A Geometry Problem Set

Transcription

1 A Geometry Problem Set Thomas Mildorf Jauary 4, 006 All aswers are itegers from 000 to 999 iclusive. Docile. Right triagle ABC with right agle C has sidelegths AC = ad BC = 4. Altitude CD is costructed, with D o the hypoteuse of ABC. The legth of CD ca be expressed as p, where p ad q are relatively prime positive itegers. Compute p + q. q Aswer: 07. Observe that m ACD = 90 m CAD = m ABC. It follows that triagles ACD ad ABC are similar. Now sice AB = 5 ad CD = BC, we have CA BA CD = 4 = I scalee triagle ABC, D is the midpoit of BC, E is the midpoit of AC, ad F is the midpoit of AB. The area of triagle DEF is 6. Compute the area of triagle ABC. Aswer: 04. Because D ad E are the midpoits of BC ad AC, DE is parallel to AB ad is half as log. It follows that each side of DEF is / as log as the correspodig side of ABC. Hece, the area of DEF is (/) = /4 of the area of ABC.. A, B, ad C are poits o a lie i that order, with BC = 6 ad AB = 4. Poits E ad F are chose o the same side of this lie such that EC = 7, AE = 6, BF = 0, ad CF = 7. Let the itersectio of BF ad CE be D. The value of the expressio [ABDE] ca be expressed as p, where p ad q are relatively prime positive itegers. [CDF ] q Compute p + q. Aswer: 00. Note that AC = AB + BC = 0, so that triagle AEC is cogruet to triagle BCF. Now [ABDE] = [AEC] [BDC] = [BCF ] [BDC] = [CDF ], so the desired ratio is. 4. Three circles are mutually exterally taget. Two of the circles have radii ad 7. If the area of the triagle formed by coectig their ceters is 84, the the area of the third circle is kπ for some iteger k. Determie k.

2 Aswer: 96. Let r deote the radius of the third circle. The the sides of the triagle are 0, + r, ad 7 + r. Usig Hero s formula ad equatig this with the give area, we have 84 = (0 + r)(r)(7)() from which r = 4, 4. Sice r is positive, it follows that the area of the third circle is 96π. 5. ABCDEF G is a regular heptago, ad P is a poit i its iterior such that ABP ( ) is p. equilateral. p ad q are relatively prime positive itegers such that m CP E = q Compute the value of p + q. Aswer: 667. Sice ABP is equilateral, BP = BA = BC, hece BCP = CP B. Let α deote the degree measure of each of the agles of ABCDEF G. The m P CB = α 60 from which m CP B = m BCP = 0 α ad m P CD = α 0. By symmetry, P lies o the agle bisector of DEF, thus m DEP = α. Fially, as m CDE = α, we have m EP C = 60 α α ( α 0 ) = 480 α. Computig α = , we fid that m EP C = = The legth of a diagoal coectig opposite vertices of a rectagular prism is 47. Determie its volume, give that oe of its dimesios is ad that the other two dimesios differ by 005. Aswer: 00. Let the other dimesios be x ad x The by the distace formula, 47 = + x + (x + 005) = x(x + 005). It follows that the desired volume is ABCDEF is a regular hexago of area 00. GHIJKL is the hexago formed by coectig adjacet midpoits of the sides of ABCDEF. Compute the area of GHIJKL. Aswer: 075. Suppose G ad H are the midpoits of AB ad BC respectively, ad let O be the ceter of ABCDEF. Note that GHIJKL is also regular ad shares ceter O. Observe that GOA is a right triagle. Sice ay two regular hexagos are similar, we have [ABCDEF ] = OA = 4. The aswer follows. [GHIJKL] OG 8. ABC is a equilateral triagle ad D is a poit o mior arc AB of the circumcircle of ABC such that BD = 005 ad CD = 006. Compute AD. Aswer: 00. Sice AB = BC = CA, Ptolemy s theorem 7. AD BC + AC BD = AB CD reduces to AD + BD = CD, which immediately gives AD. Demadig. ABCD, a rectagle with AB = ad BC = 6, is the base of pyramid P, which has a height of 4. A plae parallel to ABCD is passed through P, dividig P ito a frustum F ad a smaller pyramid P. Let X deote the ceter of the circumsphere of

