100 Geometry Problems by David Altizio, Solutions by Alvin Zou.pdf ...

Viewer
Transcript

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 of equations ⎧ ⎪ ⎨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 − φ and 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 the radius of the semicircle is 1. Since ABCD is a square, BC and EA are tangents as well. We have CX = BC = 2 and EX = EA. We can Pythagorean Theorem EDC with legs DE = 2 − EA and DC = 2 5 . with a hypotenuse CE = 2 + EA. Solving, we get EA = 12 and CE = 2 4. The diagram:

A

M

6

B

6

D

3

C

It is given that ∠AM D ∼ = ∠CM D. Since ∠AM D and ∠CDM are alternate interior angles and AB DC, ∠AM D ∼ = M C. We know that = ∠CDM −→ ∠CM D ∼ = ∠CDM . Use the Base Angle Theorem to show DC ∼ ABCD is a rectangle, so it follows that M C = 6. We notice that BM C is a 30 − 60 − 90 triangle, and ∠BM C = 30◦ . If we let x be the measure of ∠AM D, 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, EF G, GF D are congruent because of 2 equal sides and the right angle. (SsA) Let the side length of the square be a then by the Pythagorean Theorem we get 1 302 − ( a)2 = 3

a2

900 =

10 2 a 9

a2 = 810 6. First, we draw the diagram:

A E D

B

C

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 that CD = 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 √ √ a 3 b 3. a = a = 8. Drawing the diagram, we get:

C L M

N

A

B

Since quadrilateral ABLM is cyclic, ∠M LB = 180 − ∠A = 90◦ = ∠M LC. Thus ∠CM L = 90 − ∠C. We also have ∠C = ∠AN M since ALCN is cyclic. Then since ∠M AN = 180 − ∠A = 90◦ , we get ∠AM N = 90 − ∠AN M = 90 − ∠C = ∠CM L. Since the vertical angles are congruent, L, M, N are collinear. Q.E.D. 2

9. First, a diagram:

A

E I

B

X

D

Y

C

Let D and E be the feet of the altitudes from I to BC and AB respectively. Since IX = IA and ID = IE we have 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 = 12 XY, 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 prove that 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 perimeter becomes kP and its are becomes k 2 A. Thus, if we want our dilated shape to equiable, or kP = k 2 A, we should dilate the shape by a factor of k = P A . Q.E.D. 12. By simple angle chasing, we notice that triangle AEB is similar to EF C. Let the side length of the square be a, EC = x and thus BE = a − x. Because of similarity, it is x 3 a = ⇔ x = a. 4 3 4 That yields BE = 14 a. Using the Pythagorean Theorem in triangle AEB the result is 17 2 1 162 a = 16 ⇔ a2 = a2 + ( a) = 16 ⇔ 4 16 17 13. ∠ABN = ∠AM N since they are subtended from the same chord. Since ∠M XN = ∠M Y N = 90◦ , it follows that M , X, Y , and N are concyclic. Then, since M and X form angles subtended from the same chord, ∠Y XN = ∠Y M N = ∠AM N = ∠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 and we’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 ! Thus EM = M F. Now notice that (5, 12, 13) is a pythagorean triple which yields ∠BEA = 90◦ . Obviously, ∠AM B = 90◦ and √ AM = M B = 2·13 2 . Then AEBM is a cyclic quadrilateral and by Ptolemy we get √ √ √ 2 · 13 2 · 13 2 · 17 ·5+ · 12 = 13 · M E ⇔ M E = . 2 2 2 √ Therefore EF = 2 · M E = 2 · 17, so √ EF 2 = ( 2 · 17)2 = 2 · 289 = 578 .

3

Solution 2:

E

A

B

D

C

F

Extend F C, EB, EA, F D as shown until the sides intersect. The two intersection points as well as point E ∼ and F form a square, √ 2 as the newly formed triangles are congruent to triangles AEB and CF D by ASA =. So 2 then EF = (17 2) = 578 . 15. Let O be ABC’s circumcenter. Note that ∠BF E = 12 ∠BOE = 14 ∠BOC = 12 ∠A. Similarly, ∠AEF = 12 ∠B and ∠DF B = 12 ∠C. Since ∠DF E + ∠AEF = ∠DF B + ∠BF E + ∠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, we 1 ◦ have DP = 2 · 12 3 = 8. By deﬁnition, ∠DDp P = 90 so DDp P is a 30-60-90 triangle. Then DDp = 2 DP = 4, so [ABC] · DDp = (15 · 4)/3 = 20 . VABCD = 3

4

17. Here’s a colorful diagram:

P1

A3

O

P4

P2

P3

A1

Since the quadrilateral is orthodiagonal, we can write the statement we want to show as D2 = P1 P22 + P3 P42 . Let A1 and A3 be the antipodes of P1 and P3 , respectively, and let O be the center of the circle. The claim is that reﬂecting P1 P2 P3 P4 over the diameter of the circle that is parallel to P1 P3 results in the new quadrilateral A3 P4 A1 P2 . We can prove this by noting that P1 OP3 ∼ = A3 OA1 , so P1 goes to A3 and P3 goes to A1 (and obviously P2 and P4 switch places). Now, after reﬂecting, we get that A1 P2 = P3 P4 , so now we want to show that D2 = P1 P22 + A1 P22 which is true by the Pythagorean Theorem on P1 P2 A1 . 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 the midpoint of AB. Then, since this line is perpendicular to AB, we can then compute the √ radius of the sphere. 542 − 212 and First, we ﬁnd the distances from the centers of the circles to midpoint of AB. They are √ 2 − 212 . Thus, the distance from the center of the sphere to the midpoint of AB is 66 √ 54√2 − 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 of triangle XBC and XAD, respectively. We can also see that X, M, N are collinear. Hence, XN = AN and XM = BM , so we easily ﬁnd that M N = 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 that D, E, F collinear. Since line l is tangent to the circle, ∠DBC = ∠A, and therefore ∠BCD = 90 − ∠A. Notice that quadrilateral BDCF is cyclic (opposite angles 90). Thus, we have ∠BCD = ∠DF B = 90 − ∠A. However, we also have ∠EF B = ∠F BA = 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 circle shaded 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 of Cosines, 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: √ −9 + 133 r= 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 arc BC, 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 CQM B and AN P C are cyclic quadrilaterals, implying that ∠CM B = ∠CQB and ∠CN A = ∠CP A. Therefore, we can deduce that ∠M CN = 180◦ − (∠CM N + ∠M N C) = 180◦ − (90◦ − A/2 + 90◦ − B/2) = 45◦ 24. Note that since AM CN is cyclic, we have ∠AN M = ∠ACB and ∠AM N = ∠ACD = ∠CAB so M AN ∼ 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 of ABC. 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 ∠AP B = ∠CP D, so ∠AP B + ∠BP C = ∠CP D + ∠BP C, or ∠AP C = ∠BP D. We also know PB AP = , so by SAS similarity P AC ∼ P BD. Q.E.D. PC PD 27. Note that ∠OY X = ∠OXY , so ∠Y ZO = ∠OZX. Let ∠Y ZO = α, then by Law of Cosines on triangles Y ZO and OZX we get 121 + 49 − 2 · 11 · 7 cos α = 121 + 169 − 2 · 11 · 13 cos α, which simpliﬁes to cos α = 10 11 . √ 2 Plugging this back in we have OY = 170 − 140 = 30, so OY = 30 . 28. Let ∠AKF = ∠F KB = α and ∠AM H = ∠HM D = β. Also let F E ∩ HG = X. Then because ∠KAF = ∠DCB, we have ∠M F X = α + C, so ∠M XF = 180 − (α + β + C). However we also know that 2α + 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 ∠M XF = 180 − 90 = 90. From this we conclude that triangles EM X and F M X are congruent, and triangles HKX and GKX are congruent. Thus EX = XF , HX = XG, and HG ⊥ F E, so EGF H is a rhombus. Q.E.D. 29. The diagram:

