key is to construct OE and OF . Note that since E and F are midpoints, \P EO = \PFO = 90. So EP F O is cyclic. Observe that since \P EO = 90, OP is the diameter of the circumcircle of OEF . So it suffices to find the circumradius of OEF . But it is not hard to find OE = 20 and OF = 24, and it is given EF = 12. So 2 p (20)(12)(24) [4OEF ] = 2 14 and R = p = p45 . Thus OP 2 = p90 = 4050 . 4(2 14) 14 14 7 8 3 Takeaways Sometimes cyclic quadrilaterals are immediately obvious.