3 F, ad let T deote the apex of P. If the volume of P is eight times that of P, the the value of XT ca be expressed as m, where m ad are relatively prime positive itegers. Compute the value of m +. Aswer: 77. P ad P are similar; sice the volume of the former is 8 times that of the latter, if follows that the plae passes through P halfway up the pyramid P. Let Z be the apex of P, ad A, B, C, ad D the midpoits of A Z, B Z, C Z, ad D Z respectively. A B C D, the rectagular itersectio of the plae ad P, has A B = C D = 6 ad B C = D A = 8. Let O ad O deote the ceters of ABCD ad A B C D respectively. Sice the height of P is 4, OO =. By symmetry, the circumsphere of the frustum F is cetered o OO. Sice for ay poit X o OO, we have AX = BX = CX = DX ad A X = B X = C X = D X, we eed oly fid the poit X such that AX = A X. Suppose that OX = x ad XO = x. By the Pythagorea theorem i -space, we have The XT = = 507 AX = A X x = ( x) = x = x + x x = 69 4, so the aswer is = 77.. ABC is a scalee triagle. Poits D, E, ad F are selected o sides BC, CA, ad AB respectively. The cevias AD, BE, ad CF cocur at poit P. If [AF P ] = 6, [F BP ] = 6, ad [CEP ] = 4, determie the area of triagle ABC. Aswer: 5. BF [F BP ] = = F A [AF P ] CD BF AE CD = = = 48 DB F A EC DB k [ADC] [P DC] [AP C] = = 48 [ABD] [P BD] [AP B] k Sice triagles AF P ad F BP share a altitude from P, we have AE. Let [EAP ] = k. By similar reasoig,. By Ceva s theorem, = k EC 4 = [ABD] [ADC] [P DC]. Now we ote that = CD = 48. Hece, [P BD] DB k. We use the fact that [AP C] = [AP E] + [EP C] = k + 4 ad [ABP ] = [AF P ] + [F BP ] = = 89. We have 4 + k = k = k + 4k k = 4 ± = ± = ± + 6 = 08, 84 We take k = 84 sice it represets a area. Now, AE = 7 CD ad = 4. By Meelaus EC DB 7 theorem, BF AP DC = (Ceva ad Meelaus use the covetio of directed distaces, F A P D DB where XY = Y X.) This yields AP = [ABP C] from which =. Hece, [ABC] = P D [P BC] [ABP C] = ( ) = 5. ALTERNATE SOLUTION

4 Assig the weights,, ad ω to A, B, ad C. It must be that [EAP ] = 4ω, [DCP ] = k, ad [BDP ] = ωk for some k. But we have = [DCP ] [P CA] = w [BDP ] = [DCA] [BDA] [BP A] = 4(ω+) = 8(ω+). We solve this quadratic for ω = 7, 9, ad choose the former 6+ 6 sice 4ω is a area. But the weight o D is ω + so that ω+ = AP [ABP C] =. P D [P BC] Substitutig, = which implies that [P BC] = 54. Therefore, [ABC] = [P BC] [ABP C] + [P BC] = = 5.. ABCD is a cyclic quadrilateral that has a iscribed circle. The diagoals of ABCD itersect at P. If AB =, CD = 4, ad BP : DP = : 8, the the area of the iscribed circle of ABCD ca be expressed as pπ, where p ad q are relatively prime q positive itegers. Determie p + q. Aswer: 049. Because ABCD has a icircle, AD + BC = AB + CD = 5. Suppose that AD : BC = : γ. The : 8 = BP : DP = (AB BC) : (CD DA) = γ : 4. We obtai γ =, which substituted ito AD + BC = 5 gives AD =, BC =. Now, the area of ABCD ca be obtaied via Brahmagupta s formula: s = +++4 = 5, K = (s a)(s b)(s c)(s d) = 4 ad K = rs = 5r, where r is the iradius of ABCD. Thus, r = 4 from which its area 4π yields the aswer = ABC is a isosceles triagle with base AB. D is a poit o AC ad E is the poit o the extesio of BD past D such that BAE is right. If BD = 5, DE =, ad BC = 6, the CD ca be expressed as m, where m ad are relatively prime positive itegers. Determie m +. Aswer: 5. Draw i altitude CF ad deote its itersectio with BD by P. Sice ABC is isosceles, AF = F B. Now, sice BAE ad BF P are similar with a scale factor of, we have BP = BE = 7, which also yields P D = BD BP = 5 7 =. Now, applyig Meelaus to triagle ADB ad colliear poits C, P, ad F, we obtai AC DP BF CD P B F A = AC DP CD P B = = CD = AC DP ( ) P B = 6 ) = 08 7 where the mius sig was a cosequece of directed distaces. The aswer is therefore = P is a pyramid cosistig of a square base ad four slated triagular faces such that all of its edges are equal i legth. C is a cube of edge legth 6. Six pyramids similar to P are costructed by takig poits P i (all outside of C) where i =,,..., 6 ad adjoiig each to the vertices of the earest face of C, usig each face of C oce. The volume of the octahedro formed by the P i (takig the covex hull) ca be expressed A system of liear measure i which for ay poits A ad B, AB = BA. ( 7 4

