100 Geometry Problems: Solutions

Alvin Zou

April 26th, 2015

1. Let ra, rb, rc be the radii of the circles centered at A,B,C respectively. We then have the following system ofequations

BC = 9 = rb + rc

AX = ra = 6 + rb

AY = ra = 5 + rc.

Solving yields rb = 4, rc = 5, and AX = ra = 10 .

2. Denote ∠BAC = α and ∠BAD = φ, then φ is our unknown. By AC = AD we get ∠DAC = ∠CAD = α− φand thus ∠ACD = 180◦ − 2(α− φ). By that we can now get ∠ABC = 180◦ − (180◦ − 2(α− φ))−α = α− 2φ.And now using the condition we get

30◦ = ∠CAB − ∠ABC = α− (α− 2φ) = 2φ ⇔ φ = 15◦ .

3. We can draw the perpendicular from the center of the semicircle to CE (call this point X). Notice that theradius of the semicircle is 1. Since ABCD is a square, BC and EA are tangents as well. We haveCX = BC = 2 and EX = EA. We can Pythagorean Theorem △EDC with legs DE = 2− EA and DC = 2

with a hypotenuse CE = 2 + EA. Solving, we get EA = 12 and CE =

5

2.

4. The diagram:

A B

CD

M

6

6

3

It is given that ∠AMD ∼= ∠CMD. Since ∠AMD and ∠CDM are alternate interior angles and AB ‖ DC,∠AMD ∼= ∠CDM −→ ∠CMD ∼= ∠CDM . Use the Base Angle Theorem to show DC ∼= MC. We know thatABCD is a rectangle, so it follows that MC = 6. We notice that △BMC is a 30− 60− 90 triangle, and∠BMC = 30◦. If we let x be the measure of ∠AMD, then

2x+ 30 = 180

2x = 150

x = 75

1

5. From F draw a perpendicular to AD and let the foot be G. Then, we quickly notice that triangles ABE,EFG, GFD are congruent because of 2 equal sides and the right angle. (SsA) Let the side length of thesquare be a then by the Pythagorean Theorem we get

302 − (1

3a)2 = a2

900 =10

9a2

a2 = 810

6. First, we draw the diagram:

A

B C

D

E

Since ∠DEC = ∠DBC = 90, quadrilateral DEBC is cyclic, from which the result follows due to same

inscribed arcs. Q.E.D.

7. Let △ABC be an equilateral triangle with side length a, and point D be on line AC such thatCD = AC = a, and point A and D are distinct. ∠BCD = 120, and BC = CD, so △BCD is isosceles. Also,∠CBD = ∠CBD = 30, so ∠ABD = 90. Thus, △ABD is a 30-60-90 triangle, so b = a

√3, and

ba= a

√3

a=

√3 .

8. Drawing the diagram, we get:

C

A B

L

M

N

Since quadrilateral ABLM is cyclic, ∠MLB = 180− ∠A = 90◦ = ∠MLC. Thus ∠CML = 90− ∠C. We alsohave ∠C = ∠ANM since ALCN is cyclic. Then since ∠MAN = 180− ∠A = 90◦, we get

∠AMN = 90− ∠ANM = 90− ∠C = ∠CML.

Since the vertical angles are congruent, L,M,N are collinear. Q.E.D.

2

9. First, a diagram:

X

A

B C

I

YD

E

Let D and E be the feet of the altitudes from I to BC and AB respectively. Since IX = IA and ID = IE wehave that △IDX ∼= △IEA by HL congruence (both are clearly right triangles). It is also easy to show that

△IDX ∼= △IDY. So, AE = s− a = XD = 12XY, and thus XY = b+ c− a = 1400 + 1800− 2014 = 1186 .

10. Notice that, since ADBE is cyclic, we want to show that it is an isosceles trapezoid. Thus, it suffices to provethat arc DB = arc AE. However, we have arc AB = arc AC and arc AD = arc CE (since AD = CE). Thus,

we have arc DB = arc AB - arc AD = arc AC - arc CE = arc AE. Q.E.D.

11. Let P be the shape’s perimeter and A its area. Note that if we dilate the shape by a factor of k, its perimeterbecomes kP and its are becomes k2A. Thus, if we want our dilated shape to equiable, or kP = k2A, we

should dilate the shape by a factor of k = PA. Q.E.D.

12. By simple angle chasing, we notice that triangle AEB is similar to EFC. Let the side length of the square bea, EC = x and thus BE = a− x. Because of similarity, it is

a

4=

x

3⇔ x =

3

4a.

That yields BE = 14a. Using the Pythagorean Theorem in triangle AEB the result is

a2 + (1

4a) = 16 ⇔ 17

16a2 = 16 ⇔ a2 =

162

17

13. ∠ABN = ∠AMN since they are subtended from the same chord. Since ∠MXN = ∠MYN = 90◦, it followsthat M , X, Y , and N are concyclic. Then, since M and X form angles subtended from the same chord,∠Y XN = ∠YMN = ∠AMN = ∠ABN . Because BN is an extension of XN , and because

∠Y XN = ∠ABN , it follows that AB ‖ XY . Q.E.D.

14. Solution 1: Let M be the intersection point of the diagonals of the square. Then reflect A,B about M andwe’ll get C,D. Thus, the reflection of E about M which we call E′ will give us DE′ = 5 and E′C = 12, E′

must be F ! ThusEM = MF.

Now notice that (5, 12, 13) is a pythagorean triple which yields ∠BEA = 90◦. Obviously, ∠AMB = 90◦ and

AM = MB =√2·132 . Then AEBM is a cyclic quadrilateral and by Ptolemy we get

√2 · 132

· 5 +√2 · 132

· 12 = 13 ·ME ⇔ ME =

√2 · 172

.

Therefore EF = 2 ·ME =√2 · 17, so

EF 2 = (√2 · 17)2

= 2 · 289= 578 .

3

Solution 2:

A B

CD

E

F

Extend FC,EB,EA,FD as shown until the sides intersect. The two intersection points as well as point Eand F form a square, as the newly formed triangles are congruent to triangles AEB and CFD by ASA ∼=. Sothen EF 2 = (17

√2)2 = 578 .

15. Let O be △ABC’s circumcenter. Note that ∠BFE = 12∠BOE = 1

4∠BOC = 12∠A. Similarly, ∠AEF = 1

2∠Band ∠DFB = 1

2∠C. Since ∠DFE + ∠AEF = ∠DFB + ∠BFE + ∠AEF = 12 (∠A+ ∠B + ∠C) = 90◦, it

follows that DF ⊥ AE. Q.E.D.

16. Call P the projection of D onto AB, and let Dp be the projection of D onto plane ABC. Since AB = 3, wehave DP = 2 · 12

3 = 8. By definition, ∠DDpP = 90◦ so DDpP is a 30-60-90 triangle. Then DDp = 12DP = 4,

so

VABCD =[ABC] ·DDp

3= (15 · 4)/3 = 20 .

4

17. Here’s a colorful diagram:

P1

P2

P3

P4

A3

A1

O

Since the quadrilateral is orthodiagonal, we can write the statement we want to show as D2 = P1P22 + P3P

24 .

Let A1 and A3 be the antipodes of P1 and P3, respectively, and let O be the center of the circle. The claim isthat reflecting P1P2P3P4 over the diameter of the circle that is parallel to P1P3 results in the newquadrilateral A3P4A1P2. We can prove this by noting that P1OP3

∼= A3OA1, so P1 goes to A3 and P3 goes toA1 (and obviously P2 and P4 switch places). Now, after reflecting, we get that A1P2 = P3P4, so now we want

to show that D2 = P1P22 +A1P

22 which is true by the Pythagorean Theorem on △P1P2A1. Q.E.D.

18. Notice that the midpoint of AB, the centers of the two circles, and the center of the sphere form a rectangle.Thus, if we know the sides of the rectangle we can compute the distance from the center of the sphere to themidpoint of AB. Then, since this line is perpendicular to AB, we can then compute the radius of the sphere.First, we find the distances from the centers of the circles to midpoint of AB. They are

√542 − 212 and√

662 − 212. Thus, the distance from the center of the sphere to the midpoint of AB is√542 − 212 + 662 − 212. Finally, we apply pythagorean theorem once more to get the radius of the sphere to

be√542 − 212 + 662 − 212 + 212. Squaring, we get that R2 = 6831 .

19. Extend AB and CD to meet at X. We see that ∠AXD is a right angle, so XM and XN are medians oftriangle XBC and XAD, respectively. We can also see that X,M,N are collinear. Hence, XN = AN andXM = BM , so we easily find that MN = XN −XM = 1004− 500 = 504 .

20. First, we know that EF ||AB by symmetry. Now we wish to show that DF ||EF , as this would imply thatD,E, F collinear. Since line l is tangent to the circle, ∠DBC = ∠A, and therefore ∠BCD = 90−∠A. Noticethat quadrilateral BDCF is cyclic (opposite angles 90). Thus, we have ∠BCD = ∠DFB = 90− ∠A.However, we also have ∠EFB = ∠FBA = 90− ∠A by parallel lines, so lines DF and EF make the same

angle with BF . This implies that lines DF and EF are parallel, which implies D,E, F collinear. Q.E.D.

