Lesson 10: Advanced Geometry

Adithya B., Brian L., William W., Daniel X.

August 2020

Adithya B., Brian L., William W., Daniel X. Lesson 10: Advanced Geometry August 2020 1 / 31 Problem of the Week

PotW Tetrahedron ABCD satisﬁes AB = BC = CA and DA = DB = DC. Let E, F , G, H be the feet of the altitudes from A, B, C, D to their respective faces. If E, F , G, H all lie in the interiors of their respective faces, let the [EFGH] m maximum value of [ABCD] be n for relatively prime positive integers m, n. Compute m + n.

Adithya B., Brian L., William W., Daniel X. Lesson 10: Advanced Geometry August 2020 2 / 31 Problem of the Week

We consider where the plane EFG intersects segments DA, DB, DC. Suppose these are X , Y , Z. Now, note that EFG and XYZ are both equilateral triangles, but EFG 1 is the medial triangle of XYZ, so has 4 the area. If the side length of ABC is 1, then let the side length of XYZ be k. k2 Now, note that the ratio of the areas of the bases is 4 . Note that the ratio of the heights is 1 − k. Thus, we wish to k2(1−k) maximize 4 . k k q 1 2 + 2 +(1−k) 3 k k Now, note that we have 3 = 3 ≥ 2 · 2 · (1 − k) by AM-GM 1 Thus, we have that our expression is at most 27 , so our answer is 1 + 27 = 28 .

Adithya B., Brian L., William W., Daniel X. Lesson 10: Advanced Geometry August 2020 3 / 31 Angle Chasing: Inscribed Angles

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

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.

Adithya B., Brian L., William W., Daniel X. Lesson 10: Advanced Geometry August 2020 4 / 31 Angle Chasing: Cyclic Quadrilaterals

Theorem 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.

Adithya B., Brian L., William W., Daniel X. Lesson 10: Advanced Geometry August 2020 5 / 31 Angle Chasing

2016 PUMaC Geometry #7 Let ABCD be a cyclic quadrilateral with circumcircle ω and let AC and BD intersect at X . Let the line through A parallel to BD intersect line CD at E and ω at Y 6= A. If AB = 10, AD = 24, XA = 17, and XB = 21, then m the area of 4DEY can be written in simplest form as n . Find m + n.

B A

Y X

D E

Adithya B., Brian L., William W., Daniel X. Lesson 10: Advanced Geometry August 2020 6 / 31 2016 PUMaC Geometry #7

First, note that we can ﬁnd that ∠DEY = ∠CDB = ∠XAB and ∠AYD = ∠ABD. Thus, we have that DEY ∼ XAB. We also note that ABDY is an isosceles A B trapezoid, AY k BD Y X This tells us that DY = AB = 10. Note that we are given all 3 sides of XAB, so we can ﬁnd [XAB] = 84 C D Now, note that we have that the ratio of the E DY 10 triangles DEY and XAB is XB = 21 , so the [DEY ] 10 2 ratio of their areas [XAB] = 21 10 2 400 Thus, we have that [DEY ] = 84 21 = 21 , so our answer is 400 + 21 = 421

Adithya B., Brian L., William W., Daniel X. Lesson 10: Advanced Geometry August 2020 7 / 31 Angle Chasing

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 4ACD and 4EBC√satisfying a+b c DF = 2 and EF = 7. Then 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.

C B X F D

Adithya B., Brian L., William W., Daniel X. Lesson 10: Advanced Geometry August 2020 8 / 31 2019 AIME I #13

Note that ∠ACF = ∠ADF and ∠BCF = ∠BEF A Subtracting these gives ∠ACB = ∠DFE We can calculate with Law of Cosines C that cos ACB = 3 B ∠ 4 X F We can now use Law of√ Cosines on 4DEF to ﬁnd DE = 4 2. D Now, let X be the intersection of CF and AE. Let x = BX and y = DX . We have FXD ∼ AXC and E FXE ∼ BXC, so AX ·FD BX ·EF FX = AC = BC .