5 as m + p for some positive itegers m,, ad p, where p is ot divisible by the square of ay prime. Determie the value of m + + p. Aswer: 44. By a Pythagoras argumet, the legth of the altitude of each pyramid from P i to the earest face of C is. Therefore, the distace from opposite vertices of the octohedro is 6 + = 6 ( + ). Let P s ad P t be a pair of opposite vertices. The square formed by the other four vertices of the octahedro has area (6 ( + )) ( ) = 8 +. Fially, the volume is give by ( ) ( ) 6( + ) 8( + ) = Lie segmets AB ad CD itersect at P such AP = 8, BP = 4, CP =, ad DP =. Lie segmets DA ad BC are exteded past A ad C respectively util they itersect at Q. If P Q bisects BQD, the AD ca be expressed as m, where m BC ad are relatively prime positive itegers. Determie m +. Aswer: 046. Let E ad F be the projectios of P oto AD ad BC respectively. Note that the agle bisector coditio is equivalet to P E = P F. It follows that AD [AP D] AP DP = = =. BC [BP C] BP CP 7. Three spheres S, S, ad S are mutually exterally taget ad have radii 004, 507, ad 4676 respectively. Plae P is taget S, S, ad S at A, B, ad C respectively. The area of triagle ABC ca be expressed as m, where m ad are positive itegers ad is ot divisible by the square of ay prime. Determie the remaider obtaied whe m + is divided by 000. Aswer: 707. The radii are obviously deliberately chose, but the reaso is ot yet apparet. Write r i for the radius of S i, ad let O i be the ceter of S i. Let be P the projectio of O oto O B. The O P O is a right triagle with O O = r + r ad P O = r r. It follows that O P = r r = AB. Similarly, BC = r r ad CA = r r. Now we check = , so ABC is a right triagle. It follows that its area is = , which gives a aswer of ABC is a triagle i which BC =, CA = 4, AB = 5. Two cogruet circles ω ad ω are mutually exterally taget such that ω is also taget to BC ad AB while ω is also taget to AC ad AB. The radius of ω ca be expressed as m, where m ad are relatively prime positive itegers. Compute m +. Aswer: 0. Let O ad O be the ceters of ω ad ω, ad let the circles be taget to AB at T ad T respectively. Write r for the desired radius. T T O O is a rectagle, so T T = r. Now observe that BO bisects ABC. Sice cos( ABC) = /5, it follows that BT /T O = so that BT = r. Similarly, AT = r. Therefore, AB = BT + T T + T A = 7r from which r =