B

P

7

A

5

H

M

7

C

With some simple calculations we see that √ AH = 5, HM = 2, and M C = M P = 7. Also BH = 12. So by the Pythagorean theorem we have BM = 2 37. Let the foot of the altitude from P to AC be D. Then MP PD PD √42 Thus the area of P AC is √7 M B = BH . Or 2 37 = 12 . Solving we have P D = 37 1 2

· 14 ·

√42 37

=

294 √ 37

=

√ 294 37 37 .

Therefore p + q + r = 294 + 37 + 37 = 368 6

30. The complicated diagram:

E A

O

D

B

M P

D C E

Let O be the circumcentre of ABC and M be the midpoint of BC. Then its clear that P M ⊥ BC. Since ∠BDP = 90◦ = 190 − ∠P M B, quadrilateral P M BD is cyclic. So we have ∠M DP = ∠M BP . Also ∠ADM = 90 ◦ − ∠M DP = 90◦ − ∠M DP . As BP is tangent to the circumcircle of ABC we must have BC = ∠A. Therefore ∠ADM = 90◦ − ∠M DP = 90◦ − ∠A. Let D = DM ∩ AE then ADD is ∠M BP = 2 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 HHA BHC and HHA CHB are cyclic. Then ∠HC HA A = ∠HC BHB = 90 − ∠BAC and ∠AHA HB = ∠ACHC = 90 − ∠BAC. Therefore, H lies on the angle bisector of ∠HC HA HB . The same argument can be repeated for the angles HC HB HA and HA HC HB 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 = 11 and CZ = 97 + 86 = 183. Now, CB · CX = CY · CZ = 11 · 183 = 3 · 11 · 61. If BX and CX have integer lengths, this also holds for BC and since BC ≥ CX ≥ 11 (the latter holds by the Triangle Inequality) the only possible values are CX = 11, V C = 183 and CX = 33, BC = 61. But in the ﬁrst case, ACX would be degenerated which implies X ∈ AC, only possible if X = C but this contradicts AX = 86 = 97 = AC. Hence CX = 33 and BC = 61 is the only possible solution.

7

33. Let ∠BAC = A, C = 4A. ∠ABC = B, and∠BCA = C. We know that ∠BOC = 2A, so the measure of BN 4A−PQ , which means PQ = 2A, and ∠P N Q = A. Because the center of circle τ also lies on the Then A = 2 perpendicular bisector of BC, we know that ∠N OB = ∠N OC = A. Thus ∠N QC = ∠N OC = A, so ∠N QA = 180 − A, and ∠N P A = 180 − A. From this we conclude AP N Q 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 sectors of 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 1 1

·

The 4 big right triangles have a total area of 4 · 2 2 2 = 12 . Then the small external triangles are isosceles √ right triangles, and have height 2−1 2 , so their area is π 4.

√ 4· So the total area is

2−1 2

2

2

√ 3−2 2 = 2

√ π +2− 2 . 4

35. Take O the midpoint of the diameter. thus, M, O, P, S are concyclic, so ∠M P S = ∠M OS; 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 ∠A1 DC1 = ∠A1 DB + ∠BDC1 = ∠ACB + ∠BAC = 180◦ − ∠ABC = 180◦ − ∠A1 BC1 hence the quadrilateral BA1 DC1 is insciptible and ∠BAC = ∠BDA1 = ∠BC1 A1 which proves that A1 C1 //AC. Q.E.D. 27 36 37. From simple similarity we ﬁnd AE = 48 5 , EC = 5 , and DE = 5 . Now notice that quadrilateral ABDF is AE AD cyclic, so ∠AF E = ∠ABC. Thus AF E ∼ ABD, and EF = DB . It is well known that for a 13 − 14 − 15

triangle AD = 12 and BD = 5. So we have m + n = 21 .

12 5

=

48 5

EF

8

, and EF = 4. Therefore DF =

36 5

−4=

16 5 ,

so

38. Let BX = m, CX = n, and AX = d. Furthermore let the circumcircles of triangles √ AXB and AXC have 6. So the area of ABX is radii R and R respectively. We know that the area of ABC is, by Heron’s, 6 2 √ 1 √ m·6 6 n·6 6 . Similarly the area of AXC equals 7 . Then 7 R1 =

5md 4·

√ m·6 6 7

=

35d √ 24 6

From the same logic we arrive at 7d R2 = √ 6 Thus we see that the area of the circumcircles is minimized when d2 is minimized and hence when d is √ minimized. To minimize d, we have to make it the length of the altitude from A to BC. Setting 7d 2 = 6 6 we √ 19 . ﬁnd d = 127 6 , and that BX = 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, ∠KF B = 90 so BF ⊥ KL. Q.E.D. 40. Of course, a diagram would help.

C

D

E

A

B

Let AD = x and BE = y. Then, we have DC = 2x and EC = 3y. By power of a point, we have that and DC(AC) = EC(BE) =⇒ (2x)(3x) = (3y)(4y) =⇒ x2 = 2y 2 . Now, since AB is the diameter, ∠ADB √ ∠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: (16y 2 − 4x2 ) + x2 = 900 =⇒ 16y 2 − 3x2 = 900 =⇒ 16y 2 − 3(2y 2 ) = 900