Adithya B., Brian L., William W., Daniel X. Lesson 10: Advanced Geometry August 2020 9 / 31 2019 AIME I #13

x+4 7 A Thus, we find that 3 = 5 x 5 Thus, we may solve for x to get x = 4 Now, note that, by Power of a Point, C we also have that B BX · EX = CX√ · FX = AX · DX , so we X F have x(y + 4 2) = (x + 4)y √ √ D 5 2 Thus, we find that y = 2x = 4 Now, we find √ √ 5+21 2 BE = x + y + 4 2 = 4 Thus, our answer is E 5 + 21 + 2 + 4 = 32

Adithya B., Brian L., William W., Daniel X. Lesson 10: Advanced Geometry August 2020 10 / 31 Angle Chasing

2019 AIME II # 11

Triangle ABC has side lengths AB = 7, BC = 8, and CA = 9. Circle ω1 passes through B and is tangent to line AC at A. Circle ω2 passes through C and is tangent to line AB at A. Let K be the intersection of circles ω1 m and ω2 not equal to A. Then AK = n , where m and n are relatively prime positive integers. Find m + n.

A ω2

7 9

K ω1 B 8 C

Adithya B., Brian L., William W., Daniel X. Lesson 10: Advanced Geometry August 2020 11 / 31 2019 AIME II # 11

The tangency condition yields ∠ABK = ∠KAC and ∠KAB = ∠ACK. 4BKA ∼ 4AKC. AB 7 AC = 9 AK 7 Let AK = x. Then, KC = 9 , so 9 KC = 7 x. ◦ ∠AKC = 180 − ∠KAC − ∠KCA = ◦ ◦ A 180 − ∠KAC − ∠BAK = 180 − ∠A ω2 We can use the Law of Cosines on 7 9 4AKC. K 2 2 2 ω1 7 +9 −8 11 cos A = 2·7·9 = 21 B 8 C 11 cos AKC = − 21 2 81 2 9 11 x + 49 x + 2x · 7 x · 21 = 81 9 x = 2 =⇒ 011 Adithya B., Brian L., William W., Daniel X. Lesson 10: Advanced Geometry August 2020 12 / 31 Angle Chasing

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 BIB D and CIC D meet at distinct points P and D. The maximum possible area of 4BPC √ 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.

IC IB D BC P

Adithya B., Brian L., William W., Daniel X. Lesson 10: Advanced Geometry August 2020 13 / 31 2009 AIME I # 15

Let’s look at ∠BPC. BIB DP and CIC DP are cyclic A ∠BPC = ∠BPD + CPD ◦ ◦ ∠BPC = 180 − ∠BIB D + 180 − ∠CIC D ◦ ∠BIB D = 180 − ∠DBIB − ∠BDIB = ◦ 1 1 ◦ 1 180 − 2 ∠B − 2 ∠ADB = 90 + 2 ∠BAD ◦ 1 IC CIC D = 90 + CAD IB ∠ 2 ∠ BPC = 180◦ − 1 BAD − 1 CAD = D ∠ 2 ∠ 2 ∠ ◦ 1 BC 180 − 2 ∠A. P We can ﬁnd ∠A with the Law of Cosines. 102+162−142 1 ◦ cos A = 2·10·16 = 2 , so ∠A = 60 ◦ and ∠BPC = 150 . ∠BPC is ﬁxed. What is the locus of P?

Adithya B., Brian L., William W., Daniel X. Lesson 10: Advanced Geometry August 2020 14 / 31 2009 AIME I # 15

Arc of a circle! Let the center be O. What is ∠BOC? ◦ ◦ ◦ ∠BOC = 360 − 2 · 150 = 60 . ◦ Note OB = OC and ∠BOC = 60 , so BOC is equilateral. The radius is OB = BC = 14. O Area is base times height. Maximum height is at the midpoint of arc BC. √ Maximum height is 14 − 7 3. BC P The area is √ √ 1 2 · 14 · (14 − 7 3) = 98 − 49 3. 98 + 49 + 3 = 150

Adithya B., Brian L., William W., Daniel X. Lesson 10: Advanced Geometry August 2020 15 / 31 Radical Axes

Given a circle ω and a point P outside ω, let ` be any line through P Suppose ` meets ω at A and B; by Power of a Point PA · PB does not depend on ` If O is the center of ω and R is its radius, how else can we express this product? Choose ` = OP; what is PA · PB (in terms of OP and R)? Check that it’s OP2 − R2 If P is inside ω, then the answer is R2 − OP2 We deﬁne OP2 − R2 to be the power of P with respect to ω Note that the power is negative for points inside the circle

Adithya B., Brian L., William W., Daniel X. Lesson 10: Advanced Geometry August 2020 16 / 31 Radical Axes

Given two (nonconcentric) circles ω1 and ω2, what is the set of points with the same power with respect to both circles?