6 Difficult. Triagle ABC has sidelegths AB =, BC = 4, CA = 5. Triagle A B C lies outside triagle ABC ad has sides parallel to ABC with a distace of betwee correspodig sides. Compute the area of A B C. Aswer: 89. Suppose that A, B, C are ear A, B, C respectively. Let P ad Q be the feet of the perpediculars from A to A B ad A C. Sice AP = AQ =, right triagles AP A ad AQA are cogruet. Therefore, AA is the agle bisector of B A C. Sice the sides of A B C are parallel to the sides of ABC, we have that AA, BB, CC cocur at the shared iceter I of both triagles. Let ω ad ω deote the icircles of ABC ad A B C respectively. By Hero s formula, [ABC] = 84, hece, 84 = rs = r so that r = 4. It follows that the iradius of A B C is 6. Sice the homothety cetered at I that seds ω to ω seds triagle ABC to triagle A B C, we have that [A B C ] [ABC] = ( 6 4 ) = 9 4, from which [A B C ] = 89.. ABCDEF G is a regular heptago iscribed i a uit circle cetered at O. l is the lie taget to the circumcircle of ABCDEF G at A, ad P is a poit o l such that AOP is isosceles. Let p deote value of AP BP CP DP EP F P GP. Determie the value of p. Aswer:. Overlay the complex umber system with O = 0 + 0i, A = + 0i, ad P = + i. The solutios to the equatio z 7 = are precisely the seve vertices of the heptago. Lettig a, b, c, d, e, f, ad g deote the complex umbers for A, B, C, D, E, F, ad G respectively, this equatio rewrites as (z a)(z b)(z c)(z d)(z e)(z f)(z g) = z 7 = 0. The magitude of the factored product represets the product of the distaces from the arbitrary poit represeted by z. Thus, pluggig i + i, we have AP BP CP DP EP F P GP = ( + i) 7 = 8 8i = 7 8i = =. It follows that the aswer is.. ABCD is a cyclic quadrilateral with AB = 8, BC = 4, CD =, ad DA = 7. Let O ad P deote the circumceter ad itersectio of AC ad BD respectively. The value of OP ca be expressed as m, where m ad are relatively prime, positive itegers. Determie the remaider obtaied whe m + is divided by 000. Aswer: 589. Cosider D o the circumcircle of ABCD such that CD = 7 ad D A =. Let m D AB = α ad m BCD = π α. The by the Law of Cosies, cos(α) = BD = cos(π α) = cos(α) = 0 Hece D AB is a right triagle ad the circumradius of ABCD is 65. Now, by similar triagles, we have AP : BP : CP : DP = 56 : : 4 : 7. Let AP = 56x so that AC = 60x ad BD = 9x. Ptolemy s theorem applied to ABCD yields 60x 9x = = 6 from which x =. 65 6

7 Now we apply Stewart s theorem to triagle BOD ad cevia OP, obtaiig OB P D + OD BP = OP BD + BP BD P D 65 4 (x + 7x) = 9x OP + x 9x 7x = OP OP = 9 60 It follows that the aswer is = ABC is a acute triagle with perimeter 60. D is a poit o BC. The circumcircles of triagles ABD ad ADC itersect AC ad AB at E ad F respectively such that DE = 8 ad DF = 7. If EBC = BCF, the the value of AE ca be expressed as AF, where m ad are relatively prime positive itegers. Compute m +. m Aswer: 05. Sice BDEA is cyclic, EBD = EAD. Similarly, DCF = DAF. Sice we are give BCF = EBC, we have DAB = CAD. Because CD ad DF are itercepted by cogruet agles i the same circle, DF = CD = 7. Similarly, DB = 8. Now, by the agle bisector theorem, AC = 7x ad AB = 8x. Sice the perimeter of ABC is 60, 5+5x = 60 ad x =, so that AC = ad AB = 4. Now, by Power of a Poit from B, BF = BD BC = 8 5 = 5 ad CE = CD CB = 7 5 = 5. BA 4 CA Subtractig these legths from AB ad AC respectively, we fid that AF = 9 ad AE = 6. It follows that the aswer is = ABC is a scalee triagle. The circle with diameter AB itersects BC at D, ad E is the foot of the altitude from C. P is the itersectio of AD ad CE. Give that AP = 6, BP = 80, ad CP = 6, determie the circumradius of ABC. Aswer: 085. It is easily see that P = H, the orthoceter of ABC. Recall that AH = R cos(a). Thus, cos(a) = 68 40, cos(b) =, cos(c) =. Now we have R R R cos (A) + cos (B) + cos (C) + cos(a) cos(b) cos(c) = ( ) ( ) ( ) ( ) ( ) ( ) = R R R R R R R ( )R = 0 Thus, ay iteger solutio R will eed to be a factor of = Now R(cos(A) + cos(b) + cos(c)) = AH + BH + CH = 4. Sice < cos(a) + cos(b) + cos(c) 4, we establish the bouds R <. It is the a simple matter to compute R = A is the ceter of circle ω ad B is the ceter of circle ω such that A lies o ω ad B lies o ω. Let C be a poit of itersectio of ω ad ω. Γ is the circle that is 7