Thus, x =

√

2·y =

√

=⇒ 10y 2 = 900 √ =⇒ y = 3 10

√ √ 2 · 3 10 = 6 5. Plugging in our value of x, we ﬁnally get: [ABC] =

√ √ √ √ 1 (18 5)( 900 − 180) = (9 5)(12 5) = 540 . 2 9

41. The diagram:

D

D

C

P

P A

C

P

A

B

B

We seek the value of ∠DAP + ∠BCP . We begin by constructing congruent rectangles BCC B to the right and ADD A to the left of rectangle ABCD. Denote the points P and P as the points in rectangles ADD 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 and BCC B , respectively. Since P and P are just translations of P in the horizontal direction, the line P P is parallel to AB and is thus perpendicular to AD and BC. Now, note that by the ∠AP D + ∠BP C = 180◦ condition, we have that AP DP and P BP C are cyclic quadrilaterals. Then, by inscribed arcs, we have that ∠BCP = ∠BP P = ∠AP P the last equality following from congruence. Thus, we have that ∠DAP + ∠BCP = ∠DAP + ∠AP P = 180◦ − 90◦ = 90◦ 42. Let the circumcircles of triangles AP R and BP Q intersect at a point X. Because quadrilaterals AP XR and BP XQ are inscribed in circles we know ∠P XR = 180 − ∠A and ∠P XQ = 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 such that Q is adjacent to √ √ . Now draw P QR around ABCDEF 2 + 3.√ Let the equation of RP be C and D. The height of ABCDEF is 3,√so the length√of base QR is √ 3 + 2 √ 3 + 3. Solving the two y = x. Then, the equation of P Q is y = − 3(x − (2 + 3)) → y = −x √ √ √ 3+3 1 equations gives y = x = 3+3 . The area of P QR is ∗ (2 + 3) ∗ = 5 43+9 . 2 2 2 a + b + c + d = 9 + 5 + 3 + 4 = 21 44. It suﬃces to show that M B = M I = M C. First of all, the perpendicular bisector of BC so M B = M C. Now and since 1 ∠M BI = 180◦ − (∠C + ∠A + 2

since M is the midpoint of arc BC, it obviously lies on note that ∠IM B = ∠C and ∠BIM = 12 ∠A + 12 ∠B 1 1 1 ∠B) = ∠A + ∠B = ∠BIM 2 2 2

so M B = M I = M C as desired. Q.E.D. 45. Let D be the point on AB such√that T D ⊥ AB. Then AT D √ is 30 − 60 − 90 and T DB is 45 − 45 − 90, so we have T D = 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

O B

M

D

C

E

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 ∠M ED = ∠AED − ∠OEM = ∠ABD − ∠OBM = ∠ABO = ∠BAE = ∠BDE, so M ED is M -isosceles and M E = M D as desired. Q.E.D. 47. Let O be the circumcenter of ABC. Note that ADOF and ADF N are both cyclic quadrilaterals, so N D 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 N M and BP . We have M N//AB and BO//AN so AN OB is a parallelgram, therefore BO = AN = N C and BO//N C imply that BN CO is also a parallelogram. M is the middle of ON and since N P O = 90◦ then M is the circumcenter of N P O and M P = M O. Finally, since QB//M O then ∠QBP = ∠M OP = ∠M P O = ∠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 an equilateral triangle possesses. It suﬃces to show the converse: that if U V W = XY Z then ABC is equilateral. 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 triangles are similar. Next, note that ∠U W V = and that

1 1 ∠U IV = (180◦ − ∠ZAI − ∠ZBI) 2 2

∠XZY = ∠XIC = 90◦ − ∠ICX.

If the two angles ∠U W V and ∠XZY are congruent, then we must have ◦ 1 2 (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 =

43 860 · 20 = =⇒ m + n = 923 63 63

51. ∠BP Q = ∠BP A = ∠BCA so AP is the angle bisector of ∠BP C Let QD and QE be the perpendiculars √ from Q to P B and P C respectively. QD = QE = 23 P Q. Area of √ √ C) 3 3 P B·P C·sin 120 = P Q(P B + P C) Also, Area of P CB = = P CB = QD(P B+P 2 4 2 4 P B · P C. Equating the two equations for the area we get the desired result. Q.E.D. 52. The diagram:

P A A2 A1

M

C2

C1

B

C

Notice that triangles ABM and BM C are isosceles, so M A1 BC and M C1 AB. Also, A1 C1 AC because A1 and C1 are midpoints. Triangles AA1 A2 and CC1 C2 are isosceles because of equal tangents, so A1 A2 is an external angle bisector of ∠BA1 C1 and similarly C1 C2 is an external angle bisector of ∠BC1 A1 . So P is the B-excenter of BA1 C1 , 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 preliminary result: Lemma. Let ABC be a triangle, and denote by M the midpoint of segment BC. If X, Y are the orthogonal projections of the vertices B, C on the internal angle bisector of angle BAC, then M X = M Y = |b − c|/2. Proof. We shall resume to proving that M X = |b − c|/2. For this, let B be the intersection of the line BX with the sideline CA. Since the triangle ABB is isosceles, the length of segment CB is |b − c|. On other hand, X, M are the midpoints of segments BB , and BC, respectively. Thus XM = CB /2 = |b − c|/2. This proves our Lemma. Returning to the problem, let T be the midpoint of the diagonal BD. According to the Lemma, applied for the triangle BAD, we have that T B1 = T D1 = |AB − AD|/2. Similarly, according to the same Lemma, this time applied in triangle BCD, we have that T B2 = T D2 = |BC − CD|/2. On the other hand, the quadrilateral ABCD beeing circumscribed, AB + CD = AD + BC (Pithot’s theorem), and hereby, we conclude that T B1 = T B2 = T D1 = T D2 , 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 [AP B] = [AP D], and let PB and PD be the feet of the perpendiculars from P to AB and AD respectively. It’s not hard to prove that P PB : P PD = 3 : 2. Furthermore ∠PD P PB is constant (and equal to 180◦ − ∠DAB), so all such triangles PD P PB are homothetic to each other. In particular, this proves that the locus of points P is a line passing through point A. Our goal is to ﬁnd CE, where E is the projection of C onto . (This deﬁnition of E clearly minimizes said distance.) Let X = ∩ CD, and deﬁne XB and XD as before. Furthermore, let H be the foot of the altitude from A to CD. Standard trapezoid computations (i.e. dropping the altitude from A and doing Pythagorean Theorem) yield AH = XXB = 12, so XXD = 8. Now since AHD ∼ XXD D we obtain 5 5 XD AD = =⇒ XD = · 8 = 10. = XXD AH 4 4 √ √ This implies AXB = HX = XD − HD = 1, so AX = 122 + 12 = 145. Finally, since XXB ⊥ CD angles AXXB and CXE are supplementary, so AXB X ∼ XEC. Since XC = CD − XD = 18, by similar triangle ratios CE XXB 12 12 · 18 216 = =√ =⇒ EC = √ = √ . CX XA 145 145 145 56. Denote the intersection of M N and BC as X. Notice that proving that the projections of D and C onto M N implies that DX = XC. This is also equivalent to showing that triangles DM N and CM N have the same area (same base same altitude). Now for an ugly area bash: Let the area of the quadrilateral be 1, to make the calculations simpler. Then N] [BM N ] BM BN [BCM ] = [ABN ] = 21 . Notice that [BM [ABN ] = [BCM ] . This simpliﬁes to AB = BC . Let this ratio equal k. Now we can compute [BM N ] to be [BM C] ∗ k = k2 . By complimentary counting, we have that [DM N ] = [ABCD] − [ADM ] − [CDN ] − [BM N ] = 1 − (1 − k) ∗ 12 − (1 − k) ∗ [BM N ] = [DM N ], which implies that X is the midpoint of BC. QED