The answer is a line, called the radical axis of ω1 and ω2

This line is perpendicular to O1O2, where O1 and O2 are the centers of ω1 and ω2 These results can be proved with a relatively simple use of coordinates, described in the handout Note that if two circles intersect, then their radical axis is the line through their intersection points

Adithya B., Brian L., William W., Daniel X. Lesson 10: Advanced Geometry August 2020 17 / 31 Radical Axis Theorem

One useful result about radical axes is the Radical Axis Theorem:

Let ω1, ω2, ω3 be pairwise nonconcentric circles. Let `1 be the radical axis of ω2 and ω3 and deﬁne `2 and `3 similarly. Then `1, `2, and `3 are either concurrent or all parallel.

Proof: suppose `1 and `2 meet at P

Since P ∈ `1 P has equal powers with respect to ω2 and ω3

Since P ∈ `2 P has equal powers with respect to ω1 and ω3

So P has equal powers with respect to ω1 and ω2

Thus P ∈ `3 This shows that if any two radical axes intersect, then the three concur; thus they’re concurrent or all parallel If they concur, we call the concurrency point the radical center of the three circles

Adithya B., Brian L., William W., Daniel X. Lesson 10: Advanced Geometry August 2020 18 / 31 Radical Axes

2015 ARML Team # 10 ◦ −→ Let ABCD be a parallelogram with m∠A > 90 . Point E lies on DA such that BE ⊥ AD. The circumcircles of 4ABC and 4CDE intersect at points F and C. Given that AD = 35, DC = 48, and CF = 50, compute all possible values of AC.

B C

F DE A

Adithya B., Brian L., William W., Daniel X. Lesson 10: Advanced Geometry August 2020 19 / 31 2015 ARML Team # 10

CF is the radical axis of (ABC) and (CDE) Can we draw a third circle and use radical axis? B X C Want this third circle to intersect (ABC) and (CDE) in nice points M (ABE) works? Let X be foot from A to BC F AEBX rectangle =⇒ X ∈ (ABE) DE A EX = AB = CD, CX k AD =⇒ X ∈ (CDE) Radical axis theorem on (ABC), (CDE), (AEBX ) =⇒ CF , AB, EX concur at midpoint of AB Adithya B., Brian L., William W., Daniel X. Lesson 10: Advanced Geometry August 2020 20 / 31 2015 ARML Team # 10

Can now begin computation AM = BM = 24, BC = 35, CF = 50 B X C AM · BM = CM · FM 24 · 24 = CM(50 − CM) M CM = 18, 32

F Median length formula on 4ABC 2 2 2 2 DE 4CM = 2AC + 2BC − AB A 2 4CM2−2BC 2+AB2 AC = 2 BC = 35, AB = 48, CM = 18, 32 √ √ 5 23, 5 79

Adithya B., Brian L., William W., Daniel X. Lesson 10: Advanced Geometry August 2020 21 / 31 Radical Axes

2019 AIME I # 15 Let AB be a chord of a circle ω, and let P be a point on the chord AB. Circle ω1 passes through A and P and is internally tangent to ω. Circle ω2 passes through B and P and is internally tangent to ω. Circles ω1 and ω2 intersect at points P and Q. Line PQ intersects ω at X and Y . Assume 2 m that AP = 5, PB = 3, XY = 11, and PQ = n , where m and n are relatively prime positive integers. Find m + n.

AB P

Adithya B., Brian L., William W., Daniel X. Lesson 10: Advanced Geometry August 2020 22 / 31 2019 AIME I # 15

PQ radical axis of ω1, ω2 Handling the tangencies: draw X common tangent to ω, ω1 at A and ω, ω2 at B Let tangents meet at Z ZA = ZB by equal tangents to ω Q O 2 2 ZA power of Z wrt ω1, ZB power of Z wrt ω2 AB P Z on radical axis PQ Y Make another observation Is AQBZ cyclic? ∠ZQA = ∠PQA = ∠PAZ by tangency Z ∠PAZ = ∠ABZ by isosceles 4ZAB So ∠ZQA = ∠ZBA, AQBZ cyclic Adithya B., Brian L., William W., Daniel X. Lesson 10: Advanced Geometry August 2020 23 / 31 2019 AIME I # 15

X Let O be center of ω ◦ ∠OAZ = ∠OBZ = 90 (ABZ) has diameter OZ Z on (ABZ) =⇒ OQZ = 90◦ Q O ∠ So Q midpoint of chord XY XQ = YQ = 11 , PY = 11 − PQ AB P 2 2 PX · PY = PA · PB Y 11 11 2 + PQ 2 − PQ = 3 · 5 2 61 PQ = 4 65 Z