8 iterally taget to ω, ω, ad also taget to AB. If the legth of mior arc BC is, the compute the circumferece of Γ. Aswer: 7. Let Γ be taget to ω ad AB at T ad T respectively, deote its ceter by O, ad write R ad r for the radii of ω ad Γ respectively. Due to the tagecy, A, O, ad T are colliear. Thus, AO = AT T O = R r. By symmetry, AT = R/. Now, sice AT O is a right triagle, ( R ) +r = (R r) = R Rr +r, from which r = R. The circumferece of Γ is therefore /8 of the circumferece of 8 ω, which is 7 sice ABC is equilateral, so the aswer is A, B, ad C are poits o circle O such that B is the midpoit of major arc AC. D lies o mior arc AB such that BD = 65, ad E is the foot of the perpedicular from B to CD. Give that BC = 00 ad BE = 56, compute AD. Aswer: 59. Reflect A over BD to A. The m BDA = m ADB = π m BCA = π m BDC. Therefore, A lies o lie CD. It follows that CE = EA = ED+DA = ED + DA. Thus, AD = CE ED. Now Pythagoras gives AD = = 9 = ABCDE is a cyclic petago with AB = BC = CD = ad DE = EA =. The legth of AD ca be expressed as p, where p ad q are relatively prime positive itegers. q Compute p + q. Aswer: 04. Costruct A o mior arc AE such that A E = ad A B =. Now BE = EC = CA = x because each itercepts the same pair of arcs. Ptolemy o BCEA gives x = 4. Now Ptolemy o BCEA gives AC = 7/. Ad, sice AC = BD, a third Ptolemy o ABCD gives AD = 8. 4 Dracoia. ABCD is a rectagular sheet of paper. E ad F are poits o AB ad CD respectively such that BE < CF. If BCF E is folded over EF, C maps to poit C o AD ad B maps to B such that AB C = B EA. If AB = 5 ad BE =, the the area of ABCD ca be expressed as a + b c square uits, where a, b, ad c are itegers ad c is ot divisible by the square of ay prime. Compute a + b + c. Aswer: 8. By the reflectio, we have B E = BE =. Because ABCD is a rectagle, we have m C AE = m C B E = π = C AB E is cyclic with diameter C E = B C A = B EA = AB C = AB C is isosceles with AB = AC = 5. It would suffice to determie C E as this would evetually yield both sides of ABCD. Let ω deote the circumcircle of AB EC. Cosider the poit P o the mior arc B E of ω such that AP = ad P E = 5. AP EC is a isosceles trapezoid with m C AE = m C P E = π. Let C E = x. The by Pythagoras, C B = AE = x 5, but by Ptolemy s Theorem applied to this trapezoid, x + 5 = x 5 8

9 from which we fid x = 5 or. Takig C E = x = 5, we obtai AE = 65 5 = 0 6 ad C B = 5 = 4 6. Now we have AB = AE + EB = ad C B = BC = 4 6 so that the area of ABCD is , which yields a aswer of = 8.. Triagle ABC has a iradius of 5 ad a circumradius of 6. If cos B = cos A+cos C, the the area of triagle ABC ca be expressed as a b, where a, b, ad c are positive c itegers such that a ad c are relatively prime ad b is ot divisible by the square of ay prime. Compute a + b + c. Aswer: 4. It follows from cos B = cos A + cos C that cos A, cos B, cos C is a arithmetic progressio. It also follows that cos B = cos A + cos B + cos C = + r R = 6 so we may set cos A = 7 + k, cos B = 7, cos C = 7 k. We substitute these ito aother famous trig idetity, cos A + cos B + cos C + cos A cos B cos C = ( ) ( ( ) ) 7 + k k = So we have cos A =, cos B = 7 6 6, ad si C = 5 4 respectively. Fially, ( [ABC] = R si A si B si C = 6 which gives a aswer of = k + 7 (48 + 4) = 6 6 k = 058 = k = ± 48, ad cos C = 4, which imply si A =, si B = ) ( ) ( ) 5 = ABCDE is a cyclic petago with BC = CD = DE. The diagoals AC ad BE itersect at M. N is the foot of the altitude from M to AB. We have MA = 5, MD =, ad MN = 5. The area of triagle ABE ca be expressed as m where m ad are relatively prime positive itegers. Determie the remaider obtaied whe m + is divided by 000. Aswer: 77. Pythagoras gives AN = 0. We draw BD ad AD, ad costruct the altitute MP to AD, with P o AD, ad altitude MM to AE, with M o AE. Because BC = CD = DE, agles BAC, CAD, ad DAE are cogruet. Because P is 9