1 2

−

k 2

= k2 . Thus, we have

57. Denote the intersection of HM with the circumcircle to be H . Then ΔM HC is similar to ΔBM H (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 = M N ∩ AB. Since M N CB is cyclic, ∠M N B = c/2, and since ∠XBN = b/2, ∠N XB = 180 − b/2 − c/2. But, since I is the incenter, ∠BIC = 180 − b/2 − c/2 and so ∠M IB = b/2 + c/2, so M XBI 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. So AB = AM , from there we let AM = y and drop a perpendicular from A meeting M N at D. DB = DM = 2. Using pythagoras in AN D, AM D and AN M , we get: x2 + y 2 = 100 x2 − 64 = y 2 − 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 ∼ Y M H. Similarly, HJC ∼ XIA ∼ XN H. Notice that IX =

XN ·AI HN

=

AI·JH JC

and JY =

Y M ·JC HM

=

HI·JC AI .

Therefore, we have:

IX · JY = JH · HI = ZI · JZ ⇒

JH IX = IH 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 ∠O2 O1 A is half of arc AD with respect to circle (ABD). Hence, ∠AO1 O2 = ∠ABD = ∠ABC and similarly we arrive at ∠AO2 O1 = ∠ACB, so from AA similarity, we are done. Q.E.D.

13

62. First, we draw a diagram:

A

M B

D

C E

Let M be the midpoint of BC. As ABEC is a cyclic quad, ∠EBC = ∠EAC = ∠EAB = ∠ECB, so EBC is 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 AB BE AB 36 BD = =⇒ = = = 2. DE BE DE BD 18 Therefore CE = BE = 2x. Now by Pythagorean Theorem, DE 2 = DM 2 + M E 2 = DM 2 + EC 2 − M C 2 =⇒ 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 = C2 . We then have that ∠AHI + ∠CHI + ∠BIC + ∠AHB = 360, so ∠AHI = 60 + B − C2 = 3B 2 . Q.E.D. ˆ = B Fˆ C. Note 64. Let r and s meet the line CD at E and F , respectively. Angle chasing easily shows that AED that triangles ABC and ADE are similar, and also are the triangles ABD and CBF . This yields DE BC CF AD

= =

AD AB BC AB

=⇒ DE = =⇒ CF =

AD.BC AB AD.BC AB

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 have endpoints D and E, where D is on AC and E is on BC. Let the line through P parallel to AC have endpoints F 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 AF P D, HP EB, and P JCG are all parallelograms (you can angle-chase to see this). Therefore d = DP + P E = AF + HB, so HF = AB − (AF + HB) = 425 − d. Analogously, DJ = 450 − d. Let HP = x, so P J = d − x. We can see that F AP and DP J are similar to ABC (more angle-chasing can prove FA AB DJ 450−d AC 450 = 425−d = BC = 425 this), so AP x 450 and P J = d−x = BC = 510 . 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 = 510 =

510 425 d 960 450 d

−x +x

You can now add these two equations together to get 1020 =

510∗450+960∗425 425∗450

∗ d.

Now before you start multiplying stuﬀ, you can see that a bunch of 5s cancel out from the numerator and the denominator. Then you see that a couple 2s and 3s cancel out as well. After canceling stuﬀ out, you get the equation 1020 =

18+32 15 d

=

10 3 d.

Solving this is simple, and yields d = 306 . 66. Using mid-segments, we can ﬁnd that P1 P2 and P4 P3 are parallel to AC and that P1 P4 and P2 P3 are parallel to BD, so P1 P2 P3 P4 is a parallelogram. Therefore P1 P3 and P2 P4 intersect at the midpoint of P1 P3 . Again using midsegments, P1 P6 and P5 P3 are parallel to AD and P1 P5 and P6 P3 are parallel to BC so P1 P5 P6 P3 is a parallelogram, And we get P1 P3 and P5 P6 also intersect at the midpoint of P1 P3 , 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 AC respectively. Furthermore, let the line parallel to BC passing through P meet AB and AC at M and N respectively. Then it is easy to√see that AM N and ADE are both equilateral, so it is easy to compute √ 8 3 2 3 and P N = . Now the fact that P lies on the incircle implies that ∠DP E = 120◦ by that P M = 3 3 some simple angle chasing. Furthermore, it is obvious that ∠DM N = ∠M N E = 120◦ and that ∠M P D = ∠P DE, ∠N P E = ∠P ED, hence DM P ∼ DP E ∼ P N E. Let DM = EN = x. Then by similarity ratios √ √ 4 3 8 3/3 x √ =⇒ x = , = x 3 2 3/3 √ √ √ √ 14 3 28 3 10 3 4 3 + = . The side length of the triangle is twice this, or , and the so AD = N M + M D = 3 3 3 3 requested answer is 28 + 3 + 3 = 34 . 68. Let X = CQ ∩ P B. X is the A-excenter of ABC. Hence, A, I, X are collinear. Through some angle chasing, we can prove that IC ⊥ CX and that IB ⊥ BX. Hence I, B, X, C are cyclic, so . Also R, P, X, Q are cyclic, and again through some angle chasing we can ∠IXQ = ∠IXC = ∠IBC = ∠ABC 2 ﬁnd that ∠RXQ = ∠RP Q = ∠ABC . Therefore, R, I, X are collinear, and so A, R, I, X are collinear. Chasing 2 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. OP PS = 1 . Let SR 2 OQ X denote the projection of O to the segment P S. Remark that since OP = OS and OP S ∼ RQS, we have ∠OSP = ∠RSQ = ∠RQS. Furthermore, since P R is tangent to C1 and since P QRO is cyclic, we have

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

∠ROQ = ∠SP R = 12 ∠SOP = ∠SOX. Therefore SXO ∼ QRO and 1 PS QR PS SX OP PS = = = 2 =⇒ 1 = , QO SO PO QR SR OQ 2

as desired. Q.E.D.

15

70. Let ∠P AD = x, then ∠DAB = 2x, ∠ADC = 180 − 2x, and ∠ADP = 90 − x. Hence ∠AP D = 90. Now let M and N be the midpoints of BC and DA, respectively. It follows that M is the circumcenter of AP D. Then ∠P M D = 2 · ∠M AP = 2x. So M P//BC//AB. But that means that M P is the mid line of ABCD. But by symmetry N Q is the mid line as well. So M , P , Q, and D form a line. Notice that M P is equal to the circumradius of AP D, but since it is a right triangle, that value is Now it follows that 1 1 · · AD · d(AB, CD) AD [AP D] = 2 2 = [ABCD] M N · d(AB, CD) 2AB + 2CD

AD 2 .

