If the escalator stayed stationary, Billy would be able to ascend or descend in = 30 seconds. Thus, Billy can climb = 8 steps in one second.

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Alfred McKinney
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1 BMT 01 INDIVIDUAL SOLUTIONS March Billy the kid likes to play o escalators! Movig at a costat speed, he maages to climb up oe escalator i 4 secods ad climb back dow the same escalator i 40 secods. If at ay give momet the escalator cotais 48 steps, how may steps ca Billy climb i oe secod? 8 Aswer: or Solutio: If the escalator stayed statioary, Billy would be able to asced or desced i = 0 secods. Thus, Billy ca climb 48 0 = 8 steps i oe secod. 5. S-Corporatio desigs its logo by likig together 4 semicircles alog the diameter of a uit circle. Fid the perimeter of the shaded portio of the logo. Aswer: 4π Solutio: The uit circle has circumferece π ad the four semicircles cotribute π (x + (1 x)) o each side, for a total perimeter of 4π.. Two boxes cotai some umber of red, yellow, ad blue balls. The first box has red, 4 yellow, ad 5 blue balls, ad the secod box has 6 red, yellow, ad 7 blue balls. There are two ways to select a ball from these boxes; oe could first radomly choose a box ad the radomly select a ball or oe could put all the balls i the same box ad simply radomly select a ball from there. How much greater is the probability of drawig a red ball usig the secod method tha the first? Aswer: Solutio: 1 10 probability 1 The first method has a probability of ( ) = Thus, our differece is = 1, ad the secod method has 4. Let ABCD be a square with side legth, ad let a semicircle with flat side CD be draw iside the square. Of the remaiig area iside the square outside the semi-circle, the largest circle is draw. What is the radius of this circle? Aswer: 4 Solutio: Clearly, this circle will be taget to the semicircle ad taget to AB ad oe of BC or AD. Without loss of geerality, suppose it is taget to AD. The, let the ceter of this circle be O, ad let it have radius r. Drop a perpedicular from O to CD ad label it E, ad let the midpoit of CD be F. The, we have OF = 1 + r, F E = 1 r, OE = r. Applyig the Pythagorea theorem, we have ( r) + (1 r) = (1 + r), which simplifyig yields r 8r + 4 = 0. Solvig this quadratic ad gettig rid of egative solutios gives us r = 4.

2 BMT 01 INDIVIDUAL SOLUTIONS March Two positive itegers m ad satisfy max(m, ) = (m ) gcd(m, ) = mi(m, ) 6 Fid lcm(m, ). Aswer: 94 Solutio: Without loss of geerality, let m. The, m = (m ), so m = k ad = k k for some positive iteger k. Therefore, gcd(m, ) = k, so k = k k implies k = 7 6 ad m = 49, = 4. Thus, our aswer is lcm(49, 4) = Bubble Boy ad Bubble Girl live i bubbles of uit radii cetered at (0, 1) ad (0, 10) respectively. Because Bubble Boy loves Bubble Girl, he wats to reach her as quickly as possible, but he eeds to brig a gift; luckily, there are plety of gifts alog the x-axis. Assumig that Bubble Girl remais statioary, fid the legth of the shortest path Bubble Boy ca take to visit the x-axis ad the reach Bubble Girl (the bubble is a solid boudary, ad aythig the bubble ca touch, Bubble Boy ca touch too). Aswer: 7 Solutio: Cosider the reflectio of Bubble Boy ad his path over the lie y = 1. Note that Bubble Boy s ceter must touch the lie y = 1 oce, as that is the boudary for the rest of his body to touch the x-axis. The, Bubble Boy reflectio will take a path directly to Bubble Girl, ad the shortest possible path is a straight lie. Furthermore, Bubble Boy ad Bubble Girl will ed with their ceters uits apart, so as to miimize Bubble Boy s distace walked. Therefore, the value is (0 0) + ( ) (1 + 1) = Give real umbers a, b, c such that a + b c = ab bc ca = abc = 8. Fid all possible values of a. Aswer: {, ± 1} Solutio: Cosider the polyomial with roots a, b, ad c. We factor to (x )(x 6x 4) = 0 to fid a = x =, ± The three-digit prime umber p is writte i base as p ad i base 5 as p 5, ad the two represetatios share the same last digits. If the ratio of the umber of digits i p to the umber of digits i p 5 is 5 to, fid all possible values of p. Aswer: 601 Solutio: We may boud 5 p < 5 4 ad 9 p < 10 ad compute p 1 mod 100 or p 1 mod 100, givig us p = 51 ad p = 601 as possible solutios. 51 is divisible by, so our oly solutio is 601.