21. Let the center of the circle with radius 1 be A. Let the circle shaded grey have center B and let the circleshaded black that is adjacent to that grey circle be C. Let r be the radius that we want to find. Clearly,BC = 2r, AB = 4− r and AC = r + 1. We can also see that, from symmetry, ∠BAC = 60◦. From Law ofCosines, we have

4r2 = (r + 1)2 + (4− r)2 − 2(r + 1)(4− r) cos 60◦)

This reduces down to r2 + 9r − 13, and plugging into the quadratic formula:

r =−9 +

√133

2

We arrive at −9 + 133 + 2 = 126 .

5

22. ∠BAC inscribes arc BC. Assume that BD is tangent to the circle. Then, clearly, ∠DBC also inscribes arcBC, so ∠DBC = ∠BAC. However, if we let BD be not tangent, it will either increase or decrease the angle.But since we were given that the angles are equal, this is not possible. Thus, BD must be tangent to the

circle. Q.E.D.

23. Clearly, we see that CQMB and ANPC are cyclic quadrilaterals, implying that ∠CMB = ∠CQB and∠CNA = ∠CPA. Therefore, we can deduce that

∠MCN = 180◦ − (∠CMN + ∠MNC) = 180◦ − (90◦ −A/2 + 90◦ −B/2) = 45◦

24. Note that since AMCN is cyclic, we have ∠ANM = ∠ACB and ∠AMN = ∠ACD = ∠CAB so

△MAN ∼ △ABC by AA similarity. Q.E.D.

25. (a) We angle chase. Since AH is perpendicular to BC, ∠BAH = 90− ∠B. Now consider the circumcircle ofABC. We have ∠AOC = 2∠B (since it inscribes the same arc but goes through the center). Since triangle

OAC is isosceles (OA = OC), we have ∠OAC = (180− 2∠B)/2 = 90− ∠B. Q.E.D.

(b) Notice that ∠HAO = |∠A− ∠BAH − ∠OAC|. Substituting in our values from part (a), we get∠HAO = |∠A−(90−B)−(90−B)| = |∠A+2∠B−180| = |(∠A+∠B+∠C)+(∠B−∠C)−180| = |∠B−∠C|.Q.E.D.

26. We know ∠APB = ∠CPD, so ∠APB + ∠BPC = ∠CPD + ∠BPC, or ∠APC = ∠BPD. We also knowAP

PC=

PB

PD, so by SAS similarity △PAC ∼ △PBD. Q.E.D.

27. Note that ∠OYX = ∠OXY , so ∠Y ZO = ∠OZX. Let ∠Y ZO = α, then by Law of Cosines on triangles Y ZOand OZX we get 121 + 49− 2 · 11 · 7 cosα = 121 + 169− 2 · 11 · 13 cosα, which simplifies to cosα = 10

11 .

Plugging this back in we have OY 2 = 170− 140 = 30, so OY =√30 .

28. Let ∠AKF = ∠FKB = α and ∠AMH = ∠HMD = β. Also let FE ∩HG = X. Then because∠KAF = ∠DCB, we have ∠MFX = α+ C, so ∠MXF = 180− (α+ β + C). However we also know that2α+ C +D = 180 and that 2β + C +B = 180, so adding these we see that 2α+ 2β + 2C + (B +D) = 360.But we also know that B +D = 180 because quadrilateral ABCD is cyclic, so we get that α+ β + C = 90.Which means ∠MXF = 180− 90 = 90. From this we conclude that triangles EMX and FMX arecongruent, and triangles HKX and GKX are congruent. Thus EX = XF , HX = XG, and HG ⊥ FE, so

EGFH is a rhombus. Q.E.D.

29. The diagram:

5 7

7

A

B

CM

P

H

With some simple calculations we see that AH = 5, HM = 2, and MC = MP = 7. Also BH = 12. So by thePythagorean theorem we have BM = 2

√37. Let the foot of the altitude from P to AC be D. Then

MPMB

= PDBH

. Or 72√37

= PD12 . Solving we have PD = 42√

37Thus the area of △PAC is

12 · 14 · 42√

37= 294√

37= 294

√37

37 . Therefore p+ q + r = 294 + 37 + 37 = 368

6

30. The complicated diagram:

OA

B

C

P

D

E

M

E ′

D′

Let O be the circumcentre of △ABC and M be the midpoint of BC. Then its clear that PM ⊥ BC. Since∠BDP = 90◦ = 190− ∠PMB, quadrilateral PMBD is cyclic. So we have ∠MDP = ∠MBP . Also∠ADM = 90◦ − ∠MDP = 90◦ − ∠MDP . As BP is tangent to the circumcircle of △ABC we must have

∠MBP =BC⌢

2= ∠A. Therefore ∠ADM = 90◦ − ∠MDP = 90◦ − ∠A. Let D′ = DM ∩AE then ADD′ is

right. So DD′ is the D-altitude of △ADE. Similarly EE′ is the E-altitude of △ADE which means that M is

the orthocentre of △ADE. Q.E.D.

31. Note that HHABHC and HHACHB are cyclic. Then ∠HCHAA = ∠HCBHB = 90− ∠BAC and∠AHAHB = ∠ACHC = 90− ∠BAC. Therefore, H lies on the angle bisector of ∠HCHAHB . The same

argument can be repeated for the angles HCHBHA and HAHCHB to show that H is the incenter. Q.E.D.

32. Let Y and Z be the intersections of AC with the circle, w.l.o.g. CY < CZ. Then clearly CY = 97− 86 = 11and CZ = 97 + 86 = 183. Now, CB · CX = CY · CZ = 11 · 183 = 3 · 11 · 61. If BX and CX have integerlengths, this also holds for BC and since BC ≥ CX ≥ 11 (the latter holds by the Triangle Inequality) theonly possible values are CX = 11, V C = 183 and CX = 33, BC = 61. But in the first case, ACX would bedegenerated which implies X ∈ AC, only possible if X = C but this contradicts AX = 86 6= 97 = AC. HenceCX = 33 and BC = 61 is the only possible solution.

7

33. Let ∠BAC = A, ∠ABC = B, and ∠BCA = C. We know that ∠BOC = 2A, so the measure of BNC = 4A.

Then A =4A−PQ

⌢

2 , which means PQ⌢

= 2A, and ∠PNQ = A. Because the center of circle τ also lies on theperpendicular bisector of BC, we know that ∠NOB = ∠NOC = A. Thus ∠NQC = ∠NOC = A, so

∠NQA = 180−A, and ∠NPA = 180−A. From this we conclude APNQ is a parallelogram. Q.E.D.

34. First we’re going to have to visualize this.

Drawing the center of the square to each of the 8 vertices we see that we are looking for the area of 4 sectorsof a circle with central angle 45◦, 4 right triangles with length and width equal to the half the side of the

square, and 4 tiny external right triangles. The sectors form a semicircle with radius√22 , so the area of that is

π4 . The 4 big right triangles have a total area of 4 ·

1

2· 12

2 = 12 . Then the small external triangles are isosceles

right triangles, and have height√2−12 , so their area is

4 ·

(√2−12

)2

2=

3− 2√2

2

So the total area isπ

4+ 2−

√2 .

35. Take O the midpoint of the diameter. thus, M,O,P, S are concyclic, so ∠MPS = ∠MOS; since ST has

constant length, we are done. Q.E.D.

36. We have ∠BAC = ∠BDA1 since they intercept the same arc in the circumcircle of triangle ADB and∠BCA = ∠BDC1 thus∠A1DC1 = ∠A1DB+∠BDC1 = ∠ACB+∠BAC = 180◦−∠ABC = 180◦−∠A1BC1 hence the quadrilateral

BA1DC1 is insciptible and ∠BAC = ∠BDA1 = ∠BC1A1 which proves that A1C1//AC. Q.E.D.

37. From simple similarity we find AE = 485 , EC = 27

5 , and DE = 365 . Now notice that quadrilateral ABDF is

cyclic, so ∠AFE = ∠ABC. Thus △AFE ∼ △ABD, and AEEF

= ADDB

. It is well known that for a 13− 14− 15

triangle AD = 12 and BD = 5. So we have 125 =

48

5

EF, and EF = 4. Therefore DF = 36

5 − 4 = 165 , so

m+ n = 21 .

8

38. Let BX = m, CX = n, and AX = d. Furthermore let the circumcircles of triangles AXB and AXC haveradii R1 and R2 respectively. We know that the area of △ABC is, by Heron’s, 6

√6. So the area of △ABX is

m·6√6

7 . Similarly the area of △AXC equals n·6√6

7 . Then

R1 =5md

4 · m·6√6

7

=35d

24√6

From the same logic we arrive at

R2 =7d√6

Thus we see that the area of the circumcircles is minimized when d2 is minimized and hence when d isminimized. To minimize d, we have to make it the length of the altitude from A to BC. Setting 7d

2 = 6√6 we

find d = 12√6

7 , and that BX =19

7.

39. Since ∠KBL = ∠KDL = 90, KDLB is cyclic. Therefore, ∠BDL = ∠BKL But also,∠BAF = ∠BAC = ∠BDC = ∠BDL = ∠BKL = ∠BKF Therefore, BAKF is cyclic. Then since

∠KAB = 90, ∠KFB = 90 so BF ⊥ KL. Q.E.D.