We have a symmetrical result for BCQ. We can subtract those areas from the total area of the trapezoid to get the desired area. Hence 2AB + 2CD − BC − AD 2 · 11 + 2 · 19 − 5 − 7 4 [AP DCQB] = = = [ABCD] 2AB + 2CD 2 · 11 + 2 · 19 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 AB + DC 11 + 19 15 [ABCD] = = = [AB D] A D 8 4 We can ﬁnish the problem by multiplying those ratios: √ √ 4 15 = 30 3 [AP DCQB] = 10 3 · · 5 4 71. Consider the homothety that takes w to w1 . This homothety takes X to A and Y to D, so AD||XY by homothety. Similarly, we have BC||XY . Now consider a reﬂection of A and B over the line that goes through the centers of the two circles. Since this line is perpendicular to the radical axis of w1 and w2 , or XY , it is also perpendicular to AD and BC. Thus, it takes A to a point on AD, but since this point is also on w1 , it must 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 M A(M C + CD ) = M B(M D) so M A(M D ) = M B(M D). 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 the sum 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 X

Y

C B

A D 16

C

By simple angle chasing, ∠AB C = ∠AA C = ∠ABC. Using traditional chords-intercepting-arcs formulae +Y = AY . Thus AXY is isosceles and DA bisects ∠XDY . = 1 (AX we can thus write 12 AC C) =⇒ AX 2 It is well known that AB A ∼ AA C, so A A2 = AB · AC. In addition, ∠AY X = ∠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, CP Q respectively. Now from Miquel’s theorem, the three circumcircles concur at some point M . ∠O3 O1 O2 = ∠O3 O1 M + ∠M O1 O2 = 90 − ∠O1 M Q + 90 − ∠O1 M P = 180 − ∠P M Q = ∠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

O X

B A

C

H

Let X be the foot of the altitude from A to BC, and denote by H the second intersection point of AX with = CH and A H BC. This means the circumcircle of ABC. By Problem 25, ∠BAA = ∠CAH , so BA that the distances to BC from A and H are equal; in other words, [BA C] = [BH C]. The area of BH C is much easier to compute. By Law of Cosines, we have √ 52 + 82 − ( 41)2 3 AC 2 + BC 2 − AB 2 = = , cos ∠ACB = 2 · AC · BC 2·5·8 5 so XC = 3. Thus BX = 5 and by Pythagorean Theorem AX = 4. Finally, Power of a Point gives BX · CX = AX · XH =⇒ XH =

15 BX · CX = , AX 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 = 24·7·16 lies on the ray CI. Now note that r = [ABC] s 128 = 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, Q respectively onto BC. Then note that BP R ∼ BIS ∼ CQT . Thus from the ﬁrst similarity, we have IS 16 21 64 PR = =⇒ = =⇒ RB = RB SB RB 28 3 This means that SR = 28 − RB = 20 3 . Denoting the radius of circle Q as r, from the second similarity we 84−4r receive that T C = 4r 3 so T S = 28 − T C = 3 . Now consider trapezoid T QP R. Let Q denote the

17

projection from Q onto P R, and then consider right triangle QQ P . We have QP = 16 + r and P Q = 16 − r as well as QQ = T R = 104−4r . Then pythagorean theorem yields 3 2

(16 − r) +

104 − 4r 3

2

√ = (16 + r)2 =⇒ r = 44 − 6 35

so our ﬁnal answer is 254 . 77. From problem 26, P AC ∼ P BD. P M and P N are corresponding medians in these similar triangles so PA they follow the same ratio: PP M N = P B . Also because they are corresponding medians, ∠M P A = ∠N P B. So ∠N P M = ∠DP A − ∠M P A − ∠DP N = ∠DP A − ∠DP B = ∠BP A. Therefore P AB ∼ P M N ∼ P CD as desired. Q.E.D. 2 78. Let BC = 2a and let AC = 2b. Then, we can establish the following: 2(4b2 + 4a ) − 576 = 4(729) √ √ √ 3 2 2 2 2(24 + 4b ) − 4a = 4(324) Solving this system gives us a = 279 = 3 31 b = 315 2 = 2 35

Also, notice that [AF B] = 2[F EB] = 2[AF E]. Because [F EB] = 6F E sin ∠F EB = 6F E sin ∠BEC, we need to ﬁnd F E and sin ∠BEC. The former is quite easy; indeed, with power of a point, we obtain 27F E = 144 F E = 16 3 Now, we are looking for 32 sin ∠BEC. Finding sin ∠BEC is slightly more tedious; however, since we know √ BC = 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 = 243 = 38 Thus, sin ∠BEC = 855 , 648 √ and then we plug it into 32 sin BEC. 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 any point on the extension of BC past C, we have ∠AGF = ∠F CX = ∠F CG = ∠F AG, so AF G is isosceles as desired. Q.E.D. 80. Extend N M such that it intersects AB at M . This is simply an extension of the radical axis. Notice that the power of point M with respect to G2 is M B 2 , 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 that M B 2 = 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 is the midpoint of AB, M must be the midpoint of P Q. Now draw EE such that E lies on CD and EE is perpendicular to CD. Then, EE C is similar to AF C, and EE D is similar to BGD. Let CF = x, and let F E = xk. Also, let F A = h, and by similarity, HE = hk. Additionally, let GD = y, and therefore, EH EE =E AB = x + y. Since AH = F E = xk, HB = x(1 − k) + y. Then, by more similarity, we have HB D h(k+1) hk 2 = 2ky = (k + 1)(x − xk + y) xk + yk − (x + y) = 0, x(1−k)+y 2y Which is a quadratic in k. Solving for k via the quadratic formula gives us k = −1 − xy and k = 1. However, it is fairly obvious that k must be positive, and hence, we take k = 1. Hence, F E = F M = x, and M = E . Because EE was perpendicular to CD, EM must be perpendicular to CD. Now, because ∠EM C = 90 and M is the midpoint of P Q, it immediately follows that EP = EQ, which is what we wanted to prove. Q.E.D.

18

Here’s a diagram:

T P B

C

A

81. Let O be the center of ω. Let ∠T OC = θ. Then note that we have T C = 9 tan θ and AC =

9 TC − AO = −9 sin θ cos θ

Then since ACP ∼ OCT , we have 9 cos θ − 9 cos θ

9

=

AP =⇒ AP = 9 − 9 cos θ 9

Now from the Law of Cosines on P AB, 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 = 2

−162 2(−243)

=

1 3

and plugging this value in we get that

m = −27 + 54 + 405 = 432 . 82. The diagram:

D N B

OB A OA

P

M C

Let O be the midpoint of AB, the center of the semicircle (O) with the points C, D forming a convex and ON = OB cos BON . ON AD is a midline of the triangle quadrilateral ABDC. OM = OA cos AOM ABD and OM BC is a midline of the triangle ABC. Let E, F be the midpoints of PC, PD. Denote 19

α = ∠(OOA , P E) = ∠(OM, P C) and β = ∠(OOB , P F ) = ∠(ON, P D). From the trapezoid OOA EP , PE PF AB OOA = cos α and from the trapezoid OOB F P , OOB = cos β . Since OA = OB = 2 and PD P E = P2C = CD 4 = 2 = P F , we get

cos β OOA = cos AOM , OO = cos α But from the circle (O) and form AD ON, BC OM , B cos BON α = ∠(OM, P C) = ∠(BC, DC) = ∠BCD = ∠BAD = ∠BON OOA β = ∠(ON, P D) = ∠(AD, CD) = ∠ADC = ∠ABC = ∠AOM Consequently, OM ON = OOB , which means OM ON

OA OB M N . Q.E.D.