3 BMT 01 INDIVIDUAL SOLUTIONS March A at i the xy-plae is at the origi facig i the positive x-directio. The at the begis a progressio of moves, o the th of which it first walks 1 uits i the directio it is 5 facig ad the turs 60 degrees to the left. After a very large umber of moves, the at s movemets begis to coverge to a certai poit; what is the y-value of this poit? Aswer: 4 Solutio: The th move the at takes moves it 1 π( 1) si i the y-directio. The, the 5 ( 1 movemets coverge to ) ( ) ( + = ) = = If five squares of a board iitially colored white are chose at radom ad blackeed, what is the expected umber of edges betwee two squares of the same color? 16 Aswer: Solutio: There are 1 possible edges that could satisfy the give coditio. The probability the two squares touchig the edge are both black is 5 9 4, ad the probability they are 8 both white is 4 9. Thus, the expected umber of edges touched by two black squares is ( ) = Let t = (a, b, c), ad let us defie f 1 (t) = (a + b, b + c, c + a) ad f k (t) = f k 1 (f 1 (t)) for all k > 1. Furthermore, a permutatio of t has the same elemets, just i a differet order (e.g., (b, c, a)). If f 01 (s) is a permutatio of s for some s = (k, m, ), where k, m, ad are itegers such that k, m, 10, how may possible values of s are there? Aswer: 6 Solutio: Defie s(t) = a + b + c. The, s(f(t)) = a + b + b + c + c + a = s(t), ad i geeral s(f k (t)) = k s(t), so if f k (t) = t, we must have s(t) = 0. Now we show that we will always retur i at most 6 steps. Let t = (a, b, c), the f(t) = (a + b, b + c, c + a), ad f(f(t)) = (a+b+b+c, b+c+c+a, c+a+a+b) = (b, c, a). Notice that this shifts all elemets by 1, so if we repeat this operatio three more times, we will retur to (a, b, c). Thus, we have k Triagle ABC satisfies the property that A = a log x, B = a log x, ad C = a log 4x radias, for some real umbers a ad x. If the altitude to side AB has legth 8 ad the altitude to side BC has legth 9, fid the area of ABC. Aswer: 4 Solutio: Notig B = π radias, we may calculate the area [ABC] to be [ABC] = AB BC = 4 4 [ABC] [ABC] ad therefore, [ABC] = 7 = /

4 BMT 01 INDIVIDUAL SOLUTIONS March Let f() be a fuctio from itegers to itegers. Suppose f(11) = 1, ad f(a)f(b) = f(a + b) + f(a b), for all itegers a, b. Fid f(01). Aswer: Solutio: Let a = 11, b = 0, yieldig f(11)f(0) = f(11) + f(11), so f(0) =. Now, let b = 11, ad a arbitrary, which gives us f(a)f(11) = f(a + 11) + f(a 11), so f(a + 11) = f(a) f(a 11), which we may fid subsequet values of f(a) if a is a multiple of 11. I tur, we get f() = f(11) f(0) = 1, f() = f() f(11) =, f(44) = 1, f(55) = 1, f(66) =, f(77) = 1. Notice that f(0) = f(66), f(11) = f(77), ad the recurrece has depth, so we see this cycle will repeat with legth 6. Thus, f(01) = f(18 11) = f() =. 14. Triagle ABC has icircle O that is taget to AC at D. Let M be the midpoit of AC. E lies o BC so that lie AE is perpedicular to BO exteded. If AC = 01, AB = 014, DM = 49, fid CE. Aswer: 498 Solutio: Let O be taget to AB at F ad BC at G. The, we have AM = MC, CG = DC, AF = AD, BF = BG, ad GE = F A, so DC = DM + MC = CE + EG = CG, so CE + EG = CE + F A = CE + AD. Thus, we get DM + MC = CE + AD, so DM + MC = CE + AM MD, ad DM = CE, so CE = Let ABCD be a covex quadrilateral with ABD = BCD, AD = 1000, BD = 000, BC = 001, ad DC = Poit E is chose o segmet DB such that ABD = ECD. Fid AE. Aswer: Solutio: 001 ABD ECD = AED BCD. The AE = BC AD BD = Fid the sum of all possible such that is a positive iteger ad there exist a, b, c real umbers such that for every iteger m, the quatity 01m + am + bm + c is a iteger. Aswer: 9016 Solutio: Lettig a = a, b = b, ad c = c, we realize if 01 m + a m + b m + c is a iteger for every m, the we have 01 (m + 1) + a(m + 1) + b(m + 1) + c is also a iteger, so the differece betwee these two must be a iteger. The differece is 01 m + dm + e for some real d, e. We kow this quatity must be a iteger for every iteger m, so we ca apply the same trick. The differece that we get ext is 6 01 m + f, for some real f. If we apply this oce agai, we just get Thus, we wat 6 01 to be a iteger, so we just eed to fid the sum of all divisors of 6 01 = 11 61, which is (1 + )(1 + + )(1 + 11)(1 + 61) =