40. Of course, a diagram would help.

A B

C

D

E

Let AD = x and BE = y. Then, we have DC = 2x and EC = 3y. By power of a point, we have thatDC(AC) = EC(BE) =⇒ (2x)(3x) = (3y)(4y) =⇒ x2 = 2y2. Now, since AB is the diameter, ∠ADB and∠BEA are both right angles by Thales Theorem. By Pythagorean Theorem, we have that BD =

√900− x2.

Thus, the problem reduces to finding [ABC] = 12 (3x)(

√900− x2). Note, that by Pythagorean Theorem, we

also have BD =√(4y)2 − (2x)2. Using Pythag. yet again on △ABD gives:

(16y2 − 4x2) + x2 = 900

=⇒ 16y2 − 3x2 = 900

=⇒ 16y2 − 3(2y2) = 900

=⇒ 10y2 = 900

=⇒ y = 3√10

Thus, x =√2 · y =

√2 · 3

√10 = 6

√5. Plugging in our value of x, we finally get:

[ABC] =1

2(18

√5)(

√900− 180) = (9

√5)(12

√5) = 540 .

9

41. The diagram:

A B

CD

A′ B′

C ′D′

PP ′

P ′′

We seek the value of ∠DAP + ∠BCP . We begin by constructing congruent rectangles BCC ′B′ to the rightand ADD′A′ to the left of rectangle ABCD. Denote the points P ′ and P ′′ as the points in rectanglesADD′A′ and BCC ′B′, respectively, that satisfy

∠A′P ′D′ + ∠AP ′D = 180◦

and

∠BP ′′C + ∠B′P ′′C ′ = 180◦,

respectively. In other words, P ′ and P ′′ are the corresponding points to P in rectangles ADD′A′ andBCC ′B′, respectively. Since P ′ and P ′′ are just translations of P in the horizontal direction, the line P ′P ′′ isparallel to AB and is thus perpendicular to AD and BC. Now, note that by the ∠APD + ∠BPC = 180◦

condition, we have that APDP ′ and PBP ′′C are cyclic quadrilaterals. Then, by inscribed arcs, we have that

∠BCP = ∠BP ′′P = ∠APP ′

the last equality following from congruence. Thus, we have that

∠DAP + ∠BCP = ∠DAP + ∠APP ′ = 180◦ − 90◦ = 90◦

42. Let the circumcircles of triangles APR and BPQ intersect at a point X. Because quadrilaterals APXR andBPXQ are inscribed in circles we know ∠PXR = 180− ∠A and ∠PXQ = 180− ∠B. Therefore∠QXR = 360− (180− ∠A+ 180− ∠B) = ∠A+ ∠B = 180− ∠C. Thus quadrilateral QXRC is also cyclic.

Q.E.D.

43. First, find that ∠R = 45◦. Draw ABCDEF . Now draw △PQR around ABCDEF such that Q is adjacent toC and D. The height of ABCDEF is

√3, so the length of base QR is 2 +

√3. Let the equation of RP be

y = x. Then, the equation of PQ is y = −√3(x− (2 +

√3)) → y = −x

√3 + 2

√3 + 3. Solving the two

equations gives y = x =√3+32 . The area of △PQR is 1

2 ∗ (2 +√3) ∗

√3+32 = 5

√3+94 .

a+ b+ c+ d = 9 + 5 + 3 + 4 = 21

44. It suffices to show that MB = MI = MC. First of all, since M is the midpoint of arc BC, it obviously lies onthe perpendicular bisector of BC so MB = MC. Now note that ∠IMB = ∠C and ∠BIM = 1

2∠A+ 12∠B

and since

∠MBI = 180◦ − (∠C +1

2∠A+

1

2∠B) =

1

2∠A+

1

2∠B = ∠BIM

so MB = MI = MC as desired. Q.E.D.

45. Let D be the point on AB such that TD ⊥ AB. Then △ATD is 30− 60− 90 and △TDB is 45− 45− 90, sowe have TD = 12 and AD = 12

√3, so BD = 12 and AB = 12(

√3+ 1). △ACT is isosceles so AC = AT = 24.

Using the area of a triangle with sine we have the area of ABC is 12 · 24 · 12(

√3 + 1) · sin 60 = 216 + 72

√3, so

a+ b+ c = 291 .

10

46. As always, we start with a diagram:

A

B CD

E

O

M

Note that OBEM and ABED are both cyclic quadrilaterals (with diameters BO and BA respectively).Furthermore, since O is the circumcenter, △AOB is isosceles =⇒ ∠OAB = ∠OBA. Therefore

∠MED = ∠AED − ∠OEM

= ∠ABD − ∠OBM

= ∠ABO = ∠BAE = ∠BDE,

so △MED is M -isosceles and ME = MD as desired. Q.E.D.

47. Let O be the circumcenter of △ABC. Note that ADOF and ADFN are both cyclic quadrilaterals, so ND

and AO are both diameters of the circumcircle of △ADF . Q.E.D.

48. Let N be the intersection of the perpendicular bisector of AC and AC and O be the intersection of NM andBP . We have MN//AB and BO//AN so ANOB is a parallelgram, therefore BO = AN = NC andBO//NC imply that BNCO is also a parallelogram. M is the middle of ON and since NPO = 90◦ then Mis the circumcenter of NPO and MP = MO. Finally, since QB//MO then

∠QBP = ∠MOP = ∠MPO = ∠BQP so QB = QP . Q.E.D.

49. The if case (where ABC is equilateral) is easy; all one needs to do is to observe the many symmetries anequilateral triangle possesses. It suffices to show the converse: that if △UVW = △XY Z then ABC isequilateral. And to do this, we angle chase. Note that since the two triangles have the same circumradius(since all six points U, V,W,X, Y, Z are concyclic), the problem is reduced to the case where the two trianglesare similar. Next, note that

∠UWV =1

2∠UIV =

1

2(180◦ − ∠ZAI − ∠ZBI)

and that∠XZY = ∠XIC = 90◦ − ∠ICX.

If the two angles ∠UWV and ∠XZY are congruent, then we must have

12 (180

◦ − ∠ZAI − ∠ZBI) = 90◦ − ∠ICX

2∠ICX = ∠ZAI + ∠ZBI

2∠C = ∠A+ ∠B = 180◦ − ∠C

∠C = 60◦.

Applying this cyclically, we see that △ABC is equilateral. Q.E.D.

11

50. Note that DB = DI and EC = EI so the perimeter of △ADE is 43. Thus the ratio of similitude between△ADE and △ABC is 43 : 63. Hence

DE =43

63· 20 =

860

63=⇒ m+ n = 923

51. ∠BPQ = ∠BPA = ∠BCA so AP is the angle bisector of ∠BPC Let QD and QE be the perpendiculars

from Q to PB and PC respectively. QD = QE =√32 PQ. Area of

△PCB = QD(PB+PC)2 =

√34 PQ(PB + PC) Also, Area of △PCB = PB·PC·sin 120

2 =√34 PB · PC. Equating

the two equations for the area we get the desired result. Q.E.D.

52. The diagram:

A

B C

MA1

A2

C1

C2

P

Notice that triangles ABM and BMC are isosceles, so MA1 ‖ BC and MC1 ‖ AB. Also, A1C1 ‖ ACbecause A1 and C1 are midpoints. Triangles AA1A2 and CC1C2 are isosceles because of equal tangents, soA1A2 is an external angle bisector of ∠BA1C1 and similarly C1C2 is an external angle bisector of ∠BC1A1.

So P is the B-excenter of △BA1C1, and therefore it lies on the angle bisector of ∠ABC. Q.E.D.

53. Suppose that X is near to B than C. ∠KBO = ∠KXO = 90◦ thus KBXO is cyclic quadrilateral and∠OXL = ∠OCL = 90◦ thus OCLC is cyclic. From the previous result∠XKO = ∠XBO = ∠OCX = ∠OLX because OB = OC so OK = OL and since X is the projection of O

into KL then X is the middle of KL. Q.E.D.

54. Denote by B1, B2 the orthogonal projections of B on the lines IA, and IC, respectively, and similarly, let D1,D2 be the projections of D on the same lines IA, and IC, respectively. We consider the following preliminaryresult:

Lemma. Let ABC be a triangle, and denote by M the midpoint of segment BC. If X, Y are the orthogonalprojections of the vertices B, C on the internal angle bisector of angle BAC, then MX = MY = |b− c|/2.Proof. We shall resume to proving that MX = |b− c|/2. For this, let B′ be the intersection of the line BXwith the sideline CA. Since the triangle ABB′ is isosceles, the length of segment CB′ is |b− c|. On otherhand, X, M are the midpoints of segments BB′, and BC, respectively. Thus XM = CB′/2 = |b− c|/2. Thisproves our Lemma.

Returning to the problem, let T be the midpoint of the diagonal BD. According to the Lemma, applied forthe triangle BAD, we have that TB1 = TD1 = |AB −AD|/2. Similarly, according to the same Lemma, thistime applied in triangle BCD, we have that TB2 = TD2 = |BC − CD|/2. On the other hand, thequadrilateral ABCD beeing circumscribed, AB + CD = AD +BC (Pithot’s theorem), and hereby, weconclude that TB1 = TB2 = TD1 = TD2, i.e. the projections of B, and D on the lines IA, and IC lie on a

single circle (with center T , the midpoint of diagonal BD). Q.E.D.