83. This will probably help:

C

IC IB A

D

P

B

Notice that ∠BP C = ∠BP D + ∠CP D. Also, we have ∠BP D = 180 − ∠DIB B, and ∠CP D = 180 − ∠CIC D. ) and Then, ∠BP C = 360 − (∠DIB B + ∠CIC D). Also, we have ∠DIB B = 180 − ( ∠ADB+∠DBA 2 ∠ACD+∠ADC ∠ADB+∠ADC+∠ACD+∠DBA ∠CIC D = 180 − ( ). Adding, we obtain ∠DIB B + ∠CIC D = 360 − ( ) 2 2 180−(∠DCB+∠DCB)−∠CAB ∠CDB−∠CAB ∠ADB+∠ADC 360−∠CDB Notice that ∠ACD+∠DBA = = , and = . 2 2 2 2 2 Then, we add, and get ∠DIB B + ∠CIC D = 360 − 360−∠CAB . Now, it remains to ﬁnd ∠CAB. Luckily, we 2 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 = 360−60 2

1 2

∠CAB = 60

Then, ∠DIB B + ∠CIC D = 360 − = 210 and ∠BP C = 360 − (∠DIB B + ∠CIC D) = 150. Now, we have to ﬁnd the maximum area of BP C with a side length 14 and its opposite angle 150. The area is maximized when BP C is isosceles, and we can use the law of cosines to ﬁnd the length of the legs: √ √ 2r2 − 2r2 cos 150 = 196 2r2 (1 + 23 ) = r2 (2 + 3) = 196 Notice that to ﬁnd the area, it suﬃces to ﬁnd √ 2 r2 sin 150 12 = r4 . Then, A = 2+49√3 = 98 − 49 3 and our answer is 98 + 49 + 3 = 150 . 84. Let X = BM ∩ CN , and deﬁne B = AB ∩ M N , C = AC ∩ M N . First note that the condition ∠P AB = ∠BCA implies P AB ∼ ACB, and likewise ∠QAC = ∠CBA implies CAQ ∼ CBA. Therefore ∠P QA = ∠BAC = ∠AP Q and AP Q is isosceles. Next note that a homothety centered at A with scale factor 2 send ABC to AB C (due to the fact that it sends AP Q to AM N ), so additionally ACQ ∼ AC N , etc. Length ”bashing” gives AB CA C A C C AB = = = = , AM 2AP 2CQ 2C N CN which coupled with ∠P AB = ∠AC N gives BAM ∼ CC N =⇒ ∠XM A = ∠XN M . Similarly, ∠XM N = ∠XN A. Therefore 1 ∠BXC = π − (∠XN M + ∠XM N ) = π − (∠AN M + ∠AM N ) 2 1 = π − (∠BAC + ∠BAC) = π − ∠BAC, 2 establishing the desired cyclicity. Q.E.D. 20

85. WLOG AB < AC. Say the incircle meets BC at E, AB at F. Then DE 2 = 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 or = 15 + 2b . From Stewart’s a = c + 10 ⇒ c = 10. We chose c = AB < AC = b, so c = 10. Then s = a+b+c 2 2 2 2 2 theorem, we can easily derive 2x2 = b +c 9−200 . Finally, (s − a)2 = b +c 9−200 . Putting all variables in terms of 36b + 260. Solving, b √ = 10 or b = 26. Obviously, b = 26 is what we want. Moving b, we get the quadratic b2 − to Heron’s formula, Area = (28)(8)(18)(2) = 24 14. The answer is 38 . 86. We will show that < I2 I1 I4 = 90 degrees. First, we claim that A1 A2 I1 I4 is cyclic. Since < A1 I4 A2 = 90 + 1/2 < A1 A4 A2 and < A1 I1 A2 = 90 + 1/2 < A1 A3 A2 (angle bisector angle formulas), and < A1 A4 A2 =< A1 A3 A2 (cyclic quad A1 A2 A3 A4 ), we have < A1 I4 A2 =< A1 I1 A2 , which makes A1 A2 I1 I4 cyclic. By similar arguments, we show that A2 A3 I2 I1 is cyclic. Now all that remains is a basic angle chase. We have < I2 I1 I4 = 360− < A2 I1 I4 − < A2 I1 I2 = 360 − (180− < I4 A1 A2 ) − (180− < I2 A3 A2 ) = 1/2(< A4 A1 A2 + < A2 A3 A4 ) = 1/2(180) = 90. Similarly, we can show that the other three angles are 90 degrees, which makes I1 I2 I3 I4 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 want to show that the ratio M E/N F = P E/P F . Since AM = AN , we have M E/N F = (AM ∗ M E)/(N F ∗ AN ), which is the power of M wrt (with respect to) circle C divided by the power of N wrt circle C. Now, extend M P and N P to intersect circle C again at points M and N , respectively. Draw the horizontal line through P (which is also the tangent line to both circles) and let X be a point on that line to the left of P and Y be a point on that line to the right of P . Now we have < P M N =< Y P N =< XP N =< M N P (the ﬁrst and last equality follow by cyclic quads). Similarly, we get < P N M =< P N M . Therefore, triangles M N P and M N P are similar by AA similarity, so M P/N P = M P/N P . This also means that M M /N N = M P/N P . Now we can further simplify the original ratio. Since the power of M wrt circle C is also equal to M P ∗ M M , and the power of N wrt circle C is equal to N P ∗ N N , we can rewrite the ratio as (M P/N P )2 . By Law of Sines in triangle M N P , M P/N P = sin(P N M )/ sin(P N M ) = sin(AM P )/ sin(AN P ). Thus, we can write the ratio (M P/N P )2 as (M P/N P )(sin(AM P )/ sin(AN P ). By using Law of Sines once more on triangles EMP and FPN, we can rewrite this ratio as (EP/F P )(sin(AEP )/ sin(P F N )). Finally, we notice that < AEP =< AF P = 180− < P F N , so sin(AEP ) = sin(P F N ). Thus, we can simplify the ratio to EP/F P , 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 the midpoint of AA2 and A is the midpoint of AA1 , A L is the midline of triangle AA1 A2 , so A L||A1 A2 . Thus, we have that triangles BCL and BA1 A2 are similar. Similarly, we get that triangles ACL and AB1 B2 are similar. Note that O1 is inside triangle AB1 B2 and O2 is outside triangle BA1 A2 , so O1 and O2 are outside of lines AC and BC, respectively. In order to have < O1 CB1 =< O2 CA1 , we would like to show that triangles O1 CA and O2 CA1 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 like this). Notice that we have AB2 = BA2 by reﬂection, and < AB1 B2 =< ACL =< BCL =< BA1 A2 , so the circumradii of triangles AB1 B2 and BA1 A2 are equal. Thus, O1 A = O2 A1 . By reﬂection, we have CA = CA1 . Now, all we need to show is that < CAO1 =< CA1 O2 , as then we would be done by SAS congruency. < O1 AC = 180− < O1 AB1 = 180 − (90− < B1 B2 A) = 90+ < ALC, where the equality < O1 AB1 = 90− < B1 B2 A follows since O1 is inside triangle AB1 B2 . Similarly, we get < O2 CA1 = 180− < O2 A1 B = 180 − (< A1 A2 B − 90) = 90 + (180− < A1 A2 B) = 90 + (180− < CLB) = 90+ < CLA. Thus, < O1 AC =< O2 CA1 , which means triangles O1 CA and O2 CA1 are congruent by SAS, which means that < O1 CA =< O2 CA1 =< O2 CB. Q.E.D.