5 BMT 01 INDIVIDUAL SOLUTIONS March Let N 1 be a positive iteger ad k be a iteger such that 1 k N. Defie the recurrece x = x 1 + x x N for > N ad x k = 1, x 1 = x =... = x k 1 = N x k+1 =... = x N = 0. As approaches ifiity, x approaches some value. What is this value? Aswer: k N(N+1) Solutio: Defie the sequece y that satisfies the recurrece Ny + (N 1)y y N+1 = k for N with iitial coditios y 1 =... = y k 1 = y k+1 =... = y N = 0 ad y k = 1. Observe the iitial values satisfy the recurrece for y ad that y i = x i for i = 1,,..., N. Take the recurrece for y ad take > + 1. The we get for N, Ny + (N 1)y y N+1 = Ny +1 + (N 1)y y N = Ny +1 = y y N+1. Thus x, y satisfy the same recurrece. As teds to ifiity, the sequece teds to a fiite limit. Call this limit L. The N L+(N 1)L+...+L = k so L = k N(N + 1). 18. Paul ad his pet octahedro like to play games together. For this game, the octahedro radomly draws a arrow o each of its faces poitig to oe of its three edges. Paul the radomly chooses a face ad progresses from face to adjacet face, as determied by the arrows o each face, ad he wis if he reaches every face of the octahedro. What is the probability that Paul wis? Aswer: 4 Solutio: There are 18 successful cases, after choosig the placemet of the first face. Thus, the aswer is 18 7 = Equilateral triagle ABC is iscribed i a circle. Chord AD meets BC at E. If DE = 01, how may scearios exist such that both DB ad DC are itegers (two scearios are differet if AB is differet or AD is differet)? Aswer: 14 Solutio: Let e = DB, f = DC. Sice ABDC is cyclic, we have AD BC = AB f + AC e, but AB = BC = AC, so AD = e + f. Additioally, we have triagle BDE is similar to triagle ADC, so BD/AD = DE/DC, so DE = DC BD/AD. Thus, we have 01 = f e f + e, 1 so 01 = 1 f + 1. Thus, it suffices to fid the umber of iteger solutios to this equatio. e Without loss of geerality, suppose f e. Clearig demoiators, we have 01f + 01e = ef, so e = 01f/(f 01). Let k = f 01, so e = 01(k + 01)/k = /k, so if e is a iteger, the k must divide. We also have f = 01 + k x = /k, so k 01 /k, so k 01. Thus, this is the umber of divisors of 01 that are less tha or equal to 01, or (d(01 ) + 1)/ (where d represets the umber of divisors, ad because each factor 01 gets paired with a uique factor 01). Factorig 01 = 11 61, so 01 has = 7 divisors, so the umber of iteger solutios is 14.

6 BMT 01 INDIVIDUAL SOLUTIONS March A sequece a is defied such that a 0 = a k + a k k=0 ad a +1 = a for 0. Evaluate 5 + Aswer: 4 Solutio: Simplify to 1 a k + a k ad take the limit of partial products up through as. The, multiply by 1 + a + a 1 + a + a ad collapse. Remaiig is a4 0 + a =

Joe Holbrook Memorial Math Competition

Joe Holbrook Memorial Math Competitio 8th Grade Solutios October 5, 07. Sice additio ad subtractio come before divisio ad mutiplicatio, 5 5 ( 5 ( 5. Now, sice operatios are performed right to left, ( 5