12

55. Let P be an arbitrary point in the set of points such that [APB] = [APD], and let PB and PD be the feet ofthe perpendiculars from P to AB and AD respectively. It’s not hard to prove that PPB : PPD = 3 : 2.Furthermore ∠PDPPB is constant (and equal to 180◦ − ∠DAB), so all such triangles △PDPPB arehomothetic to each other. In particular, this proves that the locus of points P is a line ℓ passing throughpoint A. Our goal is to find CE, where E is the projection of C onto ℓ. (This definition of E clearlyminimizes said distance.)

Let X = ℓ ∩ CD, and define XB and XD as before. Furthermore, let H be the foot of the altitude from A toCD. Standard trapezoid computations (i.e. dropping the altitude from A and doing Pythagorean Theorem)yield AH = XXB = 12, so XXD = 8. Now since △AHD ∼ △XXDD we obtain

XD

XXD

=AD

AH=

5

4=⇒ XD =

5

4· 8 = 10.

This implies AXB = HX = XD −HD = 1, so AX =√122 + 12 =

√145.

Finally, since XXB ⊥ CD angles AXXB and CXE are supplementary, so △AXBX ∼ △XEC. SinceXC = CD −XD = 18, by similar triangle ratios

CE

CX=

XXB

XA=

12√145

=⇒ EC =12 · 18√

145=

216√145

.

56. Denote the intersection of MN and BC as X. Notice that proving that the projections of D and C onto MNimplies that DX = XC. This is also equivalent to showing that triangles DMN and CMN have the samearea (same base same altitude).

Now for an ugly area bash: Let the area of the quadrilateral be 1, to make the calculations simpler. Then

[BCM ] = [ABN ] = 12 . Notice that [BMN ]

[ABN ] = [BMN ][BCM ] . This simplifies to BM

AB= BN

BC. Let this ratio equal k.

Now we can compute [BMN ] to be [BMC] ∗ k = k2 .

By complimentary counting, we have that[DMN ] = [ABCD]− [ADM ]− [CDN ]− [BMN ] = 1− (1− k) ∗ 1

2 − (1− k) ∗ 12 − k

2 = k2 . Thus, we have

[BMN ] = [DMN ], which implies that X is the midpoint of BC. QED

57. Denote the intersection of HM with the circumcircle to be H ′. Then ∆MHC is similar to ∆BMH ′ (i.e. H ′

is the reflection of H across M), so ∠HCB = 90−B = ∠CBH ′, and ∠HCB = 90− C = ∠BCH ′. Hence A

is antipodal to H ′, and the result follows. Q.E.D.

58. Obviously I = BN ∩ CM is the incenter, and let X = MN ∩AB. Since MNCB is cyclic, ∠MNB = c/2,and since ∠XBN = b/2, ∠NXB = 180− b/2− c/2. But, since I is the incenter, ∠BIC = 180− b/2− c/2and so ∠MIB = b/2 + c/2, so MXBI is cyclic and ∠IXB = 90, implying that X is the incircle tangency

point to AB. We can replicate the argument for AC. Q.E.D.

59. Let C be the reflection of B wrt AN . Then AC = AB and AC = AM from the properties of reflection. SoAB = AM , from there we let AM = y and drop a perpendicular from A meeting MN at D. DB = DM = 2.Using pythagoras in △AND, △AMD and △ANM , we get: x2 + y2 = 100 x2 − 64 = y2 − 4 so x2 = 80

60. Let M = Y C ∩AH and N = XA ∩HC. Notice that ∠AHI = 90◦ − ∠HAI = 90◦ − ∠ZCM = ∠CY J , so△Y JC ∼ △HIA ∼ △YMH. Similarly, △HJC ∼ △XIA ∼ △XNH.

Notice that IX = XN ·AIHN

= AI·JHJC

and JY = YM ·JCHM

= HI·JCAI

. Therefore, we have:

IX · JY = JH ·HI = ZI · JZ ⇒ IX

IH=

JH

JY

and we therefore see that Y JZ ∼ ZIX, so ∠Y ZJ + ∠IZX = 90◦, so X,Y, Z are indeed collinear. Q.E.D.

61. Notice that ∠O2O1A is half of arc AD with respect to circle (ABD). Hence, ∠AO1O2 = ∠ABD = ∠ABC

and similarly we arrive at ∠AO2O1 = ∠ACB, so from AA similarity, we are done. Q.E.D.

13

62. First, we draw a diagram:

A

B CD

E

M

Let M be the midpoint of BC. As ABEC is a cyclic quad, ∠EBC = ∠EAC = ∠EAB = ∠ECB, so △EBCis isosceles. (This is well-known.)

Now let DE = x. By Angle Bisector Theorem, we can easily compute BD = 18 and DC = 22. Since∠EBC = ∠EAB, △DBE ∼ △BAE, so

BD

DE=

AB

BE=⇒ BE

DE=

AB

BD=

36

18= 2.

Therefore CE = BE = 2x. Now by Pythagorean Theorem,

DE2 = DM2 +ME2 = DM2 + EC2 −MC2

=⇒ x2 = 22 + (2x)2 − 202.

Solving this gives x2 = 132 .

63. Note that since ∠BHC = ∠BIC = 120, quadrilateral BICH is cyclic. This implies that∠CHI = ∠CBI = C

2 . We then have that ∠AHI + ∠CHI + ∠BIC + ∠AHB = 360, so

∠AHI = 60 +B − C2 = 3B

2 . Q.E.D.

64. Let r and s meet the line CD at E and F , respectively. Angle chasing easily shows that AED = BFC. Notethat triangles ABC and ADE are similar, and also are the triangles ABD and CBF . This yieldsDEBC

= ADAB

=⇒ DE = AD.BCAB

CFAD

= BCAB

=⇒ CF = AD.BCAB

That is, DE = CF , so r and s are symmetrical relative to the perpendicular bisector of CD. The conclusion

follows. Q.E.D.

65. Draw triangle ABC such that AB=425, AC=450, and BC=510. Let the line through P parallel to AB haveendpoints D and E, where D is on AC and E is on BC. Let the line through P parallel to AC have endpointsF and G, where F is on AB and G is on BC. Let the line through P parallel to BC have endpoints H and J,where H is on AB and J is on AC.

Note that AFPD, HPEB, and PJCG are all parallelograms (you can angle-chase to see this). Therefored = DP + PE = AF +HB, so HF = AB − (AF +HB) = 425− d. Analogously, DJ = 450− d. LetHP = x, so PJ = d− x. We can see that FAP and DPJ are similar to ABC (more angle-chasing can provethis), so FA

AP= 425−d

x= AB

BC= 425

450 and DJPJ

= 450−dd−x

= ACBC

= 450510 . Cross-multiplying gives the equations

425 ∗ 510 = 510d+ 425x

450 ∗ 510 = 960d− 450x

14

Dividing both sides of the first equation by 425 and both sides of the second equation by 450 gives

510 = 510425d− x

510 = 960450d+ x

You can now add these two equations together to get

1020 = 510∗450+960∗425425∗450 ∗ d.

Now before you start multiplying stuff, you can see that a bunch of 5s cancel out from the numerator and thedenominator. Then you see that a couple 2s and 3s cancel out as well. After canceling stuff out, you get theequation

1020 = 18+3215 d = 10

3 d.

Solving this is simple, and yields d = 306 .

66. Using mid-segments, we can find that P1P2 and P4P3 are parallel to AC and that P1P4 and P2P3 are parallelto BD, so P1P2P3P4 is a parallelogram. Therefore P1P3 and P2P4 intersect at the midpoint of P1P3. Againusing midsegments, P1P6 and P5P3 are parallel to AD and P1P5 and P6P3 are parallel to BC so P1P5P6P3 isa parallelogram, And we get P1P3 and P5P6 also intersect at the midpoint of P1P3, so the three lines are

concurrent. Q.E.D.

67. In equilateral △ABC, WLOG let P , the point on the incircle, be closest to A, and suppose d(P,AB) = 1,d(P,AC) = 4. Denote by D and E the tangency points of the incircle of the triangle with AB and ACrespectively. Furthermore, let the line parallel to BC passing through P meet AB and AC at M and Nrespectively. Then it is easy to see that △AMN and △ADE are both equilateral, so it is easy to compute

that PM =2√3

3and PN =

8√3

3. Now the fact that P lies on the incircle implies that ∠DPE = 120◦ by

some simple angle chasing. Furthermore, it is obvious that ∠DMN = ∠MNE = 120◦ and that∠MPD = ∠PDE, ∠NPE = ∠PED, hence △DMP ∼ △DPE ∼ △PNE. Let DM = EN = x. Then bysimilarity ratios

x

2√3/3

=8√3/3

x=⇒ x =

4√3

3,

so AD = NM +MD =10√3

3+

4√3

3=

14√3

3. The side length of the triangle is twice this, or

28√3

3, and the

requested answer is 28 + 3 + 3 = 34 .