21

89. Since F G is a tangent to the circumcircle of EDM , < EM D =< DEG =< CEF . Also, since < BCD =< BAD, triangles M AD 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 have EF/EC = M D/AM . Cross-multiplying, we get EF ∗ AM = EC ∗ M D. Also, we have < AGF = 180− < GCF − < GF C =< ADB− < GF C =< M DB, where the last equality followed from similar triangles M AD and ECF . Now we want to ﬁnd a way to express GE, since we already have a way to express EF. Thus, we look for similar triangles involving GE. Fortunately, since < CGE =< M DB and < GCE =< DBM , we have that triangles GCE and DBM are similar. Thus, we can write the ratio GE/CE = DM/BM . Cross-multiplying, we get GE ∗ BM = CE ∗ DM . Fortunately, this equals 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 get AB/AM = 1/t. Subtracting 1 from both sides, we get M B/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 ck s will be some constant. Consider the homothety H that sends w1 to w2 centered at N. It is well known that A, K, and M are collinear. Now, since arcs MB and AM are equal, < BAM =< ABM =< BN M =< AN M . Thus, triangles BN M and KBM are similar by AA. We can now write the ratio BM/KM = N M/BM , so BM 2 = N M ∗ KM . Since H takes N K to N M , N K/N M is a constant, which means that N M/KM is also a constant. Thus, we can write BM 2 = KM 2 ∗ c1 . Taking the square root, we get BM = KM ∗ c2 . Now we divide both sides by sin(AN M ). Since BM/ sin(AN M ) = 2R = constant, we have (KM/ sin(AN M )) ∗ c2 = c3 , so (KM/ sin(AN M )) = c4 . Finally, dividing by 2, we get (KM/2 sin(AN M )) = c5 , and since (KM/2 sin(AN M )) is the circumradius of triangle KBM , and it is constant, we are done. Q.E.D. 91. Extend P C past the circle and intersect it with the tangent to the circle at E. Call this point P . Now draw the other tangent from P to the circle and label the tangency point as B . Let the intersection of BE and B D be X. By Pascal’s Theorem with the conﬁguration DDB’BBE, P , X, and the intersection of DE and BB are collinear. Similarly, with the conﬁguration EEBB’B’D, P , X, and the intersection of DE and BB are collinear. Since both pairs of collinearities have two common points, they are indeed the same line. Also, since P and P are both points on the line, the line is indeed P P . 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 M N look parallel in the diagram, we will try to prove this. Since OM = ON , O lies on the perpendicular bisector of M N . Now let OA be the circumcenter of triangle AM N . Clearly, OA also lies on the perpendicular bisector of M N . Now, we want to prove that the perpendicular bisector of AQ passes through O and OA , as this would mean that AQ and M N are perpendicular to the same line, which would make them parallel. Fortunately, as AQ is the radical axis of the two circles, the perpendicular bisector of AQ must pass through O and OA , which means that AQ||M N . Also, since they are parallel, and AQM N is cyclic, AQM N is an isosceles trapezoid. If AQ is tangent to circle w, then triangles AQB and ACQ are similar (prove by PoP and SAS). Thus, we would like to prove that these triangles are similar. This is equivalent to proving that < P QB =< P CQ, as the result would follow by AA similarity. Now we do a little angle chase on < P CQ. We have < P CQ =< BCQ =< BAQ =< AM N =< P M B, where the last equality follows from parallel lines AQ and M N . Thus, we would like to prove that < BQP =< BM P , or that quadrilateral BM QP is cyclic. Since cyclic quadrilaterals have a lot of nice angle properties, we will aim to prove that < BP M =< BQM , as we can manipulate these angles the best. First we compute < BP M . < BP M =< CP N = 180− < C− < P N C = 180− < QAN − < C =< QBC− < C =< QBA+ < ABC− < C, where the equality < P N C =< QAN comes from parallel lines AQ and M N . 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 and OAN are similar, so < OQM =< OAC = 90− < ABC. Now we can substitute these values to get < BQM = (90− < QCB) − (90− < ABC) =< ABC− < QCB. Since < BP M =< BQM , quadrilateral BM QP is cyclic, which implies that < P QB =< AQN =< P CQ, which implies that triangles P QB and P CQ are similar, which implies that P Q is tangent to circle w. Q.E.D. 93. Let the midpoint of T B be E. Then, since < XET =< XM T = 90, quadrilateral XM ET is cyclic. Thus, we have < BT M =< ET M =< EXM =< EXB+ < BXM . Now we want to ﬁnd a value for < CT M . Notice that, since E and M are the midpoints of BT and BC, respectively, EM ||CT . Therefore, < CT M =< EM T . By cyclic quadrilateral XM ET , < EM T =< EXT =< EXB. Now we have that < BT M − < CT M = (< 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 G over the diameter AB, we have that HG is perpendicular to AB, so it is parallel to line l. With nothing else in mind, we angle chase a little on < HCA. < HCA =< HGA =< B F A =< DF A. Thus, we want to prove that < DF A+ < ACF = 180. Since < ACF = 180− < DCF , we want to prove that < DF A + (180− < DCF ) = 180, or < DF A =< DCF . Since triangles AF D and F CD have a shared angle at < ADF , it suﬃces to prove that triangles AF D and F CD are similar. After a (hopefully short) attempt to prove that < DAF =< DCF , as this leads nowhere, we move on to try and prove that the triangles are similar by SAS. Thus, we want to show that AD/DF = F D/DC. Cross-multiplying, we get that we want to show that AD ∗ DC = DF 2 . Now notice that AD ∗ DC = DC ∗ DA, which is the power of point D with respect to the circle. However, since DE is tangent to the circle, it is also equal to DE 2 . Thus, we want to show that DE 2 = DF 2 , or DE = DF . Finally, we can angle chase on triangle DEF . Let < DF E =< 1. Then, since AB is perpendicular to B F , < F BB = 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 =< DF E =< 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 both isosceles, AM ⊥ M O and AN ⊥ N O =⇒ AM ON is cyclic. In addition, note that angle chasing gives ∠P DE = ∠P AE ≡ ∠P AB = ∠P CB. Similarly, ∠P ED = ∠P BC, so P ED ∼ P BC. Therefore by Problem 77 P ED ∼ P M N and by Problem 26 P EM ∼ P DN . This implies ∠P M A = ∠P N A and so (AP M N ) is also cyclic. Hence A, P, M, O, N all lie one circle, implying that ∠AP O = ∠AM O = 90◦ as desired. Q.E.D. 96. It is easy to see that ABOP C is concyclic. Because BQP M is cyclic (by deﬁnition), we have ∠M BQ = ∠QP O = 90◦ . Furthermore, we have ∠BQM = ∠BP M = ∠BP O = ∠BAO, so M BQ ∼ OBA. Therefore, there is a spiral similarity about B mapping M Q to AO. Thus, M BO ∼ QBA. We therefore have OM = Similarly, we can arrive at ON = Q.E.D.