Adithya B., Brian L., William W., Daniel X. Lesson 10: Advanced Geometry August 2020 24 / 31 Radical Axis and PoP

2010 AIME II #15 In triangle ABC, AC = 13, BC = 14, and AB = 15. Points M and D lie on AC with AM = MC and ∠ABD = ∠DBC. Points N and E lie on AB with AN = NB and ∠ACE = ∠ECB. Let P be the point, other than A, of intersection of the circumcircles of 4AMN and 4ADE. Ray AP meets BQ m BC at Q. The ratio CQ can be written in the form n , where m and n are relatively prime positive integers. Find m − n.

D MN E

Q BC

Adithya B., Brian L., William W., Daniel X. Lesson 10: Advanced Geometry August 2020 25 / 31 Linearity of Power

We begin by proving a lemma: Linearity of Power

Given two circles ω1, ω2 in the plane, deﬁne

f (P) = Powω1 (P) − Powω2 (P)

Then, f is a linear function in P. Here, linear means that the function af (A)+bf (B) a b increments at a constant rate. So, a+b = f a+b A + a+b B

We use coordinates. Let P = (x, y), and ω1,ω2 have radii r1, r2 and centers (x1, y1), (x2, y2). What is f (P)? 2 2 2 2 2 2 f (P) = (x − x1) + (y − y1) − r1 − (x − x2) − (y − y2) + r2 . The quadratic terms cancel so f is linear!

Adithya B., Brian L., William W., Daniel X. Lesson 10: Advanced Geometry August 2020 26 / 31 2010 AIME II #15

How can we use linearity of power here?

f (P) = Pow(ADE)(P) − Pow(AMN)(P) Q lies on the radical axis so f (Q) = 0 A BQ How can we relate CQ with f (P)? f is linear, so QB · f (C) + QC · f (B) MN E = f (Q) D BC P BQ f (B) f (Q) = 0 so CQ = − f (C) Q BC Let’s compute f (B). What are BE, BN? AB 15 14 70 BN = 2 = 2 , BE = 13+14 · AB = 9 So, f (B) = BA · BE − BA · BN = 70 15 25 15 9 − 2 = 6

Adithya B., Brian L., William W., Daniel X. Lesson 10: Advanced Geometry August 2020 27 / 31 2010 AIME II #15

Similarly, CM = 13 , CD = 14 · AC = 182 A 2 14+15 29 182 13 f (C) = AC · (CD − CE) = 13 29 − 2 = 169 − 58 So, our ﬁnal answer is D MN E f (B) 25/6 725 P = = f (C) 169/58 507

Q BC 725 − 507 = 218

Adithya B., Brian L., William W., Daniel X. Lesson 10: Advanced Geometry August 2020 28 / 31 Radical Axis and PoP

IMO 2009/2 Let ABC be a triangle with circumcenter O. The points P and Q are interior points of the sides CA and AB respectively. Let K, L and M be the midpoints of the segments BP, CQ and PQ. respectively, and let Γ be the circle passing through K, L and M. Suppose that the line PQ is tangent to the circle Γ. Prove that OP = OQ.

Q M P

K L O

BC Adithya B., Brian L., William W., Daniel X. Lesson 10: Advanced Geometry August 2020 29 / 31 IMO 2009/2

What does OP = OQ mean? 2 2 Since Pow(ABC)(X ) = OX − R , it suﬃces to show that P, Q have the A same power with respect to (ABC)

Q How can we compute the powers of M P P, Q? We can use AQ · QB and PC · AP L K O Can we compute BQ, CP in terms of KM, ML, KL?

BC KM is the midline of PQB so QB = 2KM. Similarly, ML = 2PC Now, try to use the tangent condition to get something about the angles

Adithya B., Brian L., William W., Daniel X. Lesson 10: Advanced Geometry August 2020 30 / 31 IMO 2009/2

KM k AB, so ∠MLK = ∠KMQ = ∠AQP. Similarly, A ∠MKL = ∠APQ So, APQ ∼ MKL. What does this tell Q M P you about AQ, AP? AQ ML AP = MK L K O Putting it all together, AQ · QB ML 2MK BC = · = 1 AP · PC MK 2ML So, AQ · QB = AP · PC, as desired.

Adithya B., Brian L., William W., Daniel X. Lesson 10: Advanced Geometry August 2020 31 / 31