68. Let X = CQ ∩ PB. X is the A-excenter of △ABC. Hence, A, I,X are collinear. Through some anglechasing, we can prove that IC ⊥ CX and that IB ⊥ BX. Hence I, B,X,C are cyclic, so∠IXQ = ∠IXC = ∠IBC = ∠ABC

2 . Also R,P,X,Q are cyclic, and again through some angle chasing we can

find that ∠RXQ = ∠RPQ = ∠ABC2 . Therefore, R, I,X are collinear, and so A,R, I,X are collinear. Chasing

some more angles, we get that AC = AQ. Now draw a perpendicular from A to D on QC. Since AC = AQ,D is the midpoint of QC, and since AD ‖ IC ‖ RQ, A must be the midpoint of IR (known property of

trapezoids) and therefore AR = AI. Q.E.D.

69. Since △OPR is a right triangle with hypotense OQ, r2 = 12OQ, so it suffices to show that

PS

SR=

OP12OQ

. Let

X denote the projection of O to the segment PS. Remark that since OP = OS and △OPS ∼ △RQS, wehave ∠OSP = ∠RSQ = ∠RQS. Furthermore, since PR is tangent to C1 and since PQRO is cyclic, we have

∠ROQ = ∠SPR = 12∠SOP = ∠SOX.

Therefore △SXO ∼ △QRO and

QR

QO=

SX

SO=

12PS

PO=⇒ OP

12OQ

=PS

QR=

PS

SR,

as desired. Q.E.D.

15

70. Let ∠PAD = x, then ∠DAB = 2x, ∠ADC = 180− 2x, and ∠ADP = 90− x. Hence ∠APD = 90. Now letM and N be the midpoints of BC and DA, respectively. It follows that M is the circumcenter of △APD.Then ∠PMD = 2 · ∠MAP = 2x. So MP//BC//AB. But that means that MP is the mid line of ABCD.But by symmetry NQ is the mid line as well. So M , P , Q, and D form a line.

Notice that MP is equal to the circumradius of △APD, but since it is a right triangle, that value is AD2 .

Now it follows that[APD]

[ABCD]=

12 · 1

2 ·AD · d(AB,CD)

MN · d(AB,CD)=

AD

2AB + 2CD

We have a symmetrical result for △BCQ. We can subtract those areas from the total area of the trapezoid toget the desired area. Hence

[APDCQB]

[ABCD]=

2AB + 2CD −BC −AD

2AB + 2CD=

2 · 11 + 2 · 19− 5− 7

2 · 11 + 2 · 19 =4

5

Now construct B′ on segment CD such that AB = B′C. Since AB//B′C as well, ABCB′ is a parallelogram.Hence AB′ = 5 and DB′ = 19− 11 = 8. We can use heron’s formula to get [AB′D] = 10

√3. Now

[ABCD]

[AB′D]=

AB +DC

A′D=

11 + 19

8=

15

4

We can finish the problem by multiplying those ratios:

[APDCQB] = 10√3 · 4

5· 154

= 30√3

71. Consider the homothety that takes w to w1. This homothety takes X to A and Y to D, so AD||XY byhomothety. Similarly, we have BC||XY . Now consider a reflection of A and B over the line that goes throughthe centers of the two circles. Since this line is perpendicular to the radical axis of w1 and w2, or XY , it isalso perpendicular to AD and BC. Thus, it takes A to a point on AD, but since this point is also on w1, itmust be D. Similarly, this reflection takes B to C. Finally, since reflection preserves, lengths, we have

AD = BC. Q.E.D.

72. Note that showing < BKC =< CDB is the same as showing KBCD cyclic. Extend AC and let a point D′

be on AC with CD = CD′. Now the metric condition given becomes MA(MC + CD′) = MB(MD) soMA(MD′) = MB(MD). This implies that quadrilateral ABD′D is cyclic. Now let< CDD′ =< CD′D =< AD′D =< ABD =< 1 (by cyclic quads). Since < DCA is an exterior angle, it is thesum of < CDD′ and < CD′D, so < DCA = 2 < 1. Since CK is the angle bisector of< DCA,< DCK =< KCA =< 1. Now let X be the intersection of CK and BD. Since < XCA =< XBA,quadrilateral BAXC is cyclic as well. Now this question is just a simple angle chase. Let< CAB =< CXB =< 2. Then < CKA =< CAB− < KCA =< 2− < 1 (by exterior angles).< CDM =< CXM− < DCX =< 2− < 1 (by exterior angles again). Thus, < CKA =< CDM , so

quadrilateral KBCD is cyclic. Finally, this implies that < BKC =< CDB. Q.E.D.

73. First, a diagram:

A

B C

D

A′

B′

C′

X

Y

16

By simple angle chasing, ∠AB′C ′ = ∠AA′C ′ = ∠ABC. Using traditional chords-intercepting-arcs formulaewe can thus write 1

2 AC = 12 (AX + Y C) =⇒ AX = AY . Thus △AXY is isosceles and DA bisects ∠XDY .

It is well known that △AB′A′ ∼ △AA′C, so A′A2 = AB′ ·AC. In addition, ∠AYX = ∠AXY = ∠ACY , so△AY B′ ∼ △ACY and AY 2 = AB′ ·AC. Hence AY = AA′ = AX, so A is the circumcenter of △XA′Y and

by Fact 5 we win. Q.E.D.

74. Let O1, O2, O3 be the circumcenters of triangles AQR,BRP,CPQ respectively. Now from Miquel’s theorem,the three circumcircles concur at some point M .∠O3O1O2 = ∠O3O1M + ∠MO1O2 = 90− ∠O1MQ+ 90− ∠O1MP = 180− ∠PMQ = ∠CAB. From

symmetry, we can determine the other angles are equal as well, so the triangles are similar. Q.E.D.

75. First, a diagram:

A

B C

O

A′ H ′

X

Let X be the foot of the altitude from A to BC, and denote by H ′ the second intersection point of AX withthe circumcircle of △ABC. By Problem 25, ∠BAA′ = ∠CAH ′, so BA′ = CH ′ and A′H ′ ‖ BC. This meansthat the distances to BC from A′ and H ′ are equal; in other words, [BA′C] = [BH ′C]. The area of △BH ′Cis much easier to compute. By Law of Cosines, we have

cos∠ACB =AC2 +BC2 −AB2

2 ·AC ·BC=

52 + 82 − (√41)2

2 · 5 · 8 =3

5,

so XC = 3. Thus BX = 5 and by Pythagorean Theorem AX = 4. Finally, Power of a Point gives

BX · CX = AX ·XH ′ =⇒ XH ′ =BX · CX

AX=

15

4,

so the area of △BH ′C is 12 (BC)(XH ′) = 15 .

76. Note that the center of P lies on the ray BI where I is the incenter of △ABC, and similarly, the center of Q

lies on the ray CI. Now note that r = [ABC]s

= 24·7·16128 = 21. Now since the inradius is larger than the radius

of P , point P lies on the interior of segment BI. Now let R,S, T denote the projections from P, I,Qrespectively onto BC. Then note that △BPR ∼ △BIS ∼ △CQT . Thus from the first similarity, we have

PR

RB=

IS

SB=⇒ 16

RB=

21

28=⇒ RB =

64

3

This means that SR = 28−RB = 203 . Denoting the radius of circle Q as r, from the second similarity we

receive that TC = 4r3 so TS = 28− TC = 84−4r

3 . Now consider trapezoid TQPR. Let Q′ denote the

17

projection from Q onto PR, and then consider right triangle QQ′P . We have QP = 16 + r and PQ′ = 16− ras well as QQ′ = TR = 104−4r

3 . Then pythagorean theorem yields

(16− r)2 +

(104− 4r

3

)2

= (16 + r)2 =⇒ r = 44− 6√35

so our final answer is 254 .

77. From problem 26, △PAC ∼ △PBD. PM and PN are corresponding medians in these similar triangles sothey follow the same ratio: PM

PN= PA

PB. Also because they are corresponding medians, ∠MPA = ∠NPB. So

∠NPM = ∠DPA−∠MPA−∠DPN = ∠DPA−∠DPB = ∠BPA. Therefore △PAB ∼ △PMN ∼ △PCD

as desired. Q.E.D.

78. Let BC = 2a and let AC = 2b. Then, we can establish the following: 2(4b2 + 4a2)− 576 = 4(729)

2(242 + 4b2)− 4a2 = 4(324) Solving this system gives us a =√279 = 3

√31 b =

√3152 = 3

2

√35

Also, notice that [AFB] = 2[FEB] = 2[AFE]. Because [FEB] = 6FE sin∠FEB = 6FE sin∠BEC, we needto find FE and sin∠BEC. The former is quite easy; indeed, with power of a point, we obtain 27FE = 144FE = 16

3

Now, we are looking for 32 sin∠BEC. Finding sin∠BEC is slightly more tedious; however, since we knowBC = 2a = 6

√31, using the law of cosines to find cos∠BEC will directly allow us to find our desired value.

We then proceed as follows: 144 + 729− 648 cos∠BEC = 1116 cos∠BEC = 243648 = 3

8 Thus, sin∠BEC =√558 ,

and then we plug it into 32 sinBEC. Our final area is thus 4√55, and our final answer is 55 + 4 = 59 .