OC·AQ AC

OB · AQ AB

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

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 is tangent, but whatever.) Furthermore, since OY ⊥ AZ ⊥ BZ, where O is the center of the circle, we have KY ⊥ KB, so ALKB is cyclic. AY 2 2 Next, I claim that AK AX = ( AX ) . To prove this, note that ALY ∼ AY Z, so AY = AL · AZ. In addition, since ∠ALB = ∠AKB by cyclicity, ALX ∼ AKZ, so

2 AY AL AK AZ · AL AK = =⇒ = = , AZ AX AX AX 2 AX

as desired. Let AL = and WLOG let AB = 2. From standard altitude-to-hypotenuse calculations, it may be deduced 2 that AX = √4+ . In addition, since ∠LY O = ∠LAO = 90◦ , ALY O is cyclic, so by Ptolemy 2 AL · OY + AO · LY = AY · OL =⇒ 2 = AY

2 1 + 2 =⇒ AY = √ . 1 + 2

Therefore AK + AX

AL AB

2 =

AY AX

2 +

AL AB

2 =

2 √ 4 + 2 3 2/ 1 + 2 2 2 2 √ = =1+ + + + . 2 2 2 4 1+ 4 1+ 4 2/ 4 +

Let S denote this sum. It suﬃces to ﬁnd the maximum value of S. A way to do this without calculus is as follows: subtracting 1 and adding 14 to both sides gives √ 3 3 1 + 2 1 + 2 3 ≥ 2 = 3. + · S− = 2 2 4 1+ 4 1+ 4 Therefore S ≥

3 4

+

√

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

98. Very large diagram:

A

B

P

E

C D

AD Clearly, we have ABC ∼ ADE (they practically give that to you) so we have AB AC = AE , which rearranges to AB AC AD = AE 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 ∠P DC = ∠BDC = ∠ADC − ∠ADB = ∠AED − ∠AEC = ∠P ED Which means that CD is tangent to circle (AP DE). Similarly, CD is tangent to (ABCP ). Therefore, since AP is the radical axis between (AP DE) 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 of Ai Ai+2 with Oi+2 , taking A5 = A1 , etc. Furthermore, let P be the intersection of the diagonals A1 A3 and A2 A4 . The key observation is to note that A1 A3 is the radical axis of O2 and O4 , and that A2 A4 is the radical axis of O1 and O3 . This implies that P has equal power with respect to all four circles. One consequence of this is that P A1 · P X1 = P A4 · P A2 . Expansion gives P A1 · (P A3 + A3 X1 ) = (P X2 + X2 A4 ) · P A2 P A1 · P A3 + P A1 · A3 X1 = P X2 · P A2 + X2 A4 · P A2 . By PoP we see that P A1 · P A3 = P A2 · P X2 , so P A2 · A4 X2 + A3 X1 · P A1 = 0 =⇒

1 1 + = 0. A3 X1 · P A1 P A2 · A4 X2

Next, clever manipulation gives 1 A 1 A3 P A3 − P A1 = = A3 X1 · P A1 A1 A3 · A3 X1 · P A1 A1 A3 · A3 X1 · P A1

P A3 1 1 = − . A1 A3 A3 X1 · P A1 A3 X 1 Through the fact that P has equal power with respect to all four circles, we obtain that P X1 · P A1 = P X3 · P A3 (P A3 + A3 X1 ) · P A1 = (P A1 + A1 X3 ) · P A3 A3 X1 · P A1 = A1 X3 · P A3 . Thus 1 A1 A3

P A3 1 − A3 X1 · P A1 A3 X 1

1 1 − A1 X 3 A3 X 1 1 1 + . = A1 A 3 · A1 X 3 A3 A1 · A3 X 1 =

1 A1 A 3

1 1 1 + = . Finally, note that A1 A3 · A1 X3 is equal to the A2 A4 · A2 X 4 A4 A 2 · A4 X 2 A4 X2 · P A2 power of the point A1 with respect to O1 , which in turn is known to be equal to O1 A21 − r12 . Thus, substituting our new expressions into one of our earlier equations and replacing all of the denominators with corresponding (cyclic) Power of a Point expressions gives Similarly, we get

1 O1 A21

−

r12

+

1 O2 A22

−

r22

+

1 O3 A23

−

r32

+

1 O4 A24

− r42

= 0,

as desired. Q.E.D. 100. Let O denote the circumcenter of ABC, and WLOG let AB < AC. By simple angle-chasing, ∠BF C = ∠BDM + ∠AEF = 2∠BAD + 2∠EAC = 2∠BAC = ∠BOC, 25

so BF OC is a cyclic quadrilateral. In addition, note that the circumcircle of AP N passes through O since ∠AP O = ∠AN O = 90◦ . Therefore, if F is to lie on (AP N ), OF must be the radical axis of the circumcircles of AP N and BOC. Now let X denote the point at which the tangent to the circumcircle of ABC at A meets BC. As AP N and ABC are homothetic, their circumcircles are tangent, and as such AX is the radical axis of the two circles. Combining this with the fact that BC is the radical axis of (BF OC) and (ABC) gives that X is the radical center of all three circles. Thus what we want to prove is simpliﬁed to proving that X, F, and O are collinear. Note that by some more angle-chasing we see that ∠F OC = ∠XBF = ∠XBA + ∠ABD = ∠XAC + ∠BAD = ∠XAM + ∠A. In addition, note that ∠OM C = 12 ∠BOC = ∠A, so ∠XAM = ∠F OM . However, since ∠XAO = ∠XM O = 90◦ , quadrilateral XAOM is cyclic, so ∠XAM = ∠XOM . Therefore ∠XOM = ∠F OM , 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.

26