10 o AD, triagles MNA ad MP A are cogruet by AAS, so MP = 5 ad P A = 0, from which Pythagoras gives P D =, implyig AD =. Let α = m BAC, so m MAE = α, ad m NAE = α. Because we have si α = 5 ad cos α = 4 4 7, we compute si(α) =, ad si(α) =. We fid that MM = 4 usig si(α) = 4. By a simple Law of Sies argumet DE : EB : BD = 5 : 9 : Let [ABE] = the area of ABE. We have [ABE] = /(5 AB + 4 AE). Ptolemy o ABDE yields AB DE + AE BD = AD BE. Usig the abudace of facts that we have ascertaied previously, this gives: AB 5x + AE 40x = 9x 5AB + 40AE = 9 5AB + 4AE = 9 5 Fially, [ABE] = (5AB + 4AE) = = 77. = Therefore, the aswer is 4. I triagle ABC, we have BC =, CA = 7, ad AB = 40. Poits D, E, ad F are selected o BC, CA, ad AB respectively such that AD, BE, ad CF cocur at the circumceter of ABC. The value of AD + BE + CF ca be expressed as m where m ad are relatively prime positive itegers. Determie m +. Aswer: 59. Drop altitude AA. We have m AA B = π B, but AOB is a isosceles triagle with m AOB = C m BAO = π C. Therefore, cos DAA = cos(c B). Therefore we have AD cos(c B) = AA = AC si(c) = R si(b) si(c) so that R = cos(c B). Now, AD si(b) si(c) R + R + R AD BE R si(a) si(b) si(c) ( CF = cos(c B) + cos(a C) + cos(b A) si(b) si(c) si(c) si(a) si(a) si(b) + + ) AD BE CF = si(a) cos(b C) + si(b) cos(c A) + si(c) cos(a B) = si(a) si(b) si(c) + si(a) cos(b) cos(c) + si(b) cos(a) cos(c) + si(c) cos(a) cos(b) = si(a) si(b) si(c) + si(a + B) cos(c) + si(c) cos(a) cos(b) = si(a) si(b) si(c) + si(c) (cos(c) + cos(a) cos(b)) = si(a) si(b) si(c) + si(c) ( cos(a + B) + cos(a) cos(b)) = 4 si(a) si(b) si(c) = AD + BE + CF = R 0

11 Hero s formula yields [ABC] = = 40. We substitute this ito [ABC] = abc abc R = = 7 40 = 7. From this we fid that 4R 4 [ABC] AD + BE + CF = R = It follows that the aswer is = Circles ω ad ω are cetered o opposite sides of lie l, ad are both taget to l at P. ω passes through P, itersectig l agai at Q. Let A ad B be the itersectios of ω ad ω, ad ω ad ω respectively. AP ad BP are exteded past P ad itersect ω ad ω at C ad D respectively. If AD =, AP = 6, DP = 4, ad P Q =, the the area of triagle P BC ca be expressed as p q, where p, q, ad r are positive r itegers such that p ad r are coprime ad q is ot divisible by the square of ay prime. Determie p + q + r. Aswer: 468. We ivert about P with radius, mappig the circles ω ad ω to lies ω ad ω, each parallel to l, ad ω to a lie ω that itersects ω ad ω at A ad B respectively. Q is the itersectio of l ad ω, ad C ad D are the itersectios of the extesios of A P ad B P past P to ω ad ω respectively. We have P Q =, P A =, ad P 6 D =. The iversive distace formula gives 4 A D = R AD =. The crossed ladders theorem asserts AP DP 8 A D + B C = P Q from which B C =. However, it is clear i the iverted figure that triagles 4 C B P ad A D P are similar. Therefore, P C = ad P 8 B =. But iversio is its ow iverse trasformatio. Hece, P C = 8 ad P B =. The iversive distace formula gives BC = R B C P B P = 8 = 9. Fially, the area of P BC C may be foud via Hero s formula: K = = The aswer is therefore = I acute triagle ABC, BC = 0, CA =, ad AB = 4. ω, ω, ad ω are circles with diameters BC, CA, ad AB respectively. Let B deote the boudary of the regio iterior to the three ω i. Ω is the circle iterally taget to the three arcs of B. The radius of Ω ca be expressed as m p q, where m, p, ad are positive itegers with o commo prime divisor ad q is a positive iteger ot divisible by the square of ay prime. Compute m + + p + q. Aswer: 9. Let M A, M B, ad M C deote the midpoits of the sides opposite A, B, ad C respectively, ad write P for the ceter of Ω. Fially, let the tagets of ω, ω, ad ω with Ω be deoted by T, T, ad T. Note that M A M B = 7, M B M C = 5, ad M C M A = 6. Take poits U, U, ad U o the extesios of T M A, T M B, ad T M C past M A, M B, ad M C respectively such