79. Let the incircle of △ADC touch DC at H, and let F ′ be the A-excenter of △ADC. Since △ADC = △BCD,E and H are reflections over the midpoint M of DC, hence E is the point of contact of the A-excircle of△ADC with DC =⇒ EF ′ ⊥ CD. Since both F and F ′ lie on AX, we have F ′ ≡ F . Now letting X be anypoint on the extension of BC past C, we have

∠AGF = ∠FCX = ∠FCG = ∠FAG,

so △AFG is isosceles as desired. Q.E.D.

80. Extend NM such that it intersects AB at M ′. This is simply an extension of the radical axis. Notice that thepower of point M ′ with respect to G2 is M ′B2, and similarly, the power of M ′ with respect to G1 is M ′A2.Because every point on the radical axis has the same power with respect to both circles, we can state thatM ′B2 = M ′A2 M ′B = M ′A.

Notice that there exists a homothety centered at N which takes P to A, Q to B, and M to M ′. Since M ′ isthe midpoint of AB, M must be the midpoint of PQ. Now draw EE′ such that E′ lies on CD and EE′ isperpendicular to CD. Then, △EE′C is similar to △AFC, and △EE′D is similar to △BGD. Let CF = x,and let FE′ = xk. Also, let FA = h, and by similarity, HE = hk. Additionally, let GD = y, and therefore,AB = x+ y. Since AH = FE′ = xk, HB = x(1− k) + y. Then, by more similarity, we have EH

HB= EE′

E′Dhk

x(1−k)+y= h(k+1)

2y 2ky = (k + 1)(x− xk + y) xk2 + yk − (x+ y) = 0,

Which is a quadratic in k. Solving for k via the quadratic formula gives us k = −1− yxand k = 1. However,

it is fairly obvious that k must be positive, and hence, we take k = 1. Hence, FE′ = FM = x, and M = E′.Because EE′ was perpendicular to CD, EM must be perpendicular to CD. Now, because ∠EMC = 90 and

M is the midpoint of PQ, it immediately follows that EP = EQ, which is what we wanted to prove. Q.E.D.

18

Here’s a diagram:

BA

T

C

P

81. Let O be the center of ω. Let ∠TOC = θ. Then note that we have TC = 9 tan θ and

AC =TC

sin θ−AO =

9

cos θ− 9

Then since △ACP ∼ △OCT , we have

9cos θ − 9

9cos θ

=AP

9=⇒ AP = 9− 9 cos θ

Now from the Law of Cosines on △PAB, we have

BP 2 = 182 + (9− 9 cos θ)2 + 2(18)(9− 9 cos θ) cos θ

= −243 cos2 θ + 162 cos θ + 405

This quantity is maximized when cos θ = −1622(−243) =

13 and plugging this value in we get that

m2 = −27 + 54 + 405 = 432 .

82. The diagram:

AB

C

D

M

N

POA

OB

Let O be the midpoint of AB, the center of the semicircle (O) with the points C, D forming a convex

quadrilateral ABDC. OM = OA cos AOM and ON = OB cos BON . ON ‖ AD is a midline of the triangle△ABD and OM ‖ BC is a midline of the triangle △ABC. Let E,F be the midpoints of PC, PD. Denote

19

α = ∠(OOA, PE) = ∠(OM,PC) and β = ∠(OOB , PF ) = ∠(ON,PD). From the trapezoid OOAEP ,OOA = PE

cosα and from the trapezoid OOBFP , OOB = PFcos β . Since OA = OB = AB

2 and

PE = PC2 = CD

4 = PD2 = PF , we get

OMON

= cos AOM

cos BON, OOA

OOB= cos β

cosα But from the circle (O) and form AD ‖ ON,BC ‖ OM ,

α = ∠(OM,PC) = ∠(BC,DC) = ∠BCD = ∠BAD = ∠BONβ = ∠(ON,PD) = ∠(AD,CD) = ∠ADC = ∠ABC = ∠AOM Consequently, OM

ON= OOA

OOB, which means

OAOB ‖ MN . Q.E.D.

83. This will probably help:

PD

A B

C

IC

IB

Notice that ∠BPC = ∠BPD+∠CPD. Also, we have ∠BPD = 180−∠DIBB, and ∠CPD = 180−∠CICD.Then, ∠BPC = 360− (∠DIBB + ∠CICD). Also, we have ∠DIBB = 180− (∠ADB+∠DBA

2 ) and

∠CICD = 180− (∠ACD+∠ADC2 ). Adding, we obtain ∠DIBB +∠CICD = 360− (∠ADB+∠ADC+∠ACD+∠DBA

2 )

Notice that ∠ACD+∠DBA2 = 180−(∠DCB+∠DCB)−∠CAB

2 = ∠CDB−∠CAB2 , and ∠ADB+∠ADC

2 = 360−∠CDB2 .

Then, we add, and get ∠DIBB + ∠CICD = 360− 360−∠CAB2 . Now, it remains to find ∠CAB. Luckily, we

are given side lengths, and the law of cosines will do the trick for us:

162 + 102 − 2(10)(16) cos∠CAB = 142 cos∠CAB = 12 ∠CAB = 60

Then, ∠DIBB + ∠CICD = 360− 360−602 = 210 and ∠BPC = 360− (∠DIBB + ∠CICD) = 150. Now, we

have to find the maximum area of △BPC with a side length 14 and its opposite angle 150. The area ismaximized when △BPC is isosceles, and we can use the law of cosines to find the length of the legs:

2r2 − 2r2 cos 150 = 196 2r2(1 +√32 ) = r2(2 +

√3) = 196 Notice that to find the area, it suffices to find

r2 sin 150 12 = r2

4 . Then, A = 492+

√3= 98− 49

√3 and our answer is 98 + 49 + 3 = 150 .

84. Let X = BM ∩ CN , and define B′ = AB ∩MN , C ′ = AC ∩MN . First note that the condition∠PAB = ∠BCA implies △PAB ∼ △ACB, and likewise ∠QAC = ∠CBA implies △CAQ ∼ △CBA.Therefore ∠PQA = ∠BAC = ∠APQ and △APQ is isosceles.

Next note that a homothety centered at A with scale factor 2 send △ABC to △AB′C ′ (due to the fact thatit sends △APQ to △AMN), so additionally △ACQ ∼ △AC ′N , etc. Length ”bashing” gives

AB

AM=

AB

2AP=

CA

2CQ=

C ′A

2C ′N=

C ′C

C ′N,

which coupled with ∠PAB = ∠AC ′N gives △BAM ∼ △CC ′N =⇒ ∠XMA = ∠XNM . Similarly,∠XMN = ∠XNA. Therefore

∠BXC = π − (∠XNM + ∠XMN) = π − 1

2(∠ANM + ∠AMN)

= π − 1

2(∠BAC + ∠BAC) = π − ∠BAC,

establishing the desired cyclicity. Q.E.D.

20

85. WLOG AB < AC. Say the incircle meets BC at E,AB at F. Then DE2 = x(2x) by power of a point.Similarily DF 2 = 2x2. It’s well known we can write CE,AF = s− c, s− a, where s is the semiperimeter.Then DE = |s− c− 10|, and AF = s− a. So we have (s− a)2 = (s− c− 10)2. Solving, b = 10 ora = c+ 10 ⇒ c = 10. We chose c = AB < AC = b, so c = 10. Then s = a+b+c

2 = 15 + b2 . From Stewart’s

theorem, we can easily derive 2x2 = b2+c2−2009 . Finally, (s− a)2 = b2+c2−200

9 . Putting all variables in terms ofb, we get the quadratic b2 − 36b+ 260. Solving, b = 10 or b = 26. Obviously, b = 26 is what we want. Movingto Heron’s formula, Area =

√(28)(8)(18)(2) = 24

√14. The answer is 38 .

86. We will show that < I2I1I4 = 90 degrees. First, we claim that A1A2I1I4 is cyclic. Since< A1I4A2 = 90 + 1/2 < A1A4A2 and < A1I1A2 = 90 + 1/2 < A1A3A2 (angle bisector angle formulas), and< A1A4A2 =< A1A3A2 (cyclic quad A1A2A3A4), we have < A1I4A2 =< A1I1A2, which makes A1A2I1I4cyclic. By similar arguments, we show that A2A3I2I1 is cyclic. Now all that remains is a basic angle chase.We have < I2I1I4 = 360− < A2I1I4− < A2I1I2 = 360− (180− < I4A1A2)− (180− < I2A3A2) = 1/2(<A4A1A2+ < A2A3A4) = 1/2(180) = 90. Similarly, we can show that the other three angles are 90 degrees,

which makes I1I2I3I4 a rectangle. Q.E.D.

87. Let the line through the centers of circles C and D be vertical. WLOG, let A be to left of that line. We wantto show that the ratio ME/NF = PE/PF . Since AM = AN , we have ME/NF = (AM ∗ME)/(NF ∗AN),which is the power of M wrt (with respect to) circle C divided by the power of N wrt circle C.

Now, extend MP and NP to intersect circle C again at points M ′ and N ′, respectively. Draw the horizontalline through P (which is also the tangent line to both circles) and let X be a point on that line to the left ofP and Y be a point on that line to the right of P . Now we have < PMN =< Y PN =< XPN ′ =< M ′N ′P(the first and last equality follow by cyclic quads). Similarly, we get < PNM =< PN ′M ′. Therefore,triangles MNP and M ′N ′P are similar by AA similarity, so M ′P/N ′P = MP/NP . This also means thatMM ′/NN ′ = MP/NP .

