Advanced Geometry Adithya B., Brian L., William W., Daniel X. 8/19 §1 Introduction In recent years, the AIME has included advanced geometry problems of an \olympiad flavor." These problems, while not necessarily requiring the advanced geometric techniques of olympiads, are qualitatively different from other geometry problems. There will be a lot of geometric work at the beginning of the problem, with the (oftentimes much simpler) computation coming at the end. In these problems, it is especially important to make geometric observations before diving into calculation. We'll discuss two methods of making these observations that also see common usage in olympiad geometry, namely angle chasing and radical axes. §2 Angle Chasing To angle chase, it is useful to know a few useful theorems. Theorem 2.1 (Inscribed Angle Theorem) Let A, B, and C be inscribed in a circle O. Then, \AOB = 2\ACB. Also, angles that subtend the same arc are equal. That is, we have \ACB = \ADB in the diagram below. C D O B A Proof. Let \ACO = α and \BCO = β. Then, since OA = OC, \OAC = α and ◦ ◦ \AOC = 180 − 2α. Similarly, \BOC = 180 − 2β. Therefore, we get, ◦ \AOB = 360 − \AOC − \BOC = 2(α + β) = 2\ACB: 1 Adithya B., Brian L., William W., Daniel X. (8/19) Advanced Geometry The second part of the theorem follows directly from the fact that \ACB is independent of the position of C. Remark 2.2. It should also be noted that by continuity this result is true even when D = A or D = B. In that case, if TA is a tangent to circle O, we have \AOB = 2\T AB and \T AB = \ACB. The latter result is quite important in problems involving tangents to circles and inscribed angles. Now, with that in mind, we will move onto the most important concept in angle chasing: cyclic quadrilaterals. Theorem 2.3 (cyclic quadrilateral) Let ABCD be a convex quadrilateral. Then, if the points A; B; C; D all lie on one circle, we say ABCD is cyclic. The following three statements are equivalent: 1. ABCD is cyclic ◦ 2. \ABC + \CDA = 180 3. \ABD = \ACD B A C D Proof. The forward direction follows from the Inscribed Angle Theorem, but the converse requires more care. It cannot be stressed enough how important cyclic quadrilaterals are. They will show up in many of the examples below, and they will help to deduce important angle relationships to solve the problems. Sometimes, cyclic quadrilaterals may be given in the problem when there are 4 points on the same circle. In this case, we can use the angle properties of cyclic quadrilaterals to make other deductions. For example, cyclic quadrilaterals may lead to an important pair of similar triangles, which would help to finish the problem. Other times, we can prove that four points in our diagram is cyclic from one method, and then use the resulting angle properties to make other deductions about the diagram. Example 2.4 (2016 PUMaC Geometry #7) Let ABCD be a cyclic quadrilateral with circumcircle ! and let AC and BD intersect at X. Let the line through A parallel to BD intersect line CD at E and ! at Y =6 A. If AB = 10; AD = 24;XA = 17, and XB = 21, then the area of 4DEY can be m written in simplest form as n . Find m + n. 2 Adithya B., Brian L., William W., Daniel X. (8/19) Advanced Geometry B A Y X C D E Solution. First, using the parallel condition, we can find \DEY = \CDB = \XAB. Also, \AY D = \ABD, so DEY ∼ XAB. Now, AY k BD, so AY BD is an isosceles trapezoid. So, DY = AB = 10, and we have 102 100 400 [DEY ] = [AXB] = · 84 = 21 441 21 So, our answer is 400 + 21 = 421 . Example 2.5 (2019 AIME I #13) Triangle ABC has side lengths AB = 4, BC = 5, and CA = 6. Points D and E are on ray AB with AB < AD < AE. The point F =6 C is a point of intersection of the circumcircles of 4ACDp and 4EBC satisfying DF = 2 and EF = 7. Then a+b c BE can be expressed as d , where a, b, c, and d are positive integers such that a and d are relatively prime, and c is not divisible by the square of any prime. Find a + b + c + d. A C B X F D E Solution. Notice that \BEF = \BCF and \ADF = \ACF . Thus, \DF E = \ADF − \BEF = \ACF − \BCF = \ACB. Now, by Law of Cosines, we can find cos \ACB = 3 Adithya B., Brian L., William W., Daniel X. (8/19) Advanced Geometry q p 52+62−42 3 2 2 3 2·5·6 = 4 . Now, we can find DE = 2 + 7 − 2 · 2 · 7 · 4 = 4 2. Now, let x = BX, AC AX x+4 y = DX. Note that we have ACX ∼ F DX, so 3 = DF = FX , so FX = 3 . Now, 7 EF FX 7 note that we also have BCX ∼ F EX, so we have 5 = BC = BX , so FX = 5 x as 5 well. Setting these two values equal gives x = 4 . Now, note that,p by Power of a Point, AX · DX = CX · FX = BX · EX, so (x + 4)y = x(y + 4 2), so we can find p p p p 5 2 5+21 2 y = x 2 = 4 . Now, we can find the value of BE to be x + y + 4 2 = 4 , so our answer is 5 + 21 + 2 + 4 = 32 . Example 2.6 (2019 AIME II #11) Triangle ABC has side lengths AB = 7; BC = 8; and CA = 9: Circle !1 passes through B and is tangent to line AC at A: Circle !2 passes through C and is tangent to line AB at A: Let K be the intersection of circles !1 and !2 not equal to A: Then m AK = n ; where m and n are relatively prime positive integers. Find m + n: A ω2 7 9 K ω1 B 8 C Solution. The tangency condition yields \ABK = \KAC and \KAB = \ACK. There- AB 7 7 fore, 4BKA ∼ 4AKC. Since AC = 9 , the ratio between the two triangles is 9 . Let AK 7 9 AK = x. Then, KC = 9 , so KC = 7 x. Now, we can also note that ◦ ◦ ◦ \AKC = 180 − \KAC − \KCA = 180 − \KAC − \BAK = 180 − \A: Therefore, we can apply the Law of Cosines on AKC to solve for x. First, we can determine cos A from the Law of Cosines on 4ABC. Note that, 72 + 92 − 82 11 cos A = = : 2 · 7 · 9 21 11 Therefore, cos \AKC = − 21 . From the Law of Cosines on AKC, we find, 81 9 11 x2 + x2 + 2x · x · = 81: 49 7 21 81 66 x2 + x2 + x2 = 4x2 = 81: 49 49 9 Solving yields x = 2 , so the answer is 9 + 2 = 011 . 4 Adithya B., Brian L., William W., Daniel X. (8/19) Advanced Geometry Example 2.7 (2009 AIME I #15) In triangle ABC, AB = 10, BC = 14, and CA = 16. Let D be a point in the interior of BC. Let IB and IC denote the incenters of triangles ABD and ACD, respectively. The circumcircles of triangles BIBD and CIC D meet at distinct pointspP and D. The maximum possible area of 4BP C can be expressed in the form a − b c, where a, b, and c are positive integers and c is not divisible by the square of any prime. Find a + b + c. A IC IB D BC P Solution. Since BIBDP and CIC DP are cyclic, we have quite a few angle conditions. This motivates us to look at the angles of 4BP C. First, note that ◦ ◦ 1 1 ◦ 1 BIBD = 180 − DBIB − BDIB = 180 − B − ADB = 90 + BAD: \ \ \ 2\ 2\ 2\ ◦ 1 Similarly, \CIC D = 90 + 2 \CAD. By cyclic quadrilaterals BIBDP and CIC DP , ◦ ◦ 1 ◦ ◦ 1 \DP B = 180 −\BIBD = 90 − 2 \BAD and \DP C = 180 −\BIBD = 90 − 2 \CAD. Therefore, 1 1 1 BP C = DP B + DP C = 180◦ − BAD − CAD = 180◦ − A: \ \ \ 2\ 2\ 2\ We can find \A with the Law of Cosines on 4ABC. We have 102 + 162 − 142 1 cos A = = : 2 · 10 · 16 2 ◦ ◦ Therefore, A = 60 , so \BP C = 150 . Therefore, 4BP C has both a fixed side BC and a fixed angle \BP C. The locus of points P must be the arc BC of some circle. ◦ However, we can actually precisely state what this circle is. Since \BP C = 150 , the minor arc BC must be equal to 360◦ − 2 · 150◦ = 60◦. Also, the radius of this circle is the circumradius of BP C, which from the Law of Sines is BC 14 = = 14: 2 sin \BP C 1 Therefore, we see the center of this circle is the point O such that OBC is equilateral. We can now remove all the other points from the diagram, we just have 5 Adithya B., Brian L., William W., Daniel X. (8/19) Advanced Geometry O BC P Now, we can find the area of 4BP C with base times height. The base known to be BC = 14. To maximize the area, we have to maximize the height. The maximum heightp clearly occurs at the midpoint of minor arc BC. At this point, the height is 14 − 7 3. Therefore, the area is 1 p p [4BP C] = · 14 · (14 − 7 3) = 98 − 49 3: 2 The final answer is 98 + 49 + 3 = 150 . §3 Power of a Point and Radical Axes Recall Power of a Point: If A; B; C; D are four points on a circle and AB and CD intersect at P , then PA · PB = PC · PD.