12 that T i U i = 9 for i =,,. Because T lies o ω, T M A = 5 so that M A U = 4. Aalogously, M B U = ad M C U =. Now cosider circles ω, ω ad ω cetered at M A, M B, ad M C ad of radii 4,, ad respectively. ω, ω, ad ω are mutually exterally taget ad cotai poits U, U, ad U respectively. But the circumceter of U U U is P ; ergo, the circumcircle Ω of triagle U U U is taget to ω, ω, ad ω. Moreover, R Ω = 9 R Ω. Applyig the explicit form of the Descartes Circle Theorem for the outer circle, we fid R Ω 4 = ( + + 4) = ( + 6 6) 6 69 = from which R Ω = 9 R Ω = The aswer is therefore = Let O = (0, 0) ad A = (4, 0) deote the origi ad a poit o the positive x-axis respectively. B = (x, y) is a poit ot o the lie y = 0. These three poits determie lies l, l, ad l. Let P,..., P deote all of the poits that are equidistat from l i for i =,,. Let Q j deote the distace from P j to the l i for j =,...,. If Q + + Q = 0 Q + + Q = the the maximum possible value of x + y ca be expressed as u, where u ad v are v relatively prime positive itegers. Determie u + v. Aswer: 07. A poit that is equidistat from two lies lies o oe of the two lies that bisect agles formed at the itersectio of the two lies. Cosiderig this fact, it is clear that = 4, where P is the iceter of ABO (which we will deote I) ad P, P, ad P 4 are the three exceters. Let r deote the iradius of ABO ad r, r, ad r the three exradii. We have r + r + r + r = 4R + r = 0 r = r r r r = This pair of equatios is easily solved for r = ad R = 85, where R is the circumradius of ABO. By the Exteded Law of Sies, si B = AO = 4 = 84. Pythagoras yields R cos B = ±. The cot.5 B = 6 or Let the icircle of ABO be taget to BO, OA, ad AB at P, Q, ad R respectively. Let AQ = X ad QO = 4 X. Sice r = IQ =, we have cot.5 A = X ad

13 cot.5 O = 4 X. Sice cot.5 A cot.5 B cot.5 O = cot.5 A + cot.5 B + cot.5 O, we have cot.5 B X (4 X) = 49 cot.5 B by substitutio ad AM-GM. cot.5 B = 6 leads to 58 49, impossible, so cot.5 B = 7 7, which leads to X = 5 or 9. Thus, there are four possible B, each obtaied by 6 reflectig B over AO ad the perpedicular bisector of AO. Sice we are maximizig the sum x + y, we choose X = 5 ad assume y > 0. Now, BR = 7 ad RA = 5. Sice cos A = cos.5 A = 5 = 8, it must be that x = = 0 ad y = 7 5 = 5, which gives x + y = 5 ad a aswer of 5 + = Let G, H, ad I deote the cetroid, orthoceter, ad iceter of triagle ABC. If HI =, IG =, ad GH = 4, the the value of cos(a) + cos(b) + cos(c) ca be expressed as m, where m ad are relatively prime positive itegers. Compute m +. Aswer: 97. Examiig the Euler lie, O lies o lie HG such that GO =. Now Stewart s theorem o G, H, I, O yields 4OI + 8 = from which OI = 47. Aother famous result of Euler is that OI = R(R r). Fially, a corollary of Feuerbach s theorem gives r(r r) = IG IH = IG IH cos(hig) = HG IG IH =. It follows that cos(a) + cos(b) + cos(c) = + r = + 9 = 0. R 94 94