Now we can further simplify the original ratio. Since the power of M wrt circle C is also equal toMP ∗MM ′, and the power of N wrt circle C is equal to NP ∗NN ′, we can rewrite the ratio as (MP/NP )2.By Law of Sines in triangle MNP , MP/NP = sin(PNM)/ sin(PNM) = sin(AMP )/ sin(ANP ). Thus, wecan write the ratio (MP/NP )2 as (MP/NP )(sin(AMP )/ sin(ANP ). By using Law of Sines once more ontriangles EMP and FPN, we can rewrite this ratio as (EP/FP )(sin(AEP )/ sin(PFN)).

Finally, we notice that < AEP =< AFP = 180− < PFN , so sin(AEP ) = sin(PFN). Thus, we can simplify

the ratio to EP/FP , so the proof is complete. Q.E.D.

88. WLOG, let BC > AC. Then < CLB >< CLA. Let the projection of A onto CL be A′. Since L is themidpoint of AA2 and A′ is the midpoint of AA1, A

′L is the midline of triangle AA1A2, so A′L||A1A2. Thus,we have that triangles BCL and BA1A2 are similar. Similarly, we get that triangles ACL and AB1B2 aresimilar. Note that O1 is inside triangle AB1B2 and O2 is outside triangle BA1A2, so O1 and O2 are outside oflines AC and BC, respectively.

In order to have < O1CB1 =< O2CA1, we would like to show that triangles O1CA and O2CA1 are similar(more precisely, we would like to show that these triangles are congruent, but that is not actually necessary)(Note: this is also the importance of having a good, precise, diagram, because it helps you realize things likethis). Notice that we have AB2 = BA2 by reflection, and < AB1B2 =< ACL =< BCL =< BA1A2, so thecircumradii of triangles AB1B2 and BA1A2 are equal. Thus, O1A = O2A1. By reflection, we have CA = CA1.Now, all we need to show is that < CAO1 =< CA1O2, as then we would be done by SAS congruency.

< O1AC = 180− < O1AB1 = 180− (90− < B1B2A) = 90+ < ALC, where the equality< O1AB1 = 90− < B1B2A follows since O1 is inside triangle AB1B2. Similarly, we get < O2CA1 = 180− <O2A1B = 180− (< A1A2B − 90) = 90 + (180− < A1A2B) = 90 + (180− < CLB) = 90+ < CLA. Thus,< O1AC =< O2CA1, which means triangles O1CA and O2CA1 are congruent by SAS, which means that

< O1CA =< O2CA1 =< O2CB. Q.E.D.

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89. Since FG is a tangent to the circumcircle of EDM , < EMD =< DEG =< CEF . Also, since< BCD =< BAD, triangles MAD and ECF are similar by AA. Since EF is part of the ratio that we desire,and EF is part of triangle ECF , we want EF to be part of our ratio from this similarity. Thus, we haveEF/EC = MD/AM . Cross-multiplying, we get EF ∗AM = EC ∗MD.

Also, we have < AGF = 180− < GCF− < GFC =< ADB− < GFC =< MDB, where the last equalityfollowed from similar triangles MAD and ECF . Now we want to find a way to express GE, since we alreadyhave a way to express EF. Thus, we look for similar triangles involving GE. Fortunately, since< CGE =< MDB and < GCE =< DBM , we have that triangles GCE and DBM are similar. Thus, we canwrite the ratio GE/CE = DM/BM . Cross-multiplying, we get GE ∗BM = CE ∗DM . Fortunately, thisequals EF ∗AM , so we have GE ∗BM = EF ∗AM . Rearranging, we get GE/EF = AM/BM .

Finally, all we have to do is find AM/BM in terms of t. Taking the reciprocal, of AM/AB, we getAB/AM = 1/t. Subtracting 1 from both sides, we get MB/AM = (1− t)/t. Finally, taking the reciprocal

once more, we get GE/EF = AM/BM = t/(t− 1). Q.E.D.

90. All of the cks will be some constant. Consider the homothety H that sends w1 to w2 centered at N. It is wellknown that A, K, and M are collinear. Now, since arcs MB and AM are equal,< BAM =< ABM =< BNM =< ANM . Thus, triangles BNM and KBM are similar by AA. We can nowwrite the ratio BM/KM = NM/BM , so BM2 = NM ∗KM . Since H takes NK to NM , NK/NM is aconstant, which means that NM/KM is also a constant. Thus, we can write BM2 = KM2 ∗ c1. Taking thesquare root, we get BM = KM ∗ c2.Now we divide both sides by sin(ANM). Since BM/ sin(ANM) = 2R = constant, we have(KM/ sin(ANM)) ∗ c2 = c3, so (KM/ sin(ANM)) = c4. Finally, dividing by 2, we get(KM/2 sin(ANM)) = c5, and since (KM/2 sin(ANM)) is the circumradius of triangle KBM , and it is

constant, we are done. Q.E.D.

91. Extend PC past the circle and intersect it with the tangent to the circle at E. Call this point P ′. Now drawthe other tangent from P ′ to the circle and label the tangency point as B′. Let the intersection of BE andB′D be X.

By Pascal’s Theorem with the configuration DDB’BBE, P , X, and the intersection of DE and BB′ arecollinear. Similarly, with the configuration EEBB’B’D, P ′, X, and the intersection of DE and BB′ arecollinear.

Since both pairs of collinearities have two common points, they are indeed the same line. Also, since P andP ′ are both points on the line, the line is indeed PP ′. Thus, the intersection of BE and B′D lies on AC.Finally, we know that the entire figure is symmetric with respect to the perpendicular bisector of DE and

AC, so CX = AX, and thus BE bisects AC. Q.E.D.

92. First, draw the diagram. As, lines AQ and MN look parallel in the diagram, we will try to prove this. SinceOM = ON , O lies on the perpendicular bisector of MN . Now let OA be the circumcenter of triangle AMN .Clearly, OA also lies on the perpendicular bisector of MN . Now, we want to prove that the perpendicularbisector of AQ passes through O and OA, as this would mean that AQ and MN are perpendicular to thesame line, which would make them parallel. Fortunately, as AQ is the radical axis of the two circles, theperpendicular bisector of AQ must pass through O and OA, which means that AQ||MN . Also, since they areparallel, and AQMN is cyclic, AQMN is an isosceles trapezoid.

If AQ is tangent to circle w, then triangles AQB and ACQ are similar (prove by PoP and SAS). Thus, wewould like to prove that these triangles are similar. This is equivalent to proving that < PQB =< PCQ, asthe result would follow by AA similarity.

Now we do a little angle chase on < PCQ. We have < PCQ =< BCQ =< BAQ =< AMN =< PMB,where the last equality follows from parallel lines AQ and MN . Thus, we would like to prove that< BQP =< BMP , or that quadrilateral BMQP is cyclic.

Since cyclic quadrilaterals have a lot of nice angle properties, we will aim to prove that < BPM =< BQM ,as we can manipulate these angles the best.

First we compute < BPM . < BPM =< CPN = 180− < C− < PNC = 180− < QAN− < C =< QBC− <C =< QBA+ < ABC− < C, where the equality < PNC =< QAN comes from parallel lines AQ and MN .Finally, we have < QBA+ < ABC− < C =< QBA+ < ABC− < QCA− < QCB =< QBA+ < ABC− <QBA− < QCB =< ABC− < QCB.

22

Now we compute < BQM . < BQM =< OQB− < OQM . Since triangle OQB is isosceles with vertex O, and< QOB = 2 < QCB,< OQB = (180− 2 < QCB)/2 = 90− < QCB. By symmetry, triangles OQM andOAN are similar, so < OQM =< OAC = 90− < ABC. Now we can substitute these values to get< BQM = (90− < QCB)− (90− < ABC) =< ABC− < QCB.

Since < BPM =< BQM , quadrilateral BMQP is cyclic, which implies that < PQB =< AQN =< PCQ,which implies that triangles PQB and PCQ are similar, which implies that PQ is tangent to circle w.

Q.E.D.

93. Let the midpoint of TB be E. Then, since < XET =< XMT = 90, quadrilateral XMET is cyclic. Thus, wehave < BTM =< ETM =< EXM =< EXB+ < BXM . Now we want to find a value for < CTM .

Notice that, since E and M are the midpoints of BT and BC, respectively, EM ||CT . Therefore,< CTM =< EMT . By cyclic quadrilateral XMET , < EMT =< EXT =< EXB.

Now we have that< BTM− < CTM = (< EXB+ < BXM)− (< EXB) =< BXM =< BAM = 1/2 < BAC, which is

constant. Q.E.D.

94. First, draw the diagram, as always. Extend AB to intersect line l at a point B′.

We want to show that F,C,H collinear, or < ACH+ < ACF = 180. First of all, since H is a reflection of Gover the diameter AB, we have that HG is perpendicular to AB, so it is parallel to line l. With nothing elsein mind, we angle chase a little on < HCA. < HCA =< HGA =< B′FA =< DFA. Thus, we want to provethat < DFA+ < ACF = 180. Since < ACF = 180− < DCF , we want to prove that< DFA+ (180− < DCF ) = 180, or < DFA =< DCF . Since triangles AFD and FCD have a shared angleat < ADF , it suffices to prove that triangles AFD and FCD are similar.

After a (hopefully short) attempt to prove that < DAF =< DCF , as this leads nowhere, we move on to tryand prove that the triangles are similar by SAS. Thus, we want to show that AD/DF = FD/DC.Cross-multiplying, we get that we want to show that AD ∗DC = DF 2. Now notice thatAD ∗DC = DC ∗DA, which is the power of point D with respect to the circle. However, since DE is tangentto the circle, it is also equal to DE2. Thus, we want to show that DE2 = DF 2, or DE = DF .

Finally, we can angle chase on triangle DEF . Let < DFE =< 1. Then, since AB′ is perpendicular to B′F ,< FBB′ = 90− < 1 =< EBA. Thus, since AB is a diameter, < EBA+ < EAB = 90, so< EAB = 90− < EBA = 90− (90− < 1) =< 1. Finally, since DE is tangent to the circle,< DEF =< EAB =< 1. Since < DEF =< DFE =< 1, triangle DEF is isosceles and DE = DF , so we may

conclude. Q.E.D.

95. Let M and N be the midpoints of BE and CD respectively. Note that since △BEO and △CDO are bothisosceles, AM ⊥ MO and AN ⊥ NO =⇒ AMON is cyclic. In addition, note that angle chasing gives∠PDE = ∠PAE ≡ ∠PAB = ∠PCB. Similarly, ∠PED = ∠PBC, so △PED ∼ △PBC. Therefore byProblem 77 △PED ∼ △PMN and by Problem 26 △PEM ∼ △PDN . This implies ∠PMA = ∠PNA andso (APMN) is also cyclic. Hence A,P,M,O,N all lie one circle, implying that ∠APO = ∠AMO = 90◦ as

desired. Q.E.D.

96. It is easy to see that ABOPC is concyclic. Because BQPM is cyclic (by definition), we have∠MBQ = ∠QPO = 90◦. Furthermore, we have ∠BQM = ∠BPM = ∠BPO = ∠BAO, so MBQ ∼ OBA.Therefore, there is a spiral similarity about B mapping MQ to AO. Thus, MBO ∼ QBA. We therefore have

OM =OB ·AQ

AB

Similarly, we can arrive at ON = OC·AQAC

which is clearly equal to OM because OB = OC and AB = AC.

Q.E.D.

23

97. Note that ∠LY A = ∠Y ZA = ∠Y AZ = ∠LY A by the conditions in the problem statement. This means that△ALY is isosceles, so AL = AY . Draw in Y B. Then since

∠LAB = ∠LAY + ∠Y AZ + ∠ZAB = ∠LY A+ ∠KY Z + ∠ZY B = 180◦ − ∠AY B = 90◦,

we have AL ⊥ AB. (I’m pretty sure simply the fact that the triangle was isosceles implies that the leg istangent, but whatever.) Furthermore, since OY ⊥ AZ ⊥ BZ, where O is the center of the circle, we haveKY ⊥ KB, so ALKB is cyclic.

Next, I claim that AKAX

= (AYAX

)2. To prove this, note that △ALY ∼ △AY Z, so AY 2 = AL ·AZ. In addition,since ∠ALB = ∠AKB by cyclicity, △ALX ∼ △AKZ, so

AK

AZ=

AL

AX=⇒ AK

AX=

AZ ·AL

AX2=

(AY

AX

)2

,

as desired.

Let AL = ℓ and WLOG let AB = 2. From standard altitude-to-hypotenuse calculations, it may be deducedthat AX = 2ℓ√

4+ℓ2. In addition, since ∠LY O = ∠LAO = 90◦, ALY O is cyclic, so by Ptolemy

AL ·OY +AO · LY = AY ·OL =⇒ 2ℓ = AY√1 + ℓ2 =⇒ AY =

2ℓ√1 + ℓ2

.

Therefore

AK

AX+

(AL

AB

)2

=

(AY

AX

)2

+

(AL

AB

)2

=

(2ℓ/

√1 + ℓ2

2ℓ/√4 + ℓ2

)2

+ℓ2

4=

4 + ℓ2

1 + ℓ2+

ℓ2

4= 1 +

3

1 + ℓ2+

ℓ2

4.

Let S denote this sum. It suffices to find the maximum value of S. A way to do this without calculus is asfollows: subtracting 1 and adding 1

4 to both sides gives

S − 3

4=

3

1 + ℓ2+

1 + ℓ2

4≥ 2

√3

1 + ℓ2· 1 + ℓ2

4=

√3.

Therefore S ≥ 34 +

√3. The requested answer is 3 + 10 · 4 + 100 · 3 = 343 .

98. Very large diagram:

A

B

C

D

EP

Clearly, we have ABC ∼ ADE (they practically give that to you) so we have ABAC

= ADAE

, which rearranges toABAD

= ACAE

and because ∠DAB = ∠EAC we have ABD ∼ ACE. (Alternatively, note that there is a spiral

24

similarity about A mapping BD to CE). Now we see that ∠ABD = ∠ACE so ABCP is cyclic. Similarly,AEDP is cyclic. Furthermore, notice that

∠PDC = ∠BDC = ∠ADC − ∠ADB = ∠AED − ∠AEC = ∠PED

Which means that CD is tangent to circle (APDE). Similarly, CD is tangent to (ABCP ). Therefore, since

AP is the radical axis between (APDE) and (ABCP ), we see that AP bisects CD, as desired. Q.E.D.

99. Throughout the proof, we use directed segments. For each i, define Xi to be the second intersection point ofAiAi+2 with ⊙Oi+2, taking A5 = A1, etc. Furthermore, let P be the intersection of the diagonals A1A3 andA2A4. The key observation is to note that A1A3 is the radical axis of ⊙O2 and ⊙O4, and that A2A4 is theradical axis of ⊙O1 and ⊙O3. This implies that P has equal power with respect to all four circles.

One consequence of this is that PA1 · PX1 = PA4 · PA2. Expansion gives

PA1 · (PA3 +A3X1) = (PX2 +X2A4) · PA2

PA1 · PA3 + PA1 ·A3X1 = PX2 · PA2 +X2A4 · PA2.

By PoP we see that PA1 · PA3 = PA2 · PX2, so

PA2 ·A4X2 +A3X1 · PA1 = 0 =⇒ 1

A3X1 · PA1

+1

PA2 ·A4X2

= 0.

Next, clever manipulation gives

1

A3X1 · PA1

=A1A3

A1A3 ·A3X1 · PA1

=PA3 − PA1

A1A3 ·A3X1 · PA1

=1

A1A3

(PA3

A3X1 · PA1

− 1

A3X1

).

Through the fact that P has equal power with respect to all four circles, we obtain that

PX1 · PA1 = PX3 · PA3

(PA3 +A3X1) · PA1 = (PA1 +A1X3) · PA3

A3X1 · PA1 = A1X3 · PA3.

Thus

1

A1A3

(PA3

A3X1 · PA1

− 1

A3X1

)=

1

A1A3

(1

A1X3

− 1

A3X1

)

=1

A1A3 ·A1X3

+1

A3A1 ·A3X1

.

Similarly, we get1

A2A4 ·A2X4

+1

A4A2 ·A4X2

=1

A4X2 · PA2

. Finally, note that A1A3 ·A1X3 is equal to the

power of the point A1 with respect to ⊙O1, which in turn is known to be equal to O1A21 − r21. Thus,

substituting our new expressions into one of our earlier equations and replacing all of the denominators withcorresponding (cyclic) Power of a Point expressions gives

1

O1A21 − r21

+1

O2A22 − r22

+1

O3A23 − r23

+1

O4A24 − r24

= 0,

as desired. Q.E.D.

100. Let O denote the circumcenter of △ABC, and WLOG let AB < AC. By simple angle-chasing,

∠BFC = ∠BDM + ∠AEF

= 2∠BAD + 2∠EAC

= 2∠BAC = ∠BOC,

25

so BFOC is a cyclic quadrilateral. In addition, note that the circumcircle of △APN passes through O since∠APO = ∠ANO = 90◦. Therefore, if F is to lie on (APN), OF must be the radical axis of the circumcirclesof △APN and △BOC. Now let X denote the point at which the tangent to the circumcircle of △ABC at Ameets BC. As △APN and △ABC are homothetic, their circumcircles are tangent, and as such AX is theradical axis of the two circles. Combining this with the fact that BC is the radical axis of (BFOC) and(ABC) gives that X is the radical center of all three circles. Thus what we want to prove is simplified toproving that X,F, and O are collinear.

Note that by some more angle-chasing we see that

∠FOC = ∠XBF = ∠XBA+ ∠ABD

= ∠XAC + ∠BAD = ∠XAM + ∠A.

In addition, note that ∠OMC = 12∠BOC = ∠A, so ∠XAM = ∠FOM . However, since

∠XAO = ∠XMO = 90◦, quadrilateral XAOM is cyclic, so ∠XAM = ∠XOM . Therefore

∠XOM = ∠FOM , implying that X,F,O are collinear as desired. Q.E.D.

Acknowledgements

Thank you to everyone who posted solutions to these problems, and one big THANK YOU to David Altizio,for making this